Review of the Real
Number System
Social Security is the largest source of income for elderly Ameri
cans. It is projected that in about 30 years, however, there will be
twice as many older Americans as there are today, and the excess
revenues now accumulating in Social Security's trust funds will be
exhausted. {Source: Social Security Administration.)
To supplement their retirement incomes, more and more Ameri
cans have begun investing in mutual funds, pension plans, and other
means of savings. In Section 1.3, we relate the concepts of this chap
ter to the percent of U.S. households investing in mutual funds.
T.T Basic Concepts
T.2 Operations on Real Numbers
T .3 Exponents, Roots, and Order
of Operations
T .4 Properties of Real Numbers
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
Chapter I Review of the Real Number System
1 • 1 Basic Concepts
OBJECTIVES
Write sets using set
notation.
Use number lines.
Know the common sets
of numbers.
Find additive inverses.
Use absolute value.
Use inequality symbols.
Study Skills Workbook
Activity 2: Your Textbook
Consider the set
0,10,— ,52, 98.6 >.
' ' 10 '
(a) Which elements of the
set are natural numbers?
In this chapter we review some of the basic symbols and rules of algebra.
OBJECTIVE Q Write sets using set notation. A set is a collection of
objects called the elements or members of the set. In algebra, the elements
of a set are usually numbers. Set braces, { }, are used to enclose the ele
ments. For example, 2 is an element of the set {1, 2, 3}. Since we can count
the number of elements in the set { 1, 2, 3}, it is di finite set.
In our study of algebra, we refer to certain sets of numbers by name. The set
7V= {1,2,3,4,5,6,...}
is called the natural numbers or the counting numbers. The three dots
show that the list continues in the same pattern indefinitely. We cannot list
all of the elements of the set of natural numbers, so it is an infinite set.
When is included with the set of natural numbers, we have the set of
whole numbers, written
W= {0,1,2,3,4,5,6,...}.
A set containing no elements, such as the set of whole numbers less than 0,
is called the empty set, or null set, usually written 0.
CAUTION
Do not write {0} for the empty set; {0} is a set with one element, 0. Use
only the notation for the empty set.
(b) Which elements of the
set are whole numbers?
& List the elements in each set.
(a) {x I X is a whole number
less than 5}
Work Problem 1 at the Side.
In algebra, letters called variables are often used to represent numbers
or to define sets of numbers. For example,
(x I X is a natural number between 3 and 15}
(read "the set of all elements x such that x is a natural number between 3 and
15") defines the set
{4,5,6,7,..., 14}.
The notation {x  x is a natural number between 3 and 15} is an example
of setbuilder notation.
{ X I X has property P }
the set of all elements x such that x has a given property P
(b) {j 1 7 is a whole number
greater than 12}
Answers
1. (a) 10 and 52 (b) 0, 10, and 52
2. (a) {0,1,2,3,4} (b) {13,14,15,...}
EXAMPLE 1
Listing the Elements in Sets
List the elements in each set.
(a) {x I X is a natural number less than 4}
The natural numbers less than 4 are 1, 2, and 3. This set is {1, 2, 3}.
(b) {j 1 7 is one of the first five even natural numbers} = {2, 4, 6, 8, 10}
(c) {z I z is a natural number greater than or equal to 7}
The set of natural numbers greater than or equal to 7 is an infinite set,
written with three dots as {7, 8, 9, 10, . . . }.
Mi
Work Problem 2 at the Side.
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Section I . I Basic Concepts 3
EXAMPLE 2
Using SetBuilder Notation to Describe Sets
Use setbuilder notation to describe each set.
(a) {1,3,5,7,9}
There are often several ways to describe a set with setbuilder notation.
One way is {j  j is one of the first five odd natural numbers}.
(b) {5, 10, 15, ... }
This set can be described as {x  x is a multiple of 5 greater than 0}.
Work Problem 3 at the Side.
& Use setbuilder notation to
describe each set.
(a) {0,1,2,3,4,5}
OBJECTIVE Q Use number lines, A good way to get a picture of a set
of numbers is by using a number line. To construct a number line, choose
any point on a horizontal line and label it 0. Next, choose a point to the right
of and label it 1. The distance from to 1 establishes a scale that can be
used to locate more points, with positive numbers to the right of and nega
tive numbers to the left of 0. The number is neither positive nor negative.
A number line is shown in Figure 1 .
H \ \ \ \ h
H h
H h
54321012345
Figure 1
The set of numbers identified on the number line in Figure 1, including
positive and negative numbers and 0, is part of the set of integers, written
/={... ,3, 2,1, 0,1, 2,3,...}.
Each number on a number line is called the coordinate of the point that
it labels, while the point is the graph of the number. Figure 2 shows a num
ber line with several selected points graphed on it.
Graph of 1
\4
# • #
3
4
Coordinate'
1
/
Figure 2
Work Problem 4 at the Side.
(b) {7,14,21,28,...}
O Graph the elements of each
set.
(a) {4,2,0,2,4,6}
2 5
(b) i1,0,,
The fractions \ and , graphed on the number line in Figure 2, are
examples of rational numbers. Rational numbers can be written in decimal
form, either as terminating decimals such as f = .6,  = .125, or ^ = 2.75,
or as repeating decimals such as  = .33333. . . or ^ = .272727. ... A
repeating decimal is often written with a bar over the repeating digit(s).
Using this notation, .2727 ... is written .27.
Decimal numbers that neither terminate nor repeat are not rational, and
thus are called irrational numbers. Many square roots are irrational num
bers; for example, Vl = 1.4142136... and Vl = 2.6457513...
repeat indefinitely without pattern. (Some square roots are rational:
V 16 = 4,V 100 = 10, and so on.) Another irrational number is tt, the ratio
of the circumference of a circle to its diameter.
Some of the rational and irrational numbers discussed above are graphed
on the number line in Figure 3 on the next page. The rational numbers
together with the irrational numbers make up the set of real numbers. Every
point on a number line corresponds to a real number, and every real number
corresponds to a point on the number line.
, 16 13 29
(C) i5,y,6,y,7,
Answers
3. (a) One answer is {x  x is a whole number
less than 6}. (b) One answer is {x  x is
a multiple of 7 greater than 0}.
4. (a)
(b)
^ 4 i nnn ^
42 2 4 6
I 4 4 •! I»l
210123
(c) H — 4* 4 * f* I "
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
Chapter I Review of the Real Number System
Real numbers
[rrational
numbers
\
h
V?
I • • I
V2
4 3
Rational
numbers
2
t 1
_ 3.
.27 5
2.75
Figure 3
4X
VT6
OBJECTIVE Q Know the common sets of numbers. The sets of
numbers listed below will be used throughout the rest of this text.
Sets of Numbers
Natural numbers or
counting numbers
Whole numbers
Integers
Rational numbers
Irrational numbers
Real numbers
{1,2,3,4,5,6,...}
{0,1,2,3,4,5,6,... }
{...,3,2,1,0,1,2,3,...}
p and q are integers, q i^
Examples: f, 1.3,
9 16
"2' 8
or
2, V9 or 3, .6
{x I X is a real number that is not rational}
Examples: V 3, — v 2, tt
{x I X is represented by a point on a
number line} "^
The relationships among these sets of numbers are shown in Figure 4; in
particular, notice that the set of real numbers includes both the rational and irra
tional numbers. Every real number is either rational or irrational. Also, notice
that the integers are elements of the set of rational numbers and that whole
numbers and natural numbers are elements of the set of integers.
Real numbers
Rational numbers
% I' .6, 1.75
Integers
11,6,4
Whole
numbers
Natural
numbers
1,2,3,4,
5, 27, 45
Irrational numbers
Real
Positive integers
numbers
Integers
Zero
Rational
Negative integers
numbers
Noninteger rational numbers
Figure 4 The Real Numbers
* An example of a number that is not a coordinate of a point on a number line is V — 1 . This number,
called an imaginary number, is discussed in Section 9.7.
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Section I . I Basic Concepts 5
EXAMPLE 3
{m
Identifying Examples of Number Sets
Which numbers in
I  8,  V2,  ^, 0, .5, 1 1.12, VS, 2
are elements of each set?
(a) Integers 8, 0, and 2 are integers.
54, U, .J, 3,
(b) Rational numbers 8, ^, 0, .5, , 1.12, and 2 are rational numbers.
(c) Irrational numbers  v 2 and V 3 are irrational numbers.
(d) Real numbers
All the numbers in the given set are real numbers.
Work Problem 5 at the Side.
EXAMPLE 4
Determining Relationships between Sets
of Numbers
Decide whether each statement is true or false.
(a) All irrational numbers are real numbers.
This is true. As shown in Figure 4, the set of real numbers includes all
irrational numbers.
(b) Every rational number is an integer.
This statement is false. Although some rational numbers are integers,
other rational numbers, such as  and  1, are not.
Work Problem 6 at the Side.
& Select all the sets from the
following list that apply to
each number.
Whole number
Rational number
Irrational number
Real number
(a) 6
(b) 12
(c) .3
(d) VTs
(e) 77
22
(f)
(g) 3.14
OBJECTIVE Q Find additive inverses. Look again at the number line
in Figure 1. For each positive number, there is a negative number on the
opposite side of that lies the same distance from 0. These pairs of numbers
are called additive inverses, negatives, or opposites of each other. For
example, 5 is the additive inverse of —5, and —5 is the additive inverse of 5.
Additive Inverse
For any real number a, the number —a is the additive inverse of a.
Change the sign of a number to get its additive inverse. The sum of a num
ber and its additive inverse is always 0.
The symbol "" can be used to indicate any of the following:
1. a negative number, such as —9 or — 15;
2. the additive inverse of a number, as in "—4 is the additive inverse of 4";
3. subtraction, as in 12 — 3.
In the expression —(—5), the symbol "— " is being used in two ways: the
first  indicates the additive inverse of 5, and the second indicates a nega
tive number, —5. Since the additive inverse of —5 is 5, then —(—5) = 5. This
example suggests the following property.
For any real number a.
— {—a) = a.
© Decide whether the statement
is true ox false. If false, tell
why.
(a) All whole numbers are
integers.
(b) Some integers are whole
numbers.
(c) Every real number is
irrational.
Answers
5. (a) rational, real
(b) whole, rational, real
(c) rational, real
(d) irrational, real
(e) irrational, real
(f ) rational, real
(g) rational, real
6. (a) true (b) true
(c) false; Some real numbers are irrational,
but others are rational numbers.
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
Chapter I Review of the Real Number System
O Give the additive inverse of
each number.
(a) 9
Numbers written with positive or negative signs, such as +4, +8, —9,
and 5, are called signed numbers. A positive number can be called a
signed number even though the positive sign is usually left off. The follow
ing table shows the additive inverses of several signed numbers. Note that
is its own additive inverse.
(b) 12
Number
Additive Inverse
6
6
4
4
2
3
2
3
8.7
8.7
Work Problem 7 at the Side.
(c)
6
5
B J ECTi VE Q Use absolute value. Geometrically, the absolute
value of a number a, written  a , is the distance on the number line from to
a. For example, the absolute value of 5 is the same as the absolute value of
— 5 because each number lies five units from 0. See Figure 5. That is,
5 = 5 and — 51 = 5.
Distance is 5,
sol 5 I = 5.
Distance is 5,
so I 5 I = 5.
H \ \ \ \ \ \ \ \ \ \ h
Figure 5
CAUTION
Because absolute value represents distance, and distance is always posi
tive (or 0), the absolute value of a number is always positive (or 0).
(d)0
The formal definition of absolute value follows.
Absolute Value
i«i =
a if « is positive or
— a if « is negative
(e) 1.5
Answers
7. (a) 9 (b) 12 (c) ^ (d) (e) 1.5
The second part of this definition, \a\= —(2 if a is negative, requires careful
thought. If a is a negative number, then —a, the additive inverse or opposite
of a, is a positive number, so a is positive. For example, if a = —3, then
\a\ = — 3 = —(—3) = 3. \a\
a if a is negative.
EXAMPLE 5
Evaluating Absolute Value Expressions
Find the value of each expression.
(a) 13= 13 (b) 2= (2) = 2
(c) 0 =
Continued on Next Page
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Section I . I Basic Concepts 7
(d) 8
Evaluate the absolute value first. Then find the additive inverse.
(e)
8= (8)= 8
8
Work as in part (d):   8 1 = 8, so
8= (8)= 8.
(f) 2 + 5
Evaluate each absolute value first, and then add.
(g) 52
2 + 5 = 2 + 5 = 7
3
Work Problem 8 at the Side.
O Find the value of each
expression.
(a) 6
(b)3
(c) 1 5 1
Absolute value is useful when comparing size without regard to sign.
EXAMPLE 6
Comparing Rates of Change in Industries
The projected annual rates of employment change (in percent) in some of the
fastest growing and most rapidly declining industries from 1994 through
2005 are shown in the table.
Industry (19942005)
Percent Rate
of Change
Health services
5.7
Computer and data processing services
4.9
Child day care services
4.3
Footware, except rubber and plastic
6.7
Household audio and video equipment
4.2
Luggage, handbags, and leather products
3.3
Source: U.S. Bureau of Labor Statistics.
What industry in the list is expected to see the greatest change? the least
change?
We want the greatest change, without regard to whether the change is an
increase or a decrease. Look for the number in the list with the largest
absolute value. That number is found in footware, since  — 6.7 = 6.7. Simi
larly, the least change is in the luggage, handbags, and leather products indus
try:! 3.3 1= 3.3.
Work Problem 9 at the Side.
OBJECTIVE Q Use inequality symbols. The statement 4 + 2 = 6 is
an equation; it states that two quantities are equal. The statement 4 # 6
(read "4 is not equal to 6") is an inequality, a statement that two quantities
are not equal. When two numbers are not equal, one must be less than the
other. The symbol < means "is less than." For example,
4
8<9, 6<15, 6<l, and < .
The symbol > means "is greater than." For example,
12 > 5, 9 > 2, 4 > 6, and
> 0.
In each case, the symbol ^^points^' toward the smaller number.
(d)
(e)
(f) 6 + 3
(g)9
(h) 94
& Refer to the table in
Example 6. Of the household
audio/video equipment
industry and computer/data
processing services, which
will show the greater change
(without regard to sign)?
Answers
8. (a) 6 (b) 3 (c) 5 (d) 2 (e) 7
(f) 9 (g) 5 (h) 5
9. Computer/data processing services
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8 Chapter I Review of the Real Number System
CO) Insert < or > in each blank
to make a true statement.
(a) 3 7
The number line in Figure 6 shows the numbers 4 and 9, and we know
that 4 < 9. On the graph, 4 is to the left of 9. The lesser of two numbers is
always to the left of the other on a number line.
4 < 9
+
+
4 5 6
Figure 6
(b)9
Inequalities on a Number Line
On a number line,
a<b if a is to the left ofb; a> b if a is to the right ofb.
We can use a number line to determine order. As shown on the number
line in Figure 7, —6 is located to the left of 1. For this reason, — 6 < 1. Also,
1 > — 6. From the same number line, — 5 < — 2, or — 2 > —5.
h
h
Figure 7
(c) 4
CAUTION
Be careful when ordering negative numbers. Since — 5 is to the left of
— 2 on the number line in Figure 7, — 5 < — 2, or — 2 > —5. In each
case, the symbol points to 5, the smaller number.
i#
Work Problem 10 at the Side,
(d) 2
The following table summarizes results about positive and negative
numbers in both words and symbols.
Words
Symbols
Every negative number is less than 0.
If a is negative, then a < 0.
Every positive number is greater
than 0.
If a is positive, then a> 0.
is neither positive nor negative.
(e)
In addition to the symbols t^, <, and >, the symbols
used.
and > are often
Answers
10. (a) < (b) > (c) > (d) < (e) >
INEQUALITY SYMBOLS
Symbol
Meaning
Example
^
is not equal to
3#7
<
is less than
4< 1
>
is greater than
3>2
<
is less than or equal to
6<6
>
is greater than or equal to
8>10
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
The following table shows several inequalities and why each is true.
Inequality
Why It Is True
6<8
6<8
2< 2
2 = 2
9> 12
9> 12
3> 3
3 = 3
6 • 4 < 5(5)
24<25
Section I . I Basic Concepts 9
CD Answer true or false.
(a) 2 < 3
Notice the reason why — 2 < —2 is true. With the symbol <, Neither the <
part or the = part is true, then the inequality is true. This is also the case
with the > symbol.
In the last line, recall that the dot in 6 • 4 indicates the product 6 x 4, or
24, and 5(5) means 5 x 5, or 25. Thus, the inequality 6 • 4 < 5(5) becomes
24 < 25, which is true.
(b) 8 < 8
Work Problem 11 at the Side.
(c)
(d) 5 • 8 < 7 • 7
(e) 3(4) > 2(6)
Answers
11. (a) false
(d) true
(b) true
(e) false
(c) false
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Section 1 .2 Operations on Real Numbers I 5
L.2 Operations on Real Numbers
In this section we review the rules for adding, subtracting, mukiplying, and
dividing real numbers.
OBJECTIVE Q Add real numbers. Number lines can be used to illus
trate addition and subtraction of real numbers. To add two real numbers on a
number line, start at 0. Move right (the positive direction) to add a positive
number or left (the negative direction) to add a negative number. See Figure 8.
I 3 I 2 I
k k 1
A \ \ \ \ \ h
5 4 3 210 1
2 + (3) = 5
(a)
4
H \ h
+
H \ \ [
10 12 3 4 5 6
6 + (4) = 2
(b)
Figure 8
This procedure for adding real numbers can be generalized in the following
rules.
Adding Real Numbers
Like signs To add two numbers with the same sign, add their absolute
values. The sign of the answer (either + or — ) is the same as the sign of
the two numbers.
Unlike signs To add two numbers with different signs, subtract the
smaller absolute value from the larger. The sign of the answer is the
same as the sign of the number with the larger absolute value.
Recall that the answer to an addition problem is called the sum.
OBJECTIVES
Add real numbers.
Subtract real numbers.
Multiply real numbers.
Find the reciprocal of a
number.
Divide real numbers.
Find each sum.
(a) 2 + (7)
(b)15 + (6)
(c) 1.1 + (1.2)
EXAMPLE 1
Find each sum.
Adding Two Negative Numbers
(a) 12 + (8)
First find the absolute values.
1121= 12
and
81 = 8
Because — 12 and —8 have the same sign, add their absolute values. Both
numbers are negative, so the answer is negative.
12 + (8) = (12 + 8) = (20) = 20
(b) 6 + (3) = (6 + 3)= (6 + 3)= 9
(c) 1.2 + (.4) = (1.2 + .4) = 1.6
1\ [5 \\ (5 2\ 7
(d)
5
 +
6
3/ V6 3/ V6 6/ 6
Work Problem 1 at the Side.
3 / 1
Answers
1. (a) 9 (b) 21 (c) 2.3 (d)
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I 6 Chapter I Review of the Real Number System
O Find each sum.
(a) 12 + (1)
Animatlaj
[m\
/ideo^
(b) 3 + (7)
(c) 17 + 5
<*!4
(e) 1.5 + 3.2
Answers
2. (a) 11 (b) 4 (c) 12
»i
(e) 1.7
Copyright (Q
EXAMPLE 2
Find each sum.
Adding Numbers with Different Signs
(a) 17+11
First find the absolute values.
17=17 and 11=11
Because — 1 7 and 1 1 have different signs, subtract their absolute values.
17 11 = 6
The number — 17 has a larger absolute value than 1 1, so the answer is negative.
17+ 11 = 6
t
Negative because   17 1 > 1 1 1 1
(b)4 + (l)
Subtract the absolute values, 4 and 1 . Because 4 has the larger absolute
value, the sum must be positive.
4 + (1) = 4 1 = 3
t
Positive because 4 >  — 1 1
(c) 9 + 17 = 17  9 = 8
(d) 16+12
The absolute values are 16 and 12. Subtract the absolute values. The
negative number has the larger absolute value, so the answer is negative.
16+ 12= (16 12)= 4
4 2
(e) — + 
^ ^ 5 3
Write each number with a common denominator.
4
5 "
4
5
43
5 • 3
12 , 2
=15 ^"' 3=
2
3
• 5 10
• 5 ~ 15
2
 + 
3
12 10
= + —
15 15
/12 10 \
f has the larger
Vl5 is)
absolute value.
2
15
Subtract.
(f) 2.3 + 5.6 = 3.3
Work Problem 2 at the Side.
OBJECTIVE Q Subtract real numbers. Recall that the answer to a
subtraction problem is called the difference. Thus, the difference between
6 and 4 is 2. To see how subtraction should be defined, compare the follow
ing two statements.
64 = 2
6 + (4) = 2
The second statement is pictured on the number line in Figure 8(b) at the
beginning of this section. Similarly, 9 — 3 = 6 and 9 + (— 3) = 6 so that
9 — 3 = 9 + (— 3). These examples suggest the following rule for subtraction.
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Section 1 .2 Operations on Real Numbers I 7
Subtraction
For all real numbers a and b,
a — b = a \ (—b).
In words, change the sign of the second number (subtrahend) and add.
& Find each difference.
(a) 912
EXAMPLE 3
Subtracting Real Numbers
Find each difference.
Change to addition.
Change sign of second number (subtrahend).
(a) 6  8 = 6 + (8) = 2
— Changed
Sign changed
(b) 12  4 = 12 + (4) = 16
(c) 10 (7) =10 + [(7)]
= 10 + 7
= 3
(d) 2.4  (8.1) = 2.4 + 8.1 = 5.7
5'
This step is often omitted.
(e)5
8 5 _ jJ
3 3 ~ 3
(b) 7  2
(c) 8  (2)
Work Problem 3 at the Side.
When working a problem that involves both addition and subtraction,
add and subtract in order from left to right. Work inside brackets or paren
theses first.
(d) 6.3 (11.5)
EXAMPLE 4
Adding and Subtracting Real Numbers
Perform the indicated operations.
(a) 8 + 5  6 = (8 + 5)  6 Work from left to right.
= 3 6
= 3 + (6)
= 9
(b) 15  (3)  5  12 = (15 + 3)  5  12
= 18512
= 1312
= 1
(c) 4  (6) + 7  1 = (4 + 6) + 7  1
=2+71
= 91
= 8
Continued on Next Page
(e) 12  (5)
Answers
3. (a) 3 (b) 9 (c) 6 (d) 5.2
(e) 17
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I 8 Chapter I Review of the Real Number System
(d) 9  [8  (4)] + 6
O Perform the indicated
operations.
(a) 6 + 92
(b) 12  (4) +
(c) 6  (2)  8  1
(d) 3  [(7) + 15] + 6
= 9 [8 + 4] + 6
= 9  [4] + 6
= 9 + 4 + 6
= 5 + 6
= 1
Work inside
the brackets.
Work Problem 4 at the Side.
OBJECTIVE Q Multiply real numbers. The answer to a multiplication
problem is called the product. For example, 24 is the product of 8 and 3.
The rules for finding signs of products of real numbers are given below.
Multiplying Real Numbers
Like signs The product of two numbers with the same sign is positive.
Unlike signs The product of two numbers with different signs is negative.
& Find each product.
(a) 7(5)
(b) .9(15)
(c)
4/14
7 V 3
(d) 7(2)
(e) .8(.006)
(f) o(16)
EXAMPLE 5
Multiplying Real Numbers
Find each product.
(a) — 3(— 9) = 27 Same sign; product is positive.
(b) .5(.4) = .2
3 / 5\ 5
Wi(5J = j
(d) 6(— 9) = —54 Different signs; product is negative.
(e) .05 (.3) = .015
(f) (3) = 2
^^' 8 V13/ 26
Hi
Work Problem 5 at the Side.
OBJECTIVE Q Find the reciprocal of a number. Earlier, subtraction
was defined in terms of addition. Now, division is defined in terms of multi
plication. The definition of division depends on the idea of a multiplicative
inverse or reciprocal; two numbers are reciprocals if they have a product of 1.
(g)
■f(12)
Answers
4. (a) 1 (b) 24 (c) 13 (d) 5
5. (a) 35 (b) 13.5 (c)  (d) 14
(e) .0048 (f) 10 (g) 8
Reciprocal
The reciprocal of a nonzero number a is
1
lilll Calculator Tip Reciprocals (in decimal form) can be found with a
calculator that has a key labeled C^ or 0^ . For example, a calculator
shows that the reciprocal of 25 is .04.
Copyright ©2005 Pearson Education, Inc., pubiisiiing as Pearson AddisonWesley
The table gives several numbers and their reciprocals.
Number
Reciprocal
2
~5
5
2
6
1
6
7
11
11
7
.05
20
None
Section 1 .2 Operations on Real Numbers I 9
© Give the reciprocal of each
number.
■f(i)
111 7 ,
1
.05(20) = 1
There is no reciprocal for because there is no number that can be multi
plied by to give a product of 1 .
(a) 15
(b)
CAUTION
A number and its additive inverse have opposite signs; however, a num
ber and its reciprocal always have the same sign.
(c)
Work Problem 6 at the Side.
(d) 
OBJECTIVE Q Divide real numbers. The result of dividing one num
ber by another is called the quotient. For example, when 45 is divided by 3,
the quotient is 15. To define division of real numbers, we first write the quo
tient of 45 and 3 as y, which equals 15. The same answer will be obtained if
45 and  are multiplied, as follows.
45 1
= — = 45 . 
3 3
45
15
This suggests the following definition of division of real numbers.
Division
For all real numbers a and b (where b ¥" 0),
a — b = — = a'—.
b b
In words, multiply the first number by the reciprocal of the second
number.
(e) .125
Divide where possible.
9
(a)
(b)
There is no reciprocal for the number 0, so division by is undefined. For
example, y is undefined and  ^ is undefined.
<''x
CAUTION
Division by is undefined. However, dividing by a nonzero number
gives the quotient 0. For example,
6
— is undefined, but — = (since 06 = 0).
Be careful when is involved in a division problem.
Work Problem 7 at the Side.
(a)^
Answers
6.
(a)
1
15
(b) 
1
'l
(c)
9
8
(d)
3
(e)
7.
(a)
undefined
(b)
(c)
undefined
(d)
Copyright ©2005 Pearson Education, Inc., pubiisiiing as Pearson AddisonWesley
20 Chapter I Review of the Real Number System
O Find each quotient.
16
i»)—r
Since division is defined as multiplication by the reciprocal, the rules for
signs of quotients are the same as those for signs of products.
Dividing Real Numbers
Like signs The quotient of two nonzero real numbers with the same
sign is positive.
Unlike signs The quotient of two nonzero real numbers with different
signs is negative.
(b)
(c)
15
3
(d)
8
11
16
O Which of the following
fractions
are equal to ^?
A.
3
5
B.
3
5
C.
3
5
D.
3
5
Answers
8. (a) A
^ (b)
4 (c) 5
«^
9. B,C
Copyright ^
EXAMPLE 6
Dividing Real Numbers
Find each quotient.
(a)
12
\
a ■
6/1, ,
(b) —  = 6 I — — ) = —2 The reciprocal of 3 is 5.
30 / 1
3 2 9 6
(d) — = = 
^ ^ 5 3 5 5
15
5 ;„ 9
The reciprocal of the denominator 5 is j.
This is a complex fraction (Section 8.3), a fraction that has a fraction in
the numerator, the denominator, or both.
Work Problem 8 at the Side.
The rules for multiplication and division suggest the following results.
Equivalent Forms of a Fraction
The fractions , , and — are equal. (Assume v i^ 0.)
y y y
Example:
4 _ 4 4
^ ~ ~ 7 ~ ^'
The fractions — and are equal.
y y
4 4
Example: — = .
^77
The forms ^ and ^ are not used very often.
Every fraction has three signs: the sign of the numerator, the sign of the
denominator, and the sign of the fraction itself Changing any two of these three
signs does not change the value of the fraction. Changing only one sign, or
changing all three, does change the value.
«i<
Work Problem 9 at the Side.
Copyright ©2005 Pearson Education, Inc., pubiisiiing as Pearson AddisonWesley
Section 1 .3 Exponents, Roots, and Order of Operations 25
1 .3 Exponents, Roots, and Order of Operations
Two or more numbers whose product is a third number are factors of that
third number. For example, 2 and 6 are factors of 12 since 26=12. Other
factors of 12 are 1, 3, 4, 12, 1, 2, 3, 4, 6, and 12.
OBJECTIVE Q Use exponents. In algebra, we use exponents as a way of
writing products of repeated factors. For example, the product 2 • 2 • 2 • 2 • 2
is written
2 • 2 • 2 • 2 • 2 = 2^
^ V '
5 factors of 2
The number 5 shows that 2 is used as a factor 5 times. The number 5 is the
exponent, and 2 is the base.
25 < — Exponent
' Base
Read 2^ as "2 to the fifth power" or simply "2 to the fifth." Multiplying out
the five 2s gives
25 = 2 . 2 • 2 • 2 • 2 = 32.
OBJECTIVES
Use exponents.
Identify exponents and
bases.
Find square roots.
Use the order of
operations.
Evaluate expressions for
given values of variables.
Write each expression using
exponents.
(a) 3 • 3 • 3 • 3 • 3
Exponential Expression
If a is a real number and w is a natural number,
a" = a * a * a
a,
n factors oia
where n is the exponent, a is the base, and a^ is an exponential expres
sion. Exponents are also called powers.
(b)
2 2 2 2
l' l' l' 1
(c) (10) (10) (10)
EXAMPLE 1
Using Exponential Notation
Write each expression using exponents.
(a) 4 • 4 • 4
Here, 4 is used as a factor 3 times, so
4 . 4 . 4 = 43
3 factors of 4
Read 4^ as "4 cubed."
3 3 _ /3^2
5'5 ~ V5,
Read (f)^ as "f squared."
(c) (6) (6) (6) (6) = (6)4
(d) (.3) (.3) (.3) (.3) (.3) = (.3)5
I CI Jv Jv Jv Jv Jv Jv Jv
(b)
2 factors of ^
Work Problem 1 at the Side.
(d)(.5)(.5)
(e)jjjjjjjj
2V
Answers
1. (a) V (b) yj (c) (10)3
(d) (.5)2 (e) y^
Copyright ©2005 Pearson Education, Inc., pubiisiiing as Pearson AddisonWesley
26 Chapter I Review of the Real Number System
& Write each expression
without exponents.
(a) 53
In parts (a) and (b) of Example 1, we used the terms squared and cubed
to refer to powers of 2 and 3, respectively. The term squared comes from the
figure of a square, which has the same measure for both length and width, as
shown in Figure 9(a). Similarly, the term cubed comes from the figure of a
cube. As shown in Figure 9(b), the length, width, and height of a cube have
the same measure.
(b)34
(c) (4)5
(d) (3)
^^4
(e) (.75)3
tou i^Y ly
Video!
(a) 33 = 3 squared, or 3^ (b) 6 • 6 • 6 — 6 cubed, or 6^
Figure 9
EXAMPLE 2
Evaluating Exponential Expressions
Write each expression without exponents.
(a) 5^ = 5 • 5 = 25 5 is used as a factor 2 times.
'2V 2 2 2 8 ,
— =—. — . — = —  is use^i as a factor 3 times.
,3/ 3 3 3 27 ^
(c) 2" = 2 • 2 • 2 • 2 • 2 • 2 = 64
(d) (2)4 = (2) (2) (2) (2) =16
(e) (3)5 = (3) (3) (3) (3) (3) = 243
(b)
Parts (d) and (e) of Example 2 suggest the following generalization.
The product of an even number of negative factors is positive.
The product of an odd number of negative factors is negative.
Work Problem 2 at the Side.
(f)
2\4
5.
Answers
2. (a) 125 (b) 81 (c) 1024 (d) 81
(e) .421875 (f) ^
lilil Calculator Tip Most calculators have a key labeled (^^ or (^
that can be used to raise a number to a power. See "An Introduction to
Calculators" at the beginning of this book for more information.
OBJ ECTI VE
EXAMPLE 3
Identify exponents and bases.
Identifying Exponents and Bases
Identify the exponent and the base. Then evaluate each expression.
(a) 3^ The exponent is 6, and the base is 3.
36 = 3 • 3 • 3 • 3 • 3 • 3 = 729
(b) 5"^ The exponent is 4, and the base is 5.
54 = 5 • 5 • 5 • 5 = 625
Continued on Next Page
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Section 1 .3 Exponents, Roots, and Order of Operations 27
(c) (2)"
The exponent 6 applies
to the number 
2, so
the base is —2.
{2f = {2){2){
2)(
■2)(
2)(
2) =
= 64 The base is 2.
(d) 2*'
Since there are no parentheses,
not to —2; the base is 2.
the
exponent 6 <
ipphes only to the number 2,
26= (22
•22
•2
2) =
64
The base is 2.
O Identify the exponent and the
base. Then evaluate each
expression.
(a) 73
CAUTION
As shown in Examples 3(c) and (d), it is important to distinguish between
a'^and(a)".
— a^ = — \{a ' a ' a ' ' ' a) The base is a.
n factors of <2
{—aY = { — a) { — a) • • • { — a) The base is a.
X
n factors of —a
(b) {5f
Work Problem 3 at the Side.
OBJECTIVE Q Find square roots. As we saw in Example 2(a),
5^ = 5 • 5 = 25, so 5 squared is 25. The opposite of squaring a number is
called taking its square root. For example, a square root of 25 is 5. Another
square root of 25 is —5 since (—5)^ = 25; thus, 25 has two square roots,
5 and —5.
We write the positive ox principal square root of a number with the sym
bol V , called a radical sign. For example, the positive or principal square
root of 25 is written V 25 = 5. The negative square root of 25 is written
— V 25 = — 5. Since the square of any nonzero real number is positive, the
square root of a negative number^ such as V — 25, is not a real number.
(c)
EXAMPLE 4
Finding Square Roots
Find each square root that is a real number.
(a) V 36 = 6 since 6 is positive and 6^ = 36.
(b) Vo = since 0^
0.
3 . /3V 9
(c),/ = smcef) =.
(d) V.16 = .4 since {Af = .16. (e) V 100 = 10 since 10^ = 100.
(f) — V 100 = — 10 since the negative sign is outside the radical sign.
(g) V— 100 is not a real number since the negative sign is inside the radical
sign. No real number squared equals — 100.
Notice the difference among the expressions in parts (e), (f), and (g).
Part (e) is the positive or principal square root of 100, part (f) is the negative
square root of 100, and part (g) is the square root of — 100, which is not a
real number.
(d) (.9)5
Answers
3. (a) 3; 7; 343 (b) 4; 5; 625
(c) 4; 5; 625 (d) 5; .9; .59049
Copyright ©2005 Pearson Education, Inc., pubiisiiing as Pearson AddisonWesley
28 Chapter I Review of the Real Number System
O Find each square root that is a
real number.
(a) V9
CAUTION
The symbol v is used only for the positive square root, except that
V = 0. The symbol — v is used for the negative square root.
Work Problem 4 at the Side.
(b) V49
lilil Calculator Tip Most calculators have a square root key, usually
labeled (^J , that allows us to find the square root of a number. On
some models, the square root key must be used in conjunction with the
key marked Qnv^ or (2nd^ .
(c) V81
(d)
121
81
(e) V.25
(f)
OBJECTIVE Q Use the order of operations. To simplify an expres
sion such as 5 + 2 • 3, what should we do first — add 5 and 2, or multiply
2 and 3? When an expression involves more than one operation symbol, we
use the following order of operations.
Order of Operations
1. Work separately above and below any fraction bar.
2. If grouping symbols such as parentheses ( ), square brackets [ ],
or absolute value bars   are present, start with the innermost set
and work outward.
3. Evaluate all powers, roots, and absolute values.
4. Do any multiplications or divisions in order, working from left to
right.
5. Do any additions or subtractions in order, working from left to
right.
(g) V^T69
Simplify.
(a) 5 • 9 + 2 • 4
(b) 4  12  4 • 2
Answers
Video!
Tryi
4. (a) 3 (b) 7 (c) 9 (d) — (e) .5
(f) not a real number
(g) not a real number
5. (a) 53 (b) 2
EXAMPLE 5
Using the Order of Operations
Simplify.
(a) 5 + 23 = 5 + 6 Multiply.
= 11 Add.
(b) 24  3 • 2 + 6
Multiplications and divisions are done in the order in which they appear
from left to right, so divide first.
24^32 + 6 = 82 + 6 Divide.
= 16 + 6 Multiply
= 22 Add.
Work Problem 5 at the Side.
EXAMPLE 6
Using the Order of Operations
Simplify.
(a) 4 • 32 + 7  (2 + 8)
Work inside the parentheses first.
Continued on Next Page
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Section 1 .3 Exponents, Roots, and Order of Operations 29
4 • 32 + 7  (2 + 8) = 4 • 32 + 7  10 Add inside parentheses.
= 49 + 7— 10 Evaluate powers.
= 36 + 710 Multiply.
= 4310 Add.
= 33 Subtract.
(b) ^ • 4 + (6 ^ 3  7)
Work inside the parentheses, dividing before subtracting.
— 4 + (6 r3 — 7)= — 4 + (2— 7) Divide inside parentheses.
1
4 + (— 5) Subtract inside parentheses.
z
= 2 + (5) Muhiply.
= 3 Add.
Work Problem 6 at the Side.
EXAMPLE 7
Simplify
Using the Order of Operations
5 + 2^
6V9 92'
5 + 2" 5 + 16
6V9 92 6392
5 + 16
Evaluate powers and roots.
Multiply.
Add and subtract.
18  18
 ^
Because division by is undefined, the given expression is undefined.
Work Problem 7 at the Side.
© Simplify.
(a) (4 + 2)  32  (8  3)
(b)6 + (9) ^16
O Simplify.
10  6 + 2V9
(a)
11 • 2  3(2)^
(b)
4(8) + 6(3)
3V49   42
2
lill Calculator Tip Most calculators follow^ the order of operations given
in this section. You may want to try some of the examples to see whether
your calculator gives the same answers. Use the parentheses keys to in
sert parentheses where they are needed. To work Example 7 with a cal
culator, put parentheses around the numerator and the denominator.
OBJECTIVE Q Evaluate expressions for given values of variables.
Any collection of numbers, variables, operation symbols, and grouping sym
bols, such as
6a/?, 5m — 9n, and — 2(x^ + 4j), Algebraic expressions
is called an algebraic expression. Algebraic expressions have different
numerical values for different values of the variables. We can evaluate such
expressions by substituting given values for the variables.
Algebraic expressions are used in problem solving. For example, if
movie tickets cost $8 each, the amount in dollars you pay for x tickets can be
represented by the algebraic expression 8x. We can substitute different num
bers of tickets to get the costs to purchase those tickets.
Answers
6. (a) 8 (b) 10
7. (a) 1 (b) undefined
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30 Chapter I Review of the Real Number System
O Evaluate each expression if
w = 4, X = — 12, J = 64, and
z= 3.
(a) 5x — 2w
(b) 6{x  V^)
EXAMPLE 8
Evaluating Expressions
Evaluate each expression ifm= —4, n = 5,p = —6, and q = 25.
(a) 5m9n = 5(4)  9(5) = 20  45
= 65
Replace m with —4
and K with 5.
HI + 2« _ 4 + 2(5) _ 4 + 10 _
^"^ 4p ~ 4(6) ~ 24
6
24 ~
1
4
(c) 3m'  n\Vq) = 3{4y  {5y{\
/25)
Substitute; m = —4,
« = 5, and q = 25.
= 3(64)  25(5)
= 192  125
= 67
Evaluate powers and
roots.
Multiply.
Subtract.
(c)
5x
3 • V^
y
X  1
CAUTION
To avoid errors when evaluating expressions, use parentheses around
any negative numbers that are substituted for variables.
IK
Work Problem 8 at the Side.
(d) w^ + 2z^
& Use the expression in
Example 9 to approximate
the percent of U.S. house
holds investing in mutual
funds in 1990 and 2000.
Round answers to the
nearest tenth.
Answers
8. (a) 
9. 1990: 26.0%; 2000: 48.0%
(b) 120 (c) ^ (d)
38
EXAMPLE 9
Evaluating an Expression to Approximate Mutual
Fund Investors
An approximation of the percent of U.S. households investing in mutual
funds during the years 1980 through 2000 can be obtained by substituting a
given year for x in the expression
2.2023X  4356.6
and then evaluating. {Source: Investment Company Institute.)
(a) Approximate the percent of U.S. households investing in mutual funds
in 1980. Round to the nearest tenth.
2.2023X  4356.6
2.2023(1980)
4.0
4356.6
Letx= 1980.
Use a calculator.
Recall that the symbol — means "is approximately equal to." In 1980, about
4.0% of U.S. households invested in mutual funds.
«i
Work Problem 9 at the Side.
(b) Give the results found above and in Problem 9 at the side in a table. How
has the percent of households investing in mutual funds changed during
these years?
The table follows. The percent of U.S. households investing in mutual
funds increased dramatically during these years.
Year
Percent of U.S. Households
Investing in Mutual Funds
1980
4.0
1990
26.0
2000
48
Percent in 2000 is
twelve times
percent in 1980.
Copyright ©2005 Pearson Education, Inc., pubiisiiing as Pearson AddisonWesley
Section 1.4 Properties of Real Numbers 35
1 .4 Properties of Real Numbers
The study of any object is simplified when we know the properties of the
object. For example, a property of water is that it freezes when cooled to
0°C. Knowing this helps us to predict the behavior of water.
The study of numbers is no different. The basic properties of real
numbers reflect results that occur consistently in work with numbers, so they
have been generalized to apply to expressions with variables as well.
OBJECTIVE Q Use the distributive property. Notice that
2(3 + 5) = 28 = 16
and
2 • 3 + 2 • 5 = 6 + 10 = 16,
so
2(3 + 5) = 23 + 25.
This idea is illustrated by the divided rectangle in Figure 10.
3 5
2
Area of left part is 2 • 3 = 6.
Area of right part is 2 • 5 = 10.
Area of total rectangle is 2(3 + 5) :
Figure 10
16.
Similarly,
and
so
4[5 + (3)] = 4(2) = 8
4(5) + (4) (3) = 20 + 12 = 8,
4[5 + (3)] = 4(5) + (4)(3).
OBJECTIVES
Q Use the distributive
property.
Q Use the inverse
properties.
Q Use the identity
properties.
Q Use the commutative and
associative properties.
Q Use the multiplication
property of 0.
These arithmetic examples are generalized to all real numbers as the
distributive property of multiplication with respect to addition, or simply
the distributive property.
Distributive Property
For any real numbers a, b, and c,
a{b + c) = ab \ ac and (b + c)a = ba + ca.
The distributive property can also be written
ab \ ac = a{b + c) and ba \ ca = {b \ c)a.
It can be extended to more than two numbers as well.
a(b + c \ d) = ab \ ac \ ad
This property is important because it provides a way to rewrite di product
a(/? + c) as a sum ab + ac, or a sum as di product.
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
36 Chapter I Review of the Real Number System
' Use the distributive property
to rewrite each expression.
(a) 8(m + w)
NOTE
When we rewrite a(b + c) as ab + ac, we sometimes refer to the
process as "removing" or "clearing" parentheses.
(b) 4ip  5)
(c) 3k + 6k
(d) —6m + 2m
(e) 2r + 3s
(f) 5(4p2q + r)
EXAMPLE 1
Using the Distributive Property
Use the distributive property to rewrite each expression.
(a) 3(x + y)
Use the first form of the property.
3(x \ y) = 3x \ 3y
(b)2(5 + ^)=2(5) + (2)(^)
= 102^
(c) 4x + 8x
Use the second form of the property.
4x + 8x = (4 + 8) X = 12x
(d) 3r  7r = 3r + (7r) Definition of subtraction
= [3 + (— 7)]r Distributive property
= 4r
(e) 5p + 7q ^
Because there is no common number or variable here, we cannot use the
distributive property to rewrite the expression.
(f) 6(x + 2y 3z) = 6x + 6(2y) + 6(3z)
= 6x + I2y  18z
As illustrated in Example 1(d), the distributive property can also be used
for subtraction, so
a(b — c) = ab — ac.
& Use the distributive property
to calculate each expression.
(a) 14 • 5 + 14 • 85
(b) 78 • 33 + 22 • 33
Work Problem 1 at the Side.
The distributive property can be used to mentally perform calculations.
EXAMPLE 2
Using the Distributive Property for Calculation
Calculate 38 • 17 + 38 • 3.
38 • 17 + 38 • 3 = 38(17 + 3) Distributive property
= 38(20)
= 760
<l<
Work Problem 2 at the Side.
Answers
1. (a) 8m + 8/2 (b) Ap + 20 (c) 9k
(d) —4m (e) cannot be rewritten
(f) lOp  \0q + 5r
2. (a) 1260 (b) 3300
OBJECTIVE Q Use the inverse properties. In Section 1.1 we saw
that the additive inverse of a number a is a and that the sum of a number
and its additive inverse is 0. For example, 3 and —3 are additive inverses, as
are —8 and 8. The number is its own additive inverse.
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
In Section 1.2, we saw that two numbers with a product of 1 are recipro
cals. As mentioned there, another name for reciprocal is multiplicative inverse.
This is similar to the idea of an additive inverse. Thus, 4 and \ are multiplica
tive inverses, as are —  and — . (Recall that reciprocals have the same sign.)
We can extend these properties of arithmetic, the inverse properties of addi
tion and multiplication, to the real numbers of algebra.
Inverse Properties
For any real number a, there is a single real number —a such that
a + {—a) = and — « + « = 0.
The inverse "undoes" addition with the result 0.
For any nonzero real number a, there is a single real number \ such that
a • — = 1 and —•«=!.
a a
The inverse "undoes" multiplication with the result 1.
Section 1.4 Properties of Real Numbers 37
O Complete each statement.
(a) 4 + =
(b) 7.1 +
(c) 9 + 9 =
(d) 5
Work Problem 3 at the Side.
OBJECTIVE Q Use the identity properties. The number can be
added to any number to get that number. That is, adding leaves the identity
of a number unchanged. Thus, is the identity element for addition or the
additive identity. Similarly, multiplying by 1 leaves the identity of any num
ber unchanged, so 1 is the identity element for multiplication or the
multiplicative identity. The following identity properties summarize this
discussion and extend these properties from arithmetic to algebra.
(e)
(f ) 7 •  =
Vid€o
Video
Identity Properties
For any real number a, « + = + « = «.
Start with a number a\ add 0. The answer is "identical" to a.
Also, « • 1 = 1 • « = «.
Start with a number a\ multiply by 1. The answer is "identical" to a.
EXAMPLE 3
a = a
Using the Identity Property 1
Simplify each expression.
(a) \2m + m = \2m + \m Identity property
= (12 + \)m Distributive property
= 13m Add inside parentheses.
(b) J + J = Ij + Ij Identity property
= (1 + l)j Distributive property
= 2y Add inside parentheses.
(c) —(m — 5n) = —l(m — 5n) Identity property
= —l(m) + (—l)(—5n) Distributive property
= —m \ 5n Multiply.
Work Problem 4 at the Side.
O Simplify each expression.
(a) p3p
(b) r + r + r
(c) (3 + 4p)
(d) (k  2)
Answers
3. (a) 4 (b) 7.1 (c) (d) \
4 ^
(e)  (f) 1
4. (a) 2p (b) 3r (c) 3  4p
(d) k + 2
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
38 Chapter I Review of the Real Number System
Expressions such as 12m and 5n from Example 3 are examples oi terms.
A term is a number or the product of a number and one or more variables.
Terms with exactly the same variables raised to exactly the same powers are
called like terms. Some examples of like terms are
5p and 2\p 6x^ and 9x^. Like terms
Some examples of unlike terms are
3m and 16x 7y^ and —3y^. Unhke terms
The numerical factor in a term is called the numerical coefficient, or just
the coefficient. For example, in the term 9x^, the coefficient is 9.
OBJECTIVE Q Use the commutative and associative properties.
Simplifying expressions as in parts (a) and (b) of Example 3 is called
combining like terms. Only like terms may be combined. To combine like
terms in an expression such as
— 2m + 5m + 3 — 6m + 8,
we need two more properties. From arithmetic, we know that
3 + 9 = 12 and 9 + 3 = 12.
Also,
3 • 9 = 27 and 9 • 3 = 27.
Furthermore, notice that
(5 + 7) + (2) = 12 + (2) = 10
and
Also,
and
5 + [7 + (2)] = 5 + 5 = 10.
(5 7) (2) = 35(2)= 70
(5)[7(2)] = 5(14)=70.
These arithmetic examples can now be extended to algebra.
Commutative and Associative Properties
For any real numbers a, b, and c,
a + b = b + a
Commutative properties
and ab = ba.
Interchange the order of the two terms or factors.
Also, a \ (b \ c) = (a \ b) \ c
Associative properties
and a (be) = (ab)c.
Shift parentheses among the three terms or factors; order stays the same.
The commutative properties are used to change the order of the terms
or factors in an expression. Think of commuting from home to work and
then from work to home. The associative properties are used to regroup the
terms or factors of an expression. Remember, to associate is to be part of a
group.
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
EXAMPLE 4
Using the Commutative and Associative Properties
Simplify — 2m + 5m + 3 — 6m + 8.
—2m + 5m + 3 — 6m + 8
= {—2m + 5m) + 3 — 6m + 8 Order of operations
= (— 2 + 5)/ii + 3 — 6m + 8 Distributive property
= 3m + 3 — 6m + 8
By the order of operations, the next step would be to add 3m and 3, but they
are unlike terms. To get 3m and —6m together, use the associative and com
mutative properties. Begin by inserting parentheses and brackets according
to the order of operations.
Section 1.4 Properties of Real Numbers 39
& Simplify each expression.
(a) \2b  9b + 4b  7b + b
[(3m + 3)  6/ii] + 8
= [3m + (3  6m)] + 8
= [3m + (6m + 3)] + 8
= [(3m + [6m]) + 3] + 8
= (3m + 3) + 8
= 3m + (3 + 8)
= 3m + 11
(b) 3w + 7  8w  2
Associative property
Commutative property
Associative property
Combine like terms.
Associative property
Add.
In practice, many of the steps are not vs^ritten dovs^n, but you should real
ize that the commutative and associative properties are used v^henever the
terms in an expression are rearranged and regrouped to combine like terms.
(c) 3(6 + 20
EXAMPLE 5
Using the Properties of Real Numbers
Simplify each expression.
(a) 5ySy6y+ Uy
= (586+ 11>
= 2y
(b) 3x + 4  5(x + 1)  8
= 3x + 45jc58
= 3x5x + 458
= 2x  9
(c) 8  (3m + 2) = 8  1 (3m + 2)
= 8  3m  2
= 6 — 3m
(d) (3x)(5)(>') = [(3x)(5)]>' Order of operations
= [3 (x • 5)]y Associative property
= [3(5x)]y
= [(3 • 5)x]y
= (I5x)y
= I5(xy)
= I5xy
As previously mentioned, many of these steps are not usually w^ritten out.
Distributive property
Combine like terms.
Distributive property
Commutative property
Combine like terms.
Identity property
Distributive property
Combine like terms.
Commutative property
Associative property
Multiply.
Associative property
(d)
2(a 3) + 4 a
(e) (4m) (2«)
Work Problem 5 at the Side.
Answers
5. (a) b (b) llw + 5 (c) 186?
(d) 19  3a (e) Smn
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
40 Chapter I Review of the Real Number System
© Complete each statement.
(a) 197 • =
CAUTION
Be careful. Notice that the distributive property does not apply in
Example 5(d), because there is no addition involved.
(3x)(5)(7)#(3x)(5).(3x)(7)
OBJECTIVE Q Use the multiplication property of 0_ The additive
identity property gives a special property of 0, namely that a + = a
for any real number a. The multiplication property of gives a special
property of that involves multiplication: The product of any real number
and is 0.
Multiplication Property of
For any real number a,
« • = and • « = 0.
(b)0
Work Problem 6 at the Side.
(c)
Answers
6. (a) (b) (c) any real number
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
Linear Equations
and Applications
\
Television, first operational in the 1940s, has become the
most widespread form of communication in the world. In 2003,
106.7 million homes, 98% of all U.S. households, owned at least
one TV set, and average viewing time among all viewers exceeded
30 hours per week. Favorite primetime television programs were CSI
and Friends, which concluded a highly successful 10year run with a
finale episode on May 6, 2004. {Source: Nielsen Media Research;
Microsoft Encarta Encyclopedia 2002)
In Section 2.2 we discuss the concept of percent — one of the
most common everyday applications of mathematics — and use it in
Exercises 4550 to determine additional information about televisions
in U.S. households.
2.T Linear Equations in One
Variable
2.2 Formulas
2.3 Applications of Linear
Equations
2.4 Further Applications of Linear
Equations
Summary Exercises on Solving
Applied Problems
53
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
54 Chapter 2 Linear Equations and Applications
2 • 1 Linear Equations in One Variable
OBJECTIVES
In the previous chapter we began to use algebraic expressions. Some examples
Q Decide whether a
number is a solution of a
linear equation.
Q Solve linear equations
using the addition and
multiplication properties
of equality.
Q Solve linear equations
using the distributive
property.
Q Solve linear equations
with fractions or
decimals.
Q Identify conditional
equations, contradictions,
and identities.
of algebraic expressions are
xY
8x + 9, y 4, and . Algebraic expressions
Equations and inequalities compare algebraic expressions, just as a
balance scale compares the weights of two quantities. Many applications of
mathematics lead to equations, statements that two algebraic expressions are
equal. A linear equation in one variable involves only real numbers and one
variable raised to the first power. Examples are
X + 1 = —2, X — 3 = 5, and 2k \ 5 = 10. Linear equations
It is important to be able to distinguish between algebraic expressions and
equations. An equation always contains an equals sign, while an expression
does not.
i
Study Skills Workbook
Activity 2: Your Textbook
Are the given numbers
solutions of the given
equations?
(a) 3^= 15; 5
Linear Equation in One Variable
A linear equation in one variable can be written in the form
Ax + B = Q
where A, B, and C are real numbers, with A y^ 0.
A linear equation is also called a firstdegree equation since the greatest
power on the variable is one. Some examples of equations that are not linear
(that is, nonlinear) are
x^ + 3y = 5, =22, and Vx = 6.
Nonlinear equations
(b) r + 5 = 4; 1
OBJECTIVE Q Decide whether a number is a solution of a linear
equation. If the variable in an equation can be replaced by a real number
that makes the statement true, then that number is a solution of the equation.
For example, 8 is a solution of the equation x — 3 = 5, since replacing x
with 8 gives a true statement, 8  3 = 5. An equation is solved by finding
its solution set, the set of all solutions. The solution set of the equation
X 3 = 5 is {8}.
Hi
Work Problem 1 at the Side.
(c) 8m = 12;
Equivalent equations are equations that have the same solution set. To
solve an equation, we usually start with the given equation and replace it
with a series of simpler equivalent equations. For example,
5x + 2 = 17, 5x = 15, and x = 3 Equivalent equations
are all equivalent since each has the solution set {3}.
Answers
1. (a) yes (b) no (c) no
OBJECTIVE Q Solve linear equations using the addition and multi
plication properties of equality. Two important properties that are used
in producing equivalent equations are the addition property of equality
and the multiplication property of equality.
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
Section!. I Linear Equations in One Variable 55
Addition and Multiplication Properties of Equality
Addition Property of Equality
For all real numbers A, B, and C, the equations
A= B and A + C = B + C
are equivalent.
In words, the same number may be added to each side of an equa
tion without changing the solution set.
Multiplication Property of Equality
For all real numbers A and B, and for C 9^ 0, the equations
A= B and AC = BC
are equivalent.
In words, each side of an equation may be multiplied by the same
nonzero number without changing the solution set.
& Solve and check.
(a) 3p + 2p+ 1 = 24
(b) 3p = 2p + 4p + 5
Video
Because subtraction and division are defined in terms of addition and
multiplication, respectively, these properties can be extended:
The same number may be subtracted from each side of an equation,
and each side of an equation may be divided by the same nonzero
number, without changing the solution set.
I^r
EXAMPLE 1
Using the Addition and Multiplication Properties
to Solve a Linear Equation
Solve 4x  2x  5 = 4 + 6x + 3.
The goal is to get x alone on one side of the equation. First, combine like
terms on each side of the equation to get
2x — 5 = 7 + 6x.
Next, use the addition property to get the terms with x on the same side of
the equation and the remaining terms (the numbers) on the other side. One
way to do this is to first subtract 6x from each side.
(c) 4x + 8x = 17x  9  1
2x
Ax
5  6x
Ax 5
5+5=7+5
4x = 12
4x 12
7 + 6x — 6x
7
4 4
X = —3
Subtract 6x.
Combine like terms.
Add 5.
Combine like terms.
Divide by —4.
Proposed solution
(d) 7 + 3t9t= lit  5
Check by substituting — 3 for x in the original equation.
Check: 4x  2x  5 = 4 + 6x + 3
4(3)  2(3) 5 = 4 + 6(3) + 3
12 + 6  5 = 4 18 + 3
11 = 11
Original equation
Letx = —3.
Multiply.
True
The true statement indicates that { — 3} is the solution set.
Work Problem 2 at the Side.
Answers
2. (a) {5} (b)
. 1
(d)
(c) {2}
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
56 Chapter 2 Linear Equations and Applications
® Solve and check.
(a) 5p + 4(3  2p)
= 2+;? 10
(b) 3(z2) + 5z = 2
The steps to solve a linear equation in one variable are as follows.
Solving a Linear Equation in One Variable
Step 1 Clear fractions. Eliminate any fractions by multiplying each
side by the least common denominator.
Step 2 Simplify each side separately. Use the distributive property to
clear parentheses and combine like terms as needed.
Step 3 Isolate the variable terms on one side. Use the addition prop
erty to get all terms with variables on one side of the equation
and all numbers on the other.
Step 4 Isolate the variable. Use the multiplication property to get an
equation with just the variable (with coefficient 1) on one side.
Step 5 Check. Substitute the proposed solution into the original equation.
Video Afii mation You Try
(c) 2 + 3(x + 4) = 8x
(d) 6  (4 + m)
= 8m  2(3m + 5)
Answers
3. (a) {5} (b) {1} (c) {2} (d) {4}
OBJECTIVE Q Solve linear equations using the distributive property.
In Example 1 we did not use Step 1 or the distributive property in Step 2 as
given in the box. Many equations require one or both of these steps.
EXAMPLE 2
Using the Distributive Property to Solve a Linear
Equation
Solve 2(^ 5) + 3^ = ^+ 6.
Step 1 Since there are no fractions in this equation, Step 1 does not apply.
Step 2 Use the distributive property to simplify and combine terms on the
left side of the equation.
2(A: 5) + 3^=^+ 6
2A:  10 + 3^ = ^ + 6 2{k  5) = 2(k)  2(5) = 2^10
5k — 10 = ^+6 Combine like terms.
Step 3 Next, use the addition property of equality.
5kl0k = k+6k Subtract L
4k — 10 = 6 Combine like terms.
4^  10 + 10 = 6 + 10 Add 10.
4^ = 16 Combine like terms.
Step 4 Use the multiplication property of equality to get just k on the left.
Ak 16
— = — Divide by 4.
4 4
A: = 4
Step 5 Check by substituting 4 for k in the original equation.
Check: 2 (/r — 5) + 3A: = A: + 6 Original equation
2(4  5) + 3(4) = 4 + 6 ? Let y^ = 4.
2(l)+ 12 = 10 ?
10 = 10 True
The solution checks, so the solution set is {4}.
IK
Work Problem J at the Side.
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
Section!. I Linear Equations in One Variable 57
NOTE
Notice in Examples 1 and 2 that the equals signs are aligned in columns.
Do not use more than one equals sign in a horizontal line of work when
solving an equation.
OBJECTIVE Q Solve linear equations with fractions or decimals.
When fractions or decimals appear as coefficients in equations, our work
can be made easier if we multiply each side of the equation by the least com
mon denominator (LCD) of all the fractions. This is an application of the
multiplication property of equality, and it produces an equivalent equation
with integer coefficients.
Solve and check.
2p
(a)
P_
2
EXAMPLE 3
Solving a Linear Equation with Fractions
X + 7 2x
Solve — — +
6
Step 1 Start by eliminating the fractions. Multiply each side by the LCD, 6.
Step 2 6(^^^l + 6
/x + 7 2x  8
61^ + ^
2x  8
6(4)
6(4)
Distributive property
Multiply.
Distributive property
Multiply
Combine like terms.
6 / V 2
(x + 7) + 3(2x 8) = 24
x + 7 + 3(2x) 3(8) = 24
X + 7 + 6x  24 = 24
7x  17 = 24
7x  17 + 17 = 24 + 17 Add 17.
7x = — 7 Combine like terms.
Ix _ 1
y ~ ^
x= 1
Step 5 Check by substituting — 1 for x in the original equation
Step 3
Step 4
Divide by 7.
x + 7 2x 8
6^2 =
4
Original equation
1 + 7 2(l)  8 _
6 2
4
? Letx=1.
6 10
6^ 2 '
4
7
15 = 4 ?
—4 = —4 True
The solution checks, so the solution set is { — 1 }.
^ + 1 k + 3
(b) —r + —r
Work Problem 4 at the Side.
Answers
4. (a) {14} (b) {1}
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
58 Chapter 2 Linear Equations and Applications
& Solve and check.
(a) Mx + .06(20 x)
= .05(50)
In Sections 2.2 and 2.3 we solve problems involving interest rates and
concentrations of solutions. These problems involve percents that are
converted to decimals. The equations that are used to solve such problems
involve decimal coefficients. We can clear these decimals by multiplying by
a power of 10, such as 10^ = 10, 10^ = 100, and so on, that will allow us to
obtain integer coefficients.
(b) .10(x 6) + .05x
= .06(50)
EXAMPLE 4
Solving a Linear Equation with Decimals
Solve .06x + .09(15  x) = .07(15).
Because each decimal number is given in hundredths, multiply each side
of the equation by 100. A number can be multiplied by 100 by moving the
decimal point two places to the right.
.06x + .09(15 x) = .07(15)
.06x + .09(15 x) = .07(15)
6x + 9(15 x) = 7(15)
6x + 9(15)9(x) = 7(15)
6x + 135  9x = 105
3x + 135 = 105
3x + 135  135 = 105  135
3x = 30
3x 30
3 3
x= 10
Check by substituting 10 for x in the original equation.
Check: .06x + .09(15  x)
.06(10) + .09(15  10)
.06(10) + .09(5)
.6 + .45
Multiply by 100.
Distributive property
Multiply.
Combine like terms.
Subtract 135.
Combine like terms.
Divide by —3.
.07(15)
Original equation
.07(15)
? Letx=10.
.07(15)
7
1.05
7
1.05 = 1.05
The solution set is {10}.
Mi
True
Work Problem 5 at the Side.
Answers
5. (a) {65} (b) {24}
NOTE
Because of space limitations, we will not always show the check when
solving an equation. To be sure that your solution is correct, you
should always check your work.
OBJECTIVE Q Identify conditional equations, contradictions, and
identities. All of the preceding equations had solution sets containing one
element; for example, 2 (^ — 5) + 3^ = ^ + 6 has solution set {4}. Some linear
equations, however, have no solutions, while others have an infinite number
of solutions. The table on the next page gives the names of these types of
equations.
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
Section!. I Linear Equations in One Variable 59
Type of Linear
Equation
Number of Solutions
Indication When Solving ^^
Conditional
One
Final line is x = a number.
(See Example 5(a).)
Contradiction
None; solution set
Final line is false, such as — 15 = —20.
(See Example 5(c).)
Identity
Infinite; solution set
{all real numbers}
Final line is true, such as = 0.
(See Example 5(b).)
© Solve each equation. Decide
whether it is a conditional
equation, an identity, or a
contradiction. Give the
solution set.
(a) 5(x + 2)2(x+ 1)
= 3x+ 1
EXAMPLE 5
Recognizing Conditional Equations, Identities,
and Contradictions
Solve each equation. Decide whether it is a conditional equation, an identity,
or a contradiction.
(a) 5x  9 = 4(x  3)
5x  9 = 4x  12
5x  9  4x = 4x  12 
X 9 = 12
x9 + 9=12 + 9
x= 3
Distributive property
4x Subtract Ax.
Combine like terms.
Add 9.
9 = 4(x  3) is a
The solution set, { — 3}, has only one element, so 5x
conditional equation.
(b) 5x 15 = 5(x 3)
Use the distributive property to clear parentheses on the right side.
5x 15
5x 15 = 5(x 3)
5x— 15 = 5x— 15
5x + 15 = 5x  15 
=
Distributive property
5x + 15 Subtract 5x; add 15.
True
The final line, the true statement = 0, indicates that the solution set is {all
real numbers}, and the equation 5x  15 = 5(x  3) is an identity. (Notice
that the first step yielded 5x — 15 = 5x — 15, which is true for all values of
X. We could have identified the equation as an identity at that point.)
(c)
5x
Since the result, —15 = —20, \s false, the equation has no solution. The
solution set is j2f, so the equation 5x — 15 = 5(x — 4) is a contradiction.
Work Problem 6 at the Side.
5x
 15 = 5(x4)
5x
 15 = 5x  20
Distributive property
15
 5x = 5x  20 
5x
Subtract 5x.
15= 20
False
X + 1 2x
(b) + —
^ ^ 3 3
X +
(c) 5(3x+ 1) = x + 5
Answers
6. (a) contradiction;
(b) identity; {all real numbers}
(c) conditional; {0}
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
Section 2.2 Formulas
65
ii.ii Formulas
A mathematical model is an equation or inequality that describes a real
situation. Models for many applied problems already exist; they are called
formulas. A formula is a mathematical equation in which variables are used
to describe a relationship. Some formulas that we will be using are
d = rt, I = prt, and P = 2L + 2W. Formulas
A list of some common formulas used in algebra is given inside the covers
of this book.
OBJECTIVE Q Solve a formula for a specified variable. In some
applications, the appropriate formula may be solved for a different variable
than the one to be found. For example, the formula / = prt says that interest
on a loan or investment equals principal (amount borrowed or invested)
times rate (percent) times time at interest (in years). To determine how long
it will take for an investment at a stated interest rate to earn a predetermined
amount of interest, it would help to first solve the formula for t. This process
is called solving for a specified variable or solving a literal equation.
OBJECTIVES
Solve a formula for a
specified variable.
Solve applied problems
using formulas.
Solve percent problems.
Solve / =
variable.
(a)/p
prt for each given
The steps used in the following examples are very similar to those used
in solving linear equations from Section 2.1. When you are solving for a
specified variable^ the key is to treat that variable as if it were the only one;
treat all other variables like numbers (constants).
(b)r
EXAMPLE 1
Solving for a Specified Variable
Solve the formula / = prt for t.
We solve this formula for t by treating /, p, and r as constants (having
fixed values) and treating t as the only variable. We first write the formula
so that the variable for which we are solving, t, is on the left side. Then we
use the properties of the previous section as follows.
prt = /
{pr)t = I
{pr)t ^ l_
pr
pr
Associative property
Divide by />r.
I
t = —
pr
The result is a formula for t, time in years.
Work Problem 1 at the Side.
Answers
1. (a) p 
(b) r
pt
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
66 Chapter 2 Linear Equations and Applications
& (a) Solve the formula
P = a + b + c
for a.
To solve for a specified variable, follow these steps.
Solving for a Specified Variable
Step 1 Transform so that all terms containing the specified variable are
on one side of the equation and all terms without that variable
are on the other side.
Step 2 If necessary, use the distributive property to combine the terms
with the specified variable.* The result should be the product
of a sum or difference and the variable.
Step 3 Divide each side by the factor that is the coefficient of the
specified variable.
IMi^ B Solve the formuh. _ _ ___
^mU^ This formula gives the relationship between perimeter of a rectangle, P,
/^ V ' length of the rectangle, Z, and width of the rectangle, W. See Figure 1.
(b) Solve the formula
m = 2k + ?>b
fork.
Solving for a Specified Variable
formula^ = 2Z + 2^for W.
w
Perimeter, P, distance around a
W rectangle, is given by
P = 2L + 2W.
Figure 1
Solve the formula for Why isolating W on one side of the equals sign.
To begin, subtract 2L from each side.
P = 2L + 2W
Step 1 P  2L = 2L + 2W 2L Subtract 2L.
P 2L = 2W
Step 2 is not needed here.
P  2L 2W
= Divide by 2.
2 2
P 2L P 2L
Step 3
= W or W =
2 2
Mi
Work Problem 2 at the Side.
CAUTION
In Step 3 of Example 2, you cannot simplify the fraction by dividing 2
into the term 2L. The subtraction in the numerator must be done before
the division.
P  2L
*P  L
Answers
1. {2i) a = P h c (b) ^ =
m — ?)b
*Using the distributive property to write ab + ac diS a{b + c) is cdM^d factoring. See Chapter 7.
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
Section 2.2 Formulas 67
EXAMPLE 3
Solving a Formula with Parentheses
The formula for the perimeter of a rectangle is sometimes written in the
equivalent form P = 2(L + W). Solve this form for W.
One way to begin is to use the distributive property on the right side of the
equation to get P = IL + IW, which we would then solve as in Example 2.
Another way to begin is to divide by the coefficient 2.
P
2
P = 2(L+ W)
P
 = L+ W
2
W or W
P
2"
Divide by 2.
Subtract L.
We can show that this result is equivalent to our result in Example 2 by
multiplying L by .
^^{L) = W =l,soL = i(L).
P_2L
2~ Y
P 2L
W
w
Subtract fractions.
The final line agrees with the result in Example 2.
Work Problem 3 at the Side.
& Solve the formula
y = {x + ?>)
forx.
A rectangular solid has the shape of a box, but is solid. See Figure 2. The
labels H, W, and L represent the height, width, and length of the figure,
respectively. The surface area of any solid threedimensional figure is the
total area of its surface. For a rectangular solid, the surface area^ is
2HW + 2LW + 2LH.
EXAMPLE 4
Using the Distributive Property to Solve for a
Specified Variable
Given the surface area, height, and width of a rectangular solid, write a
formula for the length.
To solve for the length Z, treat L as the only variable and treat all other
variables as constants.
A = 2HW+ 2LW+ 2LH
A 2HW= 2LW+ 2LH
A  2HW= L(2W+ 2H)
A  2HW A  2HW
, or 1
2W+ 2H
2W + 2H
Subtract 2HW.
Distributive property
Divide by 2^ +2//.
J I
^
Figure 2
Solve the formula
A = 2HW + 2LW + 2LH
ioxW.
CAUTION
Be careful when working a problem like Example 4 to use the distributive
property correctly. We must write the expression so that the specified vari
able is ?i factor; then we can divide by its coefficient in the final step.
Work Problem 4 at the Side.
Answers
3. X = 2j  3
A  2LH
4. W=
2H + IL
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
68 Chapter 2 Linear Equations and Applications
© Solve each problem.
(a) A triangle has an area of
36 in.^ (square inches) and
a base of 12 in. Find its
height.
(b) The distance is 500 mi and
the time is 20 hr. Find the
rate.
OBJECTIVE Q Solve applied problems using formulas. The distance
formula, d = rt, relates d, the distance traveled, r, the rate or speed, and t,
the travel time.
EXAMPLE 5
Finding Average Speed
Janet Branson found that on average it took her  hr each day to drive a
distance of 15 mi to work. What was her average speed?
Find the speed r by solving d = rt for r.
d
= rt
d _
t
rt
t
Divide by t
d
t
 r
d
or r = 
t
Notice that only Step 3 was needed to solve for r in this example. Now find
the speed by substituting the given values of (i and t into this formula.
15
r =
Let J= 15, r
15
r = 20
4
3
Multiply by the reciprocal of 4.
Her average speed was 20 mph. (That is, at times she may have traveled
a little faster or slower than 20 mph, but overall her speed was 20 mph.)
m
Work Problem 5 at the Side.
J. f
Video!
(c) In 2003, Gil de Ferran won
the Indianapolis 500 (mile)
race with a speed of
156.291 mph. {Source:
World Almanac and Book
o/Fac^^, 2004.) Find his
time to the nearest
thousandth.
OBJECTIVE Q Solve percent problems. An important everyday use
of mathematics involves the concept of percent. Percent is written with the
symbol %. The word percent means "per one hundred." One percent means
"one per one hundred" or "one one hundredth."
1% = .01 or 1% =
1
100
Solving a Percent Problem
Let a represent a partial amount of/?, the base, or whole amount. Then
the following formula can be used in solving a percent problem.
amount a , ,
= — = percent (represented as a decimal)
base b
For example, if a class consists of 50 students and 32 are males, then the
percent of males in the class is
amount _ a _?>2
base b 50
= .64
= 64%.
Let a = 32,b = 50.
Answers
5. (a) 6 in.
(b) 25 mph (c) 3.199 hr
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
Section 2.2 Formulas
69
EXAMPLE 6
Solving Percent Problems
(a) A 50L mixture of acid and water contains 10 L of acid. What is the per
cent of acid in the mixture?
The given amount of the mixture is 50 L, and the part that is acid
is 10 L. Let X represent the percent of acid. Then, the percent of acid in the
mixture is
11
50
.20 or 20%.
(b) If a savings account balance of $3550 earns 8% interest in one year, how
much interest is earned?
Let X represent the amount of interest earned (that is, the part of the
whole amount invested). Since 8% = .08, the equation is
.08
= percent
3550
X = .08 (3550) Multiply by 3550.
X = 284.
The interest earned is $284.
Work Problem 6 at the Side.
EXAMPLE 7
Interpreting Percents from a Graph
In 2003, people in the United States spent an estimated $29.7 billion on their
pets. Use the graph in Figure 3 to determine how much of this amount was
spent on pet food.
Spending on Kitty and Rover
Vet care
22.6%
Grooming/
boarding
8.0%
Supplies/
medicine
25.6%
Food
43.8%
Source: American Pet Products Manufacturers
Association Inc.
Figure 3
According to the graph, 43.8% was spent on food. Let x represent this
amount in billions of dollars.
29.7
.438
43.8% = .438
X = .438 (29.7) Multiply by 29.7.
X = 13.0 Nearest tenth
Therefore, about $13.0 billion was spent on pet food.
Work Problem 7 at the Side.
© Solve each problem.
(a) A mixture of gasoline and
oil contains 20 oz, 1 oz of
which is oil. What percent
of the mixture is oil?
(b) An automobile salesman
earns an 8% commission on
every car he sells. How
much does he earn on a car
that sells for $12,000?
& Refer to Figure 3. How
much was spent on pet
supplies/medicine? Round
your answer to the nearest
tenth.
Answers
6. (a) 5% (b) $960
7. $7.6 billion
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
Section 2.3 Applications of Linear Equations 77
2.3 Applications of Linear Equations
OBJECTIVE Q Translate from words to mathematical expressions.
Producing a mathematical model of a real situation often involves translat
ing verbal statements into mathematical statements. Although the problems
we will be working with are simple ones, the methods we use will also apply
to more difficult problems later.
PROBLEMSOLVING HINT
Usually there are key words and phrases in a verbal problem that trans
late into mathematical expressions involving addition, subtraction,
multiplication, and division. Translations of some commonly used ex
pressions follow.
TRANSLATING FROM WORDS TO MATHEMATICAL EXPRESSIONS
Verbal Expression
Mathematical Expression
(where x and y are numbers)
Addition
The sum of a number and 7
6 more than a number
3 plus a number
24 added to a number
A number increased by 5
The sum of two numbers
x + 7
x + 6
3 +x
x + 24
x + 5
Subtraction
2 less than a number
12 minus a number
A number decreased by 12
A number subtracted from 10
The difference between two numbers
x2
12 X
X 12
10 X
X — y
Multiplication
1 6 times a number
A number multiplied by 6
2
— of a number (used with
fractions and percent)
Twice (2 times) a number
The product of two numbers
16x
6x
2
3^
2x
xy
Division
The quotient of 8 and a number
A number divided by 1 3
The ratio of two numbers or the
quotient of two numbers
X
U
'(y^O)
OBJECTIVES
Translate from words
to mathematical
expressions.
Write equations from
given information.
Distinguish between
expressions and
equations.
Use the six steps in solv
ing an applied problem.
Solve percent problems.
Solve investment
problems.
Solve mixture problems.
J^ Study Skills Workbook
I Activity 4: Note Taking
Translate each verbal
expression as a mathematical
expression. Use x as the
variable.
(a) 9 added to a number
(b) The difference between
7 and a number
(c) Four times a number
(d) The quotient of 7 and a
nonzero number
Work Problem 1 at the Side.
Answers
1. (a) 9 +xorx + 9 (b) 7
(c) 4x (d) 
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
78 Chapter 2 Linear Equations and Applications
® Translate each verbal
sentence into an equation.
Use X as the variable.
(a) The sum of a number and
6 is 28.
CAUTION
Because subtraction and division are not commutative operations, be
careful to correctly translate expressions involving them. For example,
"2 less than a number" is translated as x  2, not 2  x. "A number sub
tracted from 10" is expressed as 10 — x, notx — 10.
For division, the number by which we are dividing is the denomina
tor, and the number into which we are dividing is the numerator. For ex
ample, "a number divided by 13" and "13 divided into x" both translate
as f^. Similarly, "the quotient of x and j" is translated as ^.
(b) If twice a number is
decreased by 3, the result
is 17.
OBJECTIVE Q Write equations from given information. The symbol
for equality, =, is often indicated by the word is. In fact, any words that in
dicate the idea of "sameness" translate to =.
(c) The product of a number
and 7 is twice the number
plus 12.
(d) The quotient of a number
and 6, added to twice the
number, is 7.
EXAMPLE 1
Translating Words into Equations
Translate each verbal sentence into an equation.
Verbal Sentence
Equation
Twice a number, decreased by 3, is 42.
2x  3 = 42
If the product of a number and 12 is decreased
by 7, the result is 105.
12x  7 = 105
The quotient of a number and the number
^ _.o
plus 4 is 28.
— Zo
X + 4
The quotient of a number and 4, plus
the number, is 10.
^ + x= 10
4
Work Problem 2 at the Side.
OBJECTIVE Q Distinguish! between expressions and equations. An
expression translates as a phrase. An equation includes the = symbol and
translates as a sentence.
& Decide whether each is an
expression or an equation.
(a) 5x 3(x + 2) = 7
(b) 5x 3(x + 2)
EXAMPLE 2
Distinguishing between Expressions and Equations
Decide whether each is an expression or an equation.
(a) 2(3 + x)4x + 7
There is no equals sign, so this is an expression.
(b) 2(3 + x)4x + 7= 1
Because of the equals sign, this is an equation.
Note that the expression in part (a) simplifies to the expression — 2x + 13,
and the equation in part (b) has solution 7.
Hi
Work Problem 3 at the Side.
Answers
2. (a) X + 6 = 28 (b) 2x  3 = 17
(c) 7x = 2x + 12 (d) 7 + 2x = '
6
3. (a) equation (b) expression
OBJECTIVE Q Use thie six steps in solving an applied problem.
While there is no one method that will allow us to solve all types of applied
problems, the following six steps are helpful.*
* Appendix A Strategies for Problem Solving introduces additional methods and tips for solving applied
problems.
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
Section 2.3 Applications of Linear Equations 79
Solving an Applied Problem
Step 1 Read the problem, several times if necessary, until you
understand what is given and what is to be found.
Step 2 Assign a variable to represent the unknown value, using
diagrams or tables as needed. Write down what the variable
represents. Express any other unknown values in terms of
the variable.
Step 3 Write an equation using the variable expression(s).
Step 4 Solve the equation.
Step 5 State the answer to the problem. Does it seem reasonable?
Step 6 Check the answer in the words of the original problem.
O Solve the problem.
The length of a rectangle is
5 cm more than its width. The
perimeter is five times the
width. What are the dimensions
of the rectangle?
EXAMPLE 3
Solving a Geometry Problem
The length of a rectangle is 1 cm more than twice the width. The perimeter
of the rectangle is 1 10 cm. Find the length and the width of the rectangle.
Step 1 Read the problem. We must find the length and width of the rectan
gle. The length is 1 cm more than twice the width, and the perime
ter is 110 cm.
Step 2 Assign a variable. Let W = the width; then 1 + 2^ = the length.
Make a sketch, as in Figure 4.
w
1 + 2W
Figure 4
Step 3 Write an equation. The perimeter of a rectangle is given by the
formula P = 2Z + 2W.
P = 2L + 2W
110 = 2(1 + 2W) + 2W
Step 4 Solve the equation obtained in Step 3.
110 = 2 + 4^+ 2W
110 = 2 + 6W
1102 = 2 + 6W1
108 = 6W
108 _ 6W
6 ~ 6
18 = ^
LetL= 1 +2^and/^= 110.
Distributive property
Combine like terms.
Subtract 2.
Divide by 6.
Step 5 State the answer. The width of the rectangle is 18 cm and the
length is 1 + 2(18) = 37 cm.
Step 6 Check the answer by substituting these dimensions into the words
of the original problem.
Work Problem 4 at the Side.
Answers
4. width: 10 cm; length: 15 cm
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
80 Chapter 2 Linear Equations and Applications
Solve the problem.
At the end of the 2003
baseball season, Sammy Sosa
and Barry Bonds had a lifetime
total of 1 197 home runs. Bonds
had 119 more than Sosa. How
many home runs did each player
have? (Source: World Almanac
and Book of Facts, 2004.)
EXAMPLE 4
Finding Unknown Numerical Quantities
Two outstanding major league pitchers in re
cent years are Randy Johnson and Pedro
Martinez. In 2002, they combined for a total
of 573 strikeouts. Johnson had 95 more
strikeouts than Martinez. How many strike
outs did each pitcher have? {Source: World
Almanac and Book of Facts, 2004.)
Step 1 Read the problem. We are asked to
find the number of strikeouts each
pitcher had.
Step 2 Assign a variable to represent the number of strikeouts for one of
the men.
Let s = the number of strikeouts for Pedro Martinez.
We must also find the number of strikeouts for Randy Johnson.
Since he had 95 more strikeouts than Martinez,
Steps
1^
AnlniatlQf
5* + 95 = the number of strikeouts for Johnson.
Write an equation. The sum of the numbers of strikeouts is 573, so
+
Martinez's strikeouts
+
Johnson's strikeouts
(s + 95)
Total
I
573.
Step 4 Solve the equation.
^ + (^ + 95) = 573
2^ + 95 = 573
2^ + 95  95 = 573  95
Is = 478
2s _ 478
2 ~ 2
Combine like terms.
Subtract 95.
Divide by 2.
^ = 239
Step 5 State the answer. We let s represent the number of strikeouts for
Martinez, so Martinez had 239. Also,
5 + 95 = 239 + 95 = 334
is the number of strikeouts for Johnson.
Step 6 Check. 334 is 95 more than 239, and the sum of 239 and 334 is 573.
The conditions of the problem are satisfied, and our answer checks.
CAUTION
A common error in solving applied problems is forgetting to answer all
the questions asked in the problem. In Example 4, we were asked for the
number of strikeouts each player had, so there was an extra step at the
end in order to find the number Johnson had.
Answers
5. Sosa: 539; Bonds: 658
IK
Work Problem 5 at the Side.
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
Section 2.3 Applications of Linear Equations 8 I
Video
OBJECTIVE Q Solve percent problems. Recall from Section 2.2 that
percent means "per one hundred," so 5% means .05, 14% means .14, and so on.
EXAMPLE 5
Solving a Percent Problem
In 2002 there were 301 longdistance area codes in the United States. This
was an increase of 250% over the number when the area code plan originated
in 1947. How many area codes were there in 1947? (Source: SBC Telephone
Directory.)
Step 1
Read the problem. We are given that the number of area codes
increased by 250% from 1947 to 2002, and there were 301 area
codes in 2002. We must find the original number of area codes.
Step 2
Assign a variable. Let x represent the number of area codes in
1947.
250% = 250(.01) = 2.5,
so 2.5x represents the number of codes added since then.
Step 3
Write an equation from the given information.
the number in 1947 + the increase = 301
X + 2.5x = 301
Step 4
Solve the equation.
Ix + 2.5x = 301 Identity property
3.5x = 301 Combine like terms.
X = 86 Divide by 3.5.
Step 5
State the answer. There were 86 area codes in 1947.
Step 6
Check that the increase, 301  86 = 215, is 250% of 86.
CAUTION
Avoid two common errors that occur in solving problems like the one in
Example 5.
1. Do not try to find 250% of 301 and subtract that amount from 301.
The 250% should be applied to the amount in 1947, not the amount
in 2002.
Do not write the equation as
x + 2.5 = 301.
Incorrect
The percent must be multiplied by some amount; in this case, the
amount is the number of area codes in 1947, giving 2.5x.
Work Problem 6 at the Side.
© Solve each problem.
(a) A number increased by
15% is 287.5. Find the
number.
(b) Michelle Raymond was
paid $162 for a week's
work at her parttime job
after 10% deductions for
taxes. How much did she
make before the deduc
tions were made?
OBJECTIVE Q Solve investment problems. We use linear equations
to solve certain investment problems. The investment problems in this chap
ter deal with simple interest. In most realworld applications, compound
interest (covered in a later chapter) is used.
Answers
6. (a) 250 (b) $180
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
82 Chapter 2 Linear Equations and Applications
O Solve each problem.
(a) A woman invests $72,000
in two ways — some at
5% and some at 3%. Her
total annual interest in
come is $3160. Find the
amount she invests at
each rate.
EXAMPLE 6
Solving an Investment Problem
(b) A man has $34,000 to
invest. He invests some at
5% and the balance at 4%.
His total annual interest
income is $1545. Find the
amount he invests at
each rate.
After winning the state lottery, Mark LeBeau has $40,000 to invest. He will
put part of the money in an account paying 4% interest and the remainder
into stocks paying 6% interest. His accountant tells him that the total annual
income from these investments should be $2040. How much should he
invest at each rate?
Step 1
Step 2
Read the problem again. We must find the two amounts.
Assign a variable.
Let X = the amount to invest at 4%;
40,000 — X = the amount to invest at 6%.
The formula for interest is / = prt. Here the time, t, is 1 year. Make
a table to organize the given information.
Rate (as
a decimal)
Principal
Interest
.04
X
.04x
.06
40,000  X
.06(40,000  x)
40,000
2040
 Totals
Step 3 Write an equation. The last column of the table gives the equation.
interest at 4% + interest at 6%
.04x + .06(40,000 x)
= total interest
2040
Step 4 Solve the equation. We do so without clearing decimals.
.04x + .06(40,000)  .06x = 2040
Distributive property
.04x + 2400  .06x = 2040
Multiply.
.02x + 2400 = 2040
Combine like terms.
.02x= 360
Subtract 2400.
X = 18,000
Divide by .02.
Step 5 State the answer. Mark should invest $18,000 at 4%. At 6%, he
should invest $40,000  $18,000 = $22,000.
Step 6 Check by finding the annual interest at each rate; they should total
$2040.
.04 ($18,000) = $720 and .06 ($22,000) = $1320
$720 + $1320 = $2040, as required.
Videc
Hi
Work Problem 7 at the Side,
Answers
7. (a) $50,000 at 5%; $22,000 at 3%
(b) $18,500 at 5%; $15,500 at 4%
PROBLEMSOLVING HINT
In Example 6, we chose to let the variable represent the amount in
vested at 4%. Students often ask, "Can I let the variable represent the
other unknown?" The answer is yes. The equation will be different, but
in the end the two answers will be the same.
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
Section 2.3 Applications of Linear Equations 83
OBJECTIVE Q Solve mixture problems. Mixture problems involving
rates of concentration can be solved with linear equations.
EXAMPLE 7
Solving a Mixture Problem
A chemist must mix 8 L of a 40% acid solution with some 70% solution to
get a 50%) solution. How much of the 70% solution should be used?
Step 1 Read the problem. The problem asks for the amount of 70%
solution to be used.
Step 2 Assign a variable. Let x = the number of liters of 70%) solution to
be used. The information in the problem is illustrated in Figure 5.
After mixing
50% ^From 70%
From 40%
8 L Unknown (8 + x) L
number of liters, x
Figure 5
Use the given information to complete the following table.
O Solve each problem.
(a) How many liters of a
10% solution should be
mixed with 60 L of a
25% solution to get a
15% solution?
Percent
(as a decimal)
Number
of Liters
Liters of
Pure Acid
.40
8
.40(8) = 3.2
.70
X
.70x
.50
8 +x
.50(8 +x)
Step 3
Step 4
Sum must equal 1
The numbers in the right column were found by multiplying
the strengths and the numbers of liters. The number of liters of
pure acid in the 40% solution plus the number of liters of pure
acid in the 70%) solution must equal the number of liters of pure
acid in the 50%) solution.
Write an equation.
3.2 + .70x = .50(8 + x)
Solve.
3.2 + .70x = 4 + .50x
.20x = .8
x = 4
Step 5 State the answer. The chemist should use 4 L of the 70%) solution.
Step 6 Check. 8 L of 40%) solution plus 4 L of 70%) solution is
8(.40) + 4(.70) = 6L
of acid. Similarly, 8 + 4 or 12 L of 50%) solution has
12(.50) = 6L
of acid in the mixture. The total amount of pure acid is 6 L both
before and after mixing, so the answer checks.
Work Problem 8 at the Side.
Distributive property
Subtract 3.2 and .50x.
Divide by .20.
(b) How many pounds of candy
worth $8 per lb should be
mixed with 100 lb of candy
worth $4 per lb to get a
mixture that can be sold for
$7 per lb?
Answers
8. (a) 120 L (b) 3001b
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
84 Chapter 2 Linear Equations and Applications
& Solve each problem.
(a) How much pure acid
should be added to 6 L of
30% acid to increase the
concentration to 50% acid?
(b) How much water must
be added to 20 L of 50%
antifreeze solution
to reduce it to 40%)
antifreeze?
PROBLEMSOLVING HINT
When pure water is added to a solution, remember that water is 0% of
the chemical (acid, alcohol, etc.). Similarly, pure chemical is 100%
chemical.
EXAMPLE 8
Solving a Mixture Problem When One Ingredient
Is Pure
The octane rating of gasoline is a measure of its antiknock qualities. For a
standard fuel, the octane rating is the percent of isooctane. How many liters
of pure isooctane should be mixed with 200 L of 94% isooctane, referred to
as 94 octane, to get a mixture that is 98%) isooctane?
Step 1
Step 2
Read the problem. The problem asks for the amount of pure
isooctane.
Assign a variable. Let x = the number of liters of pure (100%))
isooctane. Complete a table with the given information. Recall that
100% = lOO(.Ol) = 1.
Percent
(as a decimal)
Number
of Liters
Liters of Pure
Isooctane
1
X
X
.94
200
.94(200)
.98
x + 200
.98(x + 200)
Step 3 Write an equation. The equation comes from the last column of
the table, as in Example 7.
x+ .94(200) = .98(x + 200)
Step 4 Solve.
X + .94(200) = .98x + .98(200)
X + 188 = .98x + 196
.02x = 8
x = 400
Step 5 State the answer. 400 L of isooctane are needed.
Step 6 Check by showing that 400 + .94(200) = .98(400 + 200).
Work Problem 9 at the Side.
Distributive property
Multiply.
Subtract .98x and 188
Divide by .02.
Answers
9. (a) 2.4 L (b) 5L
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
Section 2.4 Further Applications of Linear Equations 9 I
151
AudiqJ
2 . 4 Further Applications of Linear Equations
There are three common applications of linear equations that we did not
discuss in Section 2.3: money problems, uniform motion problems, and
problems involving the angles of a triangle.
OBJECTIVE Q Solve problems about different denominations of
money. These problems are very similar to the simple interest problems in
Section 2.3.
PROBLEMSOLVING HINT
In problems involving money, use the fact that
, . X V. number of monetary total monetary
denommation X .^ ^ ^, ^ = \
units of the same kind value
For example, 30 dimes have a monetary value of $.10(30)
Fifteen fivedollar bills have a value of $5(15) = $75.
$3.
Solving a Money Denomination Problem
For a bill totaling $5.65, a cashier received 25 coins consisting of nickels and
quarters. How many of each type of coin did the cashier receive?
Read the problem. The problem asks that we find the number of
nickels and the number of quarters the cashier received.
Assign a variable.
Let X represent the number of nickels;
then 25 — X represents the number of quarters.
We can organize the information in a table.
Denomination
Number of Coins
Total Value
$.05
X
.05x
$.25
25 X
.25(25  X)
F
25
5.65
Totals
Step 3
Step
Write an equation. From the last column of the table,
.05x + .25(25 x) = 5.65.
Solve.
5x + 25(25  x) = 565
5x + 625 — 25x = 565
20x = 60
x = 3
Step 5
Step 6
Multiply by 100.
Distributive property
Subtract 625; combine terms.
Divide by 20.
3 = 22
State the answer. The cashier has 3 nickels and 25
quarters.
Check. The cashier has 3 + 22 = 25 coins, and the value of the
coins is $.05(3) + $.25(22) = $5.65, as required.
Work Problem 1 at the Side.
OBJECTIVES
Solve problems about
different denominations
of money.
Solve problems about
uniform motion.
Solve problems involving
the angles of a triangle.
Solve the problem.
At the end of a day,
a cashier had 26 coins
consisting of dimes and half
dollars. The total value of
these coins was $8.60. How
many of each type did he
have?
Answers
1. 11 dimes, 15 halfdollars
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
92 Chapter 2 Linear Equations and Applications
CAUTION
Be sure that your answer is reasonable when working problems like
Example 1. Because you are dealing with a number of coins, the correct
answer can neither be negative nor a fraction.
OBJ ECTI VE
Solve problems about uniform motion.
PROBLEMSOLVING HINT
Uniform motion problems use the distance formula, d = rt. In this for
mula, when rate (or speed) is given in miles per hour, time must be
given in hours. To solve such problems, draw a sketch to illustrate
what is happening in the problem, and make a table to summarize the
given information.
EXAMPLE 2
Solving a Motion Problem (Motion in Opposite
Directions)
Two cars leave the same place at the same time, one going east and the other
west. The eastbound car averages 40 mph, while the westbound car averages
50 mph. In how many hours will they be 300 mi apart?
Step 1
Step 2
Step 3
Step 4
Step 5
Step 6
Read the problem. We must find the time it takes for the two cars to
be 300 mi apart.
Assign a variable. A sketch shows what is happening in the prob
lem: The cars are going in opposite directions. See Figure 6.
50 mph
W^
Starting
point
40 mph
^E
Total distance = 300 mi
Figure 6
Let X represent the time traveled by each car. Organize the
information in a table. Fill in each distance by multiplying rate by
time using the formula d = rt. The sum of the two distances is 300.
Rate
Time
Distance
Eastbound Car 40
X
40x
Westbound Car 50
X
50x
H
300
Write an equation.
Solve.
40x + 50x = 300
90x = 300
300
20
3
Total
Combine like terms.
Divide by 90;
lowest terms
State the answer. The cars travel y = 3 hr, or 3 hr and 20 min.
Check. The eastbound car traveled 40 (y) = ^ mi, and
for a total of
the westbound car traveled 50 (y)
^ = 300 mi, as required.
400 I 500
3 "^ 3
500
3
mi.
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
Section 2.4 Further Applications of Linear Equations 93
CAUTION
It is a common error to write 300 as the distance for each car in
Example 2. Three hundred miles is the total distance traveled.
As in Example 2, in general, the equation for a problem involving
motion in opposite directions is of the form
partial distance + partial distance = total distance.
Work Problem 2 at the Side.
& Solve the problem.
Two cars leave the same
location at the same time.
One travels north at 60 mph
and the other south at
45 mph. In how many hours
will they be 420 mi apart?
EXAMPLE 3
Solving a Motion Problem (Motion in the Same
Direction)
Jeff can bike to work in  hr. By bus, the trip takes \ hr. If the bus travels
20 mph faster than Jeff rides his bike, how far is it to his workplace?
Step 1 Read the problem. We must find the distance between Jeff's home
and his workplace.
Step 2 Assign a variable. Although the problem asks for a distance, it is
easier here to let x be Jeff's speed when he rides his bike to work.
Then the speed of the bus is x + 20. For the trip by bike.
d = rt = X
_ 3
' 4
3
and by bus.
d =
 rt= (x + 20)
1
4
4<'
Summarize this information in a table.
(x + 20).
Rate
Time
Distance
Bike ^
3
4
3
Bus X + 20
1
4
Ax + 20)
4
Same
Step 3 Write an equation. The key to setting up the correct equation is to
realize that the distance in each case is the same. See Figure 7.
Home
Workplace
'iBPElElEl' ^^ ^^^*^
pTr„=.„„
1
1
Figure 7
4 4
The distance is the same.
Solve. 4(xj = 4(j(x + 20)
Multiply by 4.
3x = X + 20
Multiply; identity property
2x = 20
Subtract x.
x= 10
Divide by 2.
Continued on Next Page
Answers
2. 4hr
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
94 Chapter 2 Linear Equations and Applications
& Solve the problem.
Elayn begins jogging at
5:00 A.M., averaging 3 mph.
Clay leaves at 5:30 a.m.,
following her, averaging
5 mph. How long will it
take him to catch up to her?
(Hint: 30 min = ^ hr.)
Step 5 State the answer. The required distance is given by
3 3 30
d = x= (10) = — = 7.5 mi.
4 4 4
Step 6 Check by finding the distance using
J = (x + 20)= (10 + 20) = — = 7.5 mi,
the same result.
As in Example 3, the equation for a problem involving motion in the
same direction is often of the form
one distance = other distance.
PROBLEMSOLVING HINT
In Example 3 it was easier to let the variable represent a quantity
other than the one that we were asked to find. This is the case in some
problems. It takes practice to learn when this approach is best,
and practice means working lots of problems!
Work Problem 3 at the Side.
O Solve the problem.
One angle in a triangle
is 15° larger than a second
angle. The third angle is
25° larger than twice the
second angle. Find the
measure of each angle.
Answers
3
3. — hr or 45 min
4
4. 35°, 50°, and 95°
OBJECTIVE Q Solve problems involving the angles of a triangle.
An important result of Euclidean geometry (the geometry of the Greek
mathematician Euclid) is that the sum of the angle measures of any triangle
is 180°. This property is used in the next example.
EXAMPLE 4
Find the
Step 1
Step 2
Step 3
Step 4
Step 5
Step 6
Finding Angle Measures
value of X, and determine the measure of each angle in Figure 8.
Read the problem. We are asked to
find the measure of each angle.
Assign a variable. Let x represent
the measure of one angle.
Write an equation. The sum of
the three measures shown in the
figure must be 180°.
X + (x + 20) + (210  3x) = 180
Solve. X + 230 = 180
x= 50
x = 50
Figure 8
Combine like terms.
Subtract 230.
Divide by — 1 .
State the answer. One angle measures 50°, another measures
X + 20 = 50 + 20 = 70°, and the third measures 210  3x =
210  3(50) = 60°.
Check. Since 50° + 70° + 60° = 180°, the answer is correct.
Work Problem 4 at the Side.
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
I I 4 Chapter 3 Linear Inequalities and Absolute Value
3.1 Linear Inequalities in One Variable
OBJECTIVES
Graph intervals on a
number line.
Solve linear inequalities
using the addition
property.
Solve linear inequalities
using the multiplication
property.
Solve linear inequalities
with three parts.
Solve applied problems
using linear inequalities.
L*
Study Skills Workbook
Activity 8: Study Cards
Solving inequalities is closely related to solving equations. In this section we
introduce properties for solving inequalities.
Inequalities are algebraic expressions related by
< "is less than,"
^ "is less than or equal to,"
> "is greater than,"
^ "is greater than or equal to."
We solve an inequality by finding all real number solutions for it. For
example, the solution set of x < 2 includes all real numbers that are less than
or equal to 2, not just the integers less than or equal to 2. For example, —2.5,
— 1 .7, — 1, ^, V 2, 1, and 2 are real numbers less than or equal to 2 and are
therefore solutions of x < 2.
OBJECTIVE Q Graph intervals on a number line. A good way to
show the solution set of an inequality is by graphing. We graph all the real
numbers satisfying x < 2 by placing a square bracket at 2 on a number line
and drawing an arrow extending from the bracket to the left (to represent the
fact that all numbers less than 2 are also part of the graph). The graph is
shown in Figure 1 .
h
h
h
h
h
^
+
+
1 1
Figure 1
The set of numbers less than or equal to 2 is an example of an interval
on the number line. To write intervals, we use interval notation. For exam
ple, using this notation, the interval of all numbers less than or equal to 2 is
written (— oo, 2]. The negative infinity symbol — oo does not indicate a num
ber. It is used to show that the interval includes all real numbers less than 2.
As on the number line, the square bracket indicates that 2 is included in the
solution set. A parenthesis is always used next to the infinity symbol. The
set of real numbers is written in interval notation as (oo, co).
EXAMPLE 1
Graphing Intervals Written in Interval Notation
on Number Lines
Write each inequality in interval notation and graph it.
(a) x> 5
The statement x > — 5 says that x can represent any number greater than
— 5, but X cannot equal —5. This interval is written (—5, oo). We show this
solution set on a number line by placing a parenthesis at —5 and drawing an
arrow to the right, as in Figure 2. The parenthesis at —5 shows that —5 is not
part of the graph.
+
^
h
h
h
h
I : »
6 5 4 3 2 1
Figure 2
Continued on Next Page
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Section 3.1 Linear Inequalities in One Variable
115
(b) 1 <x<3
This statement is read "—1 is less than or equal to x and x is less than 3."
Thus, we want the set of numbers that are between 1 and 3, with 1
included and 3 excluded. In interval notation, we write the solution set as
[—1, 3), using a square bracket at —1 because it is part of the graph and a
parenthesis at 3 because it is not part of the graph. The graph is shown in
Figure 3.
f
h
h
h
■^
1 2
Figure 3
Work Problem 1 at the Side.
Write each inequality in
interval notation and graph it.
(a) X <  1
We now summarize the various types of intervals.
Type of
Interval
Open
interval
Halfopen
interval
Closed
interval
Set
{x\a < x}
{x\a < X < b}
{x\x<b}
{xx is a real
number}
{x\a < x}
{x\a<x ^ b}
{x\a <x<Z?}
{x\x^b}
{x\a <x<Z?}
Interval
Notation
(a,b)
(^, b)
( — 00^ 00 ")
[a, 00)
(a^b]
[a.b)
(^, b]
[a,b]
Graph
An inequality says that two expressions are not equal. Solving inequali
ties is similar to solving equations.
(b)x
(c) 4 < X < 2
Linear Inequality
A linear inequality in one variable can be written in the form
Ax + B<C,
where A, B, and C are real numbers, with A i^ 0.
(Throughout this section we give definitions and rules only for <, but they
are also valid for >, <, and >.) Examples of linear inequalities include
x + 5<2, ^— 3>5, and 2k + 5 < 10. Linear inequalities
Answers
1. (a) (00,1)
C I I I ) I I I ■
4321 1 2
(b) [3,00)
— \ — ^H — \ — h^
4 3 2 1
(c) [4,2)
I [ I I ) I '
642 2 4
Copyrigfit © 2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
I I 6 Chapter 3 Linear Inequalities and Absolute Value
& Solve each inequality, check
your solutions, and graph the
solution set.
(a) ;? + 6 < 8
OBJECTIVE Q Solve linear inequalities using the addition property.
We solve an inequality by finding all numbers that make the inequality true.
Usually, an inequality has an infinite number of solutions. These solutions, like
solutions of equations, are found by producing a series of simpler equivalent
inequalities. Equivalent inequalities are inequalities with the same solution
set. We use the addition and multiplication properties of inequality to
produce equivalent inequalities.
Addition Property of Inequality
For all real numbers A, B, and C, the inequalities
A< B and A + C<B + C
are equivalent.
In words, adding the same number to each side of an inequality
does not change the solution set.
(b) 8x < 7x  6
EXAMPLE 2
Solve X — 1 <
Using the Addition Property of Inequality
12, and graph the solution set.
X 7< 12
jc7 + 7<12 + 7 Add 7.
x< 5
Check: Substitute — 5 for x in the equation x — 7 = — 12. The result should
be a true statement.
xl = 12
57= 12
? Letjc =
5.
12= 12
True
This shows that —5 is the boundary point. Now we test a number on each
side of —5 to verify that numbers less than —5 make the inequality true. We
choose —4 and —6.
7< 12
6  7 <  12 ? Letx
13 <  12 True
— 6 is in the solution set.
The check confirms that (oo, 5), graphed in Figure 4, is the solution set.
X — 1
4  7 < 12 ? Letx =
4.
11 < 12 False
4 is not in the solution set.
6.
H \ \ h
h
+H
H \ \ h
10
Figure 4
— Hi
H h
Work Problem 2 at the Side.
Answers
2. (a) (00,2)
C I I I I I ) I
321 12 3
(b) (00, 6)
C I I I I ) I I
10987654
As with equations, the addition property of inequality can be used to
subtract the same number from each side of an inequality. For example, to
solve the inequality x + 4 > 10, we subtract 4 from each side to get x > 6.
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Section 3.1 Linear Inequalities in OneVariable
117
EXAMPLE 3
Using the Addition Property of Inequality
Solve 14 + 2m ^ 3m, and graph the solution set.
14 + 2m < 3m
14 + 2m — 2m < 3m — 2m Subtract 2m.
14 < m Combine like terms.
The inequality 14 <m (14 is less than or equal to m) can also be written
m > 14 (m is greater than or equal to 14). Notice that in each case, the
inequality symbol points to the lesser number, 14.
Check: 14 + 2m = 3m
14 + 2(14) = 3(14)
42
14.
Let m
42 True
So 14 satisfies the equality part of <. Choose 10 and 15 as test points.
14 + 2/11 < 3m
14 + 2(10) < 3(10) ? Letm= 10.
34 < 30 False
10 is not in the solution set.
14 + 2(15) < 3(15) ? Letm= 15.
44 < 45 True
15 is in the solution set.
The check confirms that [14, oo) is the solution set. See Figure 5.
+
h
^^
h
h
^
10 12 14 16 18 20
Figure 5
Work Problem 3 at the Side.
O Solve Ik 5>\ + k, check,
and graph the solution set.
O Multiply both sides of each
inequality by 5. Then insert
the correct symbol, either
< or >, in the first blank.
and fill in the other blanks in
part (b).
(a) 7 < 8
35 40
CAUTION
To avoid errors, rewrite an inequality such as 14 < m as m > 14 so that
the variable is on the left, as in Example 3.
B J ECTi VE Q Solve linear inequalities using the multiplication
property. Solving an inequality such as 3x < 15 requires dividing each
side by 3 using the multiplication property of inequality. To see how this
property works, start with the true statement
2< 5.
Multiply each side by, say, 8.
2(8) < 5(8) Multiply by 8.
16<40 True
The result is true. Start again with —2 < 5, and multiply each side by —8.
2(8) < 5(8) Multiply by 8.
16 < 40 False
The result, 16 < 40, is false. To make it true, we must change the direction
of the inequality symbol to get
16 > 40.
True
Work Problem 4 at the Side.
(b) 1>
5
Answers
3. [6,00)
I I [ I I I :»
4 5 6 7 8 9
4. (a) > (b) <;20
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I I 8 Chapter 3 Linear Inequalities and Absolute Value
Solve each inequality, check,
and graph the solution set.
(a) 2x< 10
As these examples suggest, multiplying each side of an inequality by a
negative number reverses the direction of the inequality symbol. The same is
true for dividing by a negative number since division is defined in terms of
multiplication.
(b) 7^>8
Multiplication Property of Inequality
For all real numbers A, B, and C, with C t^ 0,
(a) the inequalities
A<B and AC < BC
are equivalent if C> 0;
(b) the inequalities
A<B and AC> BC
are equivalent if C < 0.
In words, each side of an inequality may be multiplied (or divided)
by ?i positive number without changing the direction of the inequality
symbol. Multiplying (or dividing) by a negative number requires that
we reverse the inequality symbol.
(c) 9m < 81
Answers
5. (a) (00, 5)
^^H \ h^ h
7 6 5 4
(b)
^H — h^^ — \ — \ — \ —
321012
(C) (9,00)
\ — \ — ^H — \ — h^
7 8 9 10 11 12
EXAMPLE 4
Using the Multiplication Property of Inequality
Solve each inequality, and graph the solution set.
(a) 5m< 30
Use the multiplication property to divide each side by 5. Since 5 > 0,
do not reverse the inequality symbol.
Divide by 5.
Check that the solution set is the interval (—0°, —6], graphed in Figure 6.
5m <
30
5m
<
5
30
5
m <
6
H ^
H \ \ h
< \ \ K
14 12 10 8 6 4 2
Figure 6
(b) 4^<32
Divide each side by —4. Since — 4 < 0, reverse the inequality symbol.
Divide by —4 and reverse the symbol.
Check the solution set. Figure 7 shows the graph of the solution set, [—8, oo).
— \ ^ — \ \ \ \ \ \ \ \ \ — ^
9 8 7 6 5 4 3 210 1
Figure 7
Work Problem 5 at the Side.
Ak
<
32
4k
4
>
32
4
k
>
8
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Section 3.1 Linear Inequalities in One Variable
119
The steps used in solving a linear inequality are given below.
Solving a Linear Inequality
Step 1 Simplify each side separately. Use the distributive property to
clear parentheses and combine like terms as needed.
Step 2 Isolate the variable terms on one side. Use the addition prop
erty of inequality to get all terms with variables on one side of
the inequality and all numbers on the other side.
Step 3 Isolate the variable. Use the multiplication property of
inequality to change the inequality to the form x < k or x > k.
CAUTION
Reverse the direction of the inequality symbol only when multiplying or
dividing each side of an inequality by a negative number.
— :»I#ilyJW*f Solving a Linear Inequality Using the Distributive
Property
Solve 3 (x + 4) + 2 > 7  X, and graph the solution set.
Step 1 3x 12 + 2>7x
Distributive property
3x 10>7x
Step 2 3x  10 + x> 7 X + jc
Addx.
2x  10 > 7
2x  10 + 10 > 7 + 10
Add 10.
2x > 17
Step 3 2  2
Divide by —2; change > to <.
17
Figure 8 shows the graph of the solution set, ( — oo, — y].
n
1 1 1 1 1 1 te
11 10 9 8 7 6
— 1 1 1 1 1 1 *"
5 4 3 2 1
Figure 8
NOTE
In Example 5, if, after distributing, we add 3x to both sides of the
inequality, we have
3x  10 + 3x > 7  X + 3jc
Add 3x.
10>2x + 7
107>2x + 77
Subtract 7.
17>2x
17
Divide by 2.
The result " y is greater than or equal to x"
means the same thing as
"x is less than or equal to y." Thus, the solution set is the same.
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120 Chapter 3 Linear Inequalities and Absolute Value
© Solve, check, and graph the
solution set of each
inequality.
(a) 53(m 1)
< 2(m + 3) + 1
1 3
(b) (m + 3) + 2<(m + 8)
EXAMPLE 6
Solving a Linear Inequality with Fractions
2 1 1
Solve (r — 3) <— (5 — r), and graph the solution set.
To clear fractions, multiply each side by the least common denominator, 6.
2. ,11,
6
ff
»41
<6
6
(r3)
/A
 6  <6
V2/
(5r)
J(5.)
Multiply by 6.
Distributive property
Step 2
4(r 3) 3<3(5  r)
4r + 12  3 < 15  3r
4r + 9 < 15  3r
4r + 9 + 3r< 15  3r + 3r
r + 9 < 15
r + 99<159
r < 6
 1 (r) >  1 (6) Multiply by  1 ; change < to >.
r > 6
Check that the solution set is (—6, oo). See Figure 9.
Distributive property
Add 3r.
Subtract 9.
Step 3
e
h
h
h
5 4 3
Figure 9
Work Problem 6 at the Side.
Answers
5'
6. (a)
5
(b)
F^H — ^ — ^ — \ — h^
12 3 4 5
13
c
2'
13
" 2
H — ^ — hf^ — ^ — h^
9 8 7 6 5 4
OBJECTIVE Q Solve linear inequalities with three parts. For some
applications, it is necessary to work with an inequality such as
3 <x + 2 < 8,
where x + 2 is between 3 and 8. To solve this inequality, we subtract 2 from
each of the three parts of the inequality, giving
32<x+22<82
1 < X < 6.
Thus, X must be between 1 and 6 so that x + 2 will be between 3 and 8. The
solution set, (1, 6), is graphed in Figure 10.
e
h
h
■^
12 3 4 5 6 7
Figure 10
CAUTION
When inequalities have three parts, the order of the parts is important.
It would be wrong to write an inequality as8<x + 2<3, since this
would imply that 8 < 3, a false statement. In general, threepart inequal
ities are written so that the symbols point in the same direction, and both
point toward the lesser number.
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Section 3.1 Linear Inequalities in OneVariable
121
EXAMPLE 7
Solving a ThreePart Inequality
Solve 2<3A:1^5, and graph the solution set.
Begin by adding 1 to each of the three parts to isolate the variable term
in the middle.
3 3
2 + 1 < 3>k 1 + 1< 5 + 1
1 < 3,k<6
^ 3/t 6
3
2 < yt< 
3
Add 1 to each part.
Divide each part by —3;
reverse the inequality symbols.
Rewrite in order based on the
number line.
Check that the solution set is [— 2,  ], as shown in Figure 11.
— \ h^ \ \ ^
4
Figure 11
O Solve, check, and graph the
solution set of each
inequality.
(a)
1
Work Problem 7 at the Side.
Examples of the types of solution sets to be expected from solving lin
ear equations and linear inequalities are shov^n belov^.
SOLUTIONS OF LINEAR EQUATIONS AND INEQUALITIES
Equation or
Inequality
Linear equation
5jc + 4 = 14
Linear inequality
5jc + 4 < 14
or
5x + 4 > 14
Threepart inequality
1 <5x + 4< 14
Typical
Solution Set
{2}
(00, 2)
(2,00)
[1,2]
Graph of
Solution Set
1
(b) 5 < 3x  4 < 9
OBJECTIVE Q Solve applied problems using linear inequalities. In
addition to the familiar "is less than" and "is greater than," other expressions
such as "is no more than" and "is at least" also indicate inequalities. The
table belov^ shoves hov^ to interpret these expressions.
Word Expression
Interpretation
a is at least b
a^b
a is no less than b
a>b
a is at most b
a^b
a is no more than b
a^b
Answers
7. (a) [2,8]
h
42 2 4
13'
I I I ] I
6 8 10
(b) 3
3
3
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122 Chapter 3 Linear Inequalities and Absolute Value
O Solve the problem.
A rental company
charges $5 to rent a leaf
blower, plus $1.75 per hr.
Dona Kenly can spend no
more than $26 to blow leaves
from her driveway and pool
deck. What is the maximum
amount of time she can use
the rented leaf blower?
& Solve the problem.
Wade has grades of 92,
90, and 84 on his first three
tests. What grade must he
make on his fourth test in
order to keep an average of at
least 90?
Answers
8. 12 hr
9. at least 94
In Examples 8 and 9, we use the six problemsolving steps from
Section 2.3, changing Step 3 to "Write an inequality" instead of "Write an
equation."
EXAMPLE 8
Using a Linear Inequality to Solve a Rental Problem
A rental company charges $15 to rent a chain saw, plus $2 per hr. Jay Jenkins
can spend no more than $35 to clear some logs from his yard. What is the
maximum amount of time he can use the rented saw?
Step 1 Read the problem again.
Step 2 Assign a variable. Let h = the number of hours he can rent the saw.
Step 3 Write an inequality. He must pay $15, plus $2A, to rent the saw for
h hours, and this amount must be no more than $35.
Cost of
is no
renting
more than
^ s/. ^
35 dollars.
V J
[5 + 2/z
<
35
2A<20
Subtract 15.
/z< 10
Divide by 2.
Step 4 Solve.
Step 5 State the answer. He can use the saw for a maximum of 10 hr. (He
may use it for less time, as indicated by the inequality A < 10.)
Step 6 Check. If Jay uses the saw for 10 hr, he will spend 15 + 2(10) =
35 dollars, the maximum amount.
Hi
Work Problem 8 at the Side.
EXAMPLE 9
Finding an Average Test Score
Helen has scores of 88, 86, and 90 on her first three algebra tests. An aver
age score of at least 90 will earn an A in the class. What possible scores on
her fourth test will earn her an A average?
Let X represent the score on the fourth test. Her average score must be
at least 90. To find the average of four numbers, add them and then divide
by 4.
Average
+ 86 + 90 + X
is at
least 90.
> 90
264 + x
90
264 + X > 360
x>96
She must score 96 or more on her fourth test.
88 + 86 + 90 + 96 360
Add the scores.
Multiply by 4.
Subtract 264.
Check:
= 90
4 4
A score of 96 or more will give an average of at least 90, as required.
Mi
Work Problem 9 at the Side.
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Section 3.2 Set Operations and Compound Inequalities
129
3.2 Set Operations and Compound Inequalities
The table shows symptoms of an overactive thyroid and an underactive thyroid.
Underactive Thyroid
Overactive Thyroid
Sleepiness, s
Insomnia, /
Dry hands, d
Moist hands, m
Intolerance of cold, c
Intolerance of heat, h
Goiter, g
Goiter, g
Source: The Merck Manual of Diagnosis and Therapy,
16th Edition, Merck Research Laboratories, 1992.
Let A^ be the set of symptoms for an underactive thyroid, and let O be the set
of symptoms for an overactive thyroid. Suppose we are interested in the set
of symptoms that are found in both sets A^ and O. In this section we discuss
the use of the words and and or as they relate to sets and inequalities.
OBJECTIVES
Ml Find the intersection of
two sets.
Q Solve compound inequali
ties with the word and.
Q Find the union of two sets.
Q Solve compound inequali
ties with the word or.
List the elements in each set.
(a) Ar\B, if A = {3, 4, 5, 6}
and^= {5,6,7}
OBJECTIVE Q Find the intersection of two sets.
two sets is defined using the word and.
The intersection of
Intersection of Sets
For any two sets A and B, the intersection of ^ and B, symbolized
^ n ^, is defined as follows:
^ n ^ = {x I X is an element of ^ and x is an element of ^}.
EXAMPLE 1
Finding the Intersection ofTwo Sets
Let^ = {1, 2, 3, 4} and^ = {2, 4, 6}. Find^ H B,
The set ^ H B contains those elements that belong to both A and B: the
numbers 2 and 4. Therefore,
^ n^= {1,2,3,4} n {2,4,6} = {2,4}.
Work Problem 1 at the Side.
(b) Nn O (Refer to the
thyroid table.)
A compound inequality consists of two inequalities linked by a connec
tive word such as and or or. Examples of compound inequalities are
and
X + 1 < 9 and x  2 > 3
2x > 4 or 3x  6 < 5.
Compound inequalities
OBJECTIVE Q Solve compound inequalities with the word and. Use
the following steps.
Solving a Compound Inequality with and
Step 1 Solve each inequality in the compound inequality individually.
Step 2 Since the inequalities are joined with and, the solution set of the
compound inequality will include all numbers that satisfy both
inequalities in Step 1 (the intersection of the solution sets).
Answers
1. (a) {5,6} (b) {g}
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130 Chapter 3 Linear Inequalities and Absolute Value
& Solve each compound
inequality, and graph the
solution set.
(a) x< 10andx>2
EXAMPLE 2
Solving a Compound Inequality with and
Solve the compound inequality x + 1 < 9 and x  2 > 3.
Step 1 Solve each inequality in the compound inequality individually.
x+l<9 and x2>3
x+ll<9l and x2 + 2>3 + 2
X < 8 and x > 5
(b) X + 3
x4
1 and
12
Step 2
Because the inequalities are joined with the word and, the solution set
will include all numbers that satisfy both inequalities in Step 1 at the
same time. Thus, the compound inequality is true whenever x < 8
andx > 5 are both true. The top graph in Figure 12 shows x < 8, and
the bottom graph shows x > 5.
x<8 ^^h
H h
H h
^^
x>5
H \ \ h
I [ I
H \ — ^
3 4 5
Figure 12
Find the intersection of the two graphs in Figure 12 to get the solution
set of the compound inequality. The intersection of the two graphs in
Figure 13 shows that the solution set in interval notation is [5, 8].
[5,8]
H h
+
+
[ I I ]
3 4 5
Figure 13
Work Problem 2 at the Side.
O Solve
EXAMPLE 3
2x > X  1 and 3x > 3 + 2x,
and graph the solution set.
Answers
2. (a) (2, 10)
I ( I I I ) I >
2 4 6 8 10 12
(b) [8, 2]
1 [ I I ] I ■
10864 2
3. [3,)
I I I I [ I I :>
10 12 3 4 5
Solving a Compound Inequality with and
Solve the compound inequality 3x  2 > 5 and 5x  1 < 21.
Step 1 Solve each inequality separately.
3x  2 > 5 and 5x  1 < 21
3x> 7 and 5x < 20
x<
 and
3
x< 4
The graphs of x <  andx < 4 are shown in Figure 14.
^— h
■^^
x<4 ^^— h
4
Stepl
5 4 3 2 1
Figure 14
Now find all values of x that satisfy both conditions; that is, the real
numbers that are less than  and also less than or equal to 4. As
shown by the graph in Figure 15, the solution set is (— °o, —4].
— ^ \ \ \ \ — ^
(00,4] ^ h
Figure 15
Work Problem 3 at the Side.
nil
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Section 3.2 Set Operations and Compound Inequalities
131
EXAMPLE 4
Solving a Compound Inequality with and
Solve X + 2 < 5 andx  10 > 2.
First solve each inequality separately.
X + 2 < 5 and x  10 > 2
X < 3 and x > 12
The graphs of x < 3 andx > 12 are shown in Figure 16.
x<3 ^^H \ h^^ \ \ \ \ \ \ \ \ \ h^
12
Jt:> 12
H — \ — \ — \ — \ — \ — \ — \ — \ — \ — \ — \ — ( I ■:3
3 6
Figure 16
12
There is no number that is both less than 3 and greater than 12, so the given
compound inequality has no solution. The solution set is 0. See Figure 17.
\ \ \ \ \ \ \ \ \ \ \ \ \ \ h^
12
Figure 17
Work Problem 4 at the Side.
O Solve.
(a) X < 5 and x > 5
(b) X + 2 > 3 and
2x + 1 < 3
OBJECTIVE Q Find the union of two sets. The union of two sets is
defined using the word or.
Union of Sets
For any two sets A and B, the union of A and B, symbolized^ U B, is
defined as follows:
^ U ^ = {xx is an element of ^ or x is an element of ^}.
& List the elements in each set.
(a) A UB.ifA = {3,4,5,6}
and^= {5,6,7}
EXAMPLE 5
Finding the Union of Two Sets
Let^ = {1, 2, 3, 4} and^ = {2, 4, 6}. Find^ U B,
Begin by listing all the elements of set ^: 1, 2, 3, 4. Then list any addi
tional elements from set B. In this case the elements 2 and 4 are already
listed, so the only additional element is 6. Therefore,
AUB= {1,2,3,4} U {2,4,6}
= {1,2,3,4,6}.
The union consists of all elements in either^ or B (or both).
(b) A^ U O from the thyroid
table at the beginning of
this section
NOTE
Although the elements 2 and 4 appeared in both sets A and B, they are
written only once in ^ U ^.
Work Problem 5 at the Side.
Answers
4. (a) (b)
5. (a) {3,4,5,6,7} (b) {s,d, c,g,i,m,h}
Copyright ©2005 Pearson Education, Inc., pubiisiiing as Pearson AddisonWesley
132 Chapter 3 Linear Inequalities and Absolute Value
© Give each solution set in both
interval and graph forms.
(a) X + 2 > 3 or
2x + 1 < 3
OBJECTIVE Q Solve compound inequalities with the word on Use
the following steps.
Solving a Compound Inequality with or
Step 1 Solve each inequality in the compound inequality individually.
Step 2 Since the inequalities are joined with or, the solution set includes
all numbers that satisfy either one of the two inequalities in
Step 1 (the union of the solution sets).
(b) X  1 > 2 or
3x + 5 < 2x + 6
EXAMPLE 6
Solving a Compound Inequality with or
Solve 6x  4 < 2x or 3x < 9.
Step 1 Solve each inequality separately.
6x  4< 2x or 3x < 9
4x< 4
X < 1 or X > 3
The graphs of these two inequalities are shown in Figure 18.
x< 1 ^^— h
h
■^
+
x>3
+
h
■^
10 12 3 4
Figure 18
Step 2 Since the inequalities are joined with or, find the union of the
two solution sets. The union is shown in Figure 19 and is written
(00, 1) U [3, 00).
H ^
(oo,l)u[3,oo) ^ h
1
h
1 2
Figure 19
CAUTION
When inequalities are used to write the solution set in Example 6, it
must be written as
X < 1 or
3,
which keeps the numbers 1 and 3 in their order on the number line.
Writing 3 < x < 1 would imply that 3 < 1, which is FALSE. There is
no other way to write the solution set of such a union.
Answers
6. (a) (^, 2) U (1, ^)
C I I ) I I ( I I ^
43210123
(b) (^, 1) U (3, ^)
C I I ) I ( I I >
10 12 3 4 5
Work Problem 6 at the Side.
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Section 3.2 Set Operations and Compound Inequalities
133
EXAMPLE 7
Solving a Compound Inequality with or
Solve 4x + 1 > 9 or 5x + 3 >  12.
First, solve each inequality separately.
4x + 1 > 9 or 5x + 3 > 12
4x> 8 or 5x> 15
X < — 2 or X > — 3
The graphs of these two inequalities are shown in Figure 20.
jc <2 ^
h
h
4
h
h
e Solve.
(a) 2x + 1 < 9 or 2x + 3 < 5
x>3
^ ^
3 2
Figure 20
By taking the union, we obtain every real number as a solution, since every real
number satisfies at least one of the two inequalities. The set of all real numbers
is written in interval notation as (— oo, oo) and graphed as in Figure 21.
(00,00) ^ — h
h
h
3 2
Figure 21
Work Problem 7 at the Side.
EXAMPLE 8
Applying Intersection and Union
The five highest domestic grossing films (adjusted for inflation) are listed in
the table.
(b) 3x
13 orx + 5
O From Example 8, list the
elements that satisfy each set.
(a) The set of films with
admissions greater than
130,000,000 and gross
income less than
$500,000,000
FIVE ALLTIME HIGHEST GROSSING FILMS
Film
Admissions
Gross Income
Gone with the Wind
200,605,313
$972,900,000
Star Wars
178,119,595
$863,900,000
The Sound of Music
142,415,376
$690,700,000
E.T.
135,987,938
$659,500,000
The Ten Commandments
131,000,000
$635,400,000
Source: New York Times Almanac, 2001.
List the elements of the following sets.
(a) The set of topfive films with admissions greater than 180,000,000 and
gross income greater than $800,000,000
The only film that satisfies both conditions is Gone with the Wind, so
the set is
{Gone with the Wind).
(b) The set of topfive films with admissions less than 170,000,000 or gross
income greater than $700,000,000
Here, a film that satisfies at least one of the conditions is in the set. This
set includes all five films:
{Gone with the Wind, Star Wars, The Sound of Music, E.T.,
The Ten Commandments] .
Work Problem 8 at the Side.
(b) The set of films with
admissions greater than
130,000,000 or gross
income less than
$500,000,000
Answers
7. (a) (00,4] (b) (a),a))
8. (a) (b) {Gone with the Wind, Star Wars,
The Sound of Music, E. T, The Ten
Commandments}
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Section 3.3 Absolute Value Equations and Inequalities I 39
3.3 Absolute Value Equations and Inequalities
In a production line, quality is controlled by randomly choosing items from
the line and checking to see how selected measurements vary from the
optimum measure. These differences are sometimes positive and sometimes
negative, so they are expressed with absolute value. For example, a machine
that fills quart milk cartons might be set to release 1 qt plus or minus 2 oz
per carton. Then the number of ounces in each carton should satisfy the
absolute value inequality \x — 32 < 2, where x is the number of ounces.
OBJECTIVE Q Use the distance definition of absolute value. In
Section 1.1 we saw that the absolute value of a number x, written x, repre
sents the distance from x to on the number line. For example, the solutions
of x = 4 are 4 and —4, as shown in Figure 22.
4 units from
\
<► h
+
+
4 units from
\
H \ ♦
4 4
X = 4 orx = 4
Figure 22
Because absolute value represents distance from 0, it is reasonable to
interpret the solutions of x > 4 to be all numbers that are more than 4 units
from 0. The set (0°, 4) U (4, oo) fits this description. Figure 23 shows
the graph of the solution set of x > 4. Because the graph consists of two
separate intervals, the solution set is described using or as x < — 4 or x > 4.
More than
4 units from
More than
4 units from
h
+
h
4 4^
x < 4 or X > 4
Figure 23
The solution set of x < 4 consists of all numbers that are less than
4 units from on the number line. Another way of thinking of this is to think
of all numbers between —4 and 4. This set of numbers is given by (—4, 4),
as shown in Figure 24. Here, the graph shows that — 4 < x < 4, which means
X > —4 andx < 4.
Less than 4 units from
h
4 < X < 4
Figure 24
h
OBJECTIVES
Use the distance defini
tion of absolute value.
Solve equations of
the form \ax + b = k,
for k > 0.
Solve inequalities of the
form \ax + b < k and of
the form \ax + b\> k, for
k>0.
Solve absolute value
equations that involve
rewriting.
Solve equations of the
form \ax + b = \cx + d.
Solve special cases of
absolute value equations
and inequalities.
Graph the solution set of each
equation or inequality.
(a) x = 3
(b) x > 3
(c) x < 3
Work Problem 1 at the Side.
The equation and inequalities just described are examples of absolute
value equations and inequalities. They involve the absolute value of a
variable expression and generally take the form
\ax \ b\ = k, \ax \ b\> k, or \ax \ b\<k,
where ^ is a positive number. From Figures 2224, we see that
x = 4 has the same solution set as x = — 4 or x = 4,
x > 4 has the same solution set as x < — 4 or x > 4,
Ixl < 4 has the same solution set as x > 4 and x < 4.
Answers
1 (a) —4—^ — \ — \ — h
(b)
(c)
hf
3210123
^ \ \ \ \ \ ^
3210123
^^ \ \ \ 1 ) >
3210123
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I 40 Chapter 3 Linear Inequalities and Absolute Value
& Solve each equation, check,
and graph the solution set.
(a) x + 2 = 3
Thus, we can solve an absolute value equation or inequality by solving the
appropriate compound equation or inequality.
Solving Absolute Value Equations and Inequalities
Let ^ be a positive real number, and/? and q be real numbers.
1. To solve \ax \ b\ = k^ solve the compound equation
ax + b = k or ax + b = —k.
The solution set is usually of the form {p, q], which includes two
numbers.
(b) 3x4= 11
2. To solve \ax \ b\> k^ solve the compound inequality
ax \ b> k or ax \ b < —k.
The solution set is of the form (—'^^p) U (q, oo), which consists of
two separate intervals.
■^
f
p q
3. To solve \ax \ b\< k^ solve the threepart inequality
—k < ax + b < k.
The solution set is of the form (p, q), a single interval.
^
OBJECTIVE Q Solve equations of the form \ax \ b\ = k, for /r > 0.
Remember that because absolute value refers to distance from the origin, an
absolute value equation will have two parts.
Answers
2. (a) {5,1}
— *H — \ — \ — h
(b)
5 4 3 2
7
1 1
I «l I I I I I I I
_ 7_
3
4 5
EXAMPLE 1
Solving an Absolute Value Equation
Solve 2x + 1 = 7.
For 2x + 1 1 to equal 7, 2x + 1 must be 7 units from on the number line.
This can happen only when 2x + 1 = 7 or 2x + 1 = — 7. This is the first case
in the preceding summary. Solve this compound equation as follows.
2x + 1 = 7 or 2x + 1 = 7
2x = 6 or 2x = — 8
X = 3 or X = —4
Check by substituting 3 and then —4 in the original absolute value equation
to verify that the solution set is { —4, 3}. The graph is shown in Figure 25.
h
+
+
1 1
Figure 25
W
Work Problem 2 at the Side.
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Section 3.3 Absolute Value Equations and Inequalities I 4 I
NOTE
Some people prefer to write the compound statements in parts 1 and 2
of the summary on the previous page as the equivalent forms
ax + b = k or —(ax + b) = k
and ax + b > k or —(ax + b)> k
These forms produce the same results.
& Solve each inequality, check,
and graph the solution set.
(a) x + 2 > 3
OBJECTIVE Q Solve inequalities of the form \ax \ b\< k and of the
form I ajir + /i I > /f, for /f > 0
EXAMPLE 2
Solving an Absolute Value Inequality with >
Solve 2x + 1> 7.
By part 2 of the summary, this absolute value inequality is rewritten as
2x + 1 > 7 or 2x + 1 < 7,
because 2x + 1 must represent a number that is more than 7 units from on
either side of the number line. Now, solve the compound inequality.
2x + 1 > 7 or 2x + 1 < 7
2x> 6 or 2x< 8
X > 3 or X < — 4
Check these solutions. The solution set is (oo, 4) U (3, oo). See Figure 26.
Notice that the graph consists of two intervals.
■^
+
+
1
Figure 26
^
(b) 3x4> 11
O Solve each inequality, check,
and graph the solution set.
(a) x + 2 < 3
Work Problem 3 at the Side.
(b) 3x4< 11
EXAMPLE 3
Solving an Absolute Value Inequality with <
Solve 2x + 1< 7.
The expression 2x + 1 must represent a number that is less than 7 units
from on either side of the number line. Another way of thinking of this is
to realize that 2x + 1 must be between —7 and 7. As part 3 of the summary
shows, this is written as the threepart inequality
7 < 2x + 1 < 7.
— 8 < 2x < 6 Subtract 1 from each part.
— 4 < X < 3 Divide each part by 2.
Check that the solution set is (4, 3), so the graph consists of the single
interval shown in Figure 27.
^
h
h
h
1 1
Figure 27
■^
Work Problem 4 at the Side.
Answers
3. (a) (^, 5)U (1,^)
^ \ \ \ \ \ ^
54321 1
7"
(b)
U [5, ^)
< l]l I I I I I I [>
20245
4. (a) (5, 1)
c )
5 4 3 210 1
(b) ,5
I [ l I I I I I I ]
4 5
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I 42 Chapter 3 Linear Inequalities and Absolute Value
(a) Solve5x + 2 9= 7.
Look back at Figures 25, 26, and 27, with the graphs of 2x + 1 = 7,
2x + 1 > 7, and 2x + 1 < 7. If we find the union of the three sets, we get
the set of all real numbers. This is because for any value of x, 2x + 1 1 will
satisfy one and only one of the following: it is equal to 7, greater than 7,
or less than 7.
(b) Solve x + 2  3 > 2, and
graph the solution set.
CAUTION
When solving absolute value equations and inequalities of the types in
Examples 1, 2, and 3, remember the following.
1. The methods described apply when the constant is alone on one side
of the equation or inequality and is positive.
2. Absolute value equations and absolute value inequalities of the form
\ax + b\> ^translate into "or" compound statements.
3. Absolute value inequalities of the form \ax + b\< ^translate into
"and" compound statements, which may be written as threepart
inequalities.
4. An "or" statement cannot be written in three parts. It would be
incorrect to use — 7>2x+l>7in Example 2, because this would
imply that 7 > 7, which is false.
OBJECTIVE Q Solve absolute value equations that involve rewriting.
Sometimes an absolute value equation or inequality requires some rewriting
before it can be set up as a compound statement, as shown in the next
example.
(c) Solve, and graph the
solution set.
3x + 2 + 4< 15
EXAMPLE 4
Solving an Absolute Value Equation That Requires
Rewriting
Solve the equation x + 3 + 5 = 12.
First, rewrite so that the absolute value expression is alone on one side
of the equals sign by subtracting 5 from each side.
x + 3 + 5  5 = 12  5 Subtract 5.
x + 3 = 7
Now use the method shown in Example 1 .
x + 3 = 7 or x + 3 = — 7
X = 4 or X = — 10
Check that the solution set is {4,  10} by substituting 4 and then  10 into
the original equation.
Answers
5. (a)
7'0
(b) (00, 7) U (3, ^)
^ I I I I I I I I I ^
(c)
13
,3
I [ l I I I I I I ]
3
2 3
We use a similar method to solve an absolute value inequality that
requires rewriting.
.IK
Work Problem 5 at the Side.
OBJECTIVE Q Solve equations of the form \ax + b\ = \cx + r/_ By
definition, for two expressions to have the same absolute value, they must
either be equal or be negatives of each other.
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Section 3.3 Absolute Value Equations and Inequalities I 43
Solving \ax + b = cx + d\
To solve an absolute value equation of the form
\ax \ b\ = \cx + rf,
solve the compound equation
ax \ b = ex \ d or ax + b = — (ex + d).
© Solve each equation.
(a) \k 1 = 5^+ 7
EXAMPLE 5
Solving an Equation with Two Absolute Values
Solve the equation z + 6 = 2z — 3.
This equation is satisfied either if z + 6 and 2z — 3 are equal to each
other, or if z + 6 and 2z — 3 are negatives of each other. Thus,
z+6 = 2z3 or z + 6= (2z  3).
Solve each equation.
z + 6 = 2z —
9 = z
or z + 6 = — 2z + 3
3z= 3
z= 1
Check that the solution set is {9,  1 }.
(b) 4r l = 3r+ 5
Work Problem 6 at the Side.
OBJECTIVE Q Solve special cases of absolute value equations and
inequalities. When a typical absolute value equation or inequality
involves a negative constant or alone on one side, as in Examples 6
and 7, we use the properties of absolute value to solve. Keep the following
in mind.
O Solve each equation.
(a) 6x + 7= 5
Special Cases for Absolute Value
1. The absolute value of an expression can never be negative: \a\
for all real numbers a.
2. The absolute value of an expression equals only when the expres
sion is equal to 0.
EXAMPLE 6
Solving Special Cases of Absolute Value Equations
Solve each equation.
(a) 5r 3= 4
See Case 1 in the preceding box. Since the absolute value of an expres
sion can never be negative, there are no solutions for this equation. The so
lution set is 0.
(b) 7x  3 =
See Case 2 in the preceding box. The expression 7x — 3 will equal
only if
7x  3 = 0.
The solution of this equation is f Thus, the solution set of the original equa
tion is HV with just one element. Check by substitution.
Work Problem 7 at the Side.
(b)
r
Answers
6. (a) {1,2} (b) {^,6
7. (a) (b) {12}
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144 Chapters Linear Inequalities and Absolute Value
O Solve.
(a) x> 1
(b) x< 5
EXAMPLE 7
Solving Special Cases of Absolute Value
Inequalities
Solve each inequality.
(a) x> 4
The absolute value of a number is always greater than or equal to 0.
Thus, x > —4 is true for all real numbers. The solution set is (— oo, oo).
(b) x + 6 3< 5
Add 3 to each side to get the absolute value expression alone on one
side.
x + 6< 2
There is no number whose absolute value is less than —2, so this inequality
has no solution. The solution set is 0.
(c) x  7 + 4 < 4
Subtracting 4 from each side gives
x 7<0.
The value of x — 7 will never be less than 0. However, \x
whenx = 7. Therefore, the solution set is {7}.
7 1 will equal
Work Problem 8 at the Side.
(c) x + 2 <
Answers
8. (a) (00,00) (b) (c) {2}
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I 66 Chapter 4 Graphs, Linear Equations, and Functions
4.1 The Rectangular Coordinate System
OBJECTIVES
Plot ordered pairs.
Find ordered pairs that
satisfy a given equation.
Graph lines.
Find X and yintercepts.
Recognize equations of
horizontal and vertical
lines.
There are many ways to present information graphically. The circle graph
(or pie chart) in Figure 1(a) shows the cost breakdown for a gallon of regular
unleaded gasoline in California. What contributes most to the cost?
Figure 1(b) shows a bar graph in which the heights of the bars represent
the Btu (British thermal units) required to cool differentsized rooms. How
many Btu are needed to cool a 1400 ft^ room?
The line graph in Figure 1(c) shows personal spending (in billions of
dollars) on medical care in the United States from 1997 through 2002. About
how much was spent on medical care in 2002?
What Goes into the Price of
A Gallon of Gas?
/
 state
sales tax
8%
Crude
^^
 State
oil cost
excise tax
24%
1
120/0
— Federal
' Refinery
excise tax
margin
12%
32%
^0^
 Dealer's
.
^^
margin
120/0
—
Choose THE Right Air
Conditioner
24,000
on c\c\c\
1 1
16,000
3
■miiiil^
^^S
¥
t
CO 12,000
^
.
8,000
_ 1
1 1
4,000
III
Source: California Energy Commission.
(a)
^^o % % % % %W
Room Size (ft^)
Source: Carey, Morris and James,
Home Improvement for Dummies,
IDG Books.
(b)
Personal Spending on
Medical Care
C ^
^ O
c
1997 1998
1999 2000 2001
Year
2002
Source: Bureau of Economic Analysis,
(c)
Figure 1
Locating a fly
on a ceiling
The line graph in Figure 1(c) presents information based on a method
for locating a point in a plane developed by Rene Descartes, a 17thcentury
French mathematician. Legend has it that Descartes, who was lying in bed
ill, was watching a fly crawl about on the ceiling near a corner of the room.
It occurred to him that the location of the fly on the ceiling could be de
scribed by determining its distances from the two adjacent walls. See the fig
ure in the margin. In this chapter we use this insight to plot points and graph
linear equations in two variables whose graphs are straight lines.
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
Section 4.1 The Rectangular Coordinate System I 67
OBJECTIVE Q Plot ordered pairs. Each of the pairs of numbers
(3, 1), (5, 6), and (4,  1) is an example of an ordered pair; that is, a pair
of numbers written within parentheses in which the order of the numbers is
important. We graph an ordered pair using two perpendicular number lines
that intersect at their points, as shown in Figure 2. The common point is
called the origin. The position of any point in this plane is determined by
referring to the horizontal number line, the xaxis, and the vertical number
line, the jaxis. The first number in the ordered pair indicates the position
relative to the xaxis, and the second number indicates the position relative
to the jaxis. The xaxis and the jaxis make up a rectangular (or
Cartesian, for Descartes) coordinate system.
Plot each point. Name the
quadrant (if any) in which
each point is located.
(a) (4, 2)
(b) (3, 2)
i
yaxis
Origin
\
i AT^axis J
• C (5, 6)
Quadrant II
Quadrant III
D (4, 5)
Quadrant I
• A (3,1)
Quadrant IV
Figure 2
Figure 3
To locate, or plot, the point on the graph that corresponds to the ordered
pair (3, 1), we move three units from to the right along the xaxis, and then
one unit up parallel to the jaxis. The point corresponding to the ordered pair
(3, 1) is labeled^ in Figure 3. Additional points are labeled ^£'. The phrase
"the point corresponding to the ordered pair (3, 1)" is often abbreviated
as "the point (3, 1)." The numbers in the ordered pairs are called compo
nents and are the coordinates of the corresponding point.
We can relate this method of locating ordered pairs to the line graph in
Figure 1(c). We move along the horizontal axis to a year, then up parallel to
the vertical axis to find medical spending for that year. Thus, we can write
the ordered pair (2002, 1370) to indicate that in 2002, personal spending on
medical care was about $1370 billion.
CAUTION
The parentheses used to represent an ordered pair are also used to
represent an open interval (introduced in Section 3.1). The context of
the discussion tells whether ordered pairs or open intervals are being
represented.
(c) (5, 6)
(d) (4, 6)
(e) (3,0)
(f) (0, 5)
The four regions of the graph, shown in Figure 3, are called quadrants
I, II, III, and IV, reading counterclockwise from the upperright quadrant.
The points on the xaxis and jaxis do not belong to any quadrant. For
example, point E in Figure 3 belongs to no quadrant.
Work Problem 1 at the Side.
Answers
1. 3
(a)
•
•
! (d)
1 1 1 f 1 1
(e)
(c) ,
•
\ • (b)
(a) II (t
(e) no qu
) IV (c) III (d) I
adrant (f ) no quadrant
Copyright ©2005 Pearson Education, Inc., pubiisiiing as Pearson AddisonWesley
I 68 Chapter 4 Graphs, Linear Equations, and Functions
& (a) Complete each ordered
pair for 3x — 4y = 12.
(0, )
( ,0)
( ,2)
OBJECTIVE Q Find ordered pairs that satisfy a given equation.
Each solution to an equation with two variables, such as 2x + 3y = 6, will
include two numbers, one for each variable. To keep track of which number
goes with which variable, we write the solutions as ordered pairs. (If x and j
are used as the variables, the xvalue is given first.) For example, we can
show that (6, —2) is a solution of 2x + 3j = 6 by substitution.
2x + 3y = 6
2(6) + 3(2) = 6
12  6 = 6
6 = 6
Letx = 6,y = —2.
True
Because the pair of numbers (6, —2) makes the equation true, it is a solution.
On the other hand, (5, 1) is not a solution of the equation 2x + 3y = 6
because
2x + 3y
2(5) + 3(1)
10 + 3
13, not 6.
To find ordered pairs that satisfy an equation, select any number for one
of the variables, substitute it into the equation for that variable, and then
solve for the other variable. Two other ordered pairs satisfying 2x + 3^ = 6
are (0, 2) and (3, 0). Since any real number could be selected for one vari
able and would lead to a real number for the other variable, linear equations
in two variables have an infinite number of solutions.
Video
(4, )
(b) Find one other ordered pair
that satisfies the equation.
Answers
2. (a) (0,3), (4,0),
2 ,(4,6)
(b) Many answers are possible; for example.
EXAMPLE 1
Completing Ordered Pairs
Complete each ordered pair for 2x + 3y = 6.
(a) (3, )
We are given x
= — 3. We substitute into the
equation
to find y
2x +
3y =
6
2(3) +
3j =
6
Letx
= 3.
6 +
3y =
3y =
y =
6
12
4
The ordered pair is
(3,4).
(b)( ,4)
Replace y with
—4 in the equation to findx.
2x +
3j =
6
2x + 3(
4) =
6
Let 7
= 4.
2x
12 =
2x =
X =
6
18
9
The ordered pair is
(9, 4).
Work Problem 2 at the Side.
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Section 4.1 The Rectangular Coordinate System I 69
OBJECTIVE Q Graph lines. The graph of an equation is the set of
points corresponding to all ordered pairs that satisfy the equation. It gives
a "picture" of the equation. Most equations in two variables are satisfied
by an infinite number of ordered pairs, so their graphs include an infinite
number of points.
To graph an equation, we plot a number of ordered pairs that satisfy the
equation until we have enough points to suggest the shape of the graph. For
example, to graph 2x + 3y = 6, we plot all the ordered pairs found in Objec
tive 2 and Example 1 on the previous page. These points, shown in a table of
values and plotted in Figure 4(a), appear to lie on a straight line. If all the or
dered pairs that satisfy the equation 2x + 3y = 6 were graphed, they would
form the straight line shown in Figure 4(b).
& Graph 3x  4y = 12. Use the
points from Problem 2 in the
margin on the previous page.
X
y
3
4
2
3
6
2
9
4
(3,4)
♦ 4
2o
42
2
4
6
2x + 3y = 6
(0,2)
(3,0)
2^4 6 8 10
(6, 2)
(9, 4)
(a)
(b)
Figure 4
Work Problem 3 at the Side.
The equation 2x + 3j = 6 is called a firstdegree equation because
it has no term with a variable to a power greater than one.
The graph of any firstdegree equation in two variables is a
straight line.
Since firstdegree equations with two variables have straightline graphs,
they are called linear equations in two variables.
Linear Equation in Two Variables
A linear equation in two variables can be written in the form
Ax + By= C,
where A, B, and C are real numbers (A and B not both 0). This form is
called standard form.
OBJECTIVE Q Find x and yintercepts. A
straight line is determined if any two different points
on the line are known, so finding two different points
is enough to graph the line. Two useful points for
graphing are the x andjintercepts. The xintercept
is the point (if any) where the line intersects the
Xaxis; likewise, the jintercept is the point (if any)
where the line intersects the j^axis.* See Figure 5.
jintercept
xintercept
Figure 5
* Some texts define an intercept as a number, not a point.
3x 4y= 12
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I 70 Chapter 4 Graphs, Linear Equations, and Functions
O Find the intercepts, and graph
2x  y = 4.
The jvalue of the point where a line intersects the xaxis is 0. Similarly,
the xvalue of the point where a line intersects the jaxis is 0. This suggests a
method for finding the x andjintercepts.
Finding Intercepts
When graphing the equation of a line,
let J = to find the xintercept;
let X = to find the jintercept.
EXAMPLE 2
Finding Intercepts
Find the x and jintercepts of4x—y =
We find the xintercept by letting y
3, and graph the equation.
0.
4x  y = 3
4x  = 3
Let 3; = 0.
4x= 3
3
xintercept is (, 0).
For the jintercept, let x = 0.
4x  y = 3
4(0) 7= 3
Letx = 0.
y=3
y = 3
jintercept is (0, 3).
The intercepts are the two points ( — , 0) and (0, 3). We show these
ordered pairs in the table next to Figure 6 and use these points to draw
the graph.
X
y
3
4
3
(
jfintercept
3;intercept
(0,3)
4xy = 3
Figure 6
Answers
4. xintercept is (2,0); jintercept is (0, —4).
3^
2x  y = 4
NOTE
While two points, such as the two intercepts in Figure 6, are sufficient
to graph a straight line, it is a good idea to use a third point to guard
against errors. Verify by substitution that (—1, — 1) also lies on the
graph of 4x — y = —3.
Work Problem 4 at the Side.
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Section 4. 1 The Rectangular Coordinate System I 7 1
OBJECTIVE Q Recognize equations of horizontal and vertical lines.
A graph can fail to have an xintercept or a jintercept, which is why the
phrase "if any" was added when discussing intercepts.
& Find the intercepts, and
graph J + 4 = 0.
EXAMPLE 3
Graphing a Horizontal Line
Graph 7 = 2.
Since y is always 2, there is no value of x corresponding to j = 0, so the
graph has no xintercept. The jintercept is (0, 2). The graph in Figure 7,
shown with a table of ordered pairs, is a horizontal line.
X
yj
1
2
2
3
2
y = 2
Horizontal
(0, 2).
line
Figure 7
Work Problem 5 at the Side.
© Find the intercepts, and
graph the line x = 2.
EXAMPLE 4
Graphing aVertical Line
Graph x + 1 = 0.
The xintercept is ( 1, 0). The standard form Ix + Oj =  1 shows that
every value of j leads to x = — 1, so no value of j makes x = 0. The only
way a straight line can have no jintercept is if it is vertical, as in Figure 8.
I«
yi
1
4
1
1
5
X + 1 =
(1,0)
Vertical
line
Figure 8
Work Problem 6 at the Side.
CAUTION
To avoid confusing equations of horizontal and vertical lines remember
that
1. An equation with only the variable x will always intersect the xaxis
and thus will be vertical.
2. An equation with only the variable y will always intersect the yaxis
and thus will be horizontal.
Answers
5. no xintercept; jintercept is (0, —4).
3^
;
1 1 1
1 1 1
1 1 1 1 1 1 ^ ^
4
: y +4=0
6. no jintercept; xintercept is (2, 0).
y
x = 2
I I I I * ^
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I 72 Chapter 4 Graphs, Linear Equations, and Functions
O Find the intercepts, and
graph the line 3x — y = 0.
Some lines have both the x andjintercepts at the origin.
EXAMPLE 5
Graphing a Line That Passes through the Origin
Graph x + 2y = 0.
Find the xintercept by letting y = 0.
Let J = 0.
xintercept is (0, 0).
x + 2y =
x + 2(0) =
x + =
x =
To find the jintercept, let x = 0.
x + 2y =
+ 2y = Letx = 0.
y = 3;intercept is (0, 0).
Both intercepts are the same ordered pair, (0, 0). (This means that the graph
goes through the origin.) To find another point to graph the line, choose any
nonzero number for x, say x = 4, and solve for j.
x + 2y =
4 + 27 = Letx = 4.
2y= 4
y=2
This gives the ordered pair (4, —2). These two points lead to the graph
shown in Figure 9. As a check, verify that (—2, 1) also lies on the line.
xintercept
and
jintercept
m
y 1
2
1
4
2
Figure 9
To find the additional point to graph, we could have chosen any number
(except 0) for J instead of x.
w
Work Problem 7 at the Side.
Answers
7. Both intercepts are (0, 0).
3x  y =
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Section 4.2 Slope I 77
4.2 Slope
Slope (steepness) is used in many practical ways. The slope of a highway
(sometimes called the grade) is often given as a percent. For example, a 10%
(or 1^ = 1^) slope means the highway rises 1 unit for every 10 horizontal
units. Stairs and roofs have slopes too, as shown in Figure 10.
1 9
/
/
7
/
/^ n
7
Slope is ~.
Slope (or □ I
pitch) is
Slope is 10.
(not to scale)
Figure 10
In each example mentioned, slope is the ratio of vertical change, or rise,
to horizontal change, or run. A simple way to remember this is to think
"slope is rise over run."
OBJECTIVE Q Find the slope of a line given two points on the line.
To obtain a formal definition of the slope of a line, we designate two differ
ent points on the line. To differentiate between the points, we write them as
(Xp y^ and {x^, y^. See Figure 11. (The small numbers 1 and 2 in these
ordered pairs are QdiWQA subscripts. Read (x^ j^) as "xsubone, jsubone.")
Change in y (rise)
yiyx
Change in x (run)
Figure 11
As we move along the line in Figure 11 from (x^ y^ to {x^, y^), the
jvalue changes (vertically) from y^ to y^, an amount equal to y^ — y^ Asy
changes from j^ to y^, the value of x changes (horizontally) from x^ to x^ by
the amount x^ — Xy The ratio of the change in j to the change in x (the rise
over the run) is called the slope of the line, with the letter m traditionally
used for slope.
OBJECTIVES
Find the slope of a line
given two points on the
line.
Find the slope of a line
given an equation of the
line.
Graph a line given its
slope and a point on the
line.
Use slopes to determine
whether two lines are
parallel, perpendicular, or
neither.
Solve problems involving
average rate of change.
Use the information given for
the walkway in the figure to
find the following.
(a) The rise
(b) The run
Slope Formula
The slope of the line through the distinct points (x^ y^) and (x^, y^ is
rise change in j j2 ~ Ji
m =
run change in x
X^ X\
(Xi 9^ X2).
(c) The slope
Work Problem 1 at the Side.
Answers
1. (a) 2 ft (b) 10 ft (c) ^ or ^
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I 78 Chapter 4 Graphs, Linear Equations, and Functions
O Find the slope of the Hne
through each pair of points.
(a) (2, 7), (4, 3)
(b) (1,2), (8, 5)
EXAMPLE 1
Finding the Slope of a Line
Find the slope of the line through the points (2, — 1) and (—5, 3).
If (2, —1) = (Xp jj) and (—5, 3) = {x^,y^, then
yiyi ^ 3 (1) ^ ^ ^ _4
X2  Xi 52 7 7'
See Figure 12. On the other hand, if the pairs are reversed so that (2, — 1)
{x^,y^ and (—5, 3) = (x^j^), the slope is
13 ^ _ _1
T~ "r
the same answer.
m
(5)
Figure 12
(c) (8, 4), (3, 2)
Example 1 suggests that the slope is the same no matter which point we
consider first. Also, using similar triangles from geometry, we can show that
the slope is the same no matter which two different points on the line we
choose.
CAUTION
When calculating slope, be careful to subtract the yvalues and the
xvalues in the same order.
yi
Correct
y\ y\
X2
or
yi
Incorrect
X]^ X2
X2
X2
Also, remember that the change in y is the numerator and the change
in X is the denominator.
Work Problem 2 at the Side.
OBJECTIVE Q Find the slope of a line given an equation of the line.
When an equation of a line is given, one way to find the slope is to use the
definition of slope by first finding two different points on the line.
Answers
2. (a) \
(b)  (c)
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EXAMPLE 2
Finding the Slope of a Line
Find the slope of the Hne 4x — y = —8.
The intercepts can be used as the two different points needed to find the
slope. Let J = to find that the xintercept is (—2, 0). Then let x = to find
that the jintercept is (0, 8). Use these two points in the slope formula. The
slope is
rise
m =
run
80
0 (2)
2
Work Problem 3 at the Side.
Section 4.2 Slope I 79
& Find the slope of each line.
(a) 2x + 7 = 6
EXAMPLE 3
Finding the Slopes of Horizontal and Vertical Lines
Find the slope of each line.
(a) 7 = 2
Figure 7 in Section 4.1 shows that the graph of j = 2 is a horizontal line.
To find the slope, select two different points on the line, such as (3, 2) and
(—1, 2), and use the slope formula.
rise 2 — 2
run 3  (  1 ) 4
m
3 (1)
In this case, the rise is 0, so the slope is 0.
(b) x= 1
As shown in Figure 8 (Section 4.1), the graph of x = — 1 (or x + 1 =0)
is a vertical line. Two points that satisfy the equation x = — 1 are (— 1, 5) and
(—1, —4). Use these two points to find the slope.
45 _ 9^
~ ~0~
m
rise
run
1  (1)
Since division by is undefined, the slope is undefined. This is why the def
inition of slope includes the restriction x^ ¥" x^.
Generalizing from Example 3, we can make the following statements
about horizontal and vertical lines.
(b) 3x  47 = 12
O Find the slope of each line.
(a) X = — 6
Slopes of Horizontal and Vertical Lines
The slope of a horizontal line is 0.
The slope of a vertical line is undefined.
(b) 7 + 5 =
Work Problem 4 at the Side.
HI"
Answers
3
3. (a) 2 (b)  4. (a) undefined (b)
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180 Chapter 4 Graphs, Linear Equations, and Functions
& Find the slope of the graph of
2x 5y = 8.
The slope of a line can also be found directly from its equation. Look
again at the equation 4x — y = —8 from Example 2. Solve this equation fory.
4x — y = — 8 Equation from Example 2
J = 4x  8 Subtract 4x.
y = 4x \ 8 Multiply by  1 .
Notice that the slope, 4, found using the slope formula in Example 2 is the
same number as the coefficient of x in the equation;; = 4x + 8. We will see
in the next section that this always happens, as long as the equation is solved
fory.
EXAMPLE 4
Finding the Slope from an Equation
Find the slope of the graph of 3x — Sj = 8.
Solve the equation fory.
3x
5y
y
3x + 8
3 8
—X
5 5
Subtract 3x.
Divide by —5.
The slope is given by the coefficient of x, so the slope is f .
Work Problem 5 at the Side.
OBJECTIVE Q Graph a line given its slope and a point on the line.
Example 5 shows how to graph a straight line by using the slope and one
point on the line.
Video
Answers
2
EXAMPLE
Using the Slope and a Point to Graph Lines
Graph each line.
(a) With slope  through the point (—1,4)
First locate the point P(— 1, 4) on a graph as shown in Figure 13. Then
use the slope to find a second point. From the slope formula,
m =
change in j
change in X 3'
so move up 2 units and then 3 units to the right to locate another point on the
graph (labeled R). The line through P(— 1, 4) and R is the required graph.
Figure 13
Continued on Next Page
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Section 4.2 Slope I 8 I
(b) Through (3,1) with slope —4
Start by locating the point P (3, 1) on a graph. Find a second point R on
the line by writing — 4 as ^ and using the slope formula.
change in J —4
change in x 1
Move down 4 units from (3, 1), and then move 1 unit to the right. Draw a line
through this second point R and P(3, 1), as shown in Figure 14.
The slope also could be written as
m
change in y
change in x
1
In this case the second point R is located up 4 units and 1 unit to the left. Verify
that this approach produces the same line.
Figure 14
Work Problem 6 at the Side.
In Example 5(a), the slope of the Hne is thQ positive number . The graph
of the line in Figure 13 goes up (rises) from left to right. The line in Exam
ple 5(b) has a negative slope, —4. As Figure 14 shows, its graph goes down
(falls) from left to right. These facts suggest the following generalization.
A positive slope indicates that the line goes up (rises) from left to right.
A negative slope indicates that the line goes down (falls) from left to right.
© Graph each line.
(a) Through (1,3);
3
m =
4
(b) Through (1,4);
m = 2
Figure 15 shows lines of positive, 0, negative, and undefined slopes.
Negative "
slope
Answers
6. (a)
Positive
slope
Undefined
slope
Figure 15
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182 Chapter 4 Graphs, Linear Equations, and Functions
OBJECTIVE Q Use slopes to determine whether two lines are
parallel, perpendicular, or neither. The slopes of a pair of parallel or per
pendicular lines are related in a special way. The slope of a line measures the
steepness of the line. Since parallel lines have equal steepness, their slopes
must be equal; also, lines with the same slope are parallel.
Slopes of Parallel Lines
Two nonvertical lines with the same slope are parallel.
Two nonvertical parallel lines have the same slope.
EXAMPLE 6
Determining whetherTwo Lines Are Parallel
Are the lines L^ through (—2, 1) and (4, 5), and L^, through (3, 0) and
(0, 2), parallel?
The slope of L^ is rui = ■
 1 _ 4 _ 2
(2) ~ 6 ~ 3 '
The slope of L^ is ^2 = ~^
Because the slopes are equal, the two lines are parallel.
2 _ 2
3 ~ 3'
To see how the slopes of perpendicular lines are related, consider a non
vertical line with slope f . If this line is rotated 90°, the vertical change and
the horizontal change are reversed and the slope is — ^, since the horizontal
change is now negative. See Figure 16. Thus, the slopes of perpendicular
lines have product — 1 and are negative reciprocals of each other. For example,
if the slopes of two lines are  and — , then the lines are perpendicular
because I ( — f) = — 1.
Slope is
Slope is I
Figure 16
Slopes of Perpendicular Lines
If neither is vertical, perpendicular lines have slopes that are negative
reciprocals; that is, their product is — 1. Also, lines with slopes that are
negative reciprocals are perpendicular.
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Section 4.2 Slope
183
EXAMPLE 7
Video
Determining whether Two Lines Are Perpendicular
Are the lines with equations 2y = 3x — 6 and 2x + 3y = —6 perpendicular?
Find the slope of each line by first solving each equation for j.
2y
= 3x 
3
= —X
2
t
Slope
2x + 3y
3j
2x — 6
7=jx2
Slope
Since the product of the slopes of the two lines is  ( —  )
perpendicular.
1, the lines are
Work Problem 7 at the Side
OBJECTIVE Q Solve problems involving average rate of change.
We know that the slope of a line is the ratio of the change in j (vertical) to
the change in x (horizontal). This idea can be applied to reallife situations.
The slope gives the average rate of change in y per unit of change in x,
where the value ofy depends on the value of x.
O Write parallel, perpendic
ular, or neither for each pair
of two distinct lines.
(a) The line through (1,2)
and (3, 5) and the line
through (4, 7) and (8, 10)
(b) The hne through (5,9)
and (3, 7) and the line
through (0, 2) and (8, 3)
EXAMPLE 8
Interpreting Slope as Average Rate of Change
The graph in Figure 17 approximates the percent of U.S. households owning
multiple personal computers in the years 19972001. Find the average rate
of change in percent per year.
(c) 2xy
2x \ y
4 and
HOMES WITH Multiple PCs
30 r
(2001,24.4)
2001
Source: The Yankee Group.
Figure 17
To determine the average rate of change, we need two pairs of data.
From the graph, if x = 1997, then j =10 and if x = 2001, then j = 24.4, so
we have the ordered pairs (1997, 10) and (2001, 24.4). By the slope formula,
. ^ u y^y^ 24.4  10 14.4
average rate of change = = = = 3.6.
^ ^ X2 xi 2001  1997 4
This means that the number of U.S. households owning multiple computers
increased hy 3.6% each year from 1997 to 2001.
Work Problem 8 at the Side,
^11
(d) 3x + 5y = 6 and
5x — 3^ = 2
O Use the ordered pairs
(1997, 10) and (2000, 20.8),
which are plotted in Figure 17,
to find the average rate of
change. How does it compare
to the average rate of change
found in Example 8?
Answers
7. (a) parallel (b) perpendicular
(c) neither (d) perpendicular
8. 3.6; It is the same.
Copyright © 2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
184 Chapter 4 Graphs, Linear Equations, and Functions
& In 1997, 36.4 percent of high
school students smoked. In
2001, 28.5 percent of high
school students smoked. Find
the average rate of change in
percent per year. (Source:
U.S. Centers for Disease
Control and Prevention.)
EXAMPLE 9
Interpreting Slope as Average Rate of Change
In 1997, sales of VCRs numbered 16.7 million. In 2002, sales of VCRs
were 13.3 million. Find the average rate of change, in millions, per year.
(Source: The Gazette, June 22, 2002.)
To use the slope formula, we need two ordered pairs. Here, if x = 1997,
then J = 16.7 and if x = 2002, then j = 13.3, which gives the ordered pairs
(1997, 16.7) and (2002, 13.3). (Note thatj is in millions.)
average rate of change
13.3  16.7 3.4
2002  1997
.68
The graph in Figure 18 confirms that the line through the ordered pairs falls
from left to right and therefore has negative slope. Thus, sales of VCRs
decreased by .68 million each year from 1997 to 2002.
Sales of VCRs
1997
1999 2001
Year
Figure 18
Work Problem 9 at the Side.
Answers
9. 1.975% per yr
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Section 4.3 Linear Equations in Two Variables
191
4.3 Linear Equations in Two Variables
OBJECTIVE Q Write an equation of a line given its slope and
yintercept. In Section 4.2 we found the slope of a line from the equation
of the line by solving the equation for y. For example, we found that the
slope of the line with equation
y = 4x + 8
is 4, the coefficient of x. What does the number 8 represent?
To find out, suppose a line has slope m and jintercept (0, b). We can
find an equation of this line by choosing another point (x, y) on the line, as
shown in Figure 19. Using the slope formula.
tfl —
X
yb
m =
X
mx = y — b
Multiply by x.
mx + b = y
AddZ?.
y — mx + b.
Rewrite.
Figure 19
This last equation is called the slopeintercept form of the equation of a line,
because we can identify the slope m and jintercept (0, Z?) at a glance. Thus,
in the line with equation
J = 4x + 8,
the number 8 indicates that the jintercept is (0, 8).
OBJECTIVES
Q Write an equation of
a line given its slope and
yintercept.
Q Graph a line using its
slope and yintercept.
Q Write an equation of
a line given its slope and
a point on the line.
Q Write an equation of
a line given two points on
the line.
Q Write an equation of
a line parallel or perpen
dicular to a given line.
Q Write an equation of
a line that models real
data.
> Write an equation in slope
intercept form for each line
with the given slope and
jintercept.
(a) Slope 2; jintercept (0, —3)
SlopeIntercept Form
The slopeintercept form of the equation of a line with slope m and
jintercept (0, /?) is
y — mx + b.
t t
Slope jintercept is (0, Z?).
EXAMPLE 1
Using the SlopeIntercept Form to Find an
Equation of a Line
Find an equation of the line with slope —\ and jintercept (0, —2).
Here m =
intercept form.
and b = —2. Substitute these values into the slope
y = mx \ b
X 2
5
Slopeintercept form
m = f;Z?= 2
Work Problem 1 at the Side.
Mr
(b) Slope ;7intercept (0, 0)
Answers
1. (a) y = 2x
3 (b) y
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I 92 Chapter 4 Graphs, Linear Equations, and Functions
& Graph each line using its
slope and jintercept.
(a) J = 2x + 3
(b) 3x + 4j = 8
Answers
2. (a) y
(b)
OBJECTIVE Q Graph a line using its slope and yintercept. If the
equation of a line is written in slopeintercept form, we can use the slope and
yintercept to obtain its graph.
jintercept to obtain its graph.
EXAMPLE 2
Graphing Lines Using Slope and /Intercept
Graph each line using its slope andjintercept.
(a) 7 = 3x  6
Here m = 3 and b = 6. Plot the jintercept (0, 6). The slope 3 can
be interpreted as
rise change in j 3
m =
run
change in x
From (0, —6), move up 3 units and to the right 1 unit, and plot a second
point at (1, —3). Join the two points with a straight line to obtain the graph
in Figure 20.
y = ix + 3
Figure 21
(b) 3y + 2x = 9
Write the equation in slopeintercept form by solving forj.
3y + 2x = 9
3y = 2x + 9 Subtract 2x.
y
Slope
= X + 3
3
Slopeintercept form
jKintercept is (0, 3).
To graph this equation, plot the jintercept (0, 3). The slope can be inter
preted as either
Y or ^. Using
3 9
move from (0, 3) down 2 units and to the
right 3 units to locate the point (3, 1). The line through these two points is
the required graph. See Figure 21 . (Verify that the point obtained using ^ as
the slope is also on this line.)
^w
Work Problem 2 at the Side.
NOTE
The slopeintercept form of a linear equation is the most useful for
several reasons. Every linear equation (of a nonvertical line) has a
unique (one and only one) slopeintercept form. In Section 4.5 we
study linear functions, which are defined using slopeintercept form.
Also, this is the form we use when graphing a line with a graphing
calculator.
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
Section 4.3 Linear Equations in Two Variables I 93
OBJECTIVE Q Write an equation of a line given its slope and a point
on the line. Let m represent the slope of a line and (x^ y^ represent
a given point on the line. Let (x, y) represent any other point on the line. See
Figure 22. Then by the slope formula,
m
y y\
m(xx^ =yy^
Multiply by x
Rewrite.
Given point
Figure 22
This last equation is thQ pointslope form of the equation of a line.
PointSlope Form
The pointslope form of the equation of a line with slope m passing
through the point (x^ y^) is
Slope
I
yy^ = m(xx^).
A A
• Given point ■
To use this form to write the equation of a line, we need to know the coordi
nates of a point (x^ y^) and the slope m of the line.
EXAMPLE 3
Using the PointSlope Form
Find an equation of the line with slope  passing through the point (2, 5).
Use the pointslope form of the equation of a line, with (x^,y^) = (—2, 5)
andm = i
jjj = m(xx^)
Pointslope form
y
1
1
[x (2)] y = 5^rn = ix. = 2
y5 = {x+2)
3y 15 = x + 2
x + 3y = 17
Multiply by 3.
Subtracts; add 15.
In Section 4.1, we defined standard form for a linear equation as
Ax\By= C,
where A, B, and C are real numbers. Most often, however. A, B, and C are
integers. In this case, let us agree that integers A, B, and C have no common
factor (except 1) and^ > 0. For example, the final equation in Example 3,
—x + 3y= 17, is written in standard form as x — 3j = — 17.
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I 94 Chapter 4 Graphs, Linear Equations, and Functions
& Write an equation of each
line in standard form.
(a) Through (2, 7); m = 3
NOTE
The definition of "standard form" is not standard from one text to
another. Any linear equation can be written in many different (all
equally correct) forms. For example, the equation 2x + Sj = 8 can be
written as 2x = 8  Sj, Sj = 8  2x, x + fj = 4, 4x + 67 = 16, and
so on. In addition to writing it in standard form Ax + By = C with
^ > 0, let us agree that the form 2x + Sj = 8 is preferred over any
multiples of each side, such as 4x + 6y = 16. (To write 4x + 6j = 16
in standard form, divide each side by 2.)
'Hi
Work Problem 3 at the Side.
(b) Through (1, 3); m
5
4
OBJECTIVE Q Write an equation of a line given two points on the
line. To find an equation of a line when two points on the line are known,
first use the slope formula to find the slope of the line. Then use the slope with
either of the given points and the pointslope form of the equation of a line.
fouTry II
Videoi
O Write an equation in standard
form for each line.
(a) Through ( 1, 2) and (5, 7)
(b) Through (2, 6) and (1, 4)
Answers
3. (a) 3x  y =
4. (a) 5x  63; =
13 (b) 5x + 4y= 17
17 (b) 2x + 3y= 14
EXAMPLE 4
Finding an Equation of a Line Given Two Points
Find an equation of the line through the points (4, 3) and (5, 7). Write
the equation in standard form.
First find the slope by using the slope formula.
m
(4)
10
9
Use either (—4, 3) or (5, —7) as (x^,y^) in the pointslope form of the equa
tion of a line. If we choose (—4, 3), then —4 = x. and 3 = j..
>'Jj=«f(xXj)
j3 = y[x(4)]
10,
y3 =  — {x + 4)
40
Pointslope form
>>, = 3,OT = f,x^
4
9y27 = lOx
10x + 9y= 13
Verify that if (5, —7) were used, the same equation would result.
Multiply by 9; distributive property
Standard form
IK
Work Problem 4 at the Side.
A horizontal line has slope 0. From the pointslope form, the equation of
a horizontal line through the point (a, b) is
yy^ = m{x
X,)
Pointslope form
yb = 0(x
a)
y^ = b,m = 0,x^ = a
y b =
Multiplication property of
y = b.
AMb.
Notice that the pointslope form does not apply to a vertical line, since
the slope of a vertical line is undefined. A vertical line through the point
(a, b) has equations = a.
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Section 4.3 Linear Equations in Two Variables I 95
In summary, horizontal and vertical lines have the following special
equations.
Equations of Horizontal and Vertical Lines
The horizontal line through the point (a, b) has equation j = b.
The vertical line through the point (a, b) has equation x = a.
& Write an equation for each
line.
(a) Through (8, 2); m =
Work Problem 5 at the Side.
>li
OBJECTIVE Q Write an equation of a line parallel or perpendicular to
a given line. As mentioned in the previous section, parallel lines have the
same slope and perpendicular lines have slopes with product — 1.
EXAMPLE 5
Finding Equations of Lines Parallel or Perpendicu
lar to a Given Line
Find the equation in slopeintercept form of the line passing through the
point (4, 5) and (a) parallel to the line 2x + Sj = 6; (b) perpendicular to
the line 2x + Sj = 6.
(a) The slope of the line 2x + Sj = 6 can be found by solving for j.
2x + Sj = 6
3j = — 2x + 6 Subtract 2x.
y
X + 2
3
Divide by 3.
i_
Slope
The slope is given by the coefficient of x, so m = — . See the figure. Since
parallel lines have the same slope, the required equation of the line through
(—4, 5) and parallel to 2x + 3j = 6 must also have slope — . To find this
equation, use the pointslope form, with (x^ y^ = (—4, 5) and m = — .
 5 =
3[x(
4)]
jj = 5, m = , Xj =
4
 5 =
(x + 4)
3
^
 5 =
2 8
3^3
Distributive property
...j.=
^
y =
2 8 1^
Add 5 = f .
i4M
%
y =
2 7
Combine like terms.
We did not clear fractions after the substitution step here because we want
the equation in slopeintercept form — that is, solved for y. Both lines are
shown in the figure.
Continued on Next Page
(b) The vertical line through
(3,5)
Answers
5. (a) y= 1 (b) X = 3
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I 96 Chapter 4 Graphs, Linear Equations, and Functions
© Write an equation in slope
intercept form of the line
passing through the point
(8, 3) and
(a) parallel to the line
2x  3j = 10.
(b) In part (a), the given line 2x + 3y = 6 was written as
^ 3
so the line has slope — . To be perpendicular to the line 2x + 3j = 6, a
line must have a slope that is the negative reciprocal of — , which is .
y
y
Use the point (—4, 5) and slope f in the pointslope form to get the equa
tion of the perpendicular line shown in the figure.
3
2
3
■[x (4)] y^ = 5,m = ix^ = 4
(x + 4)
5 = X + 6
2
y
X + 11
Distributive property
Adds.
H
Work Problem 6 at the Side.
A summary of the various forms of linear equations follows.
FORMS OF LINEAR EQUATIONS
(b) perpendicular to the line
2x3y= 10.
H^ft Equation
^^^^ Description ^^^^
When to Use ^^
y = mx + b
SlopeIntercept Form
Slope is m.
jintercept is (0, b).
The slope and j^intercept
can be easily identified
and used to quickly graph
the equation.
JJi = m(xx^)
PointSlope Form
Slope is m.
Line passes through (Xp y^).
This form is ideal for
finding the equation of a
line if the slope and a point
on the line or two points
on the line are known.
Ax\By= C
Standard Form
(A, B, and C integers, ^ > 0)
Slope is (5 ^ 0).
xintercept is (£ 0) {A i Q).
jintercept is (0, §) {B i^ 0).
The X and jintercepts
can be found quickly
and used to graph the
equation. Slope must be
calculated.
y = b
Horizontal Line
Slope is 0.
jintercept is (0, b).
If the graph intersects
only the jaxis, then y is
the only variable in the
equation.
X = a
Vertical Line
Slope is undefined,
xintercept is {a, 0).
If the graph intersects
only the xaxis, then x is
the only variable in the
equation.
Answers
2 25
6. (a) J = X + y
(b) y =
3
2
OBJECTIVE Q Write an equation of a line that models real data. We
can use the information presented in this section to write equations of lines
that mathematically describe, or model, real data if the given set of data
changes at a fairly constant rate. In this case, the data fit a linear pattern, and
the rate of change is the slope of the line.
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Section 4.3 Linear Equations in Two Variables I 97
EXAMPLE 6
Determining a Linear Equation to Describe
Real Data
Suppose it is time to fill your car with gasoline. At your local station,
89octane gas is selling for $1.80 per gal.
(a) Write an equation that describes the cost j to buy x gallons of gas.
Experience has taught you that the total price you pay is determined by
the number of gallons you buy multiplied by the price per gallon (in this
case, $1.80). As you pump the gas, two sets of numbers spin by: the number
of gallons pumped and the price for that number of gallons.
The table uses ordered pairs to illustrate this situation.
I Number of
Gallons Pumped
Price of This
Number of Gallons
0($1.80) = $0.00
1
1($1.80) = $1.80
2
2($1.80) = $3.60
3
3($1.80) = $5.40
4
4($1.80) = $7.20
%
m
^ ^h3
If we let X denote the number of gallons pumped, then the total price j in
dollars can be found by the linear equation
Total price
y= 1.80x.
Number of gallons
Theoretically, there are infinitely many ordered pairs (x, y) that satisfy this
equation, but here we are limited to nonnegative values for x, since we can
not have a negative number of gallons. There is also a practical maximum
value for x in this situation, which varies from one car to another. What
determines this maximum value?
(b) You can also get a car wash at the gas station if you pay an addi
tional $3.00. Write an equation that defines the price for gas and a car
wash.
Since an additional $3.00 will be charged, you pay 1.80x + 3.00 dollars
for X gallons of gas and a car wash, described by
y = 1.8x + 3. Delete unnecessary Os.
(c) Interpret the ordered pairs (5, 12) and (10, 21) in relation to the equation
from part (b).
The ordered pair (5, 12) indicates that the price of 5 gal of gas and a car
wash is $12.00. Similarly, (10, 21) indicates that the price of 10 gal of gas
and a car wash is $21.00.
O (a) Suppose it costs $.10 per
minute to make a long
distance call. Write an
equation to describe
the cost J to make an
xminute call.
(b) Suppose there is a flat rate
of $.20 plus a charge of
$.10 per minute to make
a call. Write an equation
that gives the costj for a
call of X minutes.
(c) Interpret the ordered pair
(15, 1.7) in relation to the
equation from part (b).
Work Problem 7 at the Side,
^11
NOTE
In Example 6(a), the ordered pair (0, 0) satisfied the equation, so the
linear equation has the form y = mx, where /? = 0. If a situation
involves an initial charge b plus a charge per unit m as in Example 6 (b),
the equation has the form j = mx + b, where b i= 0.
Answers
7. (a) y= Ax (Note: .lOx = .Ix)
(b) y= Ax+ .2
(c) The ordered pair (15,1 .7) indicates that
the price of a 15minute call is $1.70.
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I 98 Chapter 4 Graphs, Linear Equations, and Functions
O The percent of mothers of
children under 1 yr old who
participated in the U.S. labor
force is shown in the table for
selected years.
EXAMPLE 7
Year
Percent
1980
38
1984
47
1988
51
1992
54
1998
59
Source: U.S. Bureau
of the Census.
(a) Letx = represent 1980,
X = 4 represent 1984, and
so on. Use the data for
1980 and 1998 to find an
equation that models the
data.
(b) Use the equation from
part (a) to approximate the
percentage of mothers of
children under 1 yr old who
participated in the U.S.
labor force in 2000.
Finding an Equation of a Line That Models Data
Average annual tuition and fees for instate students at public 4year colleges
are shown in the table for selected years and graphed as ordered pairs
of points in the scatter diagram in Figure 23, where x = represents 1990,
X = 4 represents 1994, and so on, and j represents the cost in dollars.
Year 1 Cost (in dollars)
1990
2035
1994
2820
1996
3151
1998
3486
2000
3774
4000
3000
2000<»
o 1000
Source: U.S. National Center
for Education Statistics.
2
4 6
Year
Figure 23
10
(a) Find an equation that models the data.
Since the points in Figure 23 lie approximately on a straight line, we can
write a linear equation that models the relationship between year x and
cost J. We choose two data points, (0, 2035) and (10, 3774), to find the slope
of the line.
3774  2035 1739
m
100
10
173.9
The slope 173.9 indicates that the cost of tuition and fees for instate stu
dents at public 4year colleges increased by about $174 per year from 1990
to 2000. We use this slope, the jintercept (0, 2035), and the slopeintercept
form to write an equation of the line. Thus,
y
173.9X + 2035.
(b) Use the equation from part (a) to approximate the cost of tuition and fees
at public 4year colleges in 2002.
The value x = 12 corresponds to the year 2002, so we substitute 12 for x
in the equation.
y = 173.9X + 2035
y= 173.9(12) + 2035
7 = 4121.8
According to the model, average tuition and fees for instate students at
public 4year colleges in 2002 were about $4122.
NOTE
In Example 7, if we had chosen different data points, we would have
found a slightly different equation. However, all such equations should
yield similar results, since the data points are approximately linear.
HIK
Work Problem 8 at the Side.
Answers
8. (a) y =
1.17X + 38 (b) 61%
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Section 4.4 Linear Inequalities in Two Variables 205
4.4 Linear Inequalities in Two Variables
B J ECTi VE Q Graph linear inequalities in two variables. In
Section 3.1 we graphed linear inequalities in one variable on the number
line. We now graph linear inequalities in two variables on a rectangular
coordinate system.
Linear Inequality In Two Variables
An inequality that can be written as
Ax\ By<C or Ax + By> C,
where A, B, and C are real numbers and A and B are not both 0, is a
linear inequality in two variables.
OBJECTIVES
Graph linear inequalities
in two variables.
Graph the intersection of
two linear inequalities.
Graph the union of two
linear inequalities.
The symbols < and > may replace < and > in the definition.
Consider the graph in Figure 24. The graph of the line x \ y = 5 divides
the points in the rectangular coordinate system into three sets:
1. Those points that lie on the line itself and satisfy the equation x + j = 5
[like (0,5), (2, 3), and (5,0)];
2. Those that lie in the halfplane above the line and satisfy the inequality
X + J > 5 [like (5, 3) and (2, 4)];
3. Those that lie in the halfplane below the line and satisfy the inequality
X + J < 5 [like (0, 0) and (3,  1)].
The graph of the line x + j = 5 is called the boundary line for the
inequalities x \ y > 5 and x \ y < 5. Graphs of linear inequalities in two
variables are regions in the real number plane that may or may not include
boundary lines.
3
[■.■\x + y = 5^
..X +y > 5. ..;....;
S^(2,.4)r;r;:
rir(3,l)
\Tx+y < 5 "
(0, 0). .;.^S^
Figure 24
To graph a linear inequality in two variables, follow these steps.
Graphing a Linear Inequality
Step 1 Draw the graph of the straight line that is the boundary.
Make the line solid if the inequality involves < or >; make the
line dashed if the inequality involves < or >.
Step 2 Choose a test point. Choose any point not on the line, and
substitute the coordinates of this point in the inequality.
Step 3 Shade the appropriate region. Shade the region that includes
the test point if it satisfies the original inequality; otherwise,
shade the region on the other side of the boundary line.
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206 Chapter 4 Graphs, Linear Equations, and Functions
Graph each inequality,
(a) X + 7 < 4
(b) 3x + 7 > 6
EXAMPLE 1
Graphing a Linear Inequality
Graph 3x + 2y ^ 6.
Step 1 First graph the line 3x + 2y = 6. The graph of this line, the bound
ary of the graph of the inequality, is shown in Figure 25.
Step 2
Answers
1. (a) y
Step 3
^ '\
\
3x + 2y = 6 "I
\
2\^
::::::5::i:::::
:::j::::::Kl::::j
Figure 25
The graph of the inequality 3x + 2j > 6 includes the points of the
boundary line 3x + 2y = 6 and either the points above the line
3x + 2j = 6 or the points below that line. To decide which, select
any point not on the line 3x + 2j = 6 as a test point. The origin,
(0, 0), is often a good choice. Substitute the values from the test
point (0, 0) for x and j in the inequality.
3x + 2y> 6
3(0) + 2(0) > 6 ?
> 6 False
Because the result is false, (0, 0) does not satisfy the inequality,
and so the solution set includes all points on the other side of the
line. This region is shaded in Figure 26.
3a: + 2j > 6
Figure 26
X + y <4
(b) y
Work Problem 1 at the Side.
If the inequality is written in the form j > mx + b or y < mx + b, the
inequality symbol indicates which halfplane to shade.
If J > mx \ b, shade above the boundary line.
If J < mx + b, shade below the boundary line.
This method works only if the inequality is solved for y.
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Section 4.4 Linear Inequalities in Two Variables 207
EXAMPLE 2
Graphing a Linear Inequality
Graph x  3y < 4.
First graph the boundary line, shown in Figure 27. The points of the
boundary line do not belong to the inequality x — 3y < 4 (because the
inequality symbol is <, not <). For this reason, the line is dashed. Now
solve the inequality for j.
X  3y<4
— 3y< — X + 4 Subtracts.
X 4
J>   Multiply by
& Graph each inequality.
(a) X  y>2
3; change < to >.
Because of the is greater than symbol, shade above the line. As a check,
choose a test point not on the line, say (1, 2), and substitute for x and 7 in the
original inequality.
x3y<4
l3(2)<4 ?
5<4 True
This result agrees with the decision to shade above the line. The solution set,
graphed in Figure 27, includes only those points in the shaded halfplane
(not those on the line).
(b) 3x + 47 < 12
^
jX.3y;.<4j
1 1 •2
(1,2)1
•■■••••••i
t^fr
lO
■^^^■
:eem=a:
Figure 27
Work Problem 2 at the Side.
OBJECTIVE Q Graph the intersection of two linear inequalities. In
Section 3.2 we discussed how the words and and or are used with compound
inequalities. In that section, the inequalities had one variable. Those ideas
can be extended to include inequalities in two variables.
A pair of inequalities joined with the word and is interpreted as the in
tersection of the solution sets of the inequalities. The graph of the intersec
tion of two or more inequalities is the region of the plane where all points
satisfy all of the inequalities at the same time.
Answers
2. (a) y
I I I I I I
2 ^
i/i I I I I *" ^
xy>2
(b)
3x + 4y< 12
3
 ' N
4"^
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208 Chapter 4 Graphs, Linear Equations, and Functions
& Graph x — j < 4 and x
O Graph 7x  3j < 21 orx > 2.
Answers
y
^/K^i
lx3y< 21
J
or x>2
/I
7 1
7
EXAMPLE 3
Graphing the Intersection of Two Inequalities
Graph 2x + 4j > 5 and x > 1 .
To begin, we graph each of the two inequalities 2x + 4j > 5 andx > 1
separately. The graph of 2x + 4j > 5 is shown in Figure 28(a), and the graph
of X > 1 is shown in Figure 28(b).
3^
2
■ 2
2
\x > i;
2x + 4y> 5
and jc > 1
(a)
(b)
Figure 28
(c)
In practice, the two graphs in Figures 28(a) and 28(b) are graphed on the
same axes. Then we use heavy shading to identify the intersection of the
graphs, as shown in Figure 28(c). To check, we can use a test point from each
of the four regions formed by the intersection of the boundary lines. Verify
that only ordered pairs in the heavily shaded region satisfy both inequalities.
ii<
Work Problem 3 at the Side.
OBJECTIVE Q Graph the union of two linear inequalities. When two
inequalities are joined by the word or, we must find the union of the graphs
of the inequalities. The graph of the union of two inequalities includes all
of the points that satisfy either inequality.
EXAMPLE 4
Graphing the Union of Two Inequalities
Graph 2x + 4;; > 5 or x > 1.
The graphs of the two inequalities are shown in Figures 28(a) and 28(b)
in Example 3. The graph of the union is shown in Figure 29.
3
;
•s
^s^ ^
; LO.
2x + Ay > 5
or jc > 1
Figure 29
Work Problem 4 at the Side.
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Section 4.5 Introduction to Functions
213
4.5 Introduction to Functions
We often describe one quantity in terms of another. Consider the following.
• The amount of your paycheck if you are paid hourly depends on the number
of hours you worked.
• The cost at the gas station depends on the number of gallons of gas you
pumped into your car.
• The distance traveled by a car moving at a constant speed depends on the
time traveled.
We can use ordered pairs to represent these corresponding quantities. For
example, we indicate the relationship between the amount of your paycheck
and hours worked by writing ordered pairs in which the first number repre
sents hours worked and the second number represents paycheck amount in
dollars. Then the ordered pair (5, 40) indicates that when you work 5 hr, your
paycheck is $40. Similarly, the ordered pairs (10, 80) and (20, 160) show that
working 10 hr results in an $80 paycheck and working 20 hr results in a
$160 paycheck.
OBJECTIVES
Define and identify
relations and functions.
Find domain and range.
Identify functions defined
by graphs and equations.
Use function notation.
Identify linear functions.
O What would the ordered pair
(40, 320) in the correspon
dence between number of
hours worked and paycheck
amount (in dollars) indicate?
Work Problem 1 at the Side.
Since the amount of your paycheck depends on the number of hours
worked, your paycheck amount is called the dependent variable, and the
number of hours worked is called the independent variable. Generalizing, if
the value of the variable y depends on the value of the variable x, then j is
the dependent variable andx is the independent variable.
Independent variable — i i — Dependent variable
OBJECTIVE Q Define and identify relations and functions. Since
we can write related quantities using ordered pairs, a set of ordered pairs
such as
{(5, 40), (10, 80), (20, 160), (40, 320)}
is called a relation.
Relation
A relation is any set of ordered pairs.
A special kind of relation, called ?i function, is very important in mathemat
ics and its applications.
Function
A function is a relation in which, for each value of the first component
of the ordered pairs, there is exactly one value of the second component.
Answers
1. It indicates that when you work 40 hr, your
paycheck is $320.
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2 I 4 Chapter 4 Graphs, Linear Equations, and Functions
Determine whether each
relation defines a function.
(a) {(0,3), (1,2), (1,3)}
(b) {(2, 2), (4, 4), (6, 6)}
EXAMPLE 1
Determining whether Relations Are Functions
Tell whether each relation defines a function.
F= {(1,2), (2, 4), (3,1)}
G= {(2,1), (1,0), (0,1), (1,2), (2, 2)}
7/ ={(4,1), (2,1), (2,0)}
Relations F and G are functions, because for each different x value there
is exactly one j value. Notice that in G, the last two ordered pairs have
the same jvalue (1 is paired with 2, and 2 is paired with 2). This does not
violate the definition of function, since the first components (xvalues) are
different and each is paired with only one second component (jvalue).
In relation H, however, the last two ordered pairs have the same xvalue
paired with two different yvahxQS (—2 is paired with both 1 and 0), so 7/ is a
relation but not a function. In afunction, no two ordered pairs can have the
same first component and different second components.
H
Different 3; values
{(4,1), (2,1), (2,0)}
t t
Not a function
Same xvalue
Work Problem 2 at the Side.
(c) {(1,5), (0,5)}
In a function, there is exactly one value of the dependent variable, the
second component, for each value of the independent variable, the first
component. This is what makes functions so important in applications.
Relations and functions can also be expressed as a correspondence or
mapping from one set to another, as shown in Figure 30 for function F and
relation H from Example 1. The arrow from 1 to 2 indicates that the ordered
pair (1, 2) belongs to F — each first component is paired with exactly one
second component. In the mapping for set H, which is not a function, the first
component —2 is paired with two different second components, 1 and 0.
F is a function. H is not a function.
Figure 30
Since relations and functions are sets of ordered pairs, we can represent
them using tables and graphs. A table and graph for function F is shown in
Figure 31.
Answers
2. (a) not a function
(c) function
(b) function
X
y
1
2
2
4
3
1
(2,4),
• (1,2);
• (3,1)
Graph of F
Figure 31
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Section 4.5 Introduction to Functions
215
Finally, we can describe a relation or function using a rule that tells how
to determine the dependent variable for a specific value of the independent
variable. The rule may be given in words, such as "the dependent variable is
twice the independent variable." Usually, however, the rule is given as an
equation:
y =
2x.
f
t
;pendent
Independent
variable
variable
An equation is the most efficient way to define a relation or function.
• QL=2^ '.^^^^ (Output);)
Function machine
NOTE
Another way to think of a function relationship is to think of the indepen
dent variable as an input and the dependent variable as an output. This is
illustrated by the inputoutput (function) machine in the margin for the
function defined by 7 = 2x.
OBJECTIVE Q Find domain and range. For every relation, there are
two important sets of elements called the domain and range.
Domain and Range
In a relation, the set of all values of the independent variable (x) is the
domain. The set of all values of the dependent variable (y) is the range.
Finding Domains and Ranges of Relations
Give the domain and range of each relation. Tell whether the relation defines
a function.
(a) {(3,1), (4, 2), (4, 5), (6, 8)}
The domain, the set of xvalues, is {3, 4, 6}; the range, the set of j values,
is { — 1, 2, 5, 8}. This relation is not a function because the same xvalue 4 is
paired with two different jvalues, 2 and 5.
(c)
X
y
5
2
2
5
2
The domain of this relation is
{4,6,7,3}.
The range is
{A,B,C}.
This mapping defines a function —
each xvalue corresponds to exactly
one J value.
This is a table of ordered pairs,
so the domain is the set of xvalues,
{5, 0, 5}, and the range is the set
of jvalues, {2}. The table defines
a function because each different
xvalue corresponds to exactly
one jvalue (even though it is the
same j value).
Work Problem 3 at the Side.
& Give the domain and range of
each relation. Does the
relation define a function?
(a) {(4,0), (4,1), (4, 2)}
(c)
Year
Cell Phone
Subscribers
(in thousands)
1995
33,786
1996
44,043
1997
55,312
1998
69,209
1999
86,047
Source: Cellular
Telecommunications
Industry Association.
Answers
3. (a) domain: {4}; range: {0, 1,2}; No, the
relation does not define a function.
(b) domain: {1, 4, 7}; range: {0, 2, 3, 7};
No, the relation does not define a function.
(c) domain: {1995, 1996, 1997, 1998, 1999};
range: {33,786, 44,043, 55,312, 69,209,
86,047}; Yes, the relation defines a
function.
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2 I 6 Chapter 4 Graphs, Linear Equations, and Functions
O Give the domain and range of
each relation.
(a)
y
t (2,3)
(2,2)  •
++b
(3,1)
I I I
(3,2)
The graph of a relation gives a picture of the relation, which can be used
to determine its domain and range.
EXAMPLE 3
Finding Domains and Ranges from Graphs
Give the domain and range of each relation.
(a) y (b) Domain
t
(1, 1) _
_::(i,2);t::
(0,1)'* 1
—J4^i)
The domain is the set of xvalues,
{1,0,1,4}.
The range is the set of jvalues,
{3,1,1,2}.
The arrowheads indicate that
the line extends indefinitely left and
right, as well as up and down.
Therefore, both the domain and
the range include all real numbers,
written (—0°, ^).
The xvalues of the points on
the graph include all numbers be
tween —4 and 4, inclusive. The y
values include all numbers between
6 and 6, inclusive. Using interval
notation.
the domain is [4, 4];
the range is [—6, 6].
(d)
k^
The arrowheads indicate that
the graph extends indefinitely left
and right, as well as upward. The
domain is (—0°, 0°). Because there is
a least jvalue, —3, the range in
cludes all numbers greater than or
equal to 3, written [3, oo).
Work Problem 4 at the Side.
Since relations are often defined by equations, such as j = 2x + 3 and
y^ = X, we must sometimes determine the domain of a relation from its equa
tion. We assume the following agreement on the domain of a relation.
Agreement on Domain
The domain of a relation is assumed to be all real numbers that produce
real numbers when substituted for the independent variable.
Answers
4. (a) domain: {3, 2,2,3};
range: {2, 1,2,3}
(b) domain: [— 2, oo); range: (— oo^ oo)
(c) domain: (0°, oo); range: (oo, 0]
To illustrate this agreement, since any real number can be used as a
replacement for x inj = 2x + 3, the domain of this function is the set of all
real numbers. The function defined hy y = ^ has all real numbers except
as domain, since y is undefined if x = 0. In general, the domain of a func
tion defined by an algebraic expression is all real numbers, except those
numbers that lead to division by or an even root of a negative number.
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Section 4.5 Introduction to Functions
217
OBJECTIVE Q Identify functions defined by graphs and equations.
Since each value of x in a function corresponds to only one value of j, any
vertical line drawn through the graph of a function must intersect the graph
in at most one point. This is the vertical line test for a function.
Vertical Line Test
If every vertical line intersects the graph of a relation in no more than
one point, then the relation represents a function.
For example, the graph shown in Figure 32(a) is not the graph of a function
since a vertical line intersects the graph in more than one point. The graph in
Figure 32(b) does represent a function.
& Use the vertical line test to
decide which graphs
represent functions.
Not a function the same
xvalue corresponds to
four different J values.
(a)
^. Sz'V
B.
Function each jc value
corresponds to only one
jvalue.
(b)
Figure 32
y
/
/
/ "
EXAMPLE 4
Using the Vertical Line Test
Use the vertical line test to determine whether each relation graphed in
Example 3 is a function.
(a)
(1,1)
(0,1)
(b)
(1,2)
o(4, 3)
Function
Not a function
(d)
1
W
Function
Function
The graphs in (a), (c), and (d) represent functions. The graph of the rela
tion in (b) fails the vertical line test, since the same xvalue corresponds to
two different J values; therefore, it is not the graph of a function.
Work Problem 5 at the Side.
Answers
5. A and C are graphs of functions.
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2 I 8 Chapter 4 Graphs, Linear Equations, and Functions
© Decide whether each relation
defines a function, and give
the domain.
(a) J = 6x + 12
NOTE
Graphs that do not represent functions are still relations. Remember
that all equations and graphs represent relations and that all
relations have a domain and range.
It can be more difficult to decide whether a relation defined by an equa
tion is a function. The next example gives some hints that may help.
(b) J < 4x
(c) y
Vix
(d) y^ = 25x
Answers
6. (a) yes; (oo^ oo) (b) no; (oo^ oo)
"2
(c) yes;
EXAMPLE 5
Identifying Functions from Their Equations
Decide whether each relation defines a function and give the domain.
(a) 7 = X + 4
In the defining equation, y = x + 4, y is always found by adding 4 to x.
Thus, each value of x corresponds to just one value ofy and the relation de
fines a function; x can be any real number, so the domain is (oo, oo).
(b) y = Vlx 1
For any choice of x in the domain, there is exactly one corresponding
value for J (the radical is a nonnegative number), so this equation defines
a function. Since the equation involves a square root, the quantity under the
radical sign cannot be negative. Thus,
2x 1
2x
X ^
1
2'
and the domain of the function is [^, oo) .
(c) y^ = X
The ordered pairs (16, 4) and (16, 4) both satisfy this equation. Since
one value of x, 16, corresponds to two values ofj, 4 and —4, this equation
does not define a function. Because x is equal to the square of j, the values
of X must always be nonnegative. The domain of the relation is [0, oo).
(d) J < X  1
By definition, j is a function of x if every value of x leads to exactly one
value ofy. Here a particular value of x, say 1, corresponds to many values of
y. The ordered pairs (1, 0), (1, —1), (1, —2), (1, —3), and so on, all satisfy
the inequality. Thus, an inequality never defines a function. Any number
can be used for x, so the domain is the set of real numbers, (—0°, 0°).
(e) y = ^r
X — 1
Given any value of x in the domain, we find y by subtracting 1 , then
dividing the result into 5. This process produces exactly one value ofy for
each value in the domain, so this equation defines a function. The domain
includes all real numbers except those that make the denominator 0. We find
these numbers by setting the denominator equal to and solving for x.
X 1 =
x= 1
The domain includes all real numbers except 1, written (oo, 1) U (1, oo).
Work Problem 6 at the Side.
(d) no; [0, oo)
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Section 4.5 Introduction to Functions 2 I 9
In summary, three variations of the definition of function are given here.
Variations of the Definition of Function
1. A function is a relation in which, for each value of the first compo
nent of the ordered pairs, there is exactly one value of the second
component.
2. A function is a set of ordered pairs in which no first component is
repeated.
3. A function is a rule or correspondence that assigns exactly one
range value to each domain value.
OBJECTIVE Q Use function notation. When a function/is defined
with a rule or an equation using x and j for the independent and dependent
variables, we say "y is a function of x" to emphasize that j depends on x. We
use the notation
y=f{x),
called function notation, to express this and read/(x) as "/of x." (In this
special notation the parentheses do not indicate multiplication.) The letter/
stands ior function. For example, if j = 9x — 5, we can name this function/
and write
f{x) = 9x  5.
Note that f{x) is just another name for the dependent variable y. For
example, if j =/(x) = 9x — 5 andx = 2, then we find j, or/(2), by replac
ing x with 2.
j=/(2) = 925
= 185
= 13.
For function/, the statement "if x = 2, theny = 13" is represented by the
ordered pair (2, 13) and is abbreviated with function notation as
/(2) = 13.
Read/(2) as '/of 2" or "/at 2." Also,
/(O) = 9 •  5 = 5 and /(3) = 9(3)  5 = 32.
These ideas can be illustrated as follows.
Name of the function
\ Defining expression
y =0% = ^^^
Value of the function Name of the independent variable
CAUTION
The symbol/(x) does not indicate "/times x," but represents the jvalue
for the indicated xvalue. As just shown, /(2) is the jvalue that corre
sponds to the X value 2.
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220 Chapter 4 Graphs, Linear Equations, and Functions
e Find f (3), f(p), and
f(m + 1).
(a) f(x) = 6x  2
uT
(b)/W
3x + 5
EXAMPLE 6
Using Function Notation
Let/(x) = — x^ + 5x — 3. Find the following.
(a)/(2)
/(x) = x^ + 5x3
/(2) =22 + 523 Replace x with 2.
/(2) = 4 + 10  3
/(2) = 3
Since/(2) = 3, the ordered pair (2, 3) belongs to/
(b)/(^)
/(x) = x^ + 5x  3
f(q) = —q^ + 5q — 3 Replace x with ^.
The replacement of one variable with another is important in later courses.
Sometimes letters other than/, such as g, h, or capital letters F, G, and H
are used to name functions.
EXAMPLE 7
Using Function Notation
Let g(x) = 2x + 3. Find and simplify g (a +1).
g(x) = 2x + 3
g(a + 1) = 2(a + 1) + 3 Replace x with (2 + 1.
= 2(2 + 2 + 3
= 2a + 5
Work Problem 7 at the Side.
(c) fix) =x\
o
Answers
7. (a) 20; 6j9  2; 6m + 4
3 1 1 ,
(c) ;7/'  l\{m + 1)  lor
2 6 6
1 5
6" "6
Functions can be evaluated in a variety of ways, as shown in Example 8.
EXAMPLE 8
Using Function Notation
For each function, find/(3).
(a) fix) = 3xl
/(3) = 3(3)  7
/(3) = 97
/(3) = 2
(b)/= {(3, 5), (0,3), (3,1), (6,1)}
We want /(3), the jvalue of the
ordered pair where x = 3. As indicated
by the ordered pair (3, 1), when x = 3,
7=1, so/(3) = 1.
^ ^ Domain
Range
The domain element 3 is paired with 5 in the
range, so/(3) = 5.
Continued on Next Page
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Section 4.5 Introduction to Functions 22 I
k
;
/
40
; 2 ^
^ 4
Figure 33
To evaluate/(3), find 3 on the xaxis. See Figure 33. Then move up until
the graph of/is reached. Moving horizontally to the jaxis gives 4 for the
corresponding jvalue. Thus,/(3) = 4.
Work Problem 8 at the Side.
O For each function, find/(— 2).
(a) fix) = 4x  8
(b)/={(0,5),(l,3),
(2, 1)}
If a function/is defined by an equation with x and j, not with function
notation, use the following steps to find/(x).
Finding an Expression for f (x)
Step 1 Solve the equation for j.
Step 2 Replace y with/(x).
EXAMPLE 9
Writing Equations Using Function Notation
Rewrite each equation using function notation. Then find/(2) and/(a).
(a) J = x2 + 1
This equation is already solved fory. Since >> =/(x),
f{x) = X2 + 1.
Tofind/(2), letx= 2.
/(2) = (2)2 + 1
= 4+1
= 5
Find/(a) by letting x = a: f{a) = a^ + 1.
(h)x4y = 5
First solve x — 4>' = 5 for j. Then replace y with/(x).
X  4/ = 5
X — 5 = 4y Add 4y; subtract 5.
x5 I 5
y = ^ so fix)=x
Now find/(2) and/(a).
fi2) = '(2) '=' Letx^2.
f{a) = —a Letx = a.
4 4
Work Problem 9 at the Side.
(C) X
f{x)
4
16
2
4
2
4
4
16
O Rewrite each equation using
function notation. Then find
/(I)
(a) y = Vx + 2
(b) x2  4/ = 3
Answers
8. (a) (b) 1 (c) 4
9. (a) fix) = Vx + 2; 1
(b) fix) =^^^ or fix) 
1 2
= 4^ "
3
4
1
~ 2
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222 Chapter 4 Graphs, Linear Equations, and Functions
CO) Graph each linear function.
Give the domain and range.
(a) f{x) =7^2
OBJECTIVE B Identify linear functions. Our first twodimensional
graphing was of straight lines. Linear equations (except for vertical lines
with equations x = a) define linear functions .
Linear Function
A function that can be defined by
fix) = mx + b
for real numbers m and /? is a linear function.
Recall from Section 4.3 that m is the slope of the line and (0, b) is the
jintercept. In Example 9(b), we wrote the equation x — 4y = 5 as the linear
function defined by
/W =4^4
.^ ^.
(b) g(x) = 3
Slope ' ' jKintercept is (O,  1) .
To graph this function, plot the jintercept and use the definition of slope as
^ to find a second point on the line. Draw the straight line through the
points to obtain the graph shown in Figure 34.
fix) = ix  J
Figure 34
Answers
10. (a) y
f{x)=jx 2
domain: (—00,0°); range: (—00^ 00)
(b) 3^
g{x) = 3
>
I I I I I I " X
A linear function defined by/(jc) = b (whose graph is a horizontal line)
is sometimes called a constant function. The domain of any linear function
is (00, 00). The range of a nonconstant linear function is (00, 00), while the
range of the constant function defined by/(x) = b is {b}.
Work Problem 10 at the Side.
domain: (—00,00); range: {3}
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Section 4.6 Variation
229
4.6 Variation
C = lirr
Certain types of functions are very common, especially in business and
the physical sciences. These are functions where y depends on a multiple
of X, or J depends on a number divided by x. In such situations, y is said
to vary directly as x (in the first case) or vary inversely as x (in the second
case). For example, by the distance formula, the distance traveled varies
directly as the rate (or speed) and the time. The simple interest formula
and the formulas for area and volume are other familiar examples of
direct variation.
On the other hand, the force required to keep a car from skidding on
a curve varies inversely as the radius of the curve. Another example
of inverse variation is how travel time is inversely proportional to rate
or speed.
OBJECTIVE Q Write an equation expressing direct
variation. The circumference of a circle is given by the
formula C = lirr, where r is the radius of the circle. See the
figure. Circumference is always a constant multiple of the
radius. (C is always found by multiplying r by the constant
277.) Thus,
As the radius increases, the circumference increases.
The reverse is also true.
As the radius decreases, the circumference decreases.
Because of this, the circumference is said to vary directly as the radius.
Direct Variation
y varies directly as x if there exists some constant k such that
y = kx.
Also, y is said to be proportional to x. The number k is called the constant
of variation. In direct variation, for ^ > 0, as the value of x increases, the
value of J also increases. Similarly, as x decreases, y decreases.
OBJECTIVE Q Find thie constant of variation, and solve direct varia
tion problems. The direct variation equation j = kx defines a linear func
tion, where the constant of variation k is the slope of the line. For example,
we wrote the equation
y
\Mx
to describe the cost j to buy x gallons of gas in Example 6 of Section 4.3.
The cost varies directly as, or is proportional to, the number of gallons of gas
purchased. That is, as the number of gallons of gas increases, cost increases;
also, as the number of gallons of gas decreases, cost decreases. The constant
of variation k is 1.80, the cost of 1 gallon of gas.
OBJECTIVES
Write an equation
expressing direct
variation.
Find the constant of
variation, and solve direct
variation problems.
Solve inverse variation
problems.
Solve joint variation
problems.
Solve combined variation
problems.
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230 Chapter 4 Graphs, Linear Equations, and Functions
Find the constant of variation,
and write a direct variation
equation.
(a) Suzanne Alley is paid a
daily wage. One month she
worked 17 days and earned
$1334.50.
EXAMPLE 1
Finding the Constant of Variation and the Variation
Equation
Gina Linko is paid an hourly wage. One week she worked 43 hr and was paid
$795.50. How much does she earn per hour?
Let h represent the number of hours she works and P represent her
corresponding pay. Then, P varies directly as /z, so
P = kh.
Here ^represents Gina's hourly wage. Since P = 795.50 when h = 43,
795.50 = 43^
k = 18.50. Use a calculator.
Her hourly wage is $18.50, and P and h are related by
P = 18.50/z.
w
Work Problem 1 at the Side.
(b) Distance varies directly as
time (at a constant speed).
A car travels 100 mi at a
constant speed in 2 hr.
& The charge (in dollars) to
customers for electricity
(in kilowatthours) varies
directly as the number of
kilowatthours used. It costs
$52 to use 800 kilowatthours.
Find the cost to use 1000
kilowatthours.
Answers
1. (a) k = 78.50; Let E represent her earnings
for d days. Then E = IS.SOd.
(b) A: = 50; Let d represent the distance
traveled in h hours. Then d = 50h.
2. $65
EXAMPLE 2
Solving a Direct Variation Problem
Hooke's law for an elastic spring states that the distance a spring stretches is
proportional to the force applied. If a force of 150 newtons"^ stretches a
certain spring 8 cm, how much will a force of 400 newtons stretch the spring?
Figure 35
See Figure 35. Ifd is the distance the spring stretches and/is the force
applied, then d = kffor some constant k. Since a force of 150 newtons
stretches the spring 8 cm, we can use these values to find k.
Variation equation
LQtd= 8and/= 150.
Find k.
d = kf
8 = A • 150
4
k = —
75
Substitute ji for k in the variation equation d = kfto get
For a force of 400 newtons,
4 , . 64
</ = (400) = .
Let/= 400.
The spring will stretch y cm if a force of 400 newtons is applied.
Hi
Work Problem 2 at the Side.
*A newton is a unit of measure offeree used in physics.
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Section 4.6 Variation
231
In summary, use the following steps to solve a variation problem.
Solving aVariation Problem
Step 1 Write the variation equation.
Step 2 Substitute the initial values and solve for k.
Step 3 Rewrite the variation equation with the value of ^ from Step 2.
Step 4 Substitute the remaining values, solve for the unknown, and
find the required answer.
The direct variation equation j = ^x is a linear equation. However, other
kinds of variation involve other types of equations. For example, one
variable can be proportional to a power of another variable.
& The area of a circle varies
directly as the square of its
radius. A circle with radius
3 in. has area 28.278 in.^.
(a) Write a variation equation
and give the value of ^.
Direct Variation as a Power
y varies directly as the wth power of x if there exists a real number k
such that
y = kx"".
An example of direct variation as a power is the formula for the area of a
circle, A = irr^. Here, tt is the constant of variation, and the area varies
directly as the square of the radius.
Solving a Direct Variation Problem
The distance a body falls from rest varies directly as the square of the time it
falls (disregarding air resistance). If a skydiver falls 64 ft in 2 sec, how far
will she fall in 8 sec?
Step 1 li d represents the distance the skydiver falls and t the time it
takes to fall, then (i is a function of t, and
= ha
d = kt
for some constant k.
Step 2 To find the value of k, use the fact that the skydiver falls 64 ft
in 2 sec.
d = kt^ Variation equation
64 = k(2y Let J = 64 and r = 2.
^ = 16 Find^.
Step 3 Using 16 for k, the variation equation becomes
d = I6t\
Step 4 Let ^ = 8 to find the number of feet the skydiver will fall in 8 sec.
d= 16(8)2 = 1024 Letr=8.
The skydiver will fall 1024 ft in 8 sec.
Work Problem 3 at the Side.
(b) What is the area of a circle
with radius 4.1 in.?
Answers
3. (a) A = kr^;3.l42
(b) 52.817 in.2 (to the nearest thousandth)
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232 Chapter 4 Graphs, Linear Equations, and Functions
OBJECTIVE Q Solve inverse variation problems. In direct varia
tion, where ^ > 0, as x increases, y increases. Similarly, as x decreases,
y decreases. Another type of variation is inverse variation. With inverse
variation, where ^ > 0, as one variable increases, the other variable
decreases. For example, in a closed space, volume decreases as pressure
increases, as illustrated by a trash compactor. See Figure 36. As the com
pactor presses down, the pressure on the trash increases; in turn, the trash
occupies a smaller space.
As pressure
on trash
increases,
volume of
trash
decreases.
Figure 36
Inverse Variation
y varies inversely as x if there exists a real number k such that
k
J = •
X
Also, y varies inversely as the nth power of x if there exists a real
number k such that
•^ = 7"'
The inverse variation equation also defines a function. Since x is in the
denominator, these functions are rational functions. (See Chapter 8.)
Another example of inverse variation comes from the distance formula. In
its usual form, the formula is
d = rt.
Dividing each side by r gives
Here, t (time) varies inversely as r (rate or speed), with d (distance) serving
as the constant of variation. For example, if the distance between Chicago
and Des Moines is 300 mi, then
300
r
and the values of r and t might be any of the following.
r = 50, ^ = 6 1
r = 60,t= 5
r = 75,t = 4
As r increases,
t decreases.
r = 30, ^ = 10
r = 25,t= 12
r = 20,t= 15
As r decreases,
t increases.
If we increase the rate (speed) we drive, time decreases. If we decrease the
rate (speed) we drive, time increases.
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Section 4.6 Variation
233
EXAMPLE 4
Solving an Inverse Variation Problem
The weight of an object above Earth varies inversely as the square of its
distance from the center of Earth. A space shuttle in an elliptical orbit has a
maximum distance from the center of Earth (apogee) of 6700 mi. Its mini
mum distance from the center of Earth (perigee) is 4090 mi. See Figure 37.
If an astronaut in the shuttle weighs 57 lb at its apogee, what does the astro
naut weigh at its perigee?
Space shuttle
at perigee
Space shuttle
at apogee
^^"/jCth
Figure 37
If w is the weight and d is the distance from the center of Earth, then
k
w
d'
for some constant k. At the apogee the astronaut weighs 57 lb, and the
distance from the center of Earth is 6700 mi. Use these values to find k.
O If the temperature is constant,
the volume of a gas varies
inversely as the pressure. For
a certain gas, the volume is
10 cm^ when the pressure is
6 kg per cm^.
(a) Find the variation equation.
57
Let w = 57 Sindd= 6700.
(6700)2
k= 57(6700)2
Then the weight at the perigee with d = 4090 mi is
w =
k 57(6700)^
d'
(4090)'
153 lb. Use a calculator.
Work Problem 4 at the Side.
(b) Find the volume when the
pressure is 12 kg per cm^.
OBJECTIVE Q Solve joint variation problems. It is possible for one
variable to depend on several others. If one variable varies directly as the
product of several other variables (perhaps raised to powers), the first
variable is said to vary jointly as the others.
Joint Variation
y varies jointly as x and z if there exists a real number k such that
y = kxz>
CAUTION
Note that and in the expression "y varies jointly as x and z" translates
as the product
y = kxz.
The word and does not indicate addition here.
Answers
4. (a) V
60
(b) 5cm3
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234 Chapter 4 Graphs, Linear Equations, and Functions
& The volume of a rectangular
box of a given height is
proportional to its width and
length. A box with width 2 ft
and length 4 ft has volume
12 ft^. Find the volume of a
box with the same height that
is 3 ft wide and 5 ft long.
& The maximum load that a
cylindrical column with a
circular cross section can
hold varies directly as the
fourth power of the diameter
of the cross section and
inversely as the square of
the height. A 9m column
1 m in diameter will support
8 metric tons. How many
metric tons can be supported
by a column 12 m high and
I m in diameter?
9m^
Load = 8 metric tons
EXAMPLE 5
Solving a Joint Variation Problem
The interest on a loan or an investment is given by the formula / = prt. Here,
for a given principal p, the interest earned / varies jointly as the interest rate
r and the time t that the principal is left at interest. If an investment earns
$100 interest at 5% for 2 yr, how much interest will the same principal earn
at 4.5% for Syr?
We use the formula / = prt, where /> is the constant of variation because
it is the same for both investments. For the first investment,
/ = prt
100 =;?(.05)(2) Let/= 100, r= .05, and ^ = 2.
100 = Ap
p = 1000. Divide by .1.
Now we find /when/? = 1000, r = .045, and t = 3.
/ = 1000(.045)(3) = 135 Let;? = 1000, r = .045, and t = 3.
The interest will be $135.
m
Work Problem 5 at the Side.
OBJECTIVE Q Solve Combined variation problems. There are many
combinations of direct and inverse variation, called combined variation.
EXAMPLE 6
Solving a Combined Variation Problem
Body mass index, or BMI, is used by physicians to assess a person's level
of fatness. A BMI from 19 through 25 is considered desirable. BMI varies
directly as an individual's weight in pounds and inversely as the square of the
individual's height in inches. A person who weighs 1 18 lb and is 64 in. tall
has a BMI of 20. (The BMI is rounded to the nearest whole number.) Find
the BMI of a person who weighs 165 lb with a height of 70 in.
Let B represent the BMI, w the weight, and h the height. Then
B =
kw
h
2 •
BMI varies directly as the weight.
BMI varies inversely as the square of the height.
To find^, let^ = 20, w = 118, and/z = 64.
^(118)
20
k =
64^
20 (64")
118
k^ 694
Multiply by 64^;
divide by 118.
Use a calculator.
Now find ^ when k = 694, w = 165, and h = 70.
694(165)
B
70'
The person's BMI is 23.
23
Nearest whole
number
Work Problem 6 at the Side.
Answers
5. 22.5 ft3
6. — metric ton
9
Copyright ©2005 Pearson Education, Inc., pubiisiiing as Pearson AddisonWesley
Systems of
Linear Equations
\
51
52
53
Systems of Linear Equations
in Two Variables
Systems of Linear Equations
in Three Variables
Applications of Systems
of Linear Equations
5.4 Solving Systems of Linear
Equations by Matrix Methods
During the last decade of the twentieth century, the
number of people living with AIDS in various racial
groups in the United States followed linear patterns, as
shown in the accompanying graph. At some year during
that decade, the number of people in two of these groups
was the same. The graph can be used to determine that year
and number by finding the coordinates of the point of in
tersection of the two lines. See Exercises 1 and 2 of the
Chapter 5 Test.
The process of determining the point of intersection of
two lines is the idea behind solving a system of linear
equations in two variables. This chapter illustrates meth
ods of finding such points.
People Living with AIDS
c
(0
E
Source: U.S. Centers for Disease
Control and Prevention.
African
American
140,000
White
128,000
Hispanic
67,000
Asian
3,000
251
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
252 Chapter 5 Systems of Linear Equations
5.1 Systems of Linear Equations in Two Variables
OBJECTIVES
Solve linear systems by
graphing.
Decide whether an
ordered pair is a solution
of a linear system.
Solve linear systems (with
two equations and two
variables) by substitution.
Solve linear systems (with
two equations and two
variables) by elimination.
Solve special systems.
As technology continues to improve, the sale of digital cameras increases,
while that of conventional cameras decreases. This can be seen in Figure 1,
which illustrates this growth and decline using a graph. The two straightline
graphs intersect where the two types of cameras had the same sales.
Say Cheese!
12 3 4 5
Years since 2000
Source: Consumer Electronics
Association.
Figure 1
We could use a linear equation to model the graph of digital camera
sales and another linear equation to model the graph of conventional camera
sales. Such a set of equations is called a system of equations, in this case a
linear system of equations. The point where the graphs in Figure 1 inter
sect is a solution of each of the individual equations. It is also the solution of
the linear system of equations.
OBJECTIVE Q Solve linear systems by graphing. The solution set of
a system of equations contains all ordered pairs that satisfy all the equa
tions of the system at the same time. An example of a linear system is
X +7 = 5
2x y = A.
Linear system
of equations
VideoJ
One way to find the solution set of a linear system of equations is to graph
each equation and find the point where the graphs intersect.
EXAMPLE 1
Solving a System by Graphing
Solve the system of equations by graphing.
x+j = 5 (1)
2xy = A (2)
When we graph these linear equations as shown in Figure 2, the graph
suggests that the point of intersection is the ordered pair (3, 2).
Continued on Next Page
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
Section 5.1 Systems of Linear Equations in Two Variables 253
Figure 2
To be sure that (3, 2) is a solution of both equations, we check by substitut
ing 3 for X and 2 for j in each equation.
X + J
3 + 2
5
(1)
True
2x
2(3)
6
■y
2
2
4
4
4
4
4
(2)
True
Since (3, 2) makes both equations true, {(3, 2)} is the solution set of the system.
Work Problem 1 at the Side.
lilll Calculator Tip A graphing calculator can be used to solve a system.
Each equation must be solved for y before being entered in the calcula
tor. The point of intersection of the graphs, which is the solution of the
system, can then be displayed. Consult your owner's manual for details.
OBJECTIVE Q Decide whether an ordered pair is a solution of a
linear system. To decide if an ordered pair is a solution of a system, we
substitute the ordered pair in both equations of the system, just as we did
when we checked the solution in Example 1.
Solve each system by
graphing.
(a) X
2x
■y
■y
3
4
(1)
(2)
(b) 2x + y= 5 (1)
X + 3j = 6 (2)
EXAMPLE 2
Deciding whether an Ordered Pair Is a Solution
Decide whether the given ordered pair is a solution of the given system.
""/""'"m^ (4,2)
4x  J = 14
Replace x with 4 andj with 2 in each equation of the system.
x+j = 6 4xj=14
4 + 2 = 6 ? 4(4)  2 = 14 ?
6 = 6 True 16  2 = 14 ?
14 = 14 True
Since (4, 2) makes both equations true, (4, 2) is a solution of the system.
Continued on Next Page
Answers
1. (a) {(1,2)}
(b) {(3,1)}
(3, 1)
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
254 Chapter 5 Systems of Linear Equations
& Are the given ordered pairs
solutions of the given
systems?
(a) 2x + y= 6
x + 3y = 2
(4,2)
(b) 3x + 2y= 11
x + 5y = 36' ^"^
i,7)
3x + 2y= 11
x+ 5y = 36
3(l) + 2(7)=ll
?
1 + 5(7) = 36
7
3 + 14= 11
7
1 + 35 = 36
?
11 = 11
True
34 = 36
False
The ordered pair (—1, 7) is not a solution of the system, since it does not
make both equations true.
Work Problem 2 at the Side.
Since the graph of a linear equation is a straight line, there are three pos
sibilities for the solution set of a linear system in two variables.
(b) 9x  y= 4
4x + 3y= 11
(1,5)
Graphs of Linear Systems in Two Variables
1. The two graphs intersect in a single point. The coordinates of this
point give the only solution of the system. In this case the system is
consistent, and the equations are independent. This is the most
common case. See Figure 3(a).
2. The graphs are parallel lines. In this case the system is inconsistent;
that is, there is no solution common to both equations of the system,
and the solution set is 0. See Figure 3(b).
3. The graphs are the same line. In this case the equations are
dependent, since any solution of one equation of the system is also
a solution of the other. The solution set is an infinite set of ordered
pairs representing the points on the line. See Figure 3(c).
No
solution
Answers
2. (a) yes (b) no
OBJECTIVE Q Solve linear systems (with two equations and two
variables) by substitution. Since it can be difficult to read exact coordi
nates, especially if they are not integers, from a graph, we usually use alge
braic methods to solve systems. One such method, the substitution method,
is most useful for solving linear systems in which one equation is solved or
can be easily solved for one variable in terms of the other.
EXAMPLE 3
Solve the system.
Solving a System by Substitution
2x — y = 6
X = y + 2
(1)
(2)
Since equation (2) is solved for x, substitute j + 2 for x in equation (1).
— Continued on Next Page
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
Section 5.1 Systems of Linear Equations in Two Variables 255
2xy = e
(1)
2{y + l)y = 6
Letx = J + 2.
2y + 4y = 6
Distributive property
y + 4 = 6
Combine terms.
y = 2
Subtract 4.
We found J. Now find x by substitutin
Lg 2 for J in equation (2).
x= y + 2
=2+2=4
Thus, X = 4 and 7 = 2, giving the ordered pair (4, 2). Check this solution in
both equations of the original system.
The solution set is {(4, 2)}.
& Solve by substitution.
(a) 7x 2y= 2
y = 3x
CAUTION
Be careful! Even though we found j first in Example 3, the xcoordi
nate is always written first in the ordered pair solution of a system.
Work Problem 3 at the Side.
^11
The substitution method is summarized as follows.
Solving a Linear System by Substitution
Step 1 Solve one of the equations for either variable. If one of the
variable terms has coefficient 1 or — 1, choose it, since the sub
stitution method is usually easier this way.
Step 2 Substitute for that variable in the other equation. The result
should be an equation with just one variable.
Step 3 Solve the equation from Step 2.
Step 4 Find the other value. Substitute the result from Step 3 into the
equation from Step 1 to find the value of the other variable.
Step 5 Check the solution in both of the original equations. Then
write the solution set.
(b) 5x  Sj = 6
x = 2 y
Solve the system.
Solving a System by Substitution
Step 2
3x + 2y
Ax  y
13
1
(1)
(2)
First solve one of the equations for x or j. Since the coefficient of j
in equation (2) is — 1, it is easiest to solve for j in equation (2).
Ax
■y
y
y
1
1 Ax
1 + 4x
Substitute 1 + 4x for j in equation (1).
3x + 2j; = 13 (1)
3x + 2(1 + Ax) = 13 Letj
Continued on Next Page
(2)
Subtract Ax.
Multiply by
1 +4x.
■1.
Answers
3. (a) {(2,6)} (b) {(0,2)}
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
256 Chapter 5 Systems of Linear Equations
O Solve by substitution.
(a) 3x  y= 10
2x + 5y= I
(b) 4x 5y= 11
x + 2y = 7
Step 3 Solve for x.
3x + 2(1 + 4x) = 13
3x + 2 + 8x = 13
llx= 11
x= 1
From Step 2
Distributive property
Combine terms; subtract 2.
Divide by 1 1 .
Step 4 Now solve for j. From Step 1, j = 1 + 4x, so
J = 1 + 4(1) = 5. Letx= 1.
Step 5 Check the solution (1, 5) in both equations (1) and (2).
3a: + 2j = 13
(1)
4x — J =
1
3(1) + 2(5) =13
?
4(1)  5 =
1
7
3 + 10= 13
?
45 =
1
?
13 = 13
True
1 =
1
The solution set is
{(1,5)}.
(2)
True
Work Problem 4 at the Side.
OBJECTIVE Q Solve linear systems (with two equations and two
variables) by elimination. Another algebraic method, the elimination
method, involves combining the two equations in a system so that one vari
able is eliminated. This is done using the following logic:
If a = h and c = d, then a \ c = b \ d.
& Solve by elimination.
(a) 3x  J
2x + y
(b) 2x + 3j= 10
2x + 2j = 5
EXAMPLE 5
Solving a System by Elimination
Solve the system.
2x + 3j; = 6 (1)
4x  3 J = 6 (2)
Notice that adding the equations together will eliminate the variable y.
2x + 3y = 6 (1)
4x  3y = 6 (2)
6x =0 Add.
X = Solve for x.
To find J, substitute for x in either equation (1) or equation (2).
0.
2x+3y= 6
(1)
2(0) + 3y= 6
Letx
+ 3y= 6
3y= 6
y=2
The solution of the system is (0, 2). Check by substituting for x and 2
for J in both equations of the original system. The solution set is {(0, —2)}.
Hi
Work Problem 5 at the Side.
Answers
4. (a) {(3,1)} (b) {(1,3)}
5. (a) {(2,1)} (b) \[^,l
By adding the equations in Example 5, we eliminated the variable y
because the coefficients of the jterms were opposites. In many cases the
coefficients will not be opposites, and we must transform one or both equa
tions so that the coefficients of one pair of variable terms are opposites.
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
Section 5.1 Systems of Linear Equations in Two Variables 257
Solving a Linear System by Elimination
Step 1 Write both equations in standard form Ax + By = C.
Step 2 Make the coefficients of one pair of variable terms oppo
sites. Multiply one or both equations by appropriate numbers so
that the sum of the coefficients of either the x or jterms is 0.
Step 3 Add the new equations to eliminate a variable. The sum should
be an equation with just one variable.
Step 4 Solve the equation from Step 3 for the remaining variable.
Step 5 Find the other value. Substitute the result of Step 4 into either
of the original equations and solve for the other variable.
Step 6 Check the solution in both of the original equations. Then
write the solution set.
© Solve by elimination.
(a) X + 3j = 8
2x 5y= 17
EXAMPLE 6
Solving a System by Elimination
Solve the system.
5x2y = A (1)
2x + 3j = 13 (2)
Both equations are in standard form.
Step 1
Step 2
Step 3
Step 4
Step 5
Step 6
Suppose that you wish to eliminate the variable x. One way to do
this is to multiply equation (1) by 2 and equation (2) by 5.
lOx Ay= 8
lOx \5y= 65
Now add.
lOx Ay =
lOx \5y =
2 times each side of equation (1)
— 5 times each side of equation (2)
65
Solve for J.
\9y
y
57
Add.
Divide by
19.
To findx, substitute 3 for j in either equation (1) or (2). Substitut
ing in equation (2) gives
2x + 3j; = 13
2x + 3(3)= 13
2x + 9 = 13
2x = 4
x = 2.
(2)
Letj;
3.
Subtract 9.
Divide by 2.
The solution is (2, 3). To check, substitute 2 for x and 3 for j in both
equations (1) and (2).
5x2y = A
5(2) 2(3) = 4
106 = 4
4 = 4
(1)
True
2x + 3j
2(2) + 3(3)
4 + 9
13
13
13
13
13
(2)
True
The solution set is {(2, 3)}.
Work Problem 6 at the Side.
(b) 6x2y= 21
3x + 4>' = 36
(c) lx + 'iy= 19
3x — ly = —6
Answers
6.
(a) {(1,3)}
(c) {(5, 3)}
(b)
{(
2
17^
2,
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
258 Chapter 5 Systems of Linear Equations
O Solve each system.
1 1
(a) X  y
3x
1_
2" 6
2y= 9
EXAMPLE 7
Solving a System with Fractional Coefficients
Solve the system.
5x
ly =
= 4
1
3
13
4
(1)
(2)
If an equation in a system has fractional coefficients, as in equation (2),
first multiply by the least common denominator to clear the fractions.
.,1 3 \ ^13
\2 A^ 4
Multiply equation (2)
by the LCD, 4.
1 . 3
X + 4 • J
2
11
4
Distributive property
Equivalent to equation (2)
A'
2x + 3j = 13
The system of equations becomes
5x2y = A (1)
2x + 3j = 13, Equation (2) with fractions cleared
which is identical to the system we solved in Example 6. The solution set is
{(2, 3)}. To confirm this, check the solution in both equations (1) and (2).
Work Problem 7 at the Side.
X ly
(b)  + —
^ ^ 5 3
3x
y
NOTE
If an equation in a system contains decimal coefficients, it is best to
first clear the decimals by multiplying by 10, 100, or 1000, depending
on the number of decimal places. Then solve the system. For example,
we multiply each side of the equation
.5x + .75j
by 100 to get the equivalent equation
50x + 75j
3.25
325.
OBJECTIVE Q Solve special systems. As we saw in Figures 3(b) and (c),
some systems of linear equations have no solution or an infinite number of
solutions.
EXAMPLE 8
Solving a System of Dependent Equations
Solve the system.
2x  J = 3 (1)
6x  37 = 9 (2)
We multiply equation (1) by —3, and then add the result to equation (2).
— 6x + 3j=— 9 —3 times each side of equation (1)
6x  3j; = 9 (2)
True
Answers
7. (a) {(5,3)} (b) {(2,3)}
0=0
Continued on Next Page
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
Section 5.1 Systems of Linear Equations in Two Variables 259
Adding these equations gives the true statement = 0. In the original sys
tem, we could get equation (2) from equation (1) by multiplying equation (1)
by 3. Because of this, equations (1) and (2) are equivalent and have the same
graph, as shown in Figure 4. The equations are dependent. The solution set
is the set of all points on the line with equation 2x — y = 3, written
{(x,j) \2x  y = 3}
and read "the set of all ordered pairs (x, j), such that 2x — j = 3."
O Solve the system. Then graph
both equations.
2x + y
8x 4j
24
(1)
(2)
2xy = 3
6x 3y = 9
Same line —
infinitely
many
solutions
Figure 4
NOTE
When a system has an infinite number of solutions, as in Example 8,
either equation of the system could be used to write the solution set.
We prefer to use the equation (in standard form) with coefficients that
are integers having no common factor (except 1).
& Solve the system. Then graph
both equations.
4x  3y = S (1)
8x  6j = 14 (2)
Work Problem 8 at the Side.
EXAMPLE 9
Solving an Inconsistent System
Solve the system.
X + 3y = 4 (1)
2x  6y = 3 (2)
Multiply equation (1) by 2, and then add the result to equation (2).
2x + 6y = 8 Equation (1) multiplied by 2
2x  6y = 3 (2)
= 11 False
The result of the addition step is a false
statement, which indicates that the sys
tem is inconsistent. As shown in Figure 5,
the graphs of the equations of the system
are parallel lines. There are no ordered
pairs that satisfy both equations, so there
is no solution for the system. The solu
tion set is 0.
Figure 5
Work Problem 9 at the Side.
Answers
8. {ix,y)\2x+y = 6}
(1) and (2)
y (2)(1)
^11
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
260 Chapter 5 Systems of Linear Equations
CO) Write the equations of
Example 8 in slopeintercept
form. Use function notation.
The results of Examples 8 and 9 are generalized as follows.
Special Cases of Linear Systems
If both variables are eliminated when a system of linear equations is
solved, then
1. there is no solution if the resulting statement is false;
2. there are infinitely many solutions if the resulting statement is true.
Slopes and jintercepts can be used to decide if the graphs of a system
of equations are parallel lines or if they coincide. In Example 8, writing
each equation in slopeintercept form shows that both lines have slope 2 and
jintercept (0, —3), so the graphs are the same line and the system has an
infinite number of solutions.
[Work Problem 10 at the Side.
In Example 9, both equations have slope — , but the jintercepts are
(0, f) and (0, — ^), showing that the graphs are two distinct parallel lines.
Thus, the system has as its solution set.
[Work Problem 11 at the Side.
CD Write the equations of
Example 9 in slopeintercept
form. Use function notation.
Answers
10. Both equations are/(x) = 2x  3.
1 4 , , 1 1
ll./W = x + ;/W = x
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
Section 5.2 Systems of Linear Equations in Three Variables 267
5.2 Systems of Linear Equations in Three Variables
A solution of an equation in three variables, such as
2x + 3j;  z = 4,
is called an ordered triple and is written (x, y, z). For example, the ordered
triple (0, 1, —1) is a solution of the equation, because
2(0) + 3(1)  (1) = + 3 + 1=4.
Verify that another solution of this equation is (10, 3, 7).
In the rest of this chapter, the term linear equation is extended to equa
tions of the form
Ax + By + Cz + . . . + Dw = K,
where not all the coefficients A,B,C, . . . ,D equal 0. For example,
2x + 3j — 5z = 7 and x — 2;; — z + 3t/ — 2w = 8
are linear equations, the first with three variables and the second with five
variables.
OBJECTIVES
Understand the
geometry of systems of
three equations in three
variables.
Solve linear systems (with
three equations and three
variables) by elimination.
Solve linear systems
where some of the equa
tions have missing terms.
Solve special systems.
OBJECTIVE Q Understand the geometry of systems of three equa
tions in three variables. In this section, we discuss the solution of a
system of linear equations in three variables, such as
4x + 8 J + z = 2
X + 7j;  3z= 14
2x — 3j + 2z = 3.
Theoretically, a system of this type can be solved by graphing. However, the
graph of a linear equation with three variables is ?i plane, not a line. Since the
graph of each equation of the system is a plane, which requires three
dimensional graphing, this method is not practical. However, it does illustrate
the number of solutions possible for such systems, as shown in Figure 6.
A single solution
(a)
No points in common
(d)
Points of a line in common
All points in common
(b)
(c)
yiii
^^^
A
/ /
L^ — /\ \
A\
/ /
\/\ \
/
"^^^^^^^
^0 points in common
No points in commoi
(e)
(f)
Figure 6
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
268 Chapter 5 Systems of Linear Equations
Figure 6 on the preceding page illustrates the following cases.
Graphs of Linear Systems in Three Variables
1. The three planes may meet at a single, common point that is the
solution of the system. See Figure 6(a).
2. The three planes may have the points of a line in common so that the
infinite set of points that satisfy the equation of the line is the
solution of the system. See Figure 6(b).
3. The three planes may coincide so that the solution of the system is
the set of all points on a plane. See Figure 6(c).
4. The planes may have no points common to all three so that there is
no solution of the system. See Figures 6(d), (e), and (f).
OBJECTIVE Q Solve linear systems (with three equations and three
variables) by elimination. Is it possible to solve a system of three equa
tions in three variables such as the one that follows?
4x + 87 + z = 2
X + 7j  3z= 14
2x — 3y + 2z = 3
Graphing to find the solution set of such a system is impractical, so these
systems are solved with an extension of the elimination method from
Section 5.1, summarized as follows.
Solving a Linear System in Three Variables
Step 1 Eliminate a variable. Use the elimination method to eliminate
any variable from any two of the original equations. The result
is an equation in two variables.
Step 2 Eliminate the same variable again. Eliminate the same variable
from any other two equations. The result is an equation in the
same two variables as in Step 1.
Step 3 Eliminate a different variable and solve. Use the elimination
method to eliminate a second variable from the two equations
in two variables that result from Steps 1 and 2. The result is an
equation in one variable that gives the value of that variable.
Step 4 Find a second value. Substitute the value of the variable found
in Step 3 into either of the equations in two variables to find
the value of the second variable.
Step 5 Find a third value. Use the values of the two variables from
Steps 3 and 4 to find the value of the third variable by substi
tuting into any of the original equations.
Step 6 Check the solution in all of the original equations. Then write
the solution set.
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
Section 5.2 Systems of Linear Equations in Three Variables 269
EXAMPLE 1
Solving a System in Three Variables
An]^matjaii
Video
Solve the system.
4x + 87 + z = 2 (1)
x + 7y3z= 14 (2)
2x3y + 2z = 3 (3)
Step 1 As before, the elimination method involves eliminating a variable
from the sum of two equations. The choice of which variable to
eliminate is arbitrary. Suppose we decide to begin by eliminating z.
To do this, we multiply equation (1) by 3 and then add the result to
equation (2).
12x + 24y + 3z = 6 Multiply each side of (1) by 3.
X + 7j  3z= 14 (2)
= 8 Add
Step 2
Step 3
13x + 3ly = 8 Add. (4)
Equation (4) has only two variables. To get another equation without z,
we multiply equation (1) by 2 and add the result to equation (3). It is
essential at this point to eliminate the same variable, z.
8x  \6y  2z = 4 Multiply each side of (1) by 2.
2x  3j + 2z = 3 (3)
6x  \9y = 1 Add. (5)
Now we solve the resulting system of equations (4) and (5) for x
and J.
13x + 3\y
6x I9y
1
(4)
(5)
This step is possible only if the same variable is eliminated in Steps
1 and 2.
78x + 186j = 48
78x 2473;= 13
61
Multiply each side of (4) by 6.
Multiply each side of (5) by 13.
Add.
6ly =
Step 4 We substitute 1 for j in either equation (4) or (5). Choosing (5) gives
1.
6x 19j= 1
(5)
6x 19(1) = 1
Letj^
6x 19= 1
6x = 18
x= 3.
Step 5 We substitute — 3 for x and 1 for y in any one of the three original
equations to findz. Choosing (1) gives
4x + Sy + z = 2
4(3) + 8(l) + z = 2
4 + z = 2
z = 6.
Continued on Next Page
(1)
Letx
3 and J = 1.
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
270 Chapter 5 Systems of Linear Equations
Check that the solution
(—3, 1, 6) satisfies equations
(2) and (3) of Example 1.
(a) X + 7j 3z= 14 (2)
Does the solution satisfy
equation (2)?
Step 6
It appears that the ordered triple (3, 1, 6) is the only solution of the
system. We must check that the solution satisfies all three original
equations of the system. For equation (1),
4x + 8j + z = 2 (1)
4(3) + 8(1) + 6 = 2 ?
12 + 8 + 6 = 2 ?
2 = 2. True
Work Problem 1 at the Side.
Because (—3, 1, 6) also satisfies equations (2) and (3), the solution set is
{(3,1,6)}.
Work Problem 2 at the Side.
(b) 2x  3j + 2z = 3 (3)
OBJECTIVE Q Solve linear systems where some of the equations
have missing terms. If a linear system has an equation missing a term ^^
terms, one elimination step can be omitted.
or
Does the solution satisfy
equation (3)?
& Solve each system.
(a) X + J + z = 2
X — y + 2z = 2
—x\2y— z = \
(b) 2x + J + z = 9
— X — J + z = 1
3x  J + z = 9
EXAMPLE 2
Solving a System of Equations with MissingTerms
Solve the system.
6x  \2y = 5 (1)
8j + z = (2)
9x  z = 12 (3)
Since equation (3) is missing the variable y, a good way to begin the
solution is to eliminate y again using equations (1) and (2).
\2x  2Ay =10 Multiply each side of ( 1 ) by 2.
2Ay + 3z = Multiply each side of (2) by 3.
12x + 3z= 10 Add. (4)
= 12, to eliminate z. Multi
Use this result, together with equation (3), 9x
ply equation (3) by 3. This gives
27x — 3z
12x + 3z
36
10
39x
= 26
_ 26 _ 2
•^ " 39 " 3 •
Substituting into equation (3) gives
9x  z = 12
6 z= 12
z= 6.
Continued on Next Page
Multiply each side of (3) by 3.
(4)
Add.
12
(3)
Letx =
Answers
1. (a) yes (b) yes
2. (a) {(1,1,2)} (b) {(2,1,4)}
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
Section 5.2 Systems of Linear Equations in Three Variables 27 1
Substituting — 6 for z in equation (2) gives
Sy + z =
(2)
Sy6 =
Letz =
6
Sy = 6
3
Thus, X = I, J = I, andz = — 6. Check these values in each of the original equa
tions of the system to verify that the solution set of the system is {(3,
2 3
4'
Work Problem 3 at the Side
6)}.
II
& Solve each system.
(a) X — y = 6
2y+5z=l
3x  4z = 8
OBJECTIVE Q Solve special systems. Linear systems with three
variables may be inconsistent or may include dependent equations. The next
examples illustrate these cases.
rr^
— f22J^^53£l Solving an Inconsistent Systenn with Three
[a
Variables
Solve the system.
m
Video ^
2x  4y + 6z = 5 (1)
x+ 3y2z= I (2)
x2y + 3z=l (3)
Eliminate x by adding equations (2) and (3) to get the equation
y + z = 0.
Now, eliminate X again, using equations (1) and (3).
2x + 4y  6z = 2 Multiply each side of (3) by 2.
2x4y + 6z= 5 (1)
= 3 False
The resulting false statement indicates that equations (1) and (3) have no
common solution. Thus, the system is inconsistent and the solution set is 0.
The graph of this system would show these two planes parallel to one
another.
(b) 5x y = 26
4y+3z= 4
X + z = 5
O Solve each system.
(a) 3x  5y + 2z=l
5x+ 8j  z = 4
6x + lOj  4z = 5
NOTE
If a false statement results when adding as in Example 3, it is not
necessary to go any further with the solution. Since two of the three
planes are parallel, it is not possible for the three planes to have any
common points.
Work Problem 4 at the Side.
(b) 7x
97 + 2z =
7+ z =
8x  z =
Answers
3. (a) {(4,2,1)} (b) {(5,1,0)}
4. (a) (b) {(0,0,0)}
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
272 Chapter 5 Systems of Linear Equations
Solve the system.
X — J + z = 4
— 3x + 3j — 3z = —
2x — 23; + 2z = 8
12
EXAMPLE 4
Solving a System of Dependent Equations
with Three Variables
Solve the system.
2x 37 + 4z =
^ + ::r7
6x
2z
24
(1)
(2)
(3)
9y + 12z
Multiplying each side of equation (1) by 3 gives equation (3). Multiply
ing each side of equation (2) by —6 also gives equation (3). Because of this,
the equations are dependent. All three equations have the same graph, as
illustrated in Figure 6(c). The solution set is written
{(x,7,z)2x37 + 4z=8}.
Although any one of the three equations could be used to write the solution
set, we use the equation with coefficients that are integers with no common
factor (except 1), as we did in Section 5.1.
IK
Work Problem 5 at the Side.
Answers
5. {{x,y,z)\x  y + z = A)
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
Section 5.3 Applications of Systenns of Linear Equations 277
5 • 3 Applications of Systems of Linear Equations
Many applied problems involve more than one unknown quantity. Although
some problems with two unknowns can be solved using just one variable, it
is often easier to use two variables. To solve a problem with two unknowns,
we must write two equations that relate the unknown quantities. The system
formed by the pair of equations can then be solved using the methods of
this chapter.
Problems that can be solved by writing a system of equations have been of
interest historically. The following problem, which is given in the exercises for
this section, first appeared in a Hindu work that dates back to about a.d. 850.
The mixed price of 9 citrons [a lemonlike fruit shown in the
photo] and 7 fragrant wood apples is 107; again, the mixed price
of 7 citrons and 9 fragrant wood apples is 101. O you arithmeti
cian, tell me quickly the price of a citron and the price of a wood
apple here, having distinctly separated those prices well.
OBJECTIVES
Solve problems using two
variables.
Solve money problems
using two variables.
Solve mixture problems
using two variables.
Solve distanceratetime
problems using two
variables.
Solve problems with three
variables using a system
of three equations.
The following steps, based on the sixstep problem solving method first
introduced in Section 2.3, give a strategy for solving applied problems using
more than one variable.
Solving an Applied Problem by Writing a System of Equations
Step 1 Read the problem, several times if necessary, until you under
stand what is given and what is to be found.
Step 2 Assign variables to represent the unknown values, using
diagrams or tables as needed. Write down what each variable
represents.
Step 3 Write a system of equations that relates the unknowns.
Step 4 Solve the system of equations.
Step 5 State the answer to the problem. Does it seem reasonable?
Step 6 Check the answer in the words of the original problem.
OBJECTIVE Q Solve problems using two variables. Problems about
the perimeter of a geometric figure often involve two unknowns and can be
solved using systems of equations.
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
278 Chapter 5 Systems of Linear Equations
O Solve the problem.
The length of the
foundation of a rectangular
house is to be 6 m more than
its width. Find the length and
width of the house if the
perimeter must be 48 m.
EXAMPLE 1
Finding the Dimensions of a Soccer Field
Answers
1. length: 15 m; width: 9 m
Unlike football, where the dimensions of a playing field cannot vary, a
rectangular soccer field may have a width between 50 and 100 yd and a
length between 100 and 130 yd. Suppose that one particular field has a
perimeter of 320 yd. Its length measures 40 yd more than its width. What are
the dimensions of this field? {Source: Microsoft Encarta Encyclopedia 2002.)
Step 1 Read the problem again. We must find the dimensions of the
field.
Step 2 Assign variables. Let L = the length and W = the width. Figure 7
shows a soccer field with these variables as labels.
W
Figure 7
Step 3 Write a system of equations. Because the perimeter is 320 yd, we
find one equation by using the perimeter formula:
2L + 2W= 320.
Because the length is 40 yd more than the width, we have
L= W+ 40.
The system is, therefore,
2L + 2W= 320
L= W+ 40,
Step 4 Solve the system of equations. Since equation (2) is solved for Z,
we can use the substitution method. We substitute W ^ 40 for L in
equation (1), and solve for W.
2L + 2W= 320
2(W\ 40) + 2W= 320
2W+ 80 + 2W= 320
4^+ 80 = 320
4fF=240
W=60
Let W = 60 m the equation L =
(1)
(2)
(1)
LqXL = W+40.
Distributive property
Combine terms.
Subtract 80.
Divide by 4.
W+ 40 to find Z.
Z = 60 + 40 = 100
Step 5 State the answer. The length is 100 yd, and the width is 60 yd. Both
dimensions are within the ranges given in the problem.
Step 6 Check. The perimeter is 2(100) + 2(60) = 320 yd, and the
length, 100 yd, is indeed 40 yd more than the width, since
100  40 = 60. The answer is correct.
IK
Work Problem 1 at the Side.
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
Section 5.3 Applications of Systenns of Linear Equations 279
OBJECTIVE Q Solve money problems using two variables. Professional
sport ticket prices increase annually. Average perticket prices in three of the
four major sports (football, basketball, and hockey) now exceed $40.00.
EXAMPLE 2
Solving a Problem about Ticket Prices
It was reported in March 2004 that during the
National Hockey League and National Basketball
Association seasons, two hockey tickets and one
basketball ticket purchased at their average prices
would have cost $126.77. One hockey ticket and two
basketball tickets would have cost $128.86. What
were the average ticket prices for the two sports?
{Source: Team Marketing Report, Chicago.)
Step 1
Step 2
Step 3
Step 4
Step 5
Step 6
'^''jHB
Read the problem again. There are two un
knowns.
Assign variables. Let h represent the
average price for a hockey ticket and b represent the average price
for a basketball ticket.
Write a system of equations. Because two hockey tickets and one
basketball ticket cost a total of $126.77, one equation for the system is
2h + b= 126.77.
By similar reasoning, the second equation is
h + 2b= 128.86.
Therefore, the system is
2h+ b= 126.77 (1)
h + 2b= 128.86. (2)
Solve the system of equations. To eliminate /z, multiply equation (2)
by —2 and add.
2h+ b= 126.77
2h  4b= 257.72
3b = 130.95
b = 43.65
To find the value of A, let b
h + 2b= 128.86
h + 2(43.65) = 128.86
h + 87.30 = 128.86
A = 41.56
(1)
Multiply each side of (2) by
Add.
Divide by —3.
43.65 in equation (2).
(2)
LQtb = 43.65.
Multiply.
Subtract 87.30.
2.
State the answer. The average price for one basketball ticket was
$43.65. For one hockey ticket, the average price was $41.56.
Check that these values satisfy the conditions stated in the problem.
Work Problem 2 at the Side.
& Solve the problem.
For recent Major League
Baseball and National
Football League seasons,
based on average ticket
prices, three baseball tickets
and two football tickets
would have cost $159.50,
while two baseball tickets and
one football ticket would
have cost $89.66. What were
the average ticket prices for
the two sports? (Source:
Team Marketing Report,
Chicago.)
OBJECTIVE Q Solve mixture problems using two variables. We
solved mixture problems in Section 2.3 using one variable. For many mix
ture problems we can use more than one variable and a system of equations.
Answers
2. baseball: $19.82; football: $50.02
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
280 Chapter 5 Systems of Linear Equations
® Solve each problem.
(a) A grocer has some $4 per
lb coffee and some $8 per
lb coffee, which he will
mix to make 50 lb of
$5.60 per lb coffee. How
many pounds of each
should be used?
EXAMPLE 3
Solving a Mixture Problem
How many ounces each of 5% hydrochloric acid and 20% hydrochloric acid
must be combined to get 10 oz of solution that is 12.5% hydrochloric acid?
$4/LB I
Colombian f
Supreme r
(b) Some 40% ethyl alcohol
solution is to be mixed with
some 80% solution to get
200 Lofa 50% solution.
How many liters of each
should be used?
Step 1
Step 2
Step 3
Step 5
Step 6
Read the problem. Two solutions of different strengths are being
mixed together to get a specific amount of a solution with an "in
between" strength.
Assign variables. Let x represent the number of ounces of 5%) solu
tion and J represent the number of ounces of 20% solution. Use a
table to summarize the information from the problem.
Percent
(as a Decimal)
Ounces of
Solution
Ounces of
Pure Acid
5% = .05
X
.05x
20% = .20
y
.20j
12.5% = .125
10
(.125)10
Figure 8 also illustrates what is happening in the problem.
f^^^ f^^^ f^^^
Ounces of
solution
Ounces of
pure acid
.05x
.20j
Figure 8
10
.125(10)
Write a system of equations. When the x ounces of 5% solution
and the y ounces of 20% solution are combined, the total number of
ounces is 10, so
X + y = 10. (1)
The ounces of acid in the 5% solution (.05x) plus the ounces of acid
in the 20% solution (.20y) should equal the total ounces of acid in
the mixture, which is (.125)10, or 1.25. That is,
.05x+ .20^= 1.25. (2)
Notice that these equations can be quickly determined by reading
down in the table or using the labels in Figure 8.
Step 4 Solve the system of equations (1) and (2). Eliminate x by first mul
tiplying equation (2) by 100 to clear it of decimals and then multi
plying equation (1) by —5.
5x + 20j = 125
— 5x — 5j = — 50
15/= 75
y = 5
Because j = 5 and x + j
Multiply each side of (2) by 100.
Multiply each side of (1) by —5.
Add.
= 10, X is also 5.
State the answer. The desired mixture will require 5 oz of the
5% solution and 5 oz of the 20% solution.
Check that these values satisfy both equations of the system.
Work Problem J at the Side.
Answers
3. (a) 301bof$4;201bof$8
(b) 150Lof40%;50Lof80%
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
Section 5.3 Applications of Systenns of Linear Equations 28 I
OBJECTIVE Q Solve distanceratetime problems using two variables.
Motion problems require the distance formula, d = rt, where d is distance,
r is rate (or speed), and t is time. These applications often lead to systems of
equations.
EXAMPLE 4
I'
ideo
Solving a Motion Problem
A car travels 250 km in the same time that a truck travels 225 km. If the
speed of the car is 8 km per hr faster than the speed of the truck, find both
speeds.
Step 1
Step 2
Steps
Step 4
Read the problem again. Given the distances traveled, we need to
find the speed of each vehicle.
Assign variables.
Let X = the speed of the car
and J = the speed of the truck.
As in Example 3, a table helps organize the information. Fill in the
given information for each vehicle (in this case, distance) and use
the assigned variables for the unknown speeds (rates).
Car
d
250
Truck 225
y
To get an expression for time, solve the distance formula, d
t. Since f = t, the two times can be written as ^ and ^.
rt, for
Write a system of equations. The problem states that the car travels
8 km per hr faster than the truck. Since the two speeds are x and j,
X = J + 8.
Both vehicles travel for the same time, so from the table
250 _ 225
X y
This is not a linear equation. Multiplying each side by xy gives
2507 = 225x,
which is linear. The system is
x=3; + 8 (1)
2507 = 225x. (2)
Solve the system of equations by substitution. Replace x with j + 8
in equation (2).
2507 = 225x
250j = 225 (j + 8)
250j = 225y + 1800
25y = 1800
7 = 72
Because x = j + 8, the value of x is 72 + 8
Continued on Next Page
(2)
Letx =7 + 8.
Distributive property
Subtract 225y.
Divide by 25.
= 80.
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
282 Chapter 5 Systems of Linear Equations
O Solve the problem.
A train travels 600 mi in
the same time that a truck
travels 520 mi. Find the speed
of each vehicle if the train's
average speed is 8 mph faster
than the truck's.
Step 5 State the answer. The car's speed is 80 km per hr, and the truck's
speed is 72 km per hr.
Step 6 Check. This is especially important since one of the equations had
variable denominators.
Car: t
Truck: t
d
250
r
80
d
225
r
72
3.125 ^
3.125 <
Times are
equal.
Since 80  72 = 8, the conditions of the problem are satisfied.
Work Problem 4 at the Side.
'^
1
1
1
m
p;
l1
Y v\
kitt'.
'<^
■ J
llw^L ^
□w^r
1 .
^PVs^Di
W.5^^^
PET^
^^jIFm^
mT
\fli
■Ml
& Solve the system of
equations from Example 5.
X 3y = (1)
X z= 5 (2)
259x + 299y + 329z = 6607 (3)
Answers
4. train: 60 mph; truck: 52 mph
5. {(12,4,7)}
OBJECTIVE Q Solve problems with three variables using a system of
three equations. To solve such problems, we extend the method used for
two unknowns to three variables and three equations.
EXAMPLE 5
Solving a Problem Involving Prices
At Panera Bread, a loaf of honey wheat bread costs $2.59, a loaf of sunflower
bread costs $2.99, and a loaf of French bread costs $3.29. On a recent day,
three times as many loaves of honey wheat were sold as sunflower. The num
ber of loaves of French bread sold was 5 less than the number of loaves of
honey wheat sold. Total receipts for these breads were $66.07. How many
loaves of each type of bread were sold? (Source: Panera Bread menu.)
Step 1 Read the problem again. There are three unknowns in this problem.
Step 2 Assign variables to represent the three unknowns.
Let X = the number of loaves of honey wheat,
y = the number of loaves of sunflower,
and z = the number of loaves of French bread.
Step 3 Write a system of three equations. Since three times as many
loaves of honey wheat were sold as sunflower.
tsi
X = 3y, or x — 3y = 0. (1)
Also,
Number of loaves
5 less than the number
of French bread
equals
of loaves of honey wheat.
1
1
1
z
=
x5.
so
5. (2)
Multiplying the cost of a loaf of each kind of bread by the number
of loaves of that kind sold and adding gives the total receipts.
2.59x + 2.997 + 3.29z = 66.07
Multiply each side of this equation by 100 to clear it of decimals.
259x + 2993; + 329z = 6607 (3)
Step 4 Solve the system of three equations using the method shown in
Section 5.2.
Work Problem 5 at the Side.
Continued on Next Page
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
Section 5.3 Applications of Systenns of Linear Equations 283
Step 5 State the answer. The solution set is {(12, 4, 7)}, meaning that
12 loaves of honey wheat, 4 loaves of sunflower, and 7 loaves of
French bread were sold.
Step 6 Check. Since 12 = 3 • 4, the number of loaves of honey wheat is
three times the number of loaves of sunflower. Also, 12 — 7 = 5, so
the number of loaves of French bread is 5 less than the number of
loaves of honey wheat. Multiply the appropriate cost per loaf by the
number of loaves sold and add the results to check that total receipts
were $66.07.
Work Problem 6 at the Side.
Video
EXAMPLE 6
Solving a Business Production Problem
A company produces three color television sets, models X, Y, and Z. Each
model X set requires 2 hr of electronics work, 2 hr of assembly time, and
1 hr of finishing time. Each model Y requires 1, 3, and 1 hr of electronics,
assembly, and finishing time, respectively. Each model Z requires 3, 2,
and 2 hr of the same work, respectively. There are 100 hr available for
electronics, 100 hr available for assembly, and 65 hr available for finish
ing per week. How many of each model should be produced each week if
all available time must be used?
Step 1 Read the problem again. There are three unknowns.
Step 2 Assign variables.
Let X = the number of model X produced per week,
y = the number of model Y produced per week,
and z = the number of model Z produced per week.
Organize the information in a table.
Each
Model X
Each
Model Y
Each
Model Z
Totals
Hours of
Electronics Work 2
1
3
100
Hours of
Assembly Time 2
3
2
100
Hours of
Finishing Time 1
1
2
65
© Solve the problem.
A department store has
three kinds of perfume:
cheap, better, and best. It has
10 more bottles of cheap than
better, and 3 fewer bottles of
best than better. Each bottle
of cheap costs $8, better costs
$15, and best costs $32. The
total value of all the perfume
is $589. How many bottles of
each are there?
C$8 ] $15
Step 3 Write a system of three equations. The x model X sets require
2x hr of electronics, the y model Y sets require ly (or j) hr of elec
tronics, and the z model Z sets require 3z hr of electronics. Since
100 hr are available for electronics,
2x + J + 3z = 100. (1)
Similarly, from the fact that 100 hr are available for assembly,
2x + 3y + 2z= 100, (2)
and the fact that 65 hr are available for finishing leads to the equation
X + 7 + 2z = 65. (3)
Notice that by reading across the table, we can easily determine the
coefficients and constants in the equations of the system.
Continued on Next Page
Answers
6. 21 bottles of cheap; 11 of better; 8 of best
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
284 Chapter 5 Systems of Linear Equations
O Solve the problem.
A paper mill makes
newsprint, bond, and copy
machine paper. Each ton of
newsprint requires 3 tons of
recycled paper and 1 ton of
wood pulp. Each ton of bond
requires 2 tons of recycled
paper, 4 tons of wood pulp,
and 3 tons of rags. A ton of
copy machine paper requires
2 tons of recycled paper,
3 tons of wood pulp, and
2 tons of rags. The mill has
4200 tons of recycled paper,
5800 tons of wood pulp, and
3900 tons of rags. How much
of each kind of paper can be
made from these supplies?
Step 4 Solve the system
2x + 7 + 3z = 100
2x + 37 + 2z = 100
X + 3; + 2z = 65
to findx = 15, J = 10, andz = 20.
Step 5 State the answer. The company should produce 1 5 model X,
10 model Y, and 20 model Z sets per week.
Step 6 Check that these values satisfy the conditions of the problem.
Mi
Work Problem 7 at the Side.
Answers
7. 400 tons of newsprint; 900 tons of bond;
600 tons of copy machine paper
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
Section 5.4 Solving Systems of Linear Equations by Matrix Methods 29 I
5.4 Solving Systems of Linear Equations by Matrix Methods
OBJECTIVE Q Define a matrix. An ordered array of numbers such as
Columns
Rows
Matrix
is called a matrix. The numbers are called elements of the matrix. Matrices
(the plural of matrix) are named according to the number of rows and
columns they contain. The rows are read horizontally, and the columns are
read vertically. For example, the first row in the preceding matrix is 2 3 5
2
and the first column is . This matrix is a 2 X 3 (read "two by three") matrix
because it has 2 rows and 3 columns. The number of rows is given first, and
then the number of columns. Two other examples follow.
2X2
matrix
8
1
3
2
1
6
4X3
5
3
matrix
5
9
7
A square matrix is one that has the same number of rows as columns. The
2X2 matrix is a square matrix.
liill Calculator Tip Figure 9 shows how a graphing calculator displays the
preceding two matrices. Work with matrices is made much easier by using
technology when available. Consult your owner's manual for details.
[B]
[[8
1
3]
[2
1
6 ]
[0
5
3]
[5
9
7 ]]
OBJECTIVES
Define a matrix.
Write the augmented
matrix for a system.
Use row operations to
solve a system with two
equations.
Use row operations to
solve a system with three
equations.
Use row operations to
solve special systems.
Figure 9
In this section, we discuss a method of solving linear systems that uses
matrices. The advantage of this new method is that it can be done by a graph
ing calculator or a computer, allowing large systems of equations to be
solved easily.
OBJECTIVE Q Write the augmented matrix for a system. To begin,
we write an augmented matrix for the system. An augmented matrix
has a vertical bar that separates the columns of the matrix into two groups.
For example, to solve the system
X  3y = I
2x + J = — 5,
start with the augmented matrix
1
Augmented matrix
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
292 Chapter 5 Systems of Linear Equations
System of equations:
X — 3y = I
2x+ y = 5
Augmented matrix:
"1 3
2 1
t
Coefficients Constants
of the
variables
Place the coeflficients of the variables to the left of the bar, and the constants to
the right. The bar separates the coefficients from the constants. The matrix is
just a shorthand v^ay of v^riting the system of equations, so the rov^s of the aug
mented matrix can be treated the same as the equations of a system of equations.
We knov^ that exchanging the position of tw^o equations in a system does
not change the system. Also, multiplying any equation in a system by a
nonzero number does not change the system. Comparable changes to the
augmented matrix of a system of equations produce hqw matrices that corre
spond to systems vs^ith the same solutions as the original system.
The follovs^ing row operations produce nev^ matrices that lead to systems
having the same solutions as the original system.
Matrix Row Operations
1. Any tv^o rov^s of the matrix may be interchanged.
2. The numbers in any row may be multiplied by any nonzero real
number.
3. Any rov^ may be transformed by adding to the numbers of the row
the product of a real number and the corresponding numbers of
another row.
Examples of these row operations follow.
Row operation 1 :
2 3 9
4 83
1 7
becomes
1
4
2
8
3
7
3
9
Row operation 2:
2 3 9
4 83
1 7
becomes
6
4
1
9
8
27"
3
7_
Row operation 3:
2 3 9
4 83
1 7
becomes
4
1
3
8
5'
3
7
Interchange row 1
and row 3.
Multiply the numbers
in row 1 by 3.
Multiply the numbers in
row 3 by —2; add them
to the corresponding
numbers in row 1 .
The third row operation corresponds to the way we eliminated a variable
from a pair of equations in the previous sections.
OBJECTIVE Q Use row operations to solve a system with two
equations. Row operations can be used to rewrite a matrix. The goal is a
matrix in the form
or
a
1
b
c
d
e
1
f\
for systems with two or three equations, respectively. Notice that there are Is
down the diagonal from upper left to lower right and Os below the Is. A matrix
written this way is said to be in row echelon form. When these matrices are
rewritten as systems of equations, the value of one variable is known, and the
rest can be found by substitution. The following examples illustrate this method.
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
Section 5.4 Solving Systems of Linear Equations by Matrix Methods 293
EXAMPLE 1
Using Row Operations to Solve a System with
Two Variables
Use row operations to solve the system.
X — 3y = I
2x + y = —5
We start with the augmented matrix of the system.
Now we use the various row operations to change this matrix into one that
leads to a system that is easier to solve.
It is best to work by columns. We start with the first column and make
sure that there is a 1 in the first row, first column position. There is already a
1 in this position. Next, we get in every position below the first. To get a
in row two, column one, we use the third row operation and add to the
numbers in row two the result of multiplying each number in row one by —2.
(We abbreviate this as 2R^ + R^.) Row one remains unchanged.
1
3
1
2 +
A
K
2) 1 + 3(2)
5 + l(2)_
Original number
2 times number
from row two
from row one
1 3
r
_0 7
7
2R,+R,
The matrix now has a 1 in the first position of column one, with in every
position below the first.
Now we go to column two. A 1 is needed in row two, column two. We get
this 1 by using the second row operation, multiplying each number of row two
byi.
7R2
This augmented matrix leads to the system of equations
Ix — 3y = I X — 3y = I
Ox+lj=l ^^ 7=l.
From the second equation, y = —I. We substitute — 1 for 3; in the first
equation to get
X  3 J = 1
X 3(l)= 1
x + 3 = 1
x= 2.
The solution set of the system is {(—2, — 1)}. Check this solution by substi
tution in both equations of the system.
Work Problem 1 at the Side.
Use row operations to solve
the system.
x2y = 9
3x + y = 13
Answers
1. {(5,2)}
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
294 Chapter 5 Systems of Linear Equations
Pl Calculator Tip If the augmented matrix of the system in Example 1 is
entered as matrix A in a graphing calculator (Figure 10(a)) and the row
echelon form of the matrix is found (Figure 10(b)), the system becomes
1
2
y =
5
' 2
1.
While this system looks different from the one we obtained in Example 1,
it is equivalent, since its solution set is also {(—2, — 1)}.
MflTRIXEfl] 2 x3
C 1 5 1 ]
C £ 1 E ]
ref([n]>^Frac
[[1 1^2 5^2]
[0 1 1 ]]
(a)
(b)
Figure 10
OBJECTIVE Q Use row operations to solve a system with three
equations. As before, we use row operations to get Is down the diagonal
from left to right and all Os below each 1 .
EXAMPLE 2
Using Row Operations to Solve a System with
Three Variables
Use row operations to solve the system.
X — y \ 5z = —6
3x + 3y  z = 10
X + 3y + 2z = 5
Start by writing the augmented matrix of the system.
1
1
5
6
3
3
1
10
1
3
2
5
This matrix already has 1 in row one, column one. Next get Os in the rest of
column one. First, add to row two the results of multiplying each number of
row one by —3. This gives the matrix
1
5
6
6
16
28
3
2
5
3Rj + R2
Now add to the numbers in row three the results of multiplying each number
of row one by —1.
1
— 1
6
4
5
6
16
28
3
11
IRj +R3
Continued on Next Page
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
Section 5.4 Solving Systems of Linear Equations by Matrix Methods 295
1
1
5
6'
1
8
3
14
3
4
3
11
Get 1 in row two, column two by multiplying each number in row two by ^.
6^2
Introduce in row three, column two by adding to row three the results of
multiplying each number in row two by —4.
1 1
1
y yj 4R, + R3
Finally, obtain 1 in row three, column three by multiplying each number in
row three by
& Use row operations to solve
the system.
2x — y + z = 7
X — 3y — z = 7
X + y  5z= 9
5
61
8
3
14
3
23
3
23
3 _
23
1
1
5
6"
8
3
14
3
1
1_
23^3
This final matrix gives the system of equations
X — y \ 5z=— 6
8 14
y
3^ 3
z= 1.
Substitute — 1 for z in the second equation, j — z = y, to getj = 2. Finally,
substitute 2 for j and — 1 for z in the first equation, x — y + 5z = —6, to get
X = I. The solution set of the original system is {(1, 2, 1)}. Check by
substitution in the original system.
Work Problem 2 at the Side.
OBJECTIVE Q Use row operations to solve special systems.
EXAMPLE 3
Recognizing Inconsistent Systems or Dependent
Equations
Use row operations to solve each system.
(a) 2x  37 = 8
6x + 9y = 4
2
.6
3
9
8
4
Write the
augmented matrix.
1
6
3
2
9
4"
4
iRi
'1
3
2
4"
28
6Rj + R2
arresponding system of equa
tions is
X 
f,. = 4
= 28, False
which has no solution and is inconsistent. The solution set is 0.
Continued on Next Page
Answers
2. {(2,2,1)}
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
296 Chapter 5 Systems of Linear Equations
& Use row operations to solve
each system.
(a) X  y = 2
—2x + 2y = 2
I (b) 10x+ I2y = 30
5x — 6y = —15
10
12
30
Write the
5 6
15.
augmented matrix.
'1 f
3"
10^1
.5 6
15_
■l f
3"
.0
5Rj + Rj
The corresponding system is
X7=3
= 0, True
which has dependent equations. Using the second equation of the original
system, we write the solution set as
{(x,y)\5x  6y
Hi
■15}.
Work Problem 3 at the Side.
(b) X y = 2
2x + 2y= 4
Answers
3. (a) (b) {ix,y)\xy = 2}
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
Exponents,
Polynomials, and
Polynomial
Functions
\
In 1980 MasterCard International Incorporated first began offering
debit cards in an effort to challenge Visa, the leader in credit card
transactions at that time. Now immensely popular, debit cards draw
money from consumers' bank accounts rather than from established
lines of credit. By 2005, it is estimated that 269 million debit cards
will be in use. {Source: Microsoft Encarta Encyclopedia 2002; HSN
Consultants Inc.)
We introduced the concept of function in Section 4.5 and ex
tend our work to include polynomial functions in this chapter. In
Exercise 1 1 of Section 6.3, we use a polynomial function to model the
number of bank debit cards issued.
6.T Integer Exponents and
Scientific Notation
6.2 Adding and Subtracting
Polynomials
6.3 Polynomial Functions
6.4 Multiplying Polynomials
6.5 Dividing Polynomials
31!
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
3 I 2 Chapter 6 Exponents, Polynomials, and Polynomial Functions
6.1 Integer Exponents and Scientific Notation
OBJECTIVES
Q Use the product rule for
exponents.
Q Define and negative
exponents.
Q Use the quotient rule for
exponents.
Q Use the power rules for
exponents.
Q Simplify exponential
expressions.
Q Use the rules for exponents
with scientific notation.
Apply the product rule for
exponents, if possible, in
each case.
(a) m^ ' m^
(b) r^ ' r
(c) k^k^k^
Recall from Section 1.3 that we use exponents to write products of repeated
factors. For example,
2^ is defined as 2 • 2 • 2 • 2 • 2 = 32.
The number 5, the exponent, shows that the base 2 appears as a factor
5 times. The quantity 2^ is called an exponential or di power. We read 2^ as
"2 to the fifth power" or "2 to the fifth."
OBJECTIVE Q Use the product rule for exponents. There are sev
eral useful rules that simplify work with exponents. For example, the prod
uct 2^ • 2^ can be simplified as follows.
5 + 3 = 8
r^
~\
2523= (22222)(222) = 28
This result, that products of exponential expressions with the same base are
found by adding exponents, is generalized as the product rule for exponents.
Product Rule for Exponents
If m and n are natural numbers and a is any real number, then
In words, when multiplying powers of like bases, keep the same base
and add the exponents.
To see that the product rule is true, use the definition of an exponent.
a"^ = a ' a ' a ' ' ' a
a appears as a factor m times. a appears as a factor n times.
From this, a'^'a'^ = a'a'a'''a'a'a'a'''a
(d) m^ ' p
5 . „4
m factors
a ' a ' a ' ' ' a
(m + n) factors
n factors
(e) (4a^)(6a^)
(f) (5p')(9p')
Answers
1. (a) mi4 (b) r^ (c) k^^
(d) The product rule does not apply.
(e) 24a^ (f) 45p'^
EXAMPLE 1
Using the Product Rule for Exponents
Apply the product rule for exponents, if possible, in each case.
11
y
(a) 3" • 3^ = 34+7 = 3
(c) y^ .jS. 3,2=^3+8+2
(d)(5j2)(3j4) = 5(3)jV
= 15j2+4
(e) (7p'q)i2p'q^) = 7(2)pyqq^
(f)x2j4
(b) 535 = 53 51= 53+1 = 54
Associative and commutative properties
Multiply; product rule
= 14/'V
Because the bases are not the same, the product rule does not apply.
Work Problem 1 at the Side.
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
Section 6.1 Integer Exponents and Scientific Notation 3 I 3
CAUTION
Be careful in problems like Example 1 (a) not to multiply the bases.
Notice that 3"^ • 3^ = 3^^, not 9^^. Keep the same base and add the
exponents.
& Evaluate each expression.
(a) 290
OBJECTIVE Q Define and negative exponents. So far we have
discussed only positive exponents. Now we define as an exponent. Sup
pose we multiply 4^ by 4^. By the product rule, extended to whole numbers,
42 . 40 = 42+0 = 42^
For the product rule to hold true, 4^ must equal 1, and so we define a^ this
way for any nonzero real number a.
(b) (29)'
9Q^0
Zero Exponent
If a is any nonzero real number, then
The expression 0^ is undefined.
_/_9Q^0
(c) (29)'
EXAMPLE 2
Using as an Exponent
Evaluate each expression.
(a) 6° = 1 (b) (6)0 = 1 Base is 6.
(c) 60 = (6°) = 1 Base is 6. (d) (6)" = 1
(e) 50 + 12° = 1 + 1 = 2 (f) {Uf =1, ki0
Work Problem 2 at the Side.
(d) 290
We should define negative exponents so that the rules for exponents are
valid. With this in mind, using the product rule should give
82 . 82 = 82+(2) = 80 = 1.
This indicates that 8~^ is the reciprocal of 8^. But — is the reciprocal of 8^,
8
and a number can have only one reciprocal. Therefore, it is reasonable to
conclude that 8"^ = ^. We can generalize and make the following definition.
(e) 8^150
Negative Exponent
For any natural number n and any nonzero real number a.
1
« =
(f) il5p'f, p^Q
With this definition, the expression a" is meaningful for any integer
exponent n and any nonzero real number a.
*In advanced treatments, 0^ is called an indeterminate form.
Answers
2. (a) 1 (b) 1 (c) 1 (d) 1
(e) (f) 1
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3 I 4 Chapter 6 Exponents, Polynomials, and Polynomial Functions
& In parts (a)(f ), write the
expressions with only
positive exponents. In parts
(g) and (h), simplify each
expression.
(a) 63
CAUTION
A negative exponent does not indicate that an expression represents a
negative number. Negative exponents lead to reciprocals.
Expression
Example
a*^
3"^ = ^ =  Not negative
— a~"^
3"^ = ^= g Negative
(b)8
(c) (2x)\ x^O
(d) lr\ ri=0
(e) q\ q^O
EXAMPLE 3
Using Negative Exponents
In parts (a)(f), write the expressions with only positive exponents. In parts
(g) and (h), simplify each expression.
(a) 2 = ^
w^'^^
(c) (5z)3= z^O
/iX 5
(d)5z3 = 5 =— z^O
(e) m ^ = ,, m i=
m
ff\ ( iii^~2 _ rni L C\
{my
<^> '"'""■ = ri=. '2 +
^ _ ' 14 4.13 3
12 ~ 12 3 4 12,4 3 12
^ ^ 5 2 10
5 3
10 ~ 10
(f) {q)\ q ^
CAUTION
In Example 3(g), note that 3~^ + 4~^ ^ (3 + 4)~^. The expression on
the left is equal to ^, as shown in the solution, while the expression on
the right is 7"^ = y. Similar reasoning can be applied to part (h).
Work Problem 3 at the Side.
(g) 31 + 5
1 u ^1
(h) 41  21
Answers
1 1
3. (a) — (b)  (c) (d)
6 8 /'o^^4 ^ ^
]_
(e) , (f) ^ (g) ^ (h)
q \q) '^
15
S EXAMPLE A]
Using Negative Exponents
Evaluate each expression.
(a)^ =
1
1
1
1 2^
2'
2'
1 1 1 32 32 9
(b) 32 
1
32
" 2^ ■ 32 ~ 2^ ' 1 ~ 2^ ~ 8
Example 4 suggests the following generalizations.
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
Section 6. 1 Integer Exponents and Scientific Notation 315
Special Rules for Negative Exponents
If a i^Omdbi^ 0, then
n um
a b a
O Evaluate each expression.
(«):Fi
Work Problem 4 at the Side.
OBJECTIVE Q Use the quotient rule for exponents. A quotient, such
as —:, can be simplified in much the same way as a product. (In all quotients
a
of this type, assume that the denominator is not 0.) Using the definition of
an exponent.
a a'aaa'aaaa
a
a ' a ' a
Notice that 8 — 3 = 5. In the same way,
.3
a
a ' a ' a
= a ' a ' a ' a ' a = a^.
1
a'aaa'aaaa a
Here, 3 — 8 = — 5. These examples suggest the quotient rule for exponents.
3^
& Apply the quotient rule for
exponents, if possible, and
write each result using only
positive exponents.
(«) .6
46
(b) ^, X ^
Quotient Rule for Exponents
If a is any nonzero real number and m and n are integers, then
a
In words, when dividing powers of like bases, keep the same base and
subtract the exponent of the denominator from the exponent of the
numerator.
EXAMPLE 5
Using the Quotient Rule for Exponents
Apply the quotient rule for exponents, if possible, and write each result using
only positive exponents.
Numerator exponent
I — Denominator exponent
(a) .2
37 t
t
Minus sign
"=/'''
(b) 'J = p'' =p<, p*0
•1  12
ki=Q (d)
2'
27(3) = 2^ + 3 = 2
10
o ,  , 1
(e) ^ = 8'' = 8"^ = 
(g)
5(
Z ,
^ zi=0
")i^ = ^ = ^'"'"" = «'
(b) ^. * *
The quotient rule does not apply
because the bases are different.
Work Problem 5 at the Side.
(c)
r''
r ¥=
(d)
2'
2'
(e)
6'
6'
(f)
(g)^, t^o
(h) , 7 9^
y
Answers
4. (a) 64 (b)
5. (a) 42 (b) x9 (c) — (d) 212
(e)  (f) 82 (g) t^
(h) The quotient rule does not apply.
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
3 I 6 Chapter 6 Exponents, Polynomials, and Polynomial Functions
© Use one or more power rules
to simplify each expression.
(a) (r^)
5^4
OBJECTIVE Q Use the power rules for exponents. The expres
sion (3"*)^ can be simplified as
(3^= 3434= 34+4= 38^
where 42 = 8. This example suggests the first power rule for exponents.
The other two power rules can be demonstrated with similar examples.
(b)
Power Rules for Exponents
If a and b are real numbers and m and n are integers, then
(\m ffi
) =^ (6 9^0).
bj b
In words,
(a) to raise a power to a power, multiply exponents;
(b) to raise a product to a power, raise each factor to that power; and
(c) to raise a quotient to a power, raise the numerator and the denomi
nator to that power.
(c) (9xf
(d) (Sr*)
6\3
EXAMPLE 6
Using the Power Rules for Exponents
Use one or more power rules to simplify each expression.
(a) (;,8)3=/ 3=^,24
(ly 2' 16
^•^^ UJ = ? = ¥
(c) {}yy = 3Y = 81j^
(d) (6py = 6 V ■ 2 = ey^ = 36/ji4
2^^^ (2)3^53 (_2)3^15 8^15
(e)
M
z^
Work Problem 6 at the Side.
The reciprocal of a" is — = f — ] . Also, by definition, a^ and a " are
reciprocals since
(e)
Answers
3n
4\3
m
, m^O
27
6. (a) r20 (b) — (c) 729x3
64
(d) 125ri8 (e)
21n'
a'' a' = a'' — = 1.
Thus, since both are reciprocals of a",
IV
a " = \
Some examples of this result are
'''[1^'  (ir='
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
This discussion can be generalized as follows.
More Special Rules for Negative Exponents
Ifai^O and b i^ and n is an integer, then
F  (r=(r
Section 6. 1 Integer Exponents and Scientific Notation 317
O Write each expression with
only positive exponents and
then evaluate.
In words, any nonzero number raised to the negative nth power is equal
to the reciprocal of that number raised to the nth power.
EXAMPLE 7
Using Negative Exponents with Fractions
Write each expression with only positive exponents and then evaluate.
(a)
3^2
7N2
3.
49
9
(b)
r'
125
64
Work Problem 7 at the Side.
The definitions and rules of this section are summarized here.
Definitions and Rules for Exponents
For all integers m and n and all real numbers a and b, the following
rules apply.
Product Rule
Quotient Rule
Zero Exponent
Negative
Exponent
Power Rules
a"
a'
{ah 0)
«o = 1 (a ^ 0)
1
  — (« ^ 0)
a =
aY «"
(6^0)
(b)
Special Rules
a
(«^0)
a
bY
(a, 6^0)
OBJECTIVE Q Simplify exponential expressions. With the rules of
exponents developed so far in this section, we can simplify expressions that
involve one or more rules, as shown in Example 8 on the next page.
Answers
/4V 64 fey 36
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
3 I 8 Chapter 6 Exponents, Polynomials, and Polynomial Functions
O Simplify each expression so
that no negative exponents
appear in the final result.
Assume all variables
represent nonzero real
numbers.
(a) 54 . 56
(b) x4 . x6 . r
(c) (5^)
3^2
(d) (yy
Sexample 8M!^llJ!liJJi5a!&;
ions an(
negative
nonzero
i Rules for Exponents
Simplify each expression so that no
result. Assume all variables represent
exponents appear in the final
real numbers.
(a) 3^ • 3^ = 3^<^) = 3' = ^
or
1
27
(b)x3x4x2 = x^ + <4)+2 = x
5 ^
1
(c) (42)5 = 4(2)(5) = 4io
(d) (x
4\6 _ „(4)6 _ „24 _ ^
X ^24
—4 ? —A 7
X V" X ' y'
(e) 2 5 = 2 ■ 5
X y X y
= X42 . j;2(5)
= Xy
/
x'
(f)
(23x
2)2 = (23)2 . (x2)2
= 26x4
x' x'
= 2^ ^"^ 64
f3x'yf4x'\' 3Hxy y'
^^H J 1,0 ^ ,^ 4x3
Combination of rules
9x' y'
y^ 4x^
Power rule
^ 41 2
=  X \
4
2 _
9x
Quotient rule; (2"" = [)
31,5
(e)
a b
7b^
NOTE
There is often more than one way to simplify expressions like those in
Example 8. For instance, we could simplify Example 8(e) as follows.
x'y'
x'y'
x'x^
y a" b"
7 Use— = —; product rule
(f) i3^kr
Work Problem 8 at the Side.
«(ITr"'
OBJECTIVE Q Use the rules for exponents with scientific notation.
The number of onecelled organisms that will sustain a whale for a few
hours is 400,000,000,000,000, and the shortest wavelength of visible light is
approximately .0000004 m. It is often simpler to write these numbers using
scientific notation.
In scientific notation, a number is written with the decimal point after
the first nonzero digit and multiplied by a power of 10.
Answers
8. (a) ^or;^ (b) \ (c) 5^
5 25 X
1 b' k' k' y
(d) ^ (e)  (f) or (g) 
y a 3 y x
Scientific Notation
A number is written in scientific notation when it is expressed in the form
a X 10^
where 1 < a < 10, and n is an integer.
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
Section 6.1 Integer Exponents and Scientific Notation 3 I 9
For example, in scientific notation,
8000 = 8 X 1000 = 8 X lOl
The following numbers are not in scientific notation.
.230 X 104 46.5 X 103
.230 is less than 1. 46.5 is greater than 10.
To write a number in scientific notation, use the following steps. (If the
number is negative, ignore the negative sign, go through these steps, and
then attach a negative sign to the result.)
& Write each number in
scientific notation.
(a) 400,000
Converting to Scientific Notation
Step 1 Position the decimal point. Place a caret, ^, to the right of the
first nonzero digit, where the decimal point will be placed.
Step 2 Determine the numeral for the exponent. Count the number
of digits from the decimal point to the caret. This number gives
the absolute value of the exponent on 10.
Step 3 Determine the sign for the exponent. Decide whether multiply
ing by 10" should make the result of Step 1 larger or smaller. The
exponent should be positive to make the result larger; it should
be negative to make the result smaller.
It is helpful to remember that for w > 1, lO"'^ < 1 and 10" > 10.
EXAMPLE 9
Writing Numbers in Scientific Notation
Write each number in scientific notation.
(a) 820,000
Place a caret to the right of the 8 (the first nonzero digit) to mark the new
location of the decimal point.
8a20,000
Count from the decimal point, which is understood to be after the last 0,
to the caret.
8 20^^000. ^Decimal point
Count 5 places.
Since the number 8.2 is to be made larger, the exponent on 10 is positive.
820,000 = 8.2 X 10^
(b) .0000072
Count from left to right.
.000007.2
6 places
Since the number 7.2 is to be made smaller, the exponent on 10 is negative.
.0000072 = 7.2 X 106
Work Problem 9 at the Side.
(b) 29,800,000
(c) 6083
(d) .00172
(e) .0000000503
Answers
9. (a) 4 X 10^ (b) 2.98 X 10^
(c) 6.083 X 103 (d) 1.72 X 103
(e) 5.03 X 108
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
320 Chapter 6 Exponents, Polynomials, and Polynomial Functions
CO) Write each number in
standard notation.
(a) 4.98 X 10^
Converting from Scientific Notation
Multiplying a number by a positive power of 10 makes the number larger,
so move the decimal point to the right n places ifn is positive in 10".
Multiplying by a negative power of 10 makes a number smaller, so
move the decimal point to the left \n\ places ifn is negative.
Ifn is 0, leave the decimal point where it is.
IP
(b) 6.8 X 10^
(c) 5.372 X IQO
EXAMPLE 10
Converting from Scientific Notation
to Standard Notation
Write each number in standard notation.
(a) 6.93 X 10^
6.9300000. Attach Os as necessary.
7 places
We moved the decimal point 7 places to the right. (We had to attach five Os.)
6.93 X 10^ = 69,300,000
(b) 4.7 X 106
.000004.7
6 places
We moved the decimal point 6 places to the left.
4.7 X 106 = .0000047
(c) 1.083 X 100 = 1.083 X 1 = 1.083
Work Problem 10 at the Side.
When problems require operations with numbers that are very large and/or
very small, and a calculator is not available, we can write the numbers in
scientific notation and perform the calculations using the rules for exponents.
CD Evaluate
200,000
EXAMPLE 11
.0003
.06 X 4,000,000 '
Answers
10. (a) 498,000 (b) .00000068 (c)
11. 2.5 X 10""' or .00025
5.372
Evaluate
Using Scientific Notation in Computation
1,920,000 X .0015
.000032 X 45,000 '
1,920,000 X .0015 1.92 X 10^ X 1.5 X 10"
.000032 X 45,000
3.2 X
10"^ X 4.5 X
10^
1.92 X
1.5 X
10*^ X
103
3.2 X 4.5 X
10"^ X
10^
1.92 X
1.5 X
10^
3.2 X 4.5 X
10'
1.92 X
1.5
X
10^
3.2 X 4.5
.2 X 10^
(2 X 101) X
2 X 10^ or
104
2000
Express all numbers in
scientific notation.
Commutative property
Product rule
Quotient rule
Simplify.
Work Problem 11 at the Side.
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
Section 6.1 Integer Exponents and Scientific Notation 32 I
lilll Calculator Tip To enter numbers in scientific notation, you can use
the (ee^ or (exp^ key on a scientific calculator. For instance, to work
Example 1 1 using a popular model calculator with an (^J key, enter the
following symbols.
1.92 (eT) 6 X 1.5 (ee^ 3 (+^  (T) 3.2 (eT) 5 (+^ X 4.5 (^4 (T) =
The (exp^ key is used in exactly the same way. Notice that the negative
exponent —3 is entered by pressing 3, then (+/^ . (Keystrokes vary
among different models of calculators, so you should refer to your
owner's manual if this sequence does not apply to your particular model.)
© The distance to the sun is
9.3 X 10^ mi. How long
would it take a rocket,
traveling at 3.2 X 10^ mph,
to reach the sun?
(Hint: t = i)
EXAMPLE 12
Using Scientific Notation to Solve Problems
In 1990, the national health care expenditure was $695.6 billion. By 2000,
this figure had risen by a factor of 1.9; that is, it almost doubled in only
10 years. (Source: U.S. Centers for Medicare & Medicaid Services.)
(a) Write the 1990 health care expenditure using scientific notation.
695.6 billion = 695.6 X 10^
= (6.956 X 102) X 109
= 6.956 X 10^1
In 1990, the expenditure was $6,956 X lO^^.
Product rule
(b) What was the expenditure in 2000?
Multiply the result in part (a) by 1 .9.
(6.956 X lO^O X 1.9 = (1.9 X 6.956) X lO^^
= 13.216 X 10^1
= 1.3216 X 10^2
The 2000 expenditure was about $1,321,600,000,000 (over $1 trillion).
Commutative and
associative properties
Round to three decimal
places.
Scientific notation
Work Problem 12 at the Side.
Answers
12. approximately 2.9 X 10'^hr
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Section 6.2 Adding and Subtracting Polynomials 329
b.2 Adding and Subtracting Polynomials
OBJECTIVE Q Know the basic definitions for polynomials. Just as
whole numbers are the basis of arithmetic, polynomials are fundamental in
algebra. To understand polynomials, we must review several words from
Section 1.4. A term is a number, a variable, or the product or quotient of a
number and one or more variables raised to powers. Examples of terms include
4x,
1
m \ or
m^
7z^, 6x^z,
5
3^'
and 9. Terms
The number in the product is called the numerical coefficient, or just
the coefficient.* In the term 8x^, the coefficient is 8. In the term —4p^, it is
— 4. The coefficient of the term k is understood to be 1. The coefficient of
—r is — 1. In the term f, the coefficient is  since f
Ix
3
OBJECTIVES
Know the basic definitions
for polynomials.
Find the degree of a
polynomial.
Add and subtract
polynomials.
Identify each coefficient.
(a) 9m^
a.
Work Problem 1 at the Side.
Recall that any combination of variables or constants (numerical values)
joined by the basic operations of addition, subtraction, multiplication, and
division (except by 0), or raising to powers or taking roots is called an algebraic
expression. The simplest kind of algebraic expression is sl polynomial.
Polynomial
A polynomial is a term or a finite sum of terms in which all variables
have whole number exponents and no variables appear in denominators
or under radicals.
(b) Uyh
(c) X
(d) y
Examples of polynomials include
3x — 5, 4m^ — 5m^p + 8, and —5fs^. Polynomials
Even though the expression 3x — 5 involves subtraction, it is a sum of terms
since it could be written as 3x + (—5).
Some examples of expressions that are not polynomials are
x~^ + 3x~^, V 9 — X, and — . Not polynomials
The first of these is not a polynomial because it has negative integer expo
nents, the second because it involves a variable under a radical, and the third
because it contains a variable in the denominator.
Most of the polynomials used in this book contain only one variable. A
polynomial containing only the variable x is called a polynomial in x. A poly
nomial in one variable is written in descending powers of the variable if the
exponents on the variable decrease from left to right. For example,
x^  6x^ + 12x  5
is a polynomial in descending powers of x. The term —5 in this polynomial
can be thought of as — 5x^, since — 5x^ = — 5(1) = —5.
(e)
& Write each polynomial in
descending powers.
(a) 4 + 9j + y^
(b) 3z4 + 2z3 + z^  6z
(c) I2m^^ + 8/77^+ 10m
12
Work Problem 2 at the Side.
* More generally, any factor in a term is the coefficient of the product of the remaining factors.
For example, 3x^ is the coefficient ofy in the term 3x^y, and 3y is the coefficient of x^ in 3x^y.
Answers
1. (a) 9 (b) 12 (c) 1 (d) 1 (e)
1
2. (a) y^ + 9y4 (b) z^  3z^ + Iz^  6z
(c) 10mi2  12mio + Sm^
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
330 Chapter 6 Exponents, Polynomials, and Polynomial Functions
® Identify each polynomial as a
trinomial, binomial,
monomial, or none of these.
(a) \2m^ — 6m
(b) 6j3 + 2/  8j
Some polynomials with a specific number of terms are so common that
they are given special names. A polynomial with exactly three terms is a
trinomial, and a polynomial with exactly two terms is a binomial. A single
term polynomial is a monomial. The table that follows gives examples.
Type of Polynomial
Examples
Monomial
5x, lm\ 8, xY
Binomial
3x26, llj + 8, 5a^b + ?>a
Trinomial
y^ + ny + 6, Sp^lp^2m, 1> + 2k^ + 9z^
None of these
p^  5p^ + 2p5, 9z^ + 5c^ + 2m^ + llr^  Ir
(c) 3a5
(d) 2/tio + 2k'^  Sk^ + 2k
O Give the degree of each
polynomial.
(a) 9y4 + 8_y3  6
(b) 12m^+ 11^3 + m'^
Work Problem 3 at the Side.
OBJECTIVE Q Find the degree of a polynomial. The degree of a
term with one variable is the exponent on the variable. For example, the
degree of 2x^ is 3, the degree of — x"* is 4, and the degree of 17x (that is, 17x^)
is 1. The degree of a term in more than one variable is defined to be the sum
of the exponents on the variables. For example, the degree of Sx^j'^is 10,
because 3 + 7 = 10.
The greatest degree of any term in a polynomial is called the degree of
the polynomial. In most cases, we will be interested in finding the degree of
a polynomial in one variable. For example, 4x^ — 2x^ — 3x + 7 has degree 3,
because the greatest degree of any term is 3 (the degree of 4x^).
The table shows several polynomials and their degrees.
(c)
2k
(d)
10
(e)
3mn^ + 2m^n
Answers
3. (a) binomial
(c) monomial
4. (a) 4 (b) 9
(b) trinomial
(d) none of these
(c) 1 (d) (e) 4
Copyrights
Polynomial
Degree
9x^  5x + 8
2
llm^ + 18^14  9m^
14
5x
1, because 5x = 5x^
2
0, because —2= —2x^ (Any nonzero constant has degree 0.)
5a'b'
7, because 2 + 5 = 7
13xy4+xV+ Ixy
12, because the degrees of the terms are 5, 12, and 2; 12 is
the greatest.
NOTE
The number has no degree, since times a variable to any power is 0.
Work Problem 4 at the Side.
OBJECTIVE Q Add and subtract polynomials. We use the distribu
tive property to simplify polynomials by combining terms. For example,
x^ + 4x^ + 5x^— 1 =x^ + (4 + 5)x^ — 1 Distributive property
= x3 + 9x^  1.
On the other hand, the terms in the polynomial 4x + 5x^ cannot be
combined. As these examples suggest, only terms containing exactly the
same variables to the same powers may be combined. As mentioned in
Section 1.4, such terms are called like terms.
Copyright ©2005 Pearson Education, Inc., pubiisiiing as Pearson AddisonWesley
Section 6.2 Adding and Subtracting Polynomials 33 I
CAUTION
Remember that only like terms can be combined.
& Combine terms.
(a) llx+ \2x  7x  3x
EXAMPLE 1
Combining Like Terms
Combine terms.
(a) 5y^ + 873  3;3 = (_5 + 8 _ 1)^3 = 2y^
(b) 6x + 57  9x + 27 = 6x  9x + 5j + 2y
Associative and
commutative
properties
Combine like terms.
= 3x + ly
Since — 3x and ly are unlike terms, no further simplification is possible.
(c) 5x^y — 6xy^ + 9x^y + 13xy^ = 5x^y + 9x^y — 6xy^ + 13xy^
= Ux^y + 7x/
Work Problem 5 at the Side.
(b) llp^ + 4p^  6p^ + 8;?'
(C) 2j2z4 + 3j4 + 5j4  9j;4^2
We use the following rule to add tw^o polynomials.
Adding Polynomials
To add two polynomials, combine like terms.
Polynomials can be added horizontally or vertically.
EXAMPLE 2
Adding Polynomials
Add: (3a^  9a^ + 4a^) + (^a^ + 8^^ + 2).
Use the commutative and associative properties to rearrange the poly
nomials so that like terms are together. Then use the distributive property to
combine like terms.
(3a^  9a^ + 4a^) + (Sa^ + Ha^ + 2)
= 3«5  8«5  9a^ + 8a^ \ 4a^ + 2
= —5a^ — a^ \ 4a^ + 2 Combine like terms.
Add these same two polynomials vertically by placing like terms in columns.
3^5  9^3 + 4^2
8a^ + 8^3 +2
5(3^  a^ + 4^2 + 2
Work Problem 6 at the Side.
& Add, using both the
horizontal and vertical
methods.
(a) (Uy^ 77 + 9)
+ (4j2 llj; + 5)
(b) 6r^ + 2r3  r^
8^5 _ 2^3 + 5^2
In Section 1.2, we defined subtraction of real numbers as
a — b = a + (—b).
That is, we add the first number (minuend) and the negative (or opposite) of
the second (subtrahend). We can give a similar definition for subtraction of
polynomials by defining the negative of a polynomial as that polynomial
with the sign of every coefficient changed.
Answers
5. (a) 13x (b) I5p^ + 2p^
(c) lyh"^ + 8j4 _ gy4^2
6. (a) 8j2  I83; + 14 (b) 2r5 
4r2
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332 Chapter 6 Exponents, Polynomials, and Polynomial Functions
& Subtract, using both the
horizontal and vertical
methods.
(a) (6j3  9y^ + 8)
(2y+/ + 5)
(b) 6y^  2j2 + 5y
2/ + 8/ lly
Subtracting Polynomials
To subtract two polynomials, add the first polynomial and the negative
of the second polynomial.
EXAMPLE 3
Subtracting Polynomials
Subtract: (6m^  ?>m + 5)  (5m^ + 7m  8).
Change every sign in the second polynomial and add.
(6^2  8m + 5)  (5^2 + 7m  8)
= —6m^ — 8m + 5 + 5m^ — 7m + 8 Definition of subtraction
= — 6m^ + 5m^ — 8m — 7m + 5 + 8 Rearrange terms.
m"
15m + 13
Combine like terms.
Check by adding the sum, m^  15m + 13, to the second polynomial. The
result should be the first polynomial.
To subtract these two polynomials vertically, write the first polynomial
above the second, lining up like terms in columns.
— 6m^ — 8m + 5
— 5m^ + 7m — 8
Change all the signs in the second polynomial, and add.
— 6m^ — 8m + 5
+ 5m^ — 7m + 8 Change all signs.
— m^— 15m +13 Add in columns.
Work Problem 7 at the Side.
Answers
7. (a) 4j3  lOy^ + 3
(b) Sj^  10j;2 + ley
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Section 6.3 Polynomial Functions 335
6. 3 Polynomial Functions
OBJECTIVE Q Recognize and evaluate polynomial functions. In
Chapter 4 we studied linear (firstdegree polynomial) functions, defined
as/(x) = mx + b. Now we consider more general polynomial functions.
Polynomial Function
A polynomial function of degree n is defined by
f(x) = tt^x'' + a„_ix''~^ + • • • + ^ix + «05
for real numbers a , a ,,..., a,, and a^, where a ^0 and w is a whole
number.
Another way of describing a polynomial function is to say that it is a
function defined by a polynomial in one variable, consisting of one or more
terms. It is usually written in descending powers of the variable, and its
degree is the degree of the polynomial that defines it.
Suppose that the polynomial 3x^ — 5x + 7 defines function/ Then
f(x) = 3x2  5;^ + 7.
If X = —2, then/(x) = 3x^ — 5x + 7 takes on the value
/(2) = 3(2)25(2) + 7
= 34+10 + 7
= 29.
Letx = —2.
OBJECTIVES
Recognize and evaluate
polynonnial functions.
Use a polynomial function
to model data.
Add and subtract
polynomial functions.
Graph basic polynomial
functions.
Let/(x) =
Find each value.
(a) /(I)
+ 5x 11.
Thus,/(2) = 29 and the ordered pair (2, 29) belongs to/
(b)/(4)
EXAMPLE 1
Evaluating Polynomial Functions
Let/(x) = 4x^  x^ + 5. Find each value.
(a) /(3)
f(x) = 4x3  x2 + 5
/(3) = 43332 + 5 Substitute 3 forx.
= 427 — 9 + 5 Order of operations
= 1089 + 5
= 104
(b)/(4) =
= 4  (4)3  (4)2 +5 Letx = 4; use parentheses.
=
= 4  (—64) —16 + 5 Be careful with signs.
=
= 267
(c) /(O)
While/is the most common letter used to represent functions, recall that
other letters such as g and h are also used. The capital letter P is often used ft)r
polynomial fiinctions. Note that the fiinction defined as P(x) = 4x^  x^ + 5
yields the same ordered pairs as the function/in Example 1.
Work Problem 1 at the Side.
Answers
1. (a) 7 (b) 47 (c) 11
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
336 Chapter 6 Exponents, Polynomials, and Polynomial Functions
& Use the function in Example 2
to approximate the number of
households expected to pay at
least one bill online each
month in 2006.
OBJECTIVE Q Use a polynomial function to model data. Polynomial
functions can be used to approximate data. They are usually valid for small
intervals, and they allow us to predict (with caution) what might happen for
values just outside the intervals. These intervals are often periods of years,
as shown in Example 2.
EXAMPLE 2
Using a Polynomial Model to Approximate Data
The number of U.S. households estimated to see and pay at least one bill
online each month during the years 2000 through 2006 can be modeled by
the polynomial function defined by
P(x) = .808x2 + 2.625X + .502,
where x = corresponds to the year 2000, x = I corresponds to 2001, and
so on, and P(x) is in millions. Use this function to approximate the number
of households expected to pay at least one bill online each month in 2005.
Since x = 5 corresponds to 2005, we must find P (5).
.808x2 + 2.625X + .502
.808(5)2 + 2.625(5) + .502
33.827
P(x)
P(5)
Letx = 5.
Evaluate.
Thus, in 2005 about 33.83 million households are expected to pay at least
one bill online each month.
Work Problem 2 at the Side.
OBJECTIVE Q Add and subtract polynomial functions. The opera
tions of addition, subtraction, multiplication, and division are also defined
for functions. For example, businesses use the equation "profit equals
revenue minus cost," written using function notation as
P(x) = R(x)  C(x),
t t t
Profit Revenue Cost
function function function
where x is the number of items produced and sold. Thus, the profit function
is found by subtracting the cost function from the revenue function.
We define the following operations on functions.
Adding and Subtracting Functions
If/(x) andg(x) define functions, then
if+g)ix)=fix) + gix)
and (/  g) (x) = fix)  g(x).
Sum function
Difference function
In each case, the domain of the new function is the intersection of the
domains of/(x) andg(x).
Answers
2. 45.34 million
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
Section 6.3 Polynomial Functions 337
ifauTf
EXAMPLE 3
Video J
Adding and Subtracting Functions
For the polynomial functions defined by
f(x) = x^ — 3x + 7 and g(x) = —3x^ — 7x + 7,
find (a) the sum and (b) the difference.
(a) (f+g)(x)=fix)+g(x)
= (x^  3x + 7) + (3x2 jjc+7)
= 2x^  lOx + 14
(b)(/g)(x)=/(x)^(x)
= (x^  3x + 7)  (3x2 _ 7;^ + 7)
= (x^  3x + 7) + (3x2 + 7;^ _ 7)
Use the definition.
Substitute.
Add the polynomials.
Use the definition.
Substitute.
4x2 + 4x
Change subtraction
to addition.
Add.
Work Problem 3 at the Side.
EXAMPLE 4
Adding and Subtracting Functions
For the polynomial functions defined by
/(x) = lOx^  2x and g(x) = 2x,
find each of the following.
(a) (/+g)(2)
(/ + g) (2) = /(2) + g(2) Use the definition.
= [10(2)2  2(2)] + 2(2) Substitute.
= 40
Alternatively, we could first find (/+ g) (x).
(/ + g) (x) = /(x) + ^(x) Use the definition.
= (lOx^  2x) + 2x Substitute.
= lOx^ Combine like terms.
Then,
(/+g)(2)= 10(2)2
= 40. The result is the same.
(b)(/g)(x)and(/g)(l)
(/ — g) (x) = /(x) — ^(x) Use the definition.
= (lOx^  2x)  2x Substitute.
= lOx^ — 4x Combine like terms.
Then,
(/ g)(l) = 10(1)^  4(1) Substitute.
= 6.
Confirm that/(l) — g(l) gives the same result.
Work Problem 4 at the Side.
OLet
fix) = 3x2 + 8x  6
and g(x) = 4x2 + 4x 
Find each function.
(a) (/+g)(x)
(b)(/g)(x)
24x
OFor
fix) = 18x2
and g(x) = 3x,
find each of the following.
(a) (/+g)(x)and(/+g)(l)
(b)(/g)(x)and(/g)(l)
Answers
3. (a) x2 + 12x  14
(b) 7x2 + 4;^ + 2
4. (a) 18x221x;39
(b) 18x2  27x; 9
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
338 Chapter 6 Exponents, Polynomials, and Polynomial Functions
OBJECTIVE Q Graph basic polynomial functions. Functions were
introduced in Section 4.5. Recall that each input (or xvalue) of a function
results in one output (or jvalue). The simplest polynomial function is the
identity function, defined by/(x) = x. The domain (set of xvalues) of this
function is all real numbers, (0°, 0°), and it pairs each real number with
itself. Therefore, the range (set of jvalues) is also (—0°, ^). Its graph is a
straight line, as first seen in Chapter 4. (Notice that a linear function is
a specific kind of polynomial function.) Figure 1 shows its graph and a table
of selected ordered pairs.
m=x
■'
f{x) = X
2
2
1
1
1
1
2
2
Figure 1
Another polynomial function, defined by f(x) = x^, is the squaring
function. For this function, every real number is paired with its square. The
input can be any real number, so the domain is (oo, co). Since the square of
any real number is nonnegative, the range is [0, oo). Its graph is a. parabola.
Figure 2 shows the graph and a table of selected ordered pairs.
(2, 4)^
■'
f{x)=x2_j
2
4
1
1
1
1
2
4
Figure 2
The cubing function is defined by/(x) = x^. Every real number is
paired with its cube. The domain and the range are both (0°, ^). Its graph
is neither a line nor a parabola. See Figure 3 and the table of ordered pairs.
(Polynomial functions of degree 3 and greater are studied in detail in more
advanced courses.)
I>
f{x)=x3_,
2
8
1
1
1
1
2
8
Figure 3
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
EXAMPLE 5
Graphing Variations of the Identity, Squaring,
and Cubing Functions
Graph each function by creating a table of ordered pairs. Give the domain
and the range of each function by observing the graphs.
(a) /(x) = 2x
To find each range value, multiply the domain value by 2. Plot the points
and join them with a straight line. See Figure 4. Both the domain and the
range are {—^, ^).
Section 6.3 Polynomial Functions 339
& Graph/(x) = 2x^. Give the
domain and the range.
X
f(x) = 2x\
2
4
1
2
1
2
2
4
Figure 4
(b)/(x)=x2
For each input x, square it and then take its opposite. Plotting and joining
the points gives a parabola that opens down. See the table and Figure 5. The
domain is (oo, oo), and the range is (oo, 0].
lJL^
f (X) = X2
2
4
1
1
1
1
2
4
Figure 5
(c) fix) = x'2
For this function, cube the input and then subtract 2 from the result. The
graph is that of the cubing function shifted 2 units down. See the table and
Figure 6. The domain and the range are both (—0°, ^).
p, 6)
:: I f{x) = x^2
Domain
1 
f (x) = x3  2 1
2
10
1
3
2
1
1
2
6
(2,10)
Range
10
Figure 6
Work Problem 5 at the Side.
domain: (0°, oo); range: (°°, 0]
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
Section 6.4 Multiplying Polynomials 343
V
6.4 Multiplying Polynomials
OBJECTIVE Q Multiply terms. Recall that the product of the two
terms ?>x^ and 5x^ is found by using the commutative and associative prop
erties, along with the rules for exponents.
(3x4) (5x3) = 3 sjc^x^
= 15x4+3
= 15x7
\B\
EXAMPLE 1
Multiplying Monomials
Find each product.
(a) 4a\3,a^) = 4(3)a^ ■ a^ = Ua^
(b) 2mh\Smh^) = 2(S)m^ • m^ • z'^ • z^ = \6mh^
Work Problem 1 at the Side.
OBJECTIVE Q Multiply any two polynomials. We use the distributive
property to extend this process to find the product of any two polynomials.
OBJECTIVES
Q Multiply terms.
Q Multiply any two
polynomials.
Q Multiply binomials.
Q Find the product of the
sum and difference of two
terms.
Q Find the square of a
binomial.
Q Multiply polynomial
functions.
Find each product.
(a) 6m^{2m^)
EXAMPLE 2
Multiplying Polynomials
Find each product.
(a) 2(8x3  9x2)
2(8x3  9x2) = 2(8x3)  2(9x2) Distributive property
= 16x3+ 13^2
(b) 5x2(4x2 + 3x  2) = 5x^(4x2) + Sx^{3,x) + 5x2(2)
= 20x4 + 15x3  10x2
(c) (3x  4)(2x2 + x)
Use the distributive property to multiply each term of 2x2 + x by 3x — 4.
(3x  4) (2x2 + x) = (3x  4) (2x2) + (3x  4)(x)
"l_i
Here 3x — 4 has been treated as a single expression so that the distributive
property could be used. Now use the distributive property two more times.
= 3x(2x2) + (4) (2^2) + (3x) (x) + (4) (x)
= 6x3 _ 8^2 + 3^2 _ 4^
= 6x3 _ 5^2 _ 4^
(d) 2x2 (x + i)(jc _ 3) = 2x2 [(x + i)(_^) + (x + l)(3)]
= 2x2 [x2 + X — 3x — 3]
= 2x2 (x2  2x  3)
= 2x4 _ 4^3 _ 6^2
Work Problem 2 at the Side.
(b) U'y{9ky')
& Find each product.
(a) 2r(9r 5)
(b) l>p^{5p^ + 2p^l)
(c) {Aa  5) (3a + 6)
(d) 3x3(x + 4)(x6)
Answers
1. (a) 12m9 (b) llkY
2. (a) 18r2 + lOr (b) \5p^ + 6/  2\p^
(c) 12^2 + 9a  30 (d) 3x5 _ 5^4 _ 72x3
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344 Chapter 6 Exponents, Polynomials, and Polynomial Functions
It is often easier to multiply polynomials by writing them vertically.
O Find each product.
(a) 2m 5
3m + 4
EXAMPLE 3
Multiplying Polynomials Vertically
Find each product.
(a) (5a 2b) (3a + b)
5a — 2b
3a + b
5ab  2b^ < — b(5a  lb)
\5a^  6ab < 3a (5a  lb)
\5a^ — ab — 2b^ Combine like terms.
(b) (3m^ 2m^ + 4)(3m  5)
3m^  2m^ + 4
3m — 5
I5m^ + lOm^  20
9m^ — 6m^ + 12m
9^4  2lm^ + 10^2 + 12m  20
5(3m^  Irn^ + 4)
3m{3w? — 2m^ + 4)
Combine like terms.
Work Problem 3 at the Side.
(b) 5a^  6a^ + 2a
2a
Answers
3. (a) 6m2  7m  20
(b) 10^4  37^3 + 34^2  I6a + 15
OBJECTIVE Q Multiply binomials. When working with polynomials,
the product of two binomials occurs repeatedly. There is a shortcut method
for finding these products. Recall that a binomial has just two terms, such as
3x — 4 or 2x + 3. We can find the product of these binomials using the dis
tributive property as follows.
(3x  4)(2x + 3) = 3x(2x + 3)  4(2x + 3)
= 3x(2x) + 3x(3)  4(2x)  4(3)
= 6x^ + 9x  8x  12
Before combining like terms to find the simplest form of the answer, let us
check the origin of each of the four terms in the sum. First, 6x^ is the prod
uct of the two first terms.
(3x4)(2x+3) 3x(2x) = 6x2 First terms
To get 9x, the outer terms are multiplied.
(3x4)(2x + 3) 3x(3) = 9x Outer terms
The term — 8x comes from the inner terms.
(3x  4)(2x + 3) 4(2x) = 8x Inner terms
Finally, — 12 comes from the last terms.
(3x  4)(2x + 3) 4(3) = 12 Last terms
The product is found by combining these four results.
(3x  4)(2x + 3) = 6x^ + 9x  8x  12
= 6x2 + X  12
To keep track of the order of multiplying these terms, we use the initials
FOIL (First, Outer, Inner, Last). All the steps of the FOIL method can be
done as follows. Try to do as many of these steps as possible mentally.
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
6x2 _i2
(3x4)(2x + 3)
■ V '
8x
X Add.
EXAMPLE 4
Using the FOIL Method
Use the FOIL method to find each product.
(a) (4m  5) (3m + 1)
First terms
Outer terms
Inner terms
Last terms
(4m  5) (3m + 1)
I t
(4m  5) (3m + 1)
I ^
(4m  5) (3m + 1)
I ^
(4m  5) (3m + 1)
I ^
4m (3m) = I2m^
4m (1) = 4m
— 5 (3m) = —I5m
5(1) =5
Simplify by combining the four terms.
O
I
(4m  5) (3m + 1) = I2m^ + 4m  I5m  5
= I2m^ — 11m — 5
The procedure can be written in compact form as follows.
12^2 5
(4m  5) (3m + 1)
11m Add.
Combine these four results to get 12m^ — 1 Im — 5.
First Outer Inner Last
I \^ I I
(b) (6a  5b) (3a + 4b) = ISa^ + 24ab  \5ab  20b^
= 18^2 + 9ab  20b^
(c) (2k + 3z) (5k  3z) = 10^2 + 9^^ _ 9^2 FOIL
Section 6.4 Multiplying Polynomials 345
O Use the FOIL method to find
each product.
(a) (3z + 2)(z+ 1)
Work Problem 4 at the Side.
^11
OBJECTIVE Q Find the product of the sum and difference of two
terms. Some types of binomial products occur frequently. For example,
the product of the sum and difference of the same two terms, x and j, is
(x + y)(x  y) = x^  xy + xy  y^ FOIL
= x^  y\
(b) (5r 3)(2r 5)
(c) (4p + 5q)(3p2q)
(d)(4yz)(2y + 3z)
(e) (8r+ l)(8r 1)
Product of the Sum and Difference of Two Terms
The product of the sum and difference of the two terms x and j is
the difference of the squares of the terms.
(X+ J)(xj)=x2j2
Answers
4. (a) 3z2 + 5z + 2 (b) lOr^  31r + 15
(c) I2p^ + 7pq  10^2
(d) 872 + lOyz  3z2 (e) 64r2  1
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
346 Chapter 6 Exponents, Polynomials, and Polynomial Functions
& Find each product.
(a) (m + 5)(m  5)
(h)(x4y)(x + 4y)
EXAMPLE 5
Multiplying the Sum and Difference of Two Terms
Video I Find each product.
(a) (p + 7)(p7)=p^l^ {X + y)(,x  y) = x^  y^
(b) (2r+ 5)(2r 5) = (2r)2  5^
= 2V2  25
= 4r2  25
(c) (6m + 5n)(6m  5n) = (6my  (Sn)^
= 36m^ — 25n^
(d) 2x\x + 3)(x  3) = 2xHx^  9)
= 2x5  18x3
Work Problem 5 at the Side.
OBJECTIVE Q Find the square of a binomial. Another special bino
mial product is the square of a binomial. To find the square of x + j, or
(x + yY, multiply x \ yhy itself
(x + j) (x + j) = x^ + xj + xj + y^ FOIL
= x^ + 2xy + j^
A similar result is true for the square of a difference.
(c) {Im — 2n) (7m + 2n)
Square of a Binomial
The square of a binomial is the sum of the square of the first term,
twice the product of the two terms, and the square of the last term.
(x + yy = x^ + 2xy + y^
{x  yy = x^  2xy + y^
Video J AnimaHiigl
(d) V(_y+7)(y7)
EXAMPLE 6
Squaring Binomials
Find each product.
(a) (m + If = m^ + 2 • m • 7 + 7^ (x ^ yf = x^ + 2xy + y^
= m^ + I4m + 49
(h) (p 5y=p^ 2'p' 5 + 5^ (xyy = x^2xy^y^
= p^  I0p + 25
(c) (2p + 3vf = (IpY + 2(2/;)(3v) + (Sv)^
= Ap^ + 12;7v + 9v^
(d) (3r  5sf = (3r)2  2(3r)(5^) + (5^)^
= 9r2  30r^ + 25s^
Answers
5. (a) m2  25 (b) x^  I6y^
(c) 49^2  4n^ (d) 43;4  I96y^
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Section 6.4 Multiplying Polynomials 347
CAUTION
As the products in the formula for the square of a binomial show,
(x + yf # x^ + y^.
More generally,
(x + yf #x" +7" (w^l).
Work Problem 6 at the Side.
© Find each product.
(a) {a + If
(b) (2m  5)2
We can use the patterns for the special products with more complicated
products, as the following example shows.
EXAMPLE 7
Multiplying More Complicated Binomials
Use special products to find each product.
(a) [(3/7 2) + 5^] [(3/; 2) 5^]
= (3/7 — ly — (5qy Product of sum and difference of terms
= 9p^ — I2p \ 4 — 25q^ Square both quantities.
(b) [(2z + r)+lY
(2z + rf + 2(2z + r)(l)+ P
Square of a binomial
Square again; use the
distributive property.
(c) (x+yy = (x + y)\x+y)
= (x^ + 2xy + y^)(x + y)
= x^ + 2x^y + xy^ + x^y + 2xy^ + y^
= x^ + 3x^y + 3xy^ + y^
(d) (2a + by = (2a + by(2a + /?)2
= (4^2 + 4a/? + /?2)(4^2 + 4^^ + ^2)
= 16^4 + \6a^b + 4^2/^2 + 16^3/? + 16a2/?2
+ 4(3/?3 + Aa^b^ + 4(3Z?3 + b^
= 16^4 + 32^3/? + 24a^b^ + Sa/?^ + b^
Square x \ y.
Square 2a + b.
Work Problem 7 at the Side.
OBJECTIVE Q Multiply polynomial functions. In Section 6.3 we saw
how functions can be added and subtracted. Functions can also be multiplied.
(c) (y + 6zf
(d) (3k  In)
 1„\2
O Find each product.
(a) [(m  In)  3]
• [{m  In) + 3]
(b) [{k  5h) + 2]2
(c) (p + 2qf
Multiplying Functions
If/(x) andg(x) define functions, then
(fs)(^) — /W * S(^)' Product function
The domain of the product function is the intersection of the domains of
/(x)andg(x).
(d) (x + 2)4
Answers
6.
(a)
a^
+ 4a + 4
(b)
4m
2  20m + 25
(c)
y^
+ I2yz + 36z2
(d)
9k
2  Ukn + 4n^
7.
(a)
m2
 4mn + 4^2 _
9
(b)
k^
 \Okh + 25/^2
+ 4k
20h
(c)
P'
+ 6p^q + Upq
^ + 8g3
(d)
x'
+ 8x3 + 24x2 + 32x + 16
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
348 Chapter 6 Exponents, Polynomials, and Polynomial Functions
©For
f(x) = 2x + 7
and g(x) = x^  4,
find (/g)(x) and (/g)(2).
EXAMPLE 8
Multiplying Polynomial Functions
For/(x) = 3x + 4 and g (x) = 2x^ + x, find (fg) (x) and (/g) (  1 ).
(fg)(x) =f(x) ' g(x) Use the definition.
= (3x + 4)(2x2 + x)
= 6x^ + 3x2 + g^2 + 4_^ poiL
= 6x^ + llx^ + 4x Combine like terms.
Let X = — 1 .
Then
(/g)(l) = 6(l)3+ll(l)2 + 4(l)
= 6 + 11 4
= 1.
(Another way to find (fg)(— 1) is to find/(— 1) and g(— 1) and then multi
ply the results. Verify this by showing that /(—I) • g(— 1) equals 1. This
follows from the definition.)
Work Problem 8 at the Side.
CAUTION
Write the product/(x) • g(x) as (fg)(x), notf(g(x)), which has a differ
ent mathematical meaning as discussed in Section 12.1.
Answers
8. 2x3 + 7x2 _ 8;^ _ 28;
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Section 6.5 Dividing Polynomials 353
6. 5 Dividing Polynomials
OBJECTIVE Q Divide a polynomial by a monomial. We now discuss
polynomial division, beginning with division by a monomial. (Recall that
a monomial is a single term, such as 8x, 9m'^, or 1 \y^.)
Dividing by a Monomial
To divide a polynomial by a monomial, divide each term in the polyno
mial by the monomial, and then write each quotient in lowest terms.
OBJECTIVES
Divide a polynomial by
a monomial.
Divide a polynomial by
a polynomial of two or
more terms.
Divide polynomial
functions.
You Try I
EXAMPLE 1
Dividing a Polynomial by a Monomial
Divide.
15x'  12x + 6
(a)
\5x'
\2x 6
~ y ~ Divide each term by 3.
3 3 ^
= 5x2  4x + 2
Write in lowest terms.
Check this answer by multiplying it by the divisor, 3. You should get
ISx^ — 12x + 6 as the result.
3(5x2 _ 4^ + 2) = 15x2  12x + 6
(b)
This result is not a polynomial. (Why?) The quotient of two polynomials
need not be a polynomial.
/\ V V
t t
Divisor Quotient Original polynomial
5m^ 
 9ni + IOot STrv" 9m^ lOm
5m' ~ 5m' 5m' ' 5m'
Divide each term by 5m^.
9 2
= m  + 
5 m
Write in lowest terms.
2., 2
(c)
Sxy'  9xy + 6xY 8xy^ 9x">' ex'j
2 2
X y
2 2
X y
y
2 2
X y
+ 6
+
2 2
X y
Worl< Problem 1 at the Side.
OBJECTIVE Q Divide a polynomial by a polynomial of two or more
terms. This process is similar to that for dividing whole numbers.
O Divide.
12;? + 30
(a) ^^^7
(b)
9y^  4y^ + Sy
(c)
Sa'y  20ab
4a^b
VouTr
Video
EXAMPLE 2
Divide
Dividing a Polynomial by a Polynomial
2m^ + m — 10
m — 2
Write the problem, making sure that both polynomials are written in
descending powers of the variables.
m — 2)2m^ + m — 10
Continued on Next Page
Answers
9y 4
1. {^) 2p + 5 (b) —  2 + 
2 y
lb 5b'
(c) ^
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354 Chapter 6 Exponents, Polynomials, and Polynomial Functions
Divide.
2r^ + r
(a)
21
(b)
2k^ + \lk + 30
2^+5
Answers
2. (a) 2r + 7 (b) ^ + 6
Divide the first term of 2m^ + m — 10 by the first term of m — 2.
2m ^
Since ^^^^^— = 2m, place this result above the division line.
m
2m < — Result of ^
m — 2)2m^ + m — 10
Multiply m — 2 and 2m, and write the result below 2m^ + m — 10.
2m
m — 2)2m^ + m — 10
2m^ — 4m < 2m (m  2) = 2m^  4m
Now subtract by mentally changing the signs on 2m^ — 4m and adding.
2m
m
2)2m^ + m
2m^ — 4m
5m
10
< Subtract. The difference is 5m.
Bring down — 10 and continue by dividing 5m by m.
m
2m + 5
2)2m^ + m
10
2m — 4m
5m  10
5m  10
Bring down —10.
5(m  2) = 5m  10
Subtract. The difference is 0.
Finally, (2m ^ + m — 10) ^ (m — 2) = 2m + 5. Check by multiplying m — 2
and 2m + 5. The result should be 2m^ + m — 10.
Work Problem 2 at the Side.
EXAMPLE 3
Dividing a Polynomial with a MissingTerm
Divide 3x^ — 2x + 5 by x — 3.
Make sure that 3x^ — 2x + 5 is in descending powers of the variable.
Add a term with coefficient as a placeholder for the missing x^term.
r Missing term
3)3x^ + Ox^  2x + 5
Start with 
3x^
3x1
3x^
3x'
3x' <
X  3)3x^ + Ox^  2x + 5
3x^  9x^ < 3x2(x  3)
Subtract by mentally changing the signs on 3x^  9x^ and adding.
3x2
X  3)3x^ + Ox^  2x + 5
3x^  9x\
< Subtract.
Bring down the next term.
9x'
3x'
3)3x^ + Ox^  2x + 5
3x^  9x^_
2x
9x2
Continued on Next Page
< Bring down 2x.
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
9x2
In the next step, = 9x.
3x2 + 9^
< —
¥ = 9x
X  3)3x^ + 0x2  2x + 5
3x^  9x2
9x2 _ 2x
9x2 _ 2']x
< —
9x{x  3)
25x + 5
< —
Subtract; bring down 5.
25x
Finally, = 25.
3x2 + 9;c + 25
< —
f = 25
X  3)3x^ + 0x2  2x + 5
3x^  9x2
9x2 _ 2x
9x2 _ 2']x
25x+ 5
25x  75
< —
 25(x3)
80
< —
 Remainder
80
Write the remainder, 80, as the numerator of the fraction ^ _ 3. In summary.
= Sx" + 9x + 25 +
80
X  3
X  3
Check by mukiplying x  3 and 3x^ + 9x + 25 and adding 80. The result
should be 3x^  2x + 5.
CAUTION
. remainder
Remember to write — —. as part of the quotient.
divisor
Section 6.5 Dividing Polynomials 355
& Divide.
3P + 9^14
k2
Divide 2p^ + 7p^ + 9p + 2
by 2p + 2.
Work Problem 3 at the Side.
EXAMPLE 4
Performing a Division with a Fractional
Coefficient in the Quotient
Divide 2p^ + 5p^ + p  2hy2p + 2.
p' + ^pi
= Ip
Ip + l)lp^ + 5p^ + p  2
2p' + 2/^2
3/p2+ p
3/?2 + 3p
2/72
2/72
Since the remainder is 0, the quotient is p^ + ^p — 1.
Answers
3. 3k^ + 6k + 2\
4. p^ + p + 2 ■
28
f
k2
2
2p + 2
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
356 Chapter 6 Exponents, Polynomials, and Polynomial Functions
& Divide.
(a)
3r2  2
(b)
4x''  7x^ + X + 5
2x^ — X
EXAMPLE 5
Dividing by a Polynomial with a MissingTerm
Divide 6r4 + 9r3 + 2r2  8r + 7 by Sr^  2.
Write 3r^ — 2 as 3r^ + Or — 2 and divide as usual.
2r2 + 3r + 2
3r2 + Or  2)6r4 + 9r^ + 2r^
8r + 7
Missing term
9^3 + 6^2 _ g^
9^3 + 0^2 _ 5^
6r^  2r + 7
5^2 + Or  4
2r+ 11
Since the degree of the remainder, — 2r + 11, is less than the degree of
the divisor, 3r^ — 2, the process is now finished. The result is written
2r2 + 3r + 2 +
2r + 11
3r2  2 '
Work Problem 5 at the Side.
CAUTION
When dividing a polynomial by a polynomial of two or more terms:
1. Be sure the terms in both polynomials are in descending powers.
2. Write any missing terms with placeholders.
OBJ ECTI VE
Divide polynomial functions.
© For
and
fix)
g(x)
(f
find (t)(x) and ©(1).
2x2 + 17x + 30
2x + 5,
Dividing Functions
If/(x) and g (x) define functions, then
f\,. fix)
gJ gix)
Quotient function
The domain of the quotient function is the intersection of the domains
of/(x) and g (x), excluding any values of x for which g (x) = 0.
Answers
5. (a) r^  5r2 + 3 4
(b) 2x^ + X  3 +
3r2  2
2x + 5
2x^
6. x + 6, X ^ ; 5
2
EXAMPLE 6
Dividing Polynomial Functions
For/(x) = 2x2 + X  10 andg(x) = x  2, find (f) (x) and () (3).
A _ fix) _ 2x2 + x  10
gr' gix) x2
This quotient, found in Example 2, with x replacing m, is 2x + 5, so
A
— (x) = 2x + 5, X 9^ 2.
gJ
Then
/
(3) = 2(3) + 5 = 1. Letx
/(3) .
Kg
(Which is easier to find here: (f ) (  3) or jpy?)
3.
Work Problem 6 at the Side.
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
Factoring
11 ■ J^^^
\
7T Greatest Common Factors;
Factoring by Grouping
7.2 Factoring Trinomials
73 Special Factoring
Summary Exercises on Factoring
74 Solving Equations by Factoring
Factoring is used to solve quadratic equations, which have many
useful applications. An important one is to express the distance
a falling or propelled object travels in a specific time. Such equations
are used in astronomy and the space program to describe the motion
of objects in space.
In Section 7.4 we use the concepts of this chapter to explore how
to find the heights of objects after they are propelled or dropped.
37!
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
372 Chapter 7 Factoring
7.1 Greatest Common Factors; Factoring by Grouping
OBJECTIVES
Factor out the greatest
common factor.
Factor by grouping.
(%
Study Skills Workbook
Activity 8: Study Cards
O Factor out the greatest
common factor.
(a) 7k + 28
Writing a polynomial as the product of two or more simpler polynomials is
called factoring the polynomial. For example, the product of 3x and 5x  2 is
15x^ — 6x, and 15x^ — 6x can be factored as the product 3x(5x — 2).
3x(5x  2) = 15x^  6x Multiplying
15x^ — 6x = 3x(5x — 2) Factoring
Notice that both multiplying and factoring use the distributive property, but
in opposite directions. Factoring "undoes" or reverses multiplying.
OBJECTIVE Q Factor out the greatest common factor. The first step
in factoring a polynomial is to find the greatest common factor for the terms
of the polynomial. The greatest common factor (GCF) is the largest term
that is a factor of all terms in the polynomial. For example, the greatest com
mon factor for 8x + 12 is 4, since 4 is the largest term that is a factor of (that
is, divides into) both 8x and 12. Using the distributive property.
8x+ 12
4(2x) + 4(3)
4(2x + 3).
(b) 32m + 24
As a check, multiply 4 and 2x + 3. The result should be 8x + 12. Using the
distributive property this way is cMQd factoring out the greatest common
factor.
(c) 8a
(d) 5z + 5
EXAMPLE 1
Factoring Out the Greatest Common Factor
Factor out the greatest common factor.
(a) 9z  18
Since 9 is the GCF, factor 9 from each term.
9z 18 = 9 29 2
= 9(z2)
Check: 9(z — 2) = 9z — 18 Original polynomial
(b) 56m + ?>5p = 7(8m + 5p)
(c) 2y + 5 There is no common factor other than 1 .
(d) 12 + 24z = 12 • 1 + 12 • 2z
= 12(1 + 2z) 12 is the GCF.
Check: 12(1 + 2z) = 12(1) + 12 (2z) Distributive property
= 12 + 24z Original polynomial
Answers
1. (a) l{k + 4) (b) 8 (4m + 3)
(c) cannot be factored (d) 5 (z
CAUTION
In Example 1 (d), remember to write the factor 1 . Always check answers
by multiplying.
Work Problem 1 at the Side.
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
Section 7. 1 Greatest Common Factors; Factoring by Grouping 373
EXAMPLE 2
Factoring Out the Greatest Common Factor
Factor out the greatest common factor.
(a) 9x2 + 12x3
The numerical part of the GCF is 3, the largest number that divides into
both 9 and 12. For the variable parts, x^ and x^, use the least exponent that
appears on x; here the least exponent is 2. The GCF is Sx^.
9x2 + 12x3 = 3x^(3) + 3x2(4x)
= 3x^(3 + 4x)
(b) 32;?4 _ 24^3 + 49^5 = 8^(4;?) + »p\3) + Sp\5p^) GCF = ^p^
= Hp\4p 3 + 5p^)
(c) 3^4 _ 15^7 + 24^9 = 3^4(j _ 5^3 + g^5)
CAec^ by multiplying:
3k\l  5k' + 8P) = 3^^4(1) + 3yt4(5P) + 3^^4(8^)
= 3k^ — I5k^ + 24^^ Original polynomial
(d) 24m'n^ — ISm^n + 6m^n'
The numerical part of the GCF is 6. Here 2 is the least exponent that
appears on m, while 1 is the least exponent on n. The GCF is 6m^n.
24m'n^ — l?>m^n + 6m^n' = 6m^n{Amn) + 6m^n(—3) + 6m^n{m^n^)
= 6m^n{Amn — 3 + m^^^)
(e) 25x23;3 + 303;5 _ 15;^4^7 = 53;3(5^2 + 5^2 _ 3^ 4^ 4)
Work Problem 2 at the Side.
A greatest common factor need not be a monomial. The next example
shows a binomial greatest common factor.
EXAMPLE 3
Factoring Out a Binomial Factor
Factor out the greatest common factor.
(a) (x  5)(x + 6) + (x  5)(2x + 5)
The greatest common factor here is x — 5.
ix  5)(x + 6) + (x  5)(2x + 5) = (jc  5)[(x + 6) + (2x + 5)]
= (x 5)(x + 6 + 2x + 5)
= (x 5)(3x+ 11)
(b) z^{m + n) + x^(m + n) = (m + n) (z^ + x^)
(c) p(r + 2s)  q^(r + 2s) = (r + 2s)(j>  q^)
(d) ip  5)(p + 2)  (p  5)(3p + 4)
= (/»  5) [(/> + 2)  (3p + 4)] Factor out;?  5.
= (p5)[p + 23p4]
= (p5) [2p  2]
= (p5)[2ip+i)]
= 2(p  5)(p + I)
Be careful with signs.
Combine terms.
Look for a common factor.
Work Problem 3 at the Side.
& Factor out the greatest
common factor.
(a) 16j4 + 87
(b) I4p^  9p' + 6p^
(c) 15z2 + 45z^  60z^
(d) 4x^z  2xz + 8z2
(e) UyV + Sj^x^
(f) Sm^x^ + ISm^x^  20^^x6
& Factor out the greatest
common factor.
(a) (a + 2) (a 3)
+ (a + 2)(a + 6)
(b)(7l)(7 + 3)
(yl)(y + 4)
(c) F((2 + 5b) + m^(a + 5b)
(d) r\y + 6) + r^(y + 3)
Answers
2. (a) 8y\2y + 1) (b) p\\4 ■
(c) 15z2(l + 3z3  4z4)
9p + 6j92)
(d) 2z(2x2
■4z)
(e) 4j3x2(3j;2 + 2x)
(f) 5^4x^(1 + 3mx3  4x3)
(a) (a + 2) (la + 3)
(b) (jl)(l),orj+l
(c) (a + SZ?) (^2 + m^)
(d) r2(2j + 9)
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
374 Chapter 7 Factoring
O Factor each polynomial in
two ways.
(a) F + 3^
DuTf^
When the coefficient of the term of greatest degree is negative, it is
sometimes preferable to factor out the — 1 that is understood along with the
GCE
EXAMPLE 4
Factoring Out a Negative Common Factor
Factor —a^\ 3a^ — 5a in two ways.
First, a could be used as the common factor, giving
— a^ + ?>a^ — 5a = a{—a^) + a(3a) + a{—5) Factor out (2.
= a{—a^ \ 3a — 5).
Because of the leading negative sign, —a could be used as the common factor.
— a^ + 3a^ — 5a = —a(a^) + {—a)(—3a) + (—a)(5) Factor out a.
= —a(a^ — 3a + 5)
Either answer is correct.
(b) 6r3  5r2+ 14r
NOTE
Example 4 showed two ways of factoring a polynomial. Sometimes
there may be a reason to prefer one of these forms over the other, but
both are correct. The answer section in this book will usually give the
form where the common factor has a positive coefficient.
Work Problem 4 at the Side.
^J Factor 6p — 6q \ rp — rq.
OBJECTIVE Q Factor by grouping. Sometimes the terms of a polyno
mial have a greatest common factor of 1, but it still may be possible to factor
the polynomial by using a process cdXlQd factoring by grouping. We usually
factor by grouping when a polynomial has more than three terms. For exam
ple, to factor the polynomial
ax — ay + bx — by,
group the terms as follows.
Terms with common factors
{ax — ay) + {bx — by)
Then factor ax — ay diS a{x — y) and factor bx — by 2isb(x — y).
ax — ay + bx — by = (ax — ay) + (bx — by)
= a(x y) + b(x y)
The common factor isx — y. The final factored form is
ax — ay + bx — by = (x — y) (a + b).
Answers
4. (a) k{k + 3) or k{k  3)
(b) r(6r2 5r + 14) or
r(6r2 + 5r  14)
5. (pq){6 + r)
Hi
Work Problem 5 at the Side.
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
EXAMPLE 5
Factoring by Grouping
Factor 3x — 3y — ax + ay.
Grouping terms gives
(3x — 3y) + (—ax + ay) = 3(x — j) + a(—x + y).
There is no simple common factor here. However, if we factor out —a
instead of a in the second group of terms, we get
(3x — 3y) + (—ax + ay) = 3(x — y) — a(x — j),
= (xy)(3 a).
Check by mukiplying.
(x — y)(3 — a) = 3x — ax — 3y + ay FOIL
= 3x — 3y — ax \ ay Rearrange terms.
This final product is the original polynomial.
Work Problem 6 at the Side.
Section 7. 1 Greatest Common Factors; Factoring by Grouping 375
© Factor xj — 2j — 4x + 8.
NOTE
In Example 5, different grouping would lead to the factored form
(a  3)(y  x).
Verify by multiplying that this form is also correct.
The steps used in factoring by grouping are listed here.
O Factor 2xy + 3j + 2x + 3.
Factoring by Grouping
Step 1 Group terms. Collect the terms into groups so that each group
has a common factor.
Step 2 Factor within the groups. Factor out the common factor in
each group.
Step 3 Factor the entire polynomial. If each group now has a com
mon factor, factor it out. If not, try a different grouping.
Always check the factored form by multiplying.
EXAMPLE 6
Factoring by Grouping
Factor 6ax \ I2bx + a + 2bhy grouping.
6ax + I2bx + a + 2b = (6ax + I2bx) + (a + 2b) Group terms.
Now factor 6x from the first group, and use the identity property of multipli
cation to introduce the factor 1 in the second group.
= 6xia + 26) + 1(« + 2b)
= (a \ lb) (6x + 1) Factor out a + lb.
Again, as in Example 1(d), remember to write the 1. Check hy multiplying.
Work Problem 7 at the Side.
Answers
6. (x2)(y4)
7. (2x + 3)0 + 1)
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
376 Chapter 7 Factoring
O Factor.
(a) mn \ 6 \ 2n \ 3m
Video An i motion You
— %^^^^^^^[0 Rearranging Tei
ms before Factori
ng by Grouping
Factor />^^^ —
10  2^2 + 5^2
Neither the first two terms nor
the last two terms have a common factor
except 1 . Rearrange and group the terms as follows.
(p^q^
 2^2) + (5^2 _ 10)
Rearrange and
group the terms.
=
= ^2(^2 _ 2) + 5(/,2
2)
Factor out the
common factors.
=
= (p'2)iq' + 5)
Factor out />^ 
2.
Check: (p^
2)(q^ + 5)=p^q^
+ 5p'
 2q^  10
FOIL
= pV
10
 2q^ + 5p^
Original polynomial
(b) 4y  zx \ yx  4z
Answers
8. (a) (m + 2)(n + 3) (b) (y  z) (x + 4)
CAUTION
In Example 7, do not stop at the step
q\p^2) + 5(p^2).
This expression is not in factored form because it is a sum of two terms,
q^{p^  2) and 5(;7^  2), not a product.
ill
Work Problem 8 at the Side.
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Section 7.2 Factoring Trinomials 379
7 • 2i Factoring Trinomials
OBJECTIVE Q Factor trinomials when the coefficient of the squared
term is 1, We begin by finding the product of x + 3 andx  5.
(x + 3) (x  5) = x^  5x + 3x  15
= x2 2x 15
By this result, the factored form of x^ — 2x — 15 is (x + 3) (x — 5).
Multiplying
>
Factored form
^ (x + 3) (x  5) = x^
<
2x — 15^
Product
Factoring
Since multiplying and factoring are operations that "undo" each other, fac
toring trinomials involves using FOIL backwards. As shown here, the x^term
comes from multiplying x andx, and — 15 comes from multiplying 3 and —5.
Product of X and x is x^.
{x + 3) (x  5) = x^  2x  15
ft t
Product of 3 and 5 is 15.
We find the 2x in x^  2x  15 by multiplying the outer terms, multiply
ing the inner terms, and adding.
Outer terms: x(5) = 5x.
f
~\
(x + 3) (x  5)
Inner terms: 3 • x = 3x
Add to get —Ix.
Based on this example, follow these steps to factor a trinomial x^ + Z?x + c,
where 1 is the coefficient of the squared term.
Factoring x^ + bx + c
Step 1 Find pairs whose product is c. Find all pairs of integers
whose product is the third term of the trinomial, c.
Step 2 Find pairs whose sum is b. Choose the pair whose sum is the
coefficient of the middle term, b.
If there are no such integers, the polynomial cannot be factored. A
polynomial that cannot be factored with integer coefficients is a prime
polynomial.
Some examples of prime polynomials are
x^ + X + 2, x^ — X — 1, and 2x^ + x + 7. Prime polynomials
OBJECTIVES
Factor trinomials when
the coefficient of the
squared term is 1.
Factor trinomials when
the coefficient of the
squared term is not 1.
Use an alternative
method for factoring
trinomials.
Factor by substitution.
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
380 Chapter 7 Factoring
Factor each polynomial.
(a) p^ + 6p + 5
(b) a^ + 9a + 20
(c) k^k6
(d) b^lb+ 10
EXAMPLE 1
FactoringTrinomials in x^ + bx + c Form
Factor each polynomial.
(a) y^ + 2y 35
Write sums of those
numbers.
35 + 1 = 34
35 + (l) = 34
7 + (5) = 2 <
5 + (7) = 2
Coefficient of
the middle term
Step 1 Find pairs of numbers Step 2
whose product is —35.
35(1)
35(l)
7(5)
5(7)
The required numbers are 7 and —5, so
7^ + 27 35 = (7 + 7) (7 5).
Check hy finding the product of j + 7 and j  5.
(b) r^ + 8r + 12
Look for two numbers with a product of 12 and a sum of 8. Of all pairs
of numbers having a product of 12, only the pair 6 and 2 has a sum of 8.
Therefore,
r^ + 8r + 12 = (r + 6) (r + 2).
Because of the commutative property, it would be equally correct to write
(r + 2) (r + 6). Check by multiplying.
(e) 7^87 + 6
EXAMPLE 2
Recognizing a Prime Polynomial
Factor m^ + 6m + 7.
Look for two numbers whose product is 7 and whose sum is 6. Only two
pairs of integers, 7 and 1 and 7 and 1, give a product of 7. Neither of
these pairs has a sum of 6, so m^ + 6m + 7 cannot be factored with integer
coefficients and is prime.
Work Problem 1 at the Side.
& Factor each polynomial.
(a) m^ + 2mn — Sn^
(b) z^  Izx + 9x2
Factoring a trinomial that has more than one variable uses a similar process.
EXAMPLE 3
Factoring a Trinomial in Two Variables
fouTrjf
m\
Factor /> 2 + 6ap — I6a^.
Look for two expressions whose product is — 16^^ and whose sum is 6a.
The quantities 8a and —2a have the necessary product and sum, so
p^ + 6ap — I6a^ = (/^ + 8a) (p — 2a).
Check: (p + 8a) (p — 2a) = p^ — 2ap + 80/7 — I6a^ FOIL
/^^ + 6ap — I6a"
Original
polynomial
Work Problem 2 at the Side.
Answers
1. (a) (p+l){p + 5) (b) {a + 5){a
(c) (k3)(k+2) (d) ib5){b
(e) prime
2. (a) (m  2n) (m + 4n) (b) prime
4)
2)
Sometimes a trinomial will have a common factor that should be factored
out first.
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
Section 7.2 Factoring Trinomials 381
EXAMPLE 4
Factoring aTrinomial with a Common Factor
Factor I6y^  32y^  48j.
Start by factoring out the greatest common factor, I6y.
I6y^  32j2  48j = I6y(y^  2y  3)
To factor j^ — 2j — 3, look for two integers whose product is —3 and whose
sum is —2. The necessary integers are —3 and 1, so
16j3  32y^  48j; = I6y{y 3)(y+ 1).
& Factor 5m^ — 5m^ — lOOm^.
CAUTION
When factorings always look for a common factor first. Remember to
write the common factor as part of the answer.
Work Problem 3 at the Side.
OBJECTIVE Q Factor trinomials when the coefficient of the squared
term is not 1_ We can use a generalization of the method shown in Objec
tive 1 to factor a trinomial of the form ax^ \ bx \ c, where a 7^ \.To factor
3x^ + 7x + 2, for example, we first identify the values of a, b, and c.
ax^ + bx \ c
3x2 + 7;^ + 2
a = 3, b = 1, = 2
The product ac is 3 • 2 = 6, so we must find integers having a product of 6
and a sum of 7 (since the middle term has coefficient b = 7). The necessary
integers are 1 and 6, so we write 7x as Ix + 6x, or x + 6x. Thus,
3x2 + 7x + 2 = 3x2 + X + 6x + 2
i_
X + 6x = Ix
(3x2 \ x) + (6x + 2) Factor by grouping.
x(3jc+ 1) + 2(3x+ 1)
(3x + 1) (x + 2).
Factor each trinomial.
(a) 3^2 llj;4
(b) 6F  19^ + 10
EXAMPLE 5
Factoring aTrinomial in ax^ + bx + c Form
Factor 12r2 — 5r — 2.
Since a = 12, b = —5, and c = — 2, the product ac is —24. The two
integers whose product is —24 and whose sum is b, —5, are 3 and —8.
12r2  5r  2 = 12r2 + 3r  8r  2 Write 5r as 3r  8r.
= 3r(4r + 1) — 2(4r +1) Factor by grouping.
= (4r +1) (3r — 2) Factor out the common factor.
Work Problem 4 at the Side.
OBJECTIVE Q Use an alternative method for factoring trinomials.
An alternative approach, the method of trying repeated combinations and
using FOIL, is especially helpful when the product ac is large.
Answers
3. 5m^(m — 5) (m
4. (a) {y  4) (3;; 
4)
1) (b) (2k  5) (3k  2)
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
382 Chapter 7 Factoring
& Use the method of Example 6
to factor each trinomial.
(a) 10x2 +17^ + 3
(b) 16/2  34y  15
(c) 8^
2 _
13^+ 5
Answers
5. (a) (5x + 1) (2x + 3)
(b) (83; + 3) (2;; 5)
(c) (8^5)(^l)
EXAMPLE 6
Factoring Trinomials in ax^ + bx + c Form
Factor each polynomial.
(a) 3x2 + 7x + 2
To factor this trinomial we use an alternative method. The goal is to find
the correct numbers to fill in the blanks.
3x2 + 7x + 2 = (_
.X +
)(
_x +
)
Addition signs are used since all the signs in the trinomial indicate addition.
The first two expressions have a product of 3x2, ^^ ^^iQy must be 3x andx.
3x2 + 7x + 2 = (3_^ + ) (_^ + )
The product of the two last terms must be 2, so the numbers must be 2 and
1 . There is a choice. The 2 could be used with the 3x or with the x. Only one
of these choices can give the correct middle term, 7x Use the FOIL method
to check each one.
3x
I I
(3x + 2)(x+ 1)
6x
I I
(3x+ l)(x + 2)
6x + X = 7x
Correct middle term
2x
3x + 2x = 5x
Wrong middle term
Therefore, 3x2 + 7x + 2 = (3x + i) (^ + 2). (Compare to the method on the
preceding page.)
(b) 12r2  5r  2
To reduce the number of trials, we note that the trinomial has no
common factor (except 1). This means that neither of its factors can have
a common factor. We should keep this in mind as we choose factors. We try
4 and 3 for the two first terms.
12r2  5r  2 = (4r_
.)(3r_
)
The factors of —2 are —2 and 1 or 2 and — 1. Try both possibilities.
(4r 2)(3r+ 1)
Wrong: 4r  2 has a
common factor of 2.
This cannot be correct,
since 2 is not
a factor of
12r2  5r  2.
8r
I I
(4r l)(3r + 2)
3r
8r — 3r = 5r
Wrong middle term
The middle term on the right is 5r, instead of the — 5r that is needed. We get
— 5r by interchanging the signs in the factors.
8r
I I
(4r + 1) (3r  2)
3r
— 8r \ 3r = — 5r
Correct middle term
Thus, 12r2 — 5r — 2 = (4r + 1) (3r — 2). (Compare to Example 5.)
Work Problem 5 at the Side.
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
This alternative method of factoring a trinomial ax^ \ bx \ c, a i^ 1, is
summarized here.
Factoring ax^ + bx + c
Step 1 Find pairs whose product is a. Write all pairs of integer fac
tors of the coefficient of the squared term, a.
Step 2 Find pairs whose product is c. Write all pairs of integer fac
tors of the last term, c.
Step 3 Choose inner and outer terms. Use FOIL and various combi
nations of the factors from Steps 1 and 2 until the necessary
middle term is found.
If no such combinations exist, the trinomial is prime.
Section 7.2 Factoring Trinomials 383
© Factor each trinomial.
(a) 7p^ + 15;?^ + 2q^
(b) 6m^ + 7mn — 5n^
YouTr
EXAMPLE 7
Factoring aTrinomial in Two Variables
Factor 18m^ — I9mx — \2x^.
There is no common factor (except 1). Follow the steps to factor the tri
nomial. There are many possible factors of both 18 and  12. Try 6 and 3 for
18and3and4for12.
{6m — 3x) (3m + Ax)
Wrong: common factor
{6m + Ax) {3 m — 3x)
Wrong: common factors
Since 6 and 3 do not work in this situation, try 9 and 2 instead, with —4 and
3 as factors of — 12.
llmx
{9m + 3x) {2m  Ax)
Wrong: common factors
I I
{9m  Ax) {2m + 3x)
I9mx
— Smx
llmx + ( — 8mx)
The result on the right differs from the correct middle term only in sign, so
interchange the signs in the factors. Check by multiplying.
18^2  \9mx  12x2 = {9m + Ax) {2m  3x)
Work Problem 6 at the Side.
(c) 12z2  5zy  2y^
(d) 8^2 + 18mx  5x
 ^v2
Factor each trinomial.
(a) 6r2+ 13r+ 5
EXAMPLE 8
Factoring ax^ + bx + c, with a <
Factor 3x2 + 16x + 12.
While it is possible to factor this trinomial directly, it is helpful to first
factor out — 1. Then proceed as in the earlier examples.
3x2 + 15^ + 12 = 1(3x2  16x  12)
= l(3x + 2)(x 6)
= (3x + 2)(x 6)
This factored form can be written in other ways. Two of them are
(3x  2) (x  6) and (3x + 2) (x + 6).
Verify that these both give the original trinomial when multiplied.
Work Problem 7 at the Side.
(b) 8x2 + lox  3
Answers
6. (a) {Ip + q){p + 2q)
(b) (3m + 5/2) (2m  n)
(c) (3z2j)(4z+j;)
(d) (4m  x) (2m + 5x)
7. (a) (2r 5) (3r + 1)
(b) (4x 3)(2x 1)
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
384 Chapter 7 Factoring
O Factor each trinomial.
(a) 2m^  4m^  6m Animatti
(b) 12r4 + 6r3  90r^
EXAMPLE 9
Factoring aTrinomial with a Common Factor
Factor I6y^ + 24y^  I6y.
I6y^ + 24j2  I6y = 8y(2y^ + 3j  2) GCF = 83;
= ^y(2y — l)(y + 2) Remember the
common factor.
Work Problem 8 at the Side.
(c) 30y^  55^4  50j
J
Video I touiry
& Factor each polynomial.
(a) 6(a iy + (al)2
(b) 8(z+5)22(z+ 5) 3
(c) 15 (m 4)2
U(m4) + 2
QD Factor each trinomial.
(a) j^ + j2 _ 6
OBJECTIVE Q Factor by substitution. Sometimes we can factor a
more complicated polynomial by substituting a variable for an expression.
EXAMPLE 10
Factoring a Polynomial Using Substitution
Factor 2 (x + 3)^ + 5(x + 3)  12.
Since the binomial x + 3 appears to powers 2 and 1, we let the substitu
tion variable represent x + 3. We may choose any letter we wish except x.
We choose y to equal x + 3.
2{x + 3)2 + 5(x + 3)  12 = 2j2 + 5^;  12 Letj; = x + 3.
= (2y  3) (j + 4) Factor.
Now we replace y with x + 3 to get
2(x + 3)2 + 5(x + 3)  12 = [2(x + 3)  3] [(x + 3) + 4]
= (2x + 6  3) (x + 7)
= (2x + 3) (x + 7).
CAUTION
Remember to make the final substitution of x + 3 for j in Example 10.
Work Problem 9 at the Side.
(b) 2p' + Ip^  15 ^uTrv.
(c) 6r4  13r2 + 5
Answers
8. (a) 2m (m + l)(m  3)
(b) 6r2(r + 3) (2r  5)
(c) 5y\2y5){?>y + 2)
9. (a) {la  3) (3a  1)
(b) (4z+ 17)(2z+ 11)
(c) (3m  13) (5m 22)
10. (a) (j22)(/ + 3)
(b)(2;;2_3)(^2 + 5)
(c) (3r2  5) (2r2  1)
EXAMPLE 11
Factoring aTrinomial in ax^ + bx^ + c Form
Factor 6y^ + 7j;2  20.
The variable j appears to powers in which the larger exponent is twice
the smaller exponent. We can let a substitution variable equal the smaller
power. Here, we let m = y^.
6/ + 7/  20 = 6{y^Y + ly^  20
= 6m^ + 7m  20 Subtitute.
= (3m — 4) (2m + 5) Factor.
= (3y'4)(2y' + 5) m = y^
NOTE
Some students feel comfortable factoring polynomials like the one in
Example 1 1 directly, without using the substitution method.
[Work Problem 10 at the Side.
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Section 7.3 Special Factoring 387
7 .3 Special Factoring
OBJECTIVE Q Factor a difference of squares. The special products
introduced in Section 6.4 are used in reverse when factoring. Recall that the
product of the sum and difference of two terms leads to a difference of
squares, a pattern that occurs often when factoring.
Difference of Squares
x^  y^ = (x \ y) (x  y)
EXAMPLE 1
Factoring Differences of Squares
Factor each polynomial.
(a) 4^2 _ 64
There is a common factor of 4.
4^2 _ 64 = 4(a^  16)
= 4(« + 4)(a 4)
(b) 16^2  49p^
Factor out the common factor.
Factor the difference of squares.
A^  B^ = (A + B) {A  B)
\/ Y Y V V \/
{4mY  {IpY = (4hi + Ip) (4m  Ip)
A^  52 = (A + B) {A 
\/ \/ V Y V
■ f
= {9ky  {a + If = (9k + a^) (9k 
 [a + 2])
= (9k + a + 2) (9k 
a2)
(c) 8 IF (a + ly
We could have used the method of substitution here,
(d) x^ — 81 = (x^ + 9) (x^ — 9) Factor the difference of squares.
= (x^ + 9) (x + 3) (x  3) Factor x2  9.
Work Problem 1 at the Side.
OBJECTIVES
Factor a difference of
squares.
Factor a perfect square
trinomial.
Factor a difference of
cubes.
Factor a sum of cubes.
Factor each polynomial.
(a) 2x2  18
(b) 9a^  I6b^
(c) (m + 3)2  49z2
CAUTION
Assuming no greatest common factor except 1, it is not possible to
factor (with real numbers) a sum of squares^ such as x^ + 9 in Exam
ple 1(d). In particular, x^ + y^ ^ {x + yY, as shown next.
OBJECTIVE Q Factor a perfect square trinomial. Two other special
products from Section 6.4 lead to the following rules for factoring.
Perfect Square Trinomial
x^ + 2xy +y^ = {x+ yf
x^  2xy + j2 = (x  yy
(d) y^  16
Answers
1. (a) 2(x + 3)(x 3)
(b) (3a  4b) {3a + 4b)
(c) (m + 3 + 7z) (m + 3  Iz)
(d) (/ + 4)(j + 2)(j2)
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
388 Chapter 7 Factoring
& Identify any perfect square
trinomials.
(a) z2 + 12z + 36
(b) 2x2  4x + 4
Because the trinomial x^ + 2xy + y^ is the square of x + j, it is called a
perfect square trinomial. In this pattern, both the first and the last terms of
the trinomial must be perfect squares. In the factored form (x + j)^, twice
the product of the first and the last terms must give the middle term of the
trinomial. It is important to understand these patterns in terms of words,
since they occur with many different symbols (other than x andj).
4m^ + 20m + 25 p^  8p + 64
Perfect square trinomial Not a perfect square trinomial;
since 20m = 2 (2m) (5) middle term should be I6p or —I6p.
AH
Work Problem 2 at the Side.
(c) 9fl2 + I2ab + 16Z)2
O Factor each polynomial.
(a) 49z2 I4zk + k^
(b) 9a2 + 4Sab + 64b^
(c)
(k + mf
12(/t+m) + 36
(d) x2  2x + 1
EXAMPLE 2
Factoring Perfect Square Trinomials
Factor each polynomial.
(a) 144;?2 _ i2Qp + 25
Here 144p^ = {llpY and 25 = 5^. The sign on the middle term is — , so f^AA
if 144/7^ — 120/? + 25 is a perfect square trinomial, the factored form will l^B^
have to be Video
{Up  5)2.
Take twice the product of the two terms to see if this is correct.
2(12/;) (5)= 120/?
This is the middle term of the given trinomial, so
144/?2  120/? + 25 = (12/?  5)2.
(b) 4^2 + 20mw + 49^2
If this is a perfect square trinomial, it will equal {2m + Inf. By the
pattern described earlier, if multiplied out, this squared binomial has a
middle term of2(2m)(7n) = 28mn, which does not equal 20mn. Verify
that this trinomial cannot be factored by the methods of the previous sec
tion either. It is prime.
(c) (r + Sy + 6(r + 5) + 9 = [(r + 5) + 3]^
= (r + 8)^
since 2(r + 5) (3) = 6(r + 5), the middle term.
(d) m^  8m + 16  p^
Since there are four terms, we will use factoring by grouping. The first
three terms here form a perfect square trinomial. Group them together, and
factor as follows.
(m^  8m + 16)  p^ = (m  4)^  p^
The result is the difference of squares. Factor again to get
= (m — 4 + p) (m — 4 — p).
w
Work Problem 3 at the Side.
Answers
2. (a) perfect square trinomial
(b) not a perfect square trinomial
(c) not a perfect square trinomial
3. (a) (7z  ky (b) {3a + Sbf
(c) [{k + m)  6]2 or (^ + m  6)^
(d) {xl+y)ixly)
Perfect square trinomials, of course, can be factored using the general
methods shown earlier for other trinomials. The patterns given here provide
"shortcuts."
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
OBJECTIVE Q Factor a difference of cubes. A difference of cubes,
such as x^ — y^, can be factored as follows.
Difference of Cubes
x^  y^ = ix y) {x^ + xy + y^)
We could check this pattern by finding the product oix — y and x^ + xy + j^.
EXAMPLE 3
Factoring Differences of Cubes
Factor each polynomial.
(a) m^ 8 = m^ 2^
= (m  2){np + 2m + 2^)
= (m  2)(m^ + 2m + 4)
Check:
(m  2)(m^ + 2m + 4)
2m
1_
Opposite of the product of the cube
roots gives the middle term.
(b) 27x3 _ 8^3 = (3_^)3 _ (2j;)3
= (3x  2y) [(3xf + (3x) (2y) + (lyf]
= (3x  2y) (9x2 + 6xj + 4y^)
(c) 1000^3 _ 27„3 = (io;^)3 _ (3„)3
= (10^  3«) [(10^)2 + (10^) (3«) + (3nf]
= (\0k  3n) (100^2 + 3o^„ + 9„2)
Work Problem 4 at the Side.
Section 7.3 Special Factoring 389
O Factor each polynomial.
(a) x^  1000
(b) U^  y^
OBJECTIVE Q Factor a sum of cubes. While the binomial x^ + y^
(a sum oi squares) cannot be factored with real numbers, a sum of cubes,
such as x^ + y^, is factored as follows.
(c) 27^3  64
Sum of Cubes
x^ + y^ = {x+ y) {x^  xy + y^)
To verify this result, find the product of x + y and x^ — xj + y^. Compare
this pattern with the pattern for a difference of cubes.
NOTE
The sign of the second term in the binomial factor of a sum or differ
ence of cubes is always the same as the sign in the original polynomial.
In the trinomial factor, the first and last terms are always positive; the
sign of the middle term is the opposite o/the sign of the second term in
the binomial factor.
Answers
4. (a) (x  10) (x2 + lOx + 100)
(b) {2k  y) (4^2 + 2ky+ y^)
(c) (3m 4)(9m2+ 12m + 16)
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
390 Chapter 7 Factoring
& Factor each polynomial.
(a) 8/73 + 125
EXAMPLE 4
Factoring Sums of Cubes
Factor each polynomial.
(a) r^ + 27 = r^ + 3^
= (r + 3) (r^  3r + 3^)
= (r + 3) (r2 3r + 9)
(b) 27z3 + 125 = (3z)3 + 53
= (3z + 5)[(3z)2(3z)(5) + 52]
(b) 27m 3 + 125«
= (3z+ 5)(9z2 15Z + 25)
(c)
125/3
+ 2I656 = (5^)3 + (65^)3
= (5/ + 6^2) [(5^)2
 (50 (6^2) + (65
2)2]
= (5/ + 6*2) (25/2
 30^2 + 36*4)
(d)
3x3 + 192 = 3(x3 + 64)
= 3(x + 4)(x2  4x +
16)
CAUTION
A common error is to think that the xyterm has a coefficient of 2 when
factoring the sum or difference of cubes. Since there is no coefficient of 2,
the trinomials x^ + xy + y^ andx^  xy + y^ cannot be factored further.
w
Work Problem 5 at the Side.
The special types of factoring in this section are summarized here.
These should be memorized.
(c) 2x3 + 2000
Special Types of Factoring
Difference of Squares
Perfect Square Trinomial
Difference of Cubes
Sum of Cubes
x^ y^ = {x+ y) {x  y)
x^ + 2xy +y^ = (x+ yy
x^  2xy + y^ = (x  yY
x^  y^ = (x  y) (x^ + xy + y^)
x^ + y^ = (x + y) (x^  xy + y^)
Answers
5. (a) (2p + 5) (4p^  lOp + 25)
(b) (3m + 5n){9m^  \5mn + 25n^)
(c) 2(x+ 10) (x2 10x+ 100)
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Section 7.4 Solving Equations by Factoring 395
i .4 Solving Equations by Factoring
The equations that we have solved so far in this book have been linear equa
tions. Recall from Section 2.1 that in a linear equation, the greatest power of
the variable is 1. To solve equations of degree greater than 1, other methods
must be developed. One of these methods involves factoring.
OBJECTIVE Q Learn and use the zerofactor property. Some equa
tions can be solved by factoring. Solving equations by factoring depends on
a special property of the number 0, called the zerofactor property.
OBJECTIVES
Learn and use the
zerofactor property.
Solve applied problems
that require the
zerofactor property.
ZeroFactor Property
If two numbers have a product of 0, then at least one of the numbers
must be 0. That is, if ab = 0, then either a = or b = 0.
To prove the zerofactor property, we first assume a i^ 0. (If a does equal
0, then the property is proved already.) If a 9^ 0, then ^ exists, and each side
of ab = can be multiplied by ^ to get
1 1
' ab = '
a a
b = 0.
Thus, ifai^O, then b = 0, and the property is proved.
CAUTION
If ab = 0, then a = or b = 0. However, if ab = 6, for example, it is
not necessarily true that a = 6 or /? = 6; in fact, it is very likely that
neither a = 6 nor b = 6. The zerofactor property works only for a
product equal to 0.
YouTj
EXAMPLE 1
Using the ZeroFactor Property to Solve
an Equation
Solve (x + 6) (2x  3) = 0.
Here the product of x + 6 and 2x — 3 is 0. By the zerofactor property,
this can be true only if
X + 6 = or 2x  3 = 0.
Solve these two equations.
x + 6 = or 2x3 =
X = — 6 2x = 3
3
X = —
The solutions are x
6 or X
Continued on Next Page
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
396 Chapter 7 Factoring
Solve each equation.
(a) (3x + 5) (x + 1) =
Check the solutions 6 and  by substitution in the original equation.
If X = 6, then
(x + 6) (2x  3) =
(6 + 6)[2(6)3] = ?
0(15) = 0. True
Ifx
then
(x + 6) (2x  3) =
+ 6 2
15
?
(0) = 0. True
Both solutions check; the solution set is {— 6, }.
Work Problem 1 at the Side.
Since the product (x + 6) (2x  3) equals 2x^ + 9x  18, the equation
of Example 1 has a squared term and is an example of a. quadratic equation.
A quadratic equation has degree 2.
Quadratic Equation
An equation that can be written in the form
ax^ + 6x + c = 0,
where a 9^ 0, is a quadratic equation. This form is called standard form.
(b) (3x+ ll)(5x2) =
Quadratic equations are discussed in more detail in Chapter 10.
The steps involved in solving a quadratic equation by factoring are
summarized below.
Solving a Quadratic Equation by Factoring
Step 1 Write in standard form. Rewrite the equation if necessary so
that one side is 0.
Step 2 Factor the polynomial.
Step 3 Use the zerofactor property. Set each variable factor equal to 0.
Step 4 Find the solution(s). Solve each equation formed in Step 3.
Step 5 Check each solution in the original equation.
Answers
1. (a)
5
"3' '
(b)
' 3'5
EXAMPLE 2
Solving a Quadratic Equation by Factoring
Solve each equation.
(a) Ix^ + 3x = 2
Step 1
Step 2
Step 3 X + 2
Step 4 X
Continued on Next Page
2x2 + 3^ =
2
2x2 + 3x  2 =
Standard form
(x + 2)(2xl) =
Factor.
or 2x  1 =
Zerofactor property
2 2x =
1
Solve each equation.
X =
1
2
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Step 5 Check each solution in the original equation.
If X = — 2, then
Ix^ + 3jc = 2
2{2f + 3(2) = 2
2(4) 6 = 2
86 = 2
2 = 2.
True
If X = J, then
2x2 ^T,x = 2
 +  = 2
2 2
2 = 2.
Section 7.4 Solving Equations by Factoring 397
O Solve each equation.
True
Because both solutions check, the solution set is {—2, ^}.
(b) 4x2 = 4x  1
4x2  4x + 1 = Standard form
(2x  1)2 =
2x  1 =
2x= 1
1
" = 2
There is only one solution because the trinomial is a perfect square. The
solution set is {^}.
Factor.
Zerofactor property
Work Problem 2 at the Side.
I^r
EXAMPLE 3
Solving a Quadratic Equation with a Missing
Term
Solve 5x2 _ 25x = 0.
This quadratic equation has a missing term. Comparing it with the stan
dard form ax^ \ bx \ c = Q shows that c = 0. The zerofactor property can
still be used.
5x2  25x =
5x(x  5) =
Factor.
5jc = or x5 =
Zerofactor property
X = or X = 5
The solutions are and 5, as can be verified by substituting in the original
equation. The solution set is {0, 5}.
CAUTION
Remember to include as a solution of the equation in Example 3.
Work Problem 3 at the Side.
(a) 3x^
(b) 25x2
20x  4
& Solve each equation.
(a) x^ = 12x
(b) f\6 = Q
Answers
2. (a) 1,
'3
(b)
3. (a) {12,0} (b) {4,4}
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398 Chapter 7 Factoring
O Solve.
(x + 6)(x  2) = 8 +x
EXAMPLE 4
Solving an Equation That Requires Rewriting
Solve {2q + 1) (^ + 1) = 2(1  q) + 6.
{2q+ \){q+ 1) = 2(1  q) + 6
2q^ + 3q + I = 2  2q + 6
2q^ + 3q + \ = S  2q
2q^ + 5q 7 =
(2q + l)(ql) =
2q + 7 = or ^1 =
2q = —1 ^ = 1
7
^=2
Check that the solution set is {, 1}.
Work Problem 4 at the Side.
Multiply on each side.
Add on the right.
Standard form
Factor.
Zerofactor property
Video
The zerofactor property can be extended to solve certain polynomial
equations of degree 3 or higher, as shown in the next example.
& Solve.
EXAMPLE 5
Solving an Equation of Degree 3
= — 6x.
Solve — x^ + x^
Start by adding 6x to each side to get on the right side.
x^ + x^ + 6x =
x^ — x^ — 6x = Multiply by 1.
X (x^ — X — 6) = Factor out x.
x(x + 2) (x — 3) = Factor the trinomial.
Use the zerofactor property, extended to include the three variable factors.
x = or x + 2 = or x — 3 =
X = —2 X = 3
Check that the solution set is { 2, 0, 3}.
IK
Work Problem 5 at the Side.
OBJECTIVE Q Solve applied problems that require the zerofactor
property. An application may lead to a quadratic equation. We continue to
use the sixstep problem solving method introduced in Section 2.3.
Answers
4. {4,1}
4
5.
3'
,0, 1
x + 8
Figure 1
EXAMPLE 6
Using a Quadratic Equation in an Application
A piece of sheet metal is in the shape of a parallelogram. The longer sides of
the parallelogram are each 8 m longer than the distance between them. The
area of the parallelogram is 48 m^. Find the length of the longer sides and
the distance between them.
Step 1 Read the problem again. There will be two answers.
Step 2 Assign a variable. Let x represent the distance between the longer
sides. Then x + 8 is the length of each longer side. See Figure 1.
Continued on Next Page
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
Section 7.4 Solving Equations by Factoring 399
Step 3 Write an equation. The area of a parallelogram is given hyA = bh,
where b is the length of the longer side and h is the distance
between the longer sides. Here /? = x + 8 and h = x.
Step 4 Solve.
Step 5
Step 6
A = bh
48 = (x + 8)x
48 = x^ + 8x
LqXA
x2 + 8x  48
= (x+ 12) (x 4)
x+12 = or x4 =
X = — 12 or X = 4
48, Z? = X + 8, /z = X.
Distributive property
Standard form
Factor.
Zerofactor property
State the answer. A distance cannot be negative, so reject  12 as a
solution. The only possible solution is 4, so the distance between the
longer sides is 4 m. The length of the longer sides is 4 + 8 = 12 m.
Check. The length of the longer sides is 8 m more than the distance
between them, and the area is 4 • 12 = 48 m^, so the answer checks.
© Solve the problem.
Carl is planning to build a
rectangular deck along the
back of his house. He wants
the area of the deck to be
60 m^, and the width to be
1 m less than half the length.
What length and width
should he use?
CAUTION
A solution of the equation may not satisfy the physical requirements of
the application, as in Example 6. Reject such solutions.
Work Problem 6 at the Side.
A function defined by a quadratic polynomial is called a quadratic function.
(See Chapter 10.) The next example uses such a function.
EXAMPLE 7
Using a Quadratic Function in an Application
Quadratic functions are used to describe the height a falling object or a pro
pelled object reaches in a specific time. For example, if a small rocket
is launched vertically upward from ground level with an initial velocity of
128 ft per sec, then its height in feet after t seconds is a function defined by
h{t)= I6t^+ 128^,
if air resistance is neglected. After how many seconds will the rocket be
220 ft above the ground?
We must let h (t) = 220 and solve for t.
220= 16^2+ 128^
I6t^  128^ + 220 =
4^2  32^ + 55 =
(2t 5)(2t 11) =
2^ 5 = or 2^ 11 =
^ = 2.5 or t= 5.5
Let/z(0 = 220.
Standard form
Divide by 4.
Factor.
Zerofactor property
The rocket will reach a height of 220 ft twice: on its way up at 2.5 sec and
again on its way down at 5.5 sec.
Work Problem 7 at the Side.
O Solve the problem.
How long will it take the
rocket in Example 7 to reach
a height of 256 ft?
Answers
6. length: 12 m; width: 5 m
7. 4 sec
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Rational
Expressions and
Functions
>^ 8.1 Rational Expressions and
Functions; Multiplying and
Dividing
8.2 Adding and Subtracting
Rational Expressions
8.3 Complex Fractions
8.4 Equations with Rational
Expressions and Graphs
Summary Exercises on Rational
Expressions and Equations
8.5 Applications of Rational
Expressions
Americans have been car crazy ever since the first automobiles hit
the road early in the twentieth century. Today there are about
213.5 million vehicles in the United States driving on 3.4 million miles
of paved roadways. There is even a museum devoted exclusively to our
fourwheeled passion and its influence on our lives and culture. The
Museum of Automobile History in Syracuse, N.Y., features some 200
years of automobile memorabilia, including rare advertising pieces,
designer drawings, and Hollywood movie posters. {Source: Home and
Away, May/June 2002.)
In Exercises 67 and 68 of Section 8.2, we use a rational expres
sion to determine the cost of restoring a vintage automobile.
413
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4 I 4 Chapter 8 Rational Expressions and Functions
o.l Rational Expressions and Functions; IVIultiplying and Dividing
OBJECTIVES
Define rational
expressions.
Define rational functions
and describe their
domains.
Write rational
expressions in lowest
terms.
Multiply rational
expressions.
Find reciprocals for
rational expressions.
Divide rational
expressions.
Q
OBJECTIVE Q Define rational expressions. In arithmetic, a rational
number is the quotient of two integers, with the denominator not 0. In
algebra, a rational expression or algebraic fraction is the quotient of two
polynomials, again with the denominator not 0. For example.
X
?
m + 4 8x^  2x + 5
m
4x^ + 5x
and
or
1
Rational
expressions
are all rational expressions. In other words, rational expressions are the
elements of the set
}■
Lp and Q are polynomials, with g ^
OBJECTIVE Q Define rational functions and describe their domains.
A function that is defined by a rational expression is called a rational
function and has the form
f{x) =
P(x)
QixY
where 2 W ^ 0.
The domain of a rational function includes all real numbers except those
that make Q(x), that is, the denominator, equal to 0. For example, the do
main of
fix)
X — 5
Cannot equal
includes all real numbers except 5, because 5 would make the denominator
equal to 0.
Figure 1 shows a graph of the function defined by
/(x)
Notice that the graph does not exist when x = 5. It does not intersect the
dashed vertical line whose equation is x = 5. This line is an asymptote. We
will discuss graphs of rational functions in more detail in Section 8.4.
x = 5
Figure 1
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Section 8.1 Rational Expressions and Functions; Multiplying and Dividing 4 I 5
YouTf
EXAMPLE 1
Finding Numbers That Are Not in the Domains
of Rational Functions
Find all numbers that are not in the domain of each rational function. Then
give the domain using set notation.
3
(a) f{x) =
The only values that cannot be used are those that make the denominator
0. To find these values, set the denominator equal to and solve the resulting
equation.
7x  14 =
7x = 14 Add 14.
X = 2 Divide by 7.
The number 2 cannot be used as a replacement for x. The domain of/
includes all real numbers except 2, written using set notation as {xx ?^ 2}.
3 + X
(b) g(x) = , . ^.
X  4x + 3
Set the denominator equal to 0, and solve the equation,
x^  4x + 3 =
(x l)(x 3) = Factor.
xl=0 or x3 = Zerofactor property
X = 1 or X = 3
The domain of g includes all real numbers except 1 and 3, written {xx ?^ 1,3}.
?x + 2
(c) /z(x)
The denominator, 3, can never be 0, so the domain of /z includes all real
numbers, written (—0°, oo).
(d) fix)
x' + 4
Setting x^ + 4 equal to leads to x^ = 4. There is no real number
whose square is —4. Therefore, any real number can be used, and as in part
(c), the domain of/ includes all real numbers (co, oo).
Work Problem 1 at the Side.
OBJECTIVE Q Write rational expressions in lowest terms. In arith
metic, we write the fraction ^ in lowest terms by dividing the numerator and
denominator by 5 to get . We write rational expressions in lowest terms in a
similar way, using the fundamental property of rational numbers.
Find all numbers that are not
in the domain of each rational
function. Then give the
domain using set notation.
(a) fix)
X + 4
(b) fix)
X + 6
(c) fix)
3 + 2x
(d) fix) = ^— 
X + 1
Fundamental Property of Rational Numbers
If f is a rational number and if c is any nonzero real number, then
a
~b
ac
Vc
In words, the numerator and denominator of a rational number may
either be multiplied or divided by the same nonzero number without
changing the value of the rational number.
Answers
1. (a) 6; {xx =^ 6} (b) 2, 3; {x\x ih 2, 3}
(c) none; The domain includes all real
numbers (—0°, 0°).
(d) none; The domain includes all real
numbers (—0°, 0°).
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
4 I 6 Chapter 8 Rational Expressions and Functions
In the fundamental property, f = ff. Since ^ is equivalent to 1, the funda
mental property is based on the identity property of multiplication.
A rational expression is a quotient of two polynomials. Since the value
of a polynomial is a real number for every value of the variable for which it
is defined, any statement that applies to rational numbers will also apply to
rational expressions. We use the following steps to write rational expressions
in lowest terms.
Writing a Rational Expression in Lowest Terms
Step 1 Factor both numerator and denominator to find their greatest
common factor (GCF).
Step 2 Apply the fundamental property.
EXAMPLE 2
fouTry li
Video!'
Writing Rational Expressions in Lowest Terms
Write each rational expression in lowest terms.
k 8
2 • 8
11
Factor; apply the fundamental property.
(b)
+ k
16
The numerator cannot be factored, so this expression cannot be simplified
further and is in lowest terms.
(c)
a^ — a — 6
a^ + 5a + 6
{a  3){a \ 2)
(a + 3) (« + 2)
a — 3
Factor the numerator
and the denominator.
(d)
y
ly + 4
~ a + 3
1
_a3
~ a + 3
(y + 2){y
2)
(2 + 2
1
2(y + 2)
 2
(2 + 2
Lowest terms
Factor the difference of squares in the
numerator; factor the denominator.
Lowest terms
(e)
27 (x  3) (x^ + 3x + 9)
X — 3
= x^ + 3x + 9
Factor the difference of cubes.
Lowest terms
(f)
pr \ qr \ ps \ qs {pr + qr) + {ps + qs)
pr + qr — ps — qs {pr + qr) — (ps + qs)
^ r{p + ^) + s{p + q)
r{p + q)  s{p + q)
(p + q){r + s)
(p + q) (r
r + s
s)
r — s
Group terms.
Factor within groups.
Factor by grouping.
Lowest terms
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Section 8.1 Rational Expressions and Functions; Multiplying and Dividing 4 I 7
CAUTION
Be careful! When using the fundamental property of rational numbers,
only common factors may be divided. For example,
y2
^ y and
y2
*y 1
because the 2 in j — 2 is not di factor of the numerator. Remember to
factor before writing a fraction in lowest terms.
Work Problem 2 at the Side.
In the rational expression from Example 2(c),
a^  a  6 {a  3){a + 2)
a^ + 5a + 6'
or
(a + 3)(a + 2)'
a can take any value except —3 or —2 since these values make the denomi
nator 0. In the simplified rational expression
a + 3'
a cannot equal —3. Because of this,
a^ — a — 6
a + 5a + 6 a + 3
for all values of a except —3 or —2. From now on such statements of equal
ity will be made with the understanding that they apply only for those real
numbers that make neither denominator equal 0. We will no longer state
such restrictions.
& Write each rational
expression in lowest terms.
y' + 2y3
(a)
y' 3y + 2
(b)
3y + 9
y^  9
(c)
y + 2
y^ + 4
EXAMPLE 3
Writing Rational Expressions in Lowest Terms
Write each rational expression in lowest terms.
m — 3
(a)T
5 — m
In this rational expression, the numerator and denominator are opposites.
The given expression can be written in lowest terms by writing the denomi
nator as — 1 (m — 3), giving
m — 3
m — 3
1
= 1.
3 — m —l{m — 3) — 1
The numerator could have been rewritten instead to get the same result.
(b)
4  r
r^  16 _ (r + 4) (r  4)
4  r
(r + 4) (r  4)
l(r 4)
r + 4
1
 (r + 4) or r  4
Factor the difference of
squares in the numerator.
Write 4  r as  1 (r  4).
Fundamental property
Lowest terms
(d)
1 +/
1 +P
(e)
3x + 3y + rx + ry
5x + 5y — rx — ry
Answers
j + 3
2. (a)
(b)
3
y1  y3
(c) already in lowest tenns
3 + r
(d) 1
(e)
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
4 I 8 Chapter 8 Rational Expressions and Functions
& Write each rational
expression in lowest terms.
y2
(a)
2y
As shown in Examples 3(a) and (b), the quotient ^(a ¥= 0) can be
simplified as
a
— a
1
 1(«)  1
The following statement summarizes this result.
1.
(b)
S  b
+ b
In general, if the numerator and the denominator of a rational
expression are opposites, the expression equals — 1 .
Based on this result, the following are true:
g 7 5a + 2b
=  1 and =  1 .
1  q 5a  2b
i t
Numerator and denominator in each expression are opposites.
However, the expression
2 <n
Numerator and denominator
^ ^ 2 <^ ^^^ ^ot opposites.
cannot be simplifed further.
Work Problem 3 at the Side.
(c)
4 /
OBJECTIVE Q Multiply rational expressions. To multiply rational
expressions, follow these steps.
Multiplying Rational Expressions
Step 1 Factor all numerators and denominators as completely as
possible.
Step 2 Apply the fundamental property.
Step 3 Multiply remaining factors in the numerator and remaining
factors in the denominator. Leave the denominator in factored
form.
Step 4 Check to be sure the product is in lowest terms.
Answers
3. (a)  1 (b) already in lowest terms
1
(c) 7^
2 +;?
EXAMPLE 4
Multiplying Rational Expressions
Multiply.
5/>  5 1p^ 5(/;  1) 3/> • j9
(a)
p 10/?  10 p
_ J_ 3p
~T'y
2
Continued on Next Page
25(/;l)
Factor.
Fundamental
property
Multiply.
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
(b)
k^ + 2k 15
Section 8.1 Rational Expressions and Functions; Multiplying and Dividing
(k + 5)(k 3) k(k  1)
419
k^  4k+ 3 k^ + k20 (k3){k I) (k + 5) {k  4)
(c) (p  4)
k 4
3
5p  20
P  4 ^ p4
'Sp 20 Writer  4 as —^.
3
Factor.
1
p4
1 5(p  4)
Fundamental property;
multiply.
(d)
x^ + 2x
X + 1
1 x{x + 2) (x+ l)(x  1)
x^ + x^
(e)
X — 6
X + 1 x^(x + 1)
(x + 2)(x  1)
x(x + 1)
 3x  18
Factor.
Multiply;
lowest terms.
X — 6 (x + 3) (x — 6)
12x + 36 x^ + 7x + 12 (x  6)^ (x + 3) (x + 4)
1
X + 4
Factor.
Lowest
terms
Remember to include 1 in the numerator when all other factors are elimi
nated using the fundamental property.
Work Problem 4 at the Side.
OBJECTIVE Q Find reciprocals for rational expressions. The ration
al numbers f and § are reciprocals of each other if they have a product of 1.
The reciprocal of a rational expression is defined in the same way: Two
rational expressions are reciprocals of each other if they have a product of
1. Recall that has no reciprocal. The table shows several rational expres
sions and their reciprocals. In the first two cases, check that the product of
the rational expression and its reciprocal is 1.
Rational Expression
Reciprocal
5
k
k
5
m^ — 9m
2
2
m^ — 9m
4
undefined
The examples in the table suggest the following procedure.
Finding the Reciprocal
To find the reciprocal of a nonzero rational expression, interchange the
numerator and denominator of the expression.
Work Problem 5 at the Side.
O Multiply.
2r + 4
(a)
3r
5r
5r+ 10
(b)
c^ + 2c c^  4c + 4
(c)
m
16
1
m + 2 m + 4
(d)
X — 3
x^  25
x^ + 2x  15 x^ + 3x  40
& Find each reciprocal.
3
(a)
(b)
y +
(c)
2a  1
(d)
5
Answers
6 c  2
4. (a) — (b) (c)
25 c  1
5. (a)
(b)
y +
(c)
I + 2
2a  1
(d)
3 '~' 1 a^
(d) There is no reciprocal.
la
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
420 Chapter 8 Rational Expressions and Functions
O Divide.
(a) ^
3^
10
OBJECTIVE Q Divide rational expressions. Dividing rational expres
sions is like dividing rational numbers.
Dividing Rational Expressions
To divide two rational expressions, multiply the first (the dividend) by
the reciprocal of the second (the divisor).
(b)
5p + 1 \5p + 6
(c)
2y
y
1
>'2 + 4j + 4 j^ + J  2
EXAMPLE 5
Divide.
Dividing Rational Expressions
2z _„
— • — r Multiply by the reciprocal of the divisor.
(b)
^2z
9
_ 4_
~ Tz
8A:  16 3/t  6
111
5z^
8k
Factor.
Multiply; lowest terms
 16 4k^
3k
4r
(c)
5m^ + 17m  12
3k
Sjk  2)
3k
32k
9
5m^ + 2m
3k 6
4k^
' Hk  2)
Multiply by the reciprocal.
Factor.
Multiply; lowest terms
3m^ + 7m  20 \5n?  34m + 15
5m^ + 17m  12 I5m^  34m + 15
3m^ + 7m  20
(5m  3)(m + 4)
(m + 4) (3m  5)
5m — 3
m + I
5m^ + 2m — 3
(3m — 5) {5m 
3)
(5m  3){m + 1)
Definition of division
Factor.
Lowest terms
Work Problem 6 at the Side.
Answers
32yt 5 y
6. (a)  (b) — (c) 
3 18 ;;
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
Section 8.2 Adding and Subtracting Rational Expressions 425
o.^ Adding and Subtracting Rational Expressions
OBJECTIVE Q Add and subtract rational expressions with the same
denominator. The following steps, used to add or subtract rational num
bers, are also used to add or subtract rational expressions.
Adding or Subtracting Rational Expressions
Step 1 If the denominators are the same, add or subtract the numer
ators. Place the result over the common denominator.
If the denominators are different, first find the least common
denominator. Write all rational expressions with this LCD, and
then add or subtract the numerators. Place the result over the
common denominator.
OBJECTIVES
Add and subtract rational
expressions with the
same denominator.
Find a least common
denominator.
Add and subtract rational
expressions with different
denominators.
Step 2 Simplify. Write all answers in lowest terms.
Add or subtract.
3m 5n
(a) ^ + ^
EXAMPLE 1
Adding and Subtracting Rational Expressions
with the Same Denominator
Add or subtract as indicated.
Add the numerators.
Keep the common denominator.
3 J X iy + X
(a) \ — =
^^55 5
The denominators of these rational expressions are the same, so just add
the numerators, and place the sum over the common denominator.
(b)
2r'
11
1^'
11
2r2
4
2
Subtract the numerators; keep
the common denominator.
(c)
m
+
Lowest terms
m \ p
m
m
m
 P
m + p
(m \ p){m — p)
1
Add the numerators; keep
the common denominator.
Factor.
Lowest terms
m
(d)
f^ + 2x
+
4 + X
x^ + 2x
x^ + 2x
4 + X
(x  2)(x + 4)
1
Add.
Factor.
Lowest terms
Work Problem 1 at the Side.
( \— —
3>a 3a
2 5
(c) —  —
y y
(d)
+
a + b a + b
(e)
2y
y' +y
y" + y
Answers
3m + 5/2 17
1. (a) ^ (b) 
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
426 Chapter 8 Rational Expressions and Functions
& Find the LCD for each group
of denominators.
(a) 5k^s, 10fo4
OBJECTIVE Q Find a least common denominator. We add or subtract
rational expressions with different denominators by first writing them with a
common denominator, usually the least common denominator (LCD).
Finding the Least Common Denominator
Step 1 Factor each denominator.
Step 2 Find the least common denominator. The LCD is the product
of all different factors from each denominator, with each factor
raised to the greatest power that occurs in any denominator.
(b) 3  X, 9 X
 v2
(c) z, z + 6
(d) 2j2  3j  2,
2j2 + 3j + 1
(e) x^  2x + 1,
x^  4x + 3,
4x4
Answers
2. (a) lOytV (b) (3 +x)(3  x)
(c) z(z + 6) (d) {yl){ly+\){y+\)
(e) 4(x3)(xl)2
EXAMPLE 2
Finding Least Common Denominators
Assume that the given expressions are denominators of fractions. Find the
LCD for each group.
(a) 5x7 ^ 2x^7
Each denominator is already factored.
5xy^ = 5 • X • j^
Ix^y = 1 ' x^ ' y
LCD = 5 • 2 •
= lOxV'
x^ ' y^
Greatest exponent on x is 3.
Greatest exponent on;; is 2.
(b) ^  3, k
Each denominator is already factored. The LCD, an expression divisible
by both k  3 and k, is
k(k  3).
It is usually best to leave a least common denominator in factored form.
(c) j;2  23;  8, y' + 3y + 2
Factor the denominators.
y
2 _
2yS = (y4)iy + 2)
Factor.
Factor.
y^ + 3y + 2 = (y + 2)(y+l)
The LCD, divisible by both polynomials, is (y — 4)(y + 2)(y + 1).
(d) 8z  24, 5z2  15z
8z24 = 8(z3)
5z2  15z = 5z(z  3)
The LCD is 8 • 5z • (z  3) = 40z(z  3).
(e) m^ + 5m + 6, m^ + 4m + 4, 2m + 6
m^ + 5m + 6 = (m + 3)(m + 2)
m^ + 4m + 4 = (m + 2)^
2m + 6 = 2(m + 3)
The LCD is 2(?m + 3)(m + 2)1
Work Problem 2 at the Side.
>• Factor.
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
Section 8.2 Adding and Subtracting Rational Expressions 427
OBJECTIVE Q Add and subtract rational expressions with different
denominators. Before adding or subtracting two rational expressions, we
write each expression with the least common denominator by multiplying its
numerator and denominator by the factors needed to get the LCD. This pro
cedure is valid because we are multiplying each rational expression by a
form of 1, the identity element for multiplication.
Adding or subtracting rational expressions follows the same procedure
as that used for rational numbers. Consider the sum ^ + ^. The LCD for 15
and 12 is 60. Multiply ]^ by  (a form of 1) and multiply ^ by f (another form
of 1) so that each fraction has denominator 60. Then add the numerators.
Fundamental property
& Add or subtract.
6 1
(a)  + 
^ ^ 7 5
7
15
+
5
12
7 • 4
= +
15 • 4
28 25
60 60
28 + 25
60
53
60
5 •
12
5
• 5
Add the numerators.
(b)
EXAMPLE 3
Adding and Subtracting Rational Expressions
with Different Denominators
3k 9k
Add or subtract as indicated.
5 3
The LCD for 2p and Sp is Sp. To write the first rational expression with
a denominator of Sp, multiply by .
5
5 3
— + —
2p Sp
4 3
+ —
2p • 4 Sp
20 3_
8p 8p
20 + 3
Fundamental property
8/7
Add the numerators.
23
Sp
(c)
y y + 4
(b)
Write each rational expression with the LCD, r(r
6 5 6(r 3) r • 5
3).
r{r  3)
6r 18
r{r — 3)
6r  18
r{r  3)
5r
r{r — 3)
5r
r{r — 3)
r 18
r{r — 3)
Fundamental property
Distributive and commutative
properties
Subtract the numerators.
Combine terms in the numerator.
Work Problem 3 at the Side.
Answers
37
3. (a)
35
(b)
22
9k
(c)
y +
yiy + 4)
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
428 Chapter 8 Rational Expressions and Functions
O Subtract.
5x + 7
14
2x + 7
CAUTION
One of the most common sign errors in algebra occurs when a rational
expression with two or more terms in the numerator is being subtracted.
In this situation, the subtraction sign must be distributed to every term
in the numerator of the fraction that follows it. Study Example 4 care
fully to see how this is done.
(b)
EXAMPLE 4
Using the Distributive Property When Subtracting
Rational Expressions
Subtract.
Ix
x2
^^^ 3x + 1 3x + 1
The denominators are the same for both rational expressions. The sub
traction sign must be applied to both terms in the numerator of the second
rational expression. Notice the careful use of the distributive property here.
7x X — 2 _ Ix — (x — 2) Subtract the numerators;
3x + 1 3x + 1 ~ 3x + 1 ^^^P ^^^ common denominator.
Ix — X + 2
Distributive property;
be careful with signs.
Combine terms in the numerator.
3x +
1
Ix — X
+ 2
3x +
1
6x + 2
3x+ 1
2(3a: +
1)
3x + 1
Factor the numerator.
Lowest terms
(b)
1
q 1
q + 1
\{q + 1)
1(^1)
i.q
1)(?+1)
(q +
l)(q 1)
iq
1)(^+ 1)
q +
1^+1
(q
1)(^+ 1)
2
{q+ l)(ql)
{q 1)(^+ 1)
The LCD is (^ 1) (^ + 1);
fundamental property
Subtract.
Distributive property
Combine terms in the
numerator.
Work Problem 4 at the Side.
Answers
4. (a) 3 (b)
r^ + Ar ■
(r2)(r 1)
In some problems, rational expressions to be added or subtracted have
denominators that are opposites of each other, such as
y
+
Denominators are opposites.
y  2 2  y
The next example illustrates how to proceed in such a problem.
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
Section 8.2 Adding and Subtracting Rational Expressions 429
Video I
EXAMPLE 5
Add.
Adding Rational Expressions with Denominators
That Are Opposites
^ — +
J ~ 2 2 — y
To get a common denominator of j — 2, multiply the second expression
by  1 in both the numerator and the denominator.
8(l)
y
y
+ —
2 2
y
+
y y2 ' {2y){l)
y2 y2
yS
Add the numerators.
y
If we had used 2 — j as the common denominator and rewritten the first
expression, we would have obtained
Sy
an equivalent answer. Verify this.
Work Problem 5 at the Side.
EXAMPLE 6
Adding and SubtractingThree Rational
Expressions
Video I
s
Add and subtract as indicated.
3
X  2 X x^  2x
The denominator of the third rational expression factors as x{x — 2),
which is the LCD for the three rational expressions.
3 5 6
+
X  2 X x^  2x
3x 5{x  2)
+
x{x — 2) x{x — 2) x{x — 2)
3x + 5(x  2)  6
x(x 
2)
3x + 5x 
106
x(x 
2)
8x
 16
x(x
2)
8(a:
2)
x{x — 2)
Fundamental property
Add and subtract the numerators.
Distributive property
Combine terms in the numerator.
Factor the numerator.
Lowest terms
Work Problem 6 at the Side.
Add or subtract as indicated.
2
(a)
— +
4 4
(b)
2x  9 9  2x
© Add and subtract as
indicated.
4 2 10
+ —
X  5 X
x^ — 5x
Answers
6 6 13 13
5. (a) Tor (b) or —
x4 4x 2x9 92x
X  5
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
430 Chapters Rational Expressions and Functions
O Subtract.
— a
4a
a^ + 3a  4 a^ + 7a + U
©Add.
1
6p + 9 p^ + 2p  15
EXAMPLE 7
Subtract.
m + 4
Subtracting Rational Expressions
2m
m — 2m — 3 m
m + 4
5m + 6
2m — 3
{m — 3){m + 1) {m — 3){m — 2)
The LCD is {m  3) (m + 1) {m  2).
{m + 4){m  2) {2m  3){m + 1)
Factor each denominator.
Fundamental
{m  3)(m + \){m  2) {m  3){m  2){m + 1) Property
{m + 4)(m  2)  {2m  3)(m + 1)
{m — 3)(m + \){m — 2)
m^ + 2m — 8 — {2m^ — m — 3)
Subtract.
{m — 3)(m + \){m — 2)
m^ + 2m — 8 — 2m^ + m + 3
{m — 3)(m + \){m — 2)
— m^ + 3m — 5
{m — 3)(m + \){m — 2)
Muhiply in the numerator.
Distributive property;
be careful with signs.
Combine terms in the
numerator.
If we try to factor the numerator, we find that this rational expression is in
lowest terms.
Work Problem 7 at the Side.
EXAMPLE 8
Add.
Adding Rational Expressions
+
x^ + lOx + 25 x^ + 7x + 10
5 2
+
(x + 5)2 ■ (x + 5)(x + 2)
The LCD is (x + 5)2(x + 2).
5 (x + 2) 2 (x + 5)
+
(x + 5)2(x + 2) (x + 5Y(x + 2)
5(x + 2) + 2(x + 5)
(x + 5)2(x + 2)
5x + 10 + 2x + 10
(x + SYix + 2)
7x + 20
(x + 5f{x + 2)
Factor each denominator.
Fundamental property
Add.
Distributive property
Combine terms in the
numerator.
Work Problem 8 at the Side.
Answers
5a' + a
{a + 4) (fl  1) {a + 3)
5p + 17
{p  ?>)\p + 5)
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
Section 8.3 Complex Fractions 435
8.3 Complex Fractions
A complex fraction is an expression having a fraction in the numerator,
denominator, or both. Examples of complex fractions include
1 4 m^ 9
1 +  
X y ^ m + I
and 1 — . Complex fractions
6 
/w + 3
m^  1
OBJECTIVE Q Simplify complex fractions by simplifying the numera
tor and denominator. (Method 1) There are two different methods for
simplifying complex fractions.
Simplifying a Complex Fraction: Method 1
Step 1 Simplify the numerator and denominator separately.
Step 2 Divide by multiplying the numerator by the reciprocal of the
denominator.
Step 3 Simplify the resulting fraction, if possible.
In Step 2, we are treating the complex fraction as a quotient of two
rational expressions and dividing. Before performing this step, be sure that
both the numerator and denominator are single fractions.
OBJECTIVES
Simplify complex
fractions by simplifying
the numerator and
denominator. (Method 1)
Simplify complex
fractions by multiplying by
a common denominator.
(Method 2)
Compare the two
methods of simplifying
complex fractions.
Simplify rational
expressions with negative
exponents.
EXAMPLE 1
Simplifying Complex Fractions by Method 1
Use Method 1 to simplify each complex fraction.
X + 1
(a)
X  1
2x
Both the numerator and the denominator are already simplified, so
divide by multiplying the numerator by the reciprocal of the denominator.
X + 1
X
X + 1
X
jc 1
X 1
Ix
2x
x+ 1
2x
X
X 1
2x{x 4
1)
x{x —
1)
2(x +
1)
X  1
Continued on Next Page
Write as a division problem.
Multiply by the reciprocal of ^^
Multiply.
Simplify.
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
436 Chapters Rational Expressions and Functions
Use Method 1 to simplify
each complex fraction.
(a)
(2 + 2
5a
a — 3
la
(b)
2 + 
y
y
j; y_
3j _ 2
y y
2y+ 1
3j
y_ Simplify the numerator and denominator.
 2 (Step 1)
3j
y
2y+ 1
2;^+ 1
3J2
J
Write as a division problem.
y Multiply by the reciprocal of
2v^.(Step2)
Multiply and simplify.
(Step 3)
Work Problem 1 at the Side.
OBJECTIVE Q Simplify complex fractions by multiplying by a common
denominator. (Method 2) The second method for simplifying complex
fractions uses the identity property of multiplication.
2 +
(b)
^
I
Simplifying a Complex Fraction: Method 2
Step 1 Multiply the numerator and denominator of the complex fraction
by the least common denominator of the fractions in the numera
tor and the fractions in the denominator of the complex fraction.
Step 2 Simplify the resulting fraction, if possible.
(c)
r'  A
1 + ^
r
Answers
l{a + 2)
5{a  3)
(b)
2k
2k 1
1 ,r(r2)
(c) ,
EXAMPLE 2
Simplifying Complex Fractions by Method 2
Use Method 2 to simplify each complex fraction.
1
2 +
(a)
y
y
Multiply the numerator and denominator by the LCD of all the fractions
in the numerator and denominator of the complex fraction. (This is the same
as multiplying by 1 .) Here the LCD is y.
2 +
1
2 +
]_
y
2 +
1
y
2
y
2
y
■ y
2. + 
y
y
2y + 1
• y
Multiply the numerator and
denominator by j^, since  = 1.
(Step 1)
Distributive property
372
Compare this method with that used in Example 1(b).
Continued on Next Page
Simplify (Step 2)
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
Section 8.3 Complex Fractions 437
Ip +
(b)
P 1
^P
2p +
P 1
p(pl)
^p^yp{pi)
2p[pip  1)] +
P 1
pip  1)
3p[p(pl)] pip I)
2p[p(p  D] + 5p
3p[pip l)]2{p 1)
2p^  2/ + 5p
3p^  3p^ 2p + 2
Multiply the
numerator and
denominator
by the LCD,
Pip I)
Distributive
property
Multiply; lowest
terms
Work Problem 2 at the Side.
& Use Method 2 to simplify
each complex fraction.
+ 6
(a)
1
OBJECTIVE Q Compare the two methods of simplifying complex
fractions. Choosing whether to use Method 1 or Method 2 to simplify a
complex fraction is usually a matter of preference. Some students prefer one
method over the other, vs^hile other students feel comfortable v^ith both meth
ods and rely on practice vs^ith many examples to determine vs^hich method
they v^ill use on a particular problem.
In the next example, wq illustrate hov^ to simplify a complex fraction us
ing both methods so that you can observe the processes and decide for your
self the pros and cons of each method.
EXAMPLE 3
Simplifying Complex Fractions Using Both Methods
Use both Method 1 and Method 2 to simplify each complex fraction.
Method 1
(a)
X 
 3
5
2
X
 9
2
X 3
5
ix
3)(x +
3)
2
5
X —
3 ■ (x
3)(x
+ 3)
2
ix
3)(x
+ 3)
X — 3 5
2{x + 3)
5
— Continued on Next Page
Method 2
(a)
x^  9
X 3
(x  3)(x + 3)
(x  3)(x + 3)
2{x + 3)
(x  3){x + 3)
(b)
1 1
 +
y y  I
y y  I
Answers
15
2. (a)
8  3 J
I83; 2y
(b) ^
1 1
—or —
2y
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
438 Chapters Rational Expressions and Functions
& Use both methods to simplify
each complex fraction.
(a)
y + 2
3
(b)
1
1
a
b
1
1
a'
b'
Method 1
(b)
1 1
 + 
X y
1
1
^ + ^
xy
xy
y'
x'
xY
x'r
I
y + X
xy

/x^
2 2
X y
y + X
y'
x'
xy
2
X
2
y
y + X
xY
xy
{y
x)(y
xy
y
Method 2
(b)
1 1
 + 
X y
1
_1_
7
'^xY + (^ ]xY
xy^ + x^y
2 2
y^  x^
xyjy + x)
xy
y  X
Work Problem 3 at the Side.
O Simplify each expression,
using only positive exponents
in the answer.
(a)
4r~^ + s~^
(b)
b'
+ 2
Answers
3. (Both methods give the same answers.)
5(3; 2) _ ab
(a)
4. (a)
3
4rs'
(b)
(b)
b + a
b
1 + 2b^
OBJECTIVE Q Simplify rational expressions with negative expo
nents. Rational expressions and complex fractions sometimes involve
negative exponents. To simplify such expressions, we begin by rewriting the
expressions with only positive exponents.
EXAMPLE 4
Simplifying a Rational Expression with Negative
Exponents
Simplify
m
' +p'
, using only positive exponents in the answer.
2m — p
First write the expression with only positive exponents.
m ^ + p
2m~^ — p
1 1
m n^
m
]_
P
Definition of negative exponent
Note that the 2 in 2m ^ is not raised to the —2 power (since m is the base
for the exponent —2), so 2m~
m
Simplify the complex fraction using
Method 2, multiplying numerator and denominator by the LCD, m^p^.
1 1
m v^
2 2
m p
1
m
I 2 2
+ m p
1
m
P
22/^
mp 2
^m
m^p^
m
m^p^
]_
P
mp^ + m^
m^p
ip'
Work Problem 4 at the Side.
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
Section 8.4 Equations with Rational Expressions and Graphs 44 I
8.4 Equations with Rational Expressions and Graphs
In Section 8.1, we defined the domain of a rational expression as the set of
all possible values of the variable. Any value that makes the denominator
is excluded.
OBJECTIVE Q Determine the domain of a rational equation. The
domain of a rational equation is the intersection (overlap) of the domains
of the rational expressions in the equation.
EXAMPLE 1
Determining the Domains of Rational Equations
Find the domain of each equation.
(a)
2 _ 3 _ ^
X 2 2x
The domains of the three rational terms of the equation are, in order,
{x\x 7^ 0}, (00, oo), and {x\x t^ 0}. The intersection of these three
domains is all real numbers except 0, which may be written {x\x ^ 0}.
(b)
12
x3 x + 3 x^9
The domains of these three terms are, respectively, {x\x ^3},
{x\x 7^—3}, and {x\x ^ ±3}. The domain of the equation is the intersection
of the three domains, all real numbers except 3 and —3, written {x\x ^ ±3}.
Work Problem 1 at the Side.
OBJECTIVES
Determine the domain of
a rational equation.
Solve rational equations.
Recognize the graph of a
rational function.
Find the domain of each
equation.
3 1 5
(a)  +  = —
^ 2 6x
OBJECTIVE Q Solve rational equations. The easiest way to solve
most equations involving rational expressions is to multiply all terms in the
equation by the least common denominator. This step will clear the equation
of all denominators. We can do this only with equations, not expressions.
CAUTION
When each side of an equation is multiplied by a variable expression,
the resulting "solutions" may not satisfy the original equation. You must
either determine and observe the domain or check all potential solu
tions in the original equation. It is wise to do both.
(b)
X — 5 X + 5
25
EXAMPLE 2
Solve
Solving an Equation with Rational Expressions
2 _ 3 _ ^
X 2 2x
The domain, which excludes 0, was found in Example 1(a).
i^i'\^t.ii
2x
 2x
,x 2
^2
4  3x = 7
3x = 3
x= 1
Continued on Next Page
 <l)
Multiply by the LCD, 2x.
Distributive property
Multiply.
Subtract 4.
Divide by —3.
Answers
1. (a) {xx^O} (b) {xx^±5}
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442 Chapter 8 Rational Expressions and Functions
3 2 5
& Solve +  = — .
20 X 4x
Check: Replace x with — 1 in the original equation.
2_3__J_
X 1 Ix
2 3 _ 7
1 2 ~ 2(l)
. 3 7
2 =  ?
2 2
_7 _ _7
~2 ~ ~2
The solution set is { — 1 1 .
Original equation
Let X = — 1 .
True
Work Problem 2 at the Side.
O Solve each equation.
3 1
(a)
X + 1
1
(b)
+
X + 3
EXAMPLE 3
Solve
Solving an Equation with No Solution
3 12
X — 3 x + 3 X— 9
Using the result from Example 1(b), we know that the domain excludes
3 and 3, since these values make one or more of the denominators in the
equation equal 0. Multiply each side by the LCD, (x + 3) (x  3).
{x + 3) (x  3)
(jc + 3)(jc 3)
X  3
2
X + 3
(x + 3) (x  3)
12
x'
X
(x + 3)(x 3)
X + 3
(x + 3)(x  3). ,
vX
12
Distributive
property
2(x + 3) 3(x 3)= 12
2x + 6  3x + 9 = 12
x+ 15 = 12
x= 3
X = 3
Multiply.
Distributive property
Combine terms.
Subtract 15.
Divide by — 1 .
Since 3 is not in the domain, it cannot be a solution of the equation. Substi
tuting 3 in the original equation shows why.
Check:
12
X — 3 X + 3
2 3
12
Original equation
? Letx = 3.
33 3 + 3 3^9
2 _ 3 _ 22
6 ~ '
Since division by is undefined, the given equation has no solution, and the
solution set is 0.
«f<
Work Problem 3 at the Side.
Answers
2. {5}
3. (a) (b)
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Section 8.4 Equations with Rational Expressions and Graphs 443
juTr
Afiimatiocil
EXAMPLE 4
Solve
Solving an Equation witli Rational Expressions
1 7
p" + p 2 p"  \ lip" + 3p + 1)
Factor each denominator to find the LCD, l{p — \){p + 2){p + 1).
The domain excludes 1, 2, and  1. Multiply each side by the LCD.
2{p  Dip + 2){p + \)
1
{p + l){p\) {p+\){p
= lip  l){p + 2){p + \)
1)
2Xp+\)lip + l) = lip\)
6p + 6  2p  4 = 7p  7
4p + 2 = lpl
9 = 3p
3=p
2(p + 2)ip+ 1)
Distributive property
Distributive property
Combine terms
Subtract 4p; add 7.
Divide by 3.
Note that 3 is in the domain; substitute 3 forp in the original equation to
check that the solution set is {3}.
Work Problem 4 at the Side.
O Solve
X + X
6
2
x^ + 5x + 6
Ani
Video
EXAMPLE 5
Solving an Equation That Leads to a Quadratic
Equation
Solve
6x
I
3x + 1 ~ X 3x + r
Since the denominator 3x + 1 cannot equal 0,
\ is excluded from the
domain, as is 0. Multiply each side by the LCD, x(3x +1).
x(3x + 1)
x(3x + 1)
3x + 1
2
3x + 1
x(3x + 1)
x(3x + 1)
6x
3x + 1
 x(3x + 1)
6x
3x + 1.
Distributive property
2x = 3x + 1 — 6x^
Write this quadratic equation in standard form with on the right side.
6x2  3x + 2x  1 =
6x2  X  1 =
(3x+ l)(2x 1) =
3x + 1 = or 2x  1 =
1 1
x= or x = —
3 2
Standard form
Factor.
Zerofactor property
Because  is not in the domain of the equation, it is not a solution. Check
that the solution set is {^}.
Work Problem 5 at the Side.
& Solve
1 X
X + 4
+
16
Answers
4. {9}
5. {1}
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
444 Chapter 8 Rational Expressions and Functions
© Graph each rational function,
and give the equations of the
vertical and horizontal
asymptotes.
(a) /(x) = 
1
OBJECTIVE Q Recognize the graph of a rational function. As men
tioned in Section 8.1, a function defined by a rational expression is a
rational function. Because one or more values of x may be excluded from
the domain of most rational functions, their graphs are often discontinuous.
That is, there will be one or more breaks in the graph. For example, we use
point plotting and observing the domain to graph the simple rational func
tion defined by
m = i
The domain of this function includes all real numbers except 0. Thus,
there will be no point on the graph with x = 0. The vertical line with equation
X = is called a vertical asymptote of the graph. The horizontal line with
equation J = is called a horizontal asymptote. We show some typical or
dered pairs in the table for both negative and positive xvalues.
X
3
2
1
1
2
1
4
1
10
1
10
1
4
1
2
1
2
3
y
1
3
1
2
1
2
4
10
10
4
2
1
1
2
1
3
(b) f{x) =
X + 3
Answers
6. (a) vertical asymptote: x = 0; horizontal
asymptote:;; =
'm^ '
(b) vertical asymptote: x = — 3; horizontal
asymptote: J =
Notice that the closer positive values of x are to 0, the larger j is. Similarly,
the closer negative values of x are to 0, the smaller (more negative) y is.
Using this observation, excluding from the domain, and plotting the points
in the table, we obtain the graph in Figure 2.
3
;
t
•vvrrrvv10
:: Vertical:...]...:...
EXrHZIZZn
Asymptote V^
x = o rt5
i...l :....l...i....
Im=I=h
3'">"2'"i'"ft
:mH:i: :::;;:] ^
■■■'''■ "HsLo
■■■[■■■ritt2'A<3i
L Li Horizontal
:::::::: :::::t:5:
:::::::: ::::::::[:]l
;~4:]:Asymptote
::::;:::: ::::::::::::;:::i:r
iiJ4:j. = 0H
' ri\
Zio;L;::i:Li;i:i
Figure 2
The graph of
Figure 3
gix) =
2
X — 3
is shown in Figure 3. Some ordered pairs are shown in the table.
X
2
1
1
2
2.5
2.75
3.25
3.5
4
5
6
y
2
5
1
2
2
3
1
2
4
8
8
4
2
1
2
3
There is no point on the graph for x = 3 because 3 is excluded from the
domain. The dashed line x = 3 represents the vertical asymptote and is not
part of the graph. As suggested by the points from the table, the graph gets
closer to the vertical asymptote as the xvalues get closer to 3. Again, y =
is a horizontal asymptote.
^H(
Work Problem 6 at the Side.
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
Section 8.5 Applications of Rational Expressions 45 I
Video
8 .5 Applications of Rational Expressions
OBJECTIVE Q Find the value of an unknown variable in a formula.
In this section, we work with formulas that contain rational expressions.
EXAMPLE 1
Finding the Value of aVariable in a Fornnula
In physics, the focal length,/, of a lens is given by the formula
i  i i
7 " ^ ^ ?
where p is the distance from the object to the lens and q is the distance from
the lens to the image. See Figure 4. Find q if/? = 20 cm and/= 10 cm.
#
^ko
Focal Length of Camera Lens
Figure 4
Replace/with 10 and/> with 20.
i i i
10
1 1
— + 
20 q
20q
— = (— + 
10 ~ ^V20 ^ q
Idq
1
+ 20q
To "'"V20
2q = q + 20
q = 20
The distance from the lens to the image is 20 cm.
Let/= 10, p = 20.
Multiply by the LCD, 20q.
a Distributive property.
Multiply.
Subtract q.
OBJECTIVES
Find the value of an
unknown variable in a
formula.
Solve a formula for a
specified variable.
Solve applications using
proportions.
Solve applications about
distance, rate, and time.
Solve applications about
work rates.
O Use the formula given in
Example 1 to answer each
part.
(a) Find;? if /= 15 and
q = 25.
(b) Findfifp
q = 9.
6 and
Work Problem 1 at the Side.
{M
OBJECTIVE Q Solve a formula for a specified variable. The goal in
solving for a specified variable is to isolate it on one side of the equals sign.
EXAMPLE 2
Solving a Formula for a Specified Variable
1 1 1
Solve — = — \ — for p.
f P ^
]_ _
f~
1
1 1
 + 
p q
pq =fy ^fp
Continued on Next Page
Multiply by the LCD Jpq.
Distributive property
(c) Find q if/= 12 and
p= 16.
Answers
75 18
1. (a) y (b) — (c) 48
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452 Chapters Rational Expressions and Functions
3 3 5
& Solve — \ — = — for ^.
p q r
Transform the equation so that the terms with p (the specified variable) are
on the same side. One way to do this is to subtract^ from each side.
PqfP=fy Subtract^.
p{q — f) =fq Factor out;?.
fq
P =
qf
Divide by q — f.
Work Problem 2 at the Side.
EXAMPLE 3
& Solve A
Rr
R + r
fori?.
Solve / =
nE
Solving a Formula for a Specified Variable
for n.
R + nr
R + nr
{R + nr)I = (R + nr) 
nE
R + nr
Multiply hy R + nr.
RI + nrl = nE
RI= nE  nrl
Subtract nrl.
RI= n(Erl)
Factor out n.
RI
Err"
Divide by E — rl.
CAUTION
Refer to the steps in Examples 2 and 3 that factor out the desired
variable. This is a step that often gives students difficulty. Remember
that the variable for which you are solving must be a factor on only
one side of the equation. Then each side can be divided by the remain
ing factor in the last step.
Work Problem 3 at the Side.
We can now solve problems that translate into equations with rational
expressions. To do so, we continue to use the sixstep problemsolving
method from Section 2.3.
OBJECTIVE Q Solve applications using proportions. A ratio is a
comparison of two quantities. The ratio of a to Z? may be written in any of the
following ways:
Answers
2. q
3. R
3rp
5p — 3r
Ar
ovR
— 3rp
3r — 5p
Ar
aiob, a : b, or — . Ratio of a to b
b
Ratios are usually written as quotients in algebra. A proportion is a state
ment that two ratios are equal, such as
a c
— = — . Proportion
b a
Proportions are a useful and important type of rational equation.
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
Section 8.5 Applications of Rational Expressions 453
EXAMPLE 4
Solving a Proportion
In 2002, about 15 of every 100 Americans had no health insurance coverage.
The population at that time was about 288 million. How many million Amer
icans had no health insurance? {Source: U.S. Bureau of the Census.)
Step 1 Read the problem.
Step 2 Assign a variable. Let x = the number (in millions) who had no
health insurance.
Step 3 Write an equation. To get an equation, set up a proportion. The
ratio 15 to 100 should equal the ratio x to 288.
Step 4 Solve. 28,800
15
100
X
288
Write a proportion.
(") =
viooy
= ^^•«°°(2"88)
Multiply by a common
denominator.
4320 =
= lOOx
Simplify
X =
= 43.2
Divide by 100.
State the answer. There were 43.2 million Americans with no
health insurance in 2002.
Step 5
Step 6 Check that the ratio of 43.2 million to 288 million equals
100
O Solve the problem.
In 2002, approximately
1 1.6% (that is, 1 1 .6 of every
100) of the 73,500,000
children under 18 yr of age
in the United States had
no health insurance. How
many such children were
uninsured? (Source: U.S.
Bureau of the Census.)
Work Problem 4 at the Side.
EXAMPLE 5
Solving a Proportion Involving Rates
Marissa's car uses 10 gal of gas to travel 210 mi. She has 5 gal of gas in the
car, and she still needs to drive 640 mi. If we assume the car continues to use
gas at the same rate, how many more gallons will she need?
Step 1 Read the problem.
Step 2 Assign a variable. Let x = the additional number of gallons of gas.
Step 3 Write an equation. To get an equation, set up a proportion.
gallons — > 10 _ 5 + X < — gallons
Step 4
10 • 21 • 64
miles ^210 640
Solve. The LCD is 10 2164.
10 \ /5 + x
64  10 = 21(5 +x)
640 = 105 + 21x
535 = 21x
25.5 « X
miles
Distributive property
Subtract 105.
Divide by 21; round to the nearest tenth.
Step 5 State the answer. Marissa will need about 25.5 more gallons of gas.
Step 6 Check. The 25.5 gal plus the 5 gal equals 30.5 gal.
30.5 ^_ , 10
— « .047 and —
640 210
.047
Since the ratios are equal, the answer is correct.
& Solve the problem.
In a recent year, the average
American family spent 8.2 of
every 100 dollars on health
care. This amounted to $3665
per family. To the nearest
dollar, what was the average
family income at that time?
(Source: U.S. Health Care
Financing Administration,
U.S. Bureau of the Census.)
Work Problem 5 at the Side.
Answers
4. 8,526,000
5. $44,695
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454 Chapters Rational Expressions and Functions
OBJECTIVE Q Solve applications about distance, rate, and time.
The next examples use the distance formula d = rt introduced in Section
2.2. A familiar example of a rate is speed, which is the ratio of distance to
time, or r = f.
rouTr
EXAMPLE 6
Solving a Problem about Distance, Rate, and Time
Step 1
Step 2
A tour boat goes 10 mi against the current in a small river in the same time
that it goes 15 mi with the current. If the speed of the current is 3 mph, find
the speed of the boat in still water.
Read the problem. We must find the speed of the boat in still water.
Assign a variable.
Letx = the speed of the boat in still water.
When the boat is traveling against the current, the current slows the
boat down, and the speed of the boat is the difference between its
speed in still water and the speed of the current, that is, x — 3 mph.
When the boat is traveling with the current, the current speeds
the boat up, and the speed of the boat is the sum of its speed in still
water and the speed of the current, that is, x + 3 mph.
Thus, X — 3 = the speed of the boat against the current,
and X + 3 = the speed of the boat with the current.
Because the time is the same going against the current as with
the current, find time in terms of distance and rate (speed) for each
situation. Start with the distance formula.
and divide each side by r to get t =
distance is 10 mi and the rate is x
rt,
f. Going against the current, the
3, giving
10
_d_
r X  3'
Going with the current, the distance is 15 mi and the rate is x + 3, so
r X + 3'
This information is summarized in the following table.
Distance
Rate
Time
Against
Current
10
x3
10
x3
With
Current
15
x + 3
15
x + 3
Times
are equal.
Step 3 Write an equation. Because the times are equal,
10 15
X — 3
Continued on Next Page
x + 3'
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Section 8.5 Applications of Rational Expressions 455
Step 4 Solve. The LCD is (x + 3) (x
10 15
3).
X  3 X + 3
{x + 3) (x 
3)(/_^3)(^ + 3)(x
10(x + 3)= 15(x3)
10x + 30= 15X45
30 = 5x  45
75 = 5x
15 = x
3)
vx + 3/
Multiply by
the LCD.
Multiply.
Distributive property
Subtract lOx.
Add 45.
Divide by 5.
Step 5 State the answer. The speed of the boat in still water is 15 mph.
10 15
Step 6 Check the answer:
15
IS true.
15 + 3
Work Problem 6 at the Side.
© Solve the problem.
A plane travels 100 mi
against the vs^ind in the same
time that it takes to travel
120 mi v^ith the v^ind. The
w^ind speed is 20 mph.
(a) Complete this table.
d
r
t
Against , „„
Wind ^""
x20
With
Wind ^^"
x + 20
EXAMPLE 7
Solving a Problem about Distance, Rate, and Time
At O'Hare Airport, Cheryl and Bill are vs^alking to the gate (at the same
speed) to catch their flight to Akron, Ohio. Since Bill vs^ants a vs^indovs^ seat,
he steps onto the moving sidevs^alk and continues to v^alk v^hile Cheryl uses
the stationary sidev^alk. If the sidev^alk moves at 1 m per sec and Bill saves
50 sec covering the 300m distance, v^hat is their v^alking speed?
Step 1
Step 2
Step 3
Read the problem. We must find their v^alking speed.
Assign a variable. Let x represent their v^alking speed in meters per
second. Thus Cheryl travels at x meters per second and Bill travels at
X + 1 meters per second. Express their times in terms of the knov^n
distances and the variable rates. As in Example 6, start v^ith d = rt and
divide each side by r to get t = 7. For Cheryl, the distance is 300 m
and the rate is x, so Cheryl's time is
d
r
300
X
Bill travels 300 m at a rate of x + 1, so his time is
d 300
^ = ~ = 7
This information is summarized in the foUov^ing table.
Distance
Rate
Time
Cheryl 300
X
300
X
Bill 300
x+ 1
300
x + 1
Write an equation using the times from the table.
Bill's time is Cheryl's time less 50 seconds.
300 300
X + 1 X
Continued on Next Page
50
(b) Find the speed of the plane
in still air.
Answers
6. (a)
100
120
20 ' X + 20
(b) 220 mph
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
456 Chapters Rational Expressions and Functions
O Solve the problem.
Dona Kenly drove 300 mi
north from San Antonio,
mostly on the freeway. She
usually averaged 55 mph, but
an accident slowed her speed
through Dallas to 15 mph. If
her trip took 6 hr, how many
miles did she drive at reduced
speed?
Normal
Speed
Reduced
Speed
300
r
55
15
Step 4 Solve.
300
300
X + 1 X
, / 300 \ , , /300 \
x{x + 1) =x{x+ I) 50
\x + IJ \ X J
Multiply by the
LCD,x(x+ 1).
.J 300 \ , .y300\
x{x +1) ^,]= x{x+ 1) ^ x{x +
\X + \J \ X J
1)(50)
Distributive property
300x = 300(x + 1)  50x(x + 1)
Multiply.
300x = 300x + 300  SOx^  50x
Distributive property
50x2 + 50;^  300 =
Standard form
x2 + X  6 =
Divide by 50.
(x + 3)(x2) =
Factor.
x + 3 = or x2 =
Zerofactor property
X = — 3 or X = 2
Discard the negative answer, since speed cannot be negative.
Step 5 State the answer. Their walking speed is 2 m per sec.
Step 6 Check the answer in the words of the original problem.
Work Problem 7 at the Side.
OBJECTIVE Q Solve applications about work rates.
work are closely related to distance problems.
Problems about
PROBLEMSOLVING HINT
People work at different rates. If the letters r, t, and^ represent the rate
at which the work is done, the time required, and the amount of work
accomplished, respectively, then A = rt. Notice the similarity to the
distance formula, d = rt.
Amount of work can be measured in terms of jobs accomplished.
Thus, if 1 job is completed, A = I, and the formula gives the rate as
1 = rt
_ ]_
t'
To solve a work problem, we begin by using the following fact to express all
rates of work.
Rate of Work
If a job can be accomplished in t units of time, then the rate of work is
 job per unit of time.
Answers
1
7. 11 mi
4
See if you can identify the six problemsolving steps in the next example.
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
Section 8.5 Applications of Rational Expressions 457
EXAMPLE 8
Solving a Problem about Work
Video
Letitia and Kareem are working on a neighborhood cleanup. Kareem can
clean up all the trash in the area in 7 hr, while Letitia can do the same job in
5 hr. How long will it take them if they work together?
Let X = the number of hours it will take the two people working to
gether. Just as we made a table for the distance formula, d = rt, make a table
here for^ = rt, with^ = 1. Since A = \, the rate for each person will be 7,
where t is the time it takes the person to complete the job alone. For exam
ple, since Kareem can clean up all the trash in 7 hr, his rate is \ of the job per
hour. Similarly, Letitia 's rate is \ of the job per hour.
Rate
Time Working
Together
Fractional Part
of the Job Done
Kareem —
X
1
Letitia 
X
1
5"
Since together they complete 1 job, the sum of the fractional parts
accomplished by them should equal 1 .
Part done
Part done 1 whole
jy Kareem
+ by Letitia is job.
1
1
+ X = \
5
35 X + X =351 The LCD is 35
V7 5 /
5x + 7x = 35
12x = 35
35
"=12
35 
Working together, Kareem and Letitia can do the entire job in 72 hr, or 2 hr
and 55 min. Check this result in the original problem.
Work Problem 8 at the Side.
O Solve each problem.
(a)
Stan needs 45 min to do the
dishes, while Deb can do
them in 30 min. How long
will it take them if they
work together?
Rate
Time
Working
Together
Fractional
Part of the
Job Done
Stan ^
X
'^'^ h
X
(b) Suppose it takes Stan
35 min to do the dishes, and
together they can do them in
15 min. How long will it
take Deb to do them alone?
There is another way to approach problems about work. For instance, in
Example 8, x represents the number of hours it will take the two people
working together to complete the entire job. In one hour, \ of the entire job
will be completed. Kareem completes ^ of the job in one hour, and Letitia
completes \ of the job, so the sum of their rates should equal \. Thus,
i i i
7 5~ x'
Multiplying each side of this equation by 35x gives 5x + 7x = 35. This is the
same equation we got in Example 8 in the third line from the bottom. Thus
the solution of the equation is the same using either approach.
Answers
1
8. (a) 18 mm (b) 26 mm
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
Roots, Radicals,
and Root
Functions
\
Tom Skilling is the chief meteorologist for the Chicago Tribune.
He writes a column titled "Ask Tom Why," where readers
question him on a variety of topics. Reader Ted Fleischaker wrote: "I
cannot remember the formula to calculate the distance to the horizon.
I have a stunning view from my 14th floor condo, 150 feet above the
ground. How far can I see?" (See Exercise 125 in Section 9.3.)
In Skilling's answer, he explained the formula for finding the
distance d to the horizon in miles,
d= 1.224V^,
where h is the height in feet. Square roots such as this one are often
found in formulas. This chapter deals with roots and radicals.
9.T Radical Expressions and
Graphs
9.2 Rational Exponents
9.3 Simplifying Radical
Expressions
9.4 Adding and Subtracting
Radical Expressions
9.5 Multiplying and Dividing
Radical Expressions
Summary Exercises on Operations
witti Radicals and Rational Exponents
9.6 Solving Equations with
Radicals
9.7 Complex Numbers
479
480 Chapter 9 Roots, Radicals, and Root Functions
9.1 Radical Expressions and Graplis
OBJECTIVES
Find roots of numbers.
Find principal roots.
Graph functions defined
by radical expressions.
Find nth roots of
nth powers.
Use a calculator to
find roots.
Simplify.
(a) ^
OBJECTIVE Q Find roots of numbers.
square roots of positive numbers such as
In Section 1.3 we found
36 = 6, because 6 • 6 = 36 and Vl44 = 12, because 12 • 12 = 144.
We now extend our discussion of roots to cube roots, fourth roots, and higher
roots. In general, y a is a number whose wth power equals a. That is,
y a — b means b^ = a.
The number a is the radicand, n is the index or order, and the expression
Va is a radical.
Index
Radical
sign
Radicand
(b) ^1000
EXAMPLE 1
Simplify.
Simplifying Higher Roots
(a) V27 = 3, because 3^ = 27.
(b) ^^125 = 5, because 5^ = 125.
(c) V 16 = 2, because 2"^ =16.
^' 5 —
(d) V32 = 2, because 25 = 32.
Hi
Work Problem 1 at the Side.
(c) V81
OBJECTIVE Q Find principal roots. Ifn is even, positive numbers
have two nth roots. For example, both 4 and —4 are square roots of 16, and
2 and —2 are fourth roots of 16. In such cases, the notation va represents
the positive root, called the principal root.
(d) V64
rAudioJ
nth Root
Ifn is even and a is positive or 0, then
va represents the principal nth root of a, and
— va represents the negative nth root of a.
Ifn is even and a is negative, then
Va is not a real number.
Ifn is oddy then
there is exactly one nth root of a, written va.
If n is even, then the two nth roots of a are often written together as
± V «, with ± read "positive or negative."
Answers
1. (a) 2 (b) 10 (c) 3 (d) 2
Section 9.1 Radical Expressions and Graphs 48 I
EXAMPLE 2
Find each root.
Finding Roots
(a) VlOO = 10
Because the radicand is positive, there are two square roots, 10 and — 10.
We want the principal root, which is 10.
(b) VTOO = 10
Here, we want the negative square root, — 10.
(c) ^81 = 3
(d) V^
The index is even and the radicand is negative, so this is not a real number.
(e)
8 = —2, because (—2)'
Work Problem 2 at the Side.
® Find each root.
(a) V4
(b) V27
OBJECTIVE Q Graph functions defined by radical expressions. A
radical expression is an algebraic expression that contains radicals. For
example,
3 — V X, V X, and V 2x — 1 Radical expressions
are radical expressions.
In earlier chapters we graphed functions defined by polynomial and
rational expressions. Now we examine the graphs of functions defined by
the radical expressions f{x) = vx and f{x) = vx.
Figure 1 shows the graph of the square root function with a table of
selected points.
X
fix) = Vx
1
1
4
2
9
3
Figure 1
Only nonnegative values can be used for x, so the domain is [0, oo). Because v x
is the principal square root of x, it always has a nonnegative value, so the range
is also [0, co).
Figure 2 shows the graph of the cube root function and a table of
selected points.
I«
{{x)=Vx
8
2
1
1
1
1
8
2
3
8
Figure 2
Since any real number (positive, negative, or 0) can be used for x in the cube
root function, v x can be positive, negative, or 0. Thus both the domain and
the range of the cube root function are (— °o, ^).
(c) V36
(d) V625
(e) if^
(f) ^^
Answers
2. (a) 2 (b) 3 (c) 6 (d) 5
(e) 2 (f) not a real number
482 Chapter 9 Roots, Radicals, and Root Functions
& Graph each function by
creating a table of values.
Give the domain and range.
(a) fix) = Vx + 2
(b) /(x) = ^^^
EXAMPLE 3
Graphing Functions Defined with Radicals
Graph each function by creating a table of values. Give the domain and
the range.
(a) fix) = V^^
A table of values is shown. The x values were chosen in such a way that the
function values are all integers. For the radicand to be nonnegative, we must
have X  3 > 0, or X > 3. Therefore, the domain is [3, co). Again, function
values are positive or 0, so the range is [0, oo). The graph is shown in Figure 3.
X
f(x) = Vx^
3
V3  3 =
4
V4  3 = 1
7
Vl 3 = 2
12
ii
• f •ri/U) = Vr3i
:0.
:i±i::t:l:i±tn'
Figure 3
(b) fix) = ^ + 2
See the table and Figure 4. Both the domain and the range are (oo oo).
Wx
f(x)=^ + 2~
8
^y^ + 2 =
1
^/^ + 2 = 1
^ + 2 = 2
1
>^ + 2 = 3
8
Vs +2 = 4
\
Hiii
8
1
;;l;T;;i;;j;;j;;;i;;;i;;8;;i;;;i
;j/(x)=A^i2j
«<
Figure 4
Work Problem 3 at the Side.
OBJECTIVE Q Find nth roots of nth powers. A square root of a^
(where (3 t^ 0) is a number that can be squared to give a^. This number is
either a or —a. Since the symbol y a^ represents the nonnegative square
root, we must write y a^ with absolute value bars, as a , because a may be a
negative number.
Answers
3. (a) domain: [0, 0°); range: [2, oo)
5
:^_^^
2
r"^^
1 1 IQ
:i 4 9
: fix) = v^+ 2

(b) domain: (—00,00); range: (—00, 00)
y
_
2
;^—
>
^7
, , 1^
9
: fix) = v^
^

SEXAMPLE 4]
Simplifying
root that is a
= 7
7 = 7
Square Roots
real number.
Using
Absolute Value
Find each square
(a) V72= 7
(b) V(7)2 =
(c) Vf= 1^1
(d) V(/t)2 =
Section 9.1 Radical Expressions and Graphs 483
Work Problem 4 at the Side.
We can generalize this idea to any nth root.
Ifn is an even positive integer, va"" = « I,
and ifn is an odd positive integer, va" = a.
In words, use absolute value when n is even; do not use absolute value
when n is odd.
EXAMPLE 5
Simplifying Higlier Roots Using Absolute Value
Simplify each root.
(a) v( — 3)^= — 3= 3 « is even; use absolute value.
(b) V(4)= = 4 wis odd.
(c) ^{9}' = 9 = 9
(d)
/ 4 12 1 2
m = — \m \ = —m
(e) 
No absolute value bars are needed here because m^ is nonnegative for
any real number value of m.
(f)
(g)
a^ because a^^
.3 1
(ay.
X = \x ^
We use absolute value bars to guarantee that the result is not negative
(because x^ can be either positive or negative, depending on x). If desired,
\x^ I can be written as x^ • \x 1.
Work Problem 5 at the Side.
OBJECTIVE Q Use a calculator to find roots. While numbers such as
V 9 and v — 8 are rational, radicals are often irrational numbers. To find
approximations of roots such as VI 5, v 10, and v 2, we usually use scien
tific or graphing calculators. Using a calculator, we find
Vis « 3.872983346, ^10 « 2.15443469, and Vl « 1.189207115,
where the symbol « means "is approximately equal to." In this book we will
usually show approximations rounded to three decimal places. Thus, we
would write
15 « 3.873, ^10 « 2.154, and ^2 « 1.189.
liiil Calculator Tip The methods for finding approximations differ
among makes and models, and you should always consult your owner's
manual for keystroke instructions. Be aware that graphing calculators
often differ from scientific calculators in the order in which keystrokes
are made.
O Find each square
root that is
a real number.
(a)
V^
(b)
Vi
(c)
V(6)2
(d)
V?
& Simplify.
(a)
Vm
(b)
^6
(c)
3/2I6
Vl25
(d)
^y243
(e)
^{Pf
(f)
Vj24
Answers
4. (a) 7
(b) J (c) 6
(d) \r\
5. (a) 2
(b) 2 (c) ^
(d) :
! (e) \p\ (f) 
y'
484 Chapter 9 Roots, Radicals, and Root Functions
& Use a calculator to
approximate each radical to
three decimal places.
(a) Vl7
(b)
Figure 5 shows how the preceding approximations are displayed on a TI83
Plus or TI84 graphing calculator. In Figure 5 (a), eight or nine decimal places
are shown, while in Figure 5 (b), the number of decimal places is fixed at three.
4 1 ' 4
i ■ 
(a)
(b)
Figure 5
There is a simple way to check that a calculator approximation is "in the
ballpark." Because 16 is a little larger than 15, v 16 = 4 should be a little
larger than V 15. Thus, 3.873 is a reasonable approximation for v 15.
(c) ^9482
EXAMPLE 6
Finding Approximations for Roots
Use a calculator to verify that each approximation is correct.
(a) V39 « 6.245 (b) V72 « 8.485
(c) V93 « 4.531
(d) ^^39 « 2.499
Work Problem 6 at the Side.
(d) ^^6825
O Use the formula in
Example 7 to approximate/
to the nearest thousand if
L = 6X 105
and C = 4 X 10"".
EXAMPLE 7
Using Roots to Calculate Resonant Frequency
In electronics, the resonant frequency/ of a circuit may be found by the formula
1
/ =
IttVLC
where/ is in cycles per second, L is in henrys, and C is in farads.* Find the
resonant frequency/ if Z = 5 X 10"* henrys and C = 3 X 10'° farads.
Give your answer to the nearest thousand.
Find the value of/ when i = 5 X lO"'' and C = 3 X lO""^.
/ =
1
IttVLC
1
Given formula
Substitute for L and C.
277V(5 X 10"^) (3 X 10""')
= 411,000 Usea calculator.
The resonant frequency/ is approximately 41 1,000 cycles per sec.
Work Problem 7 at the Side.
Answers
6. (a) 4.123 (b) 19.026
(c) 21.166 (d) 9.089
7. 325,000 cycles per sec
*Henrys and farads are units of measure in electronics.
Section 9.2 Rational Exponents 489
9.2 Rational Exponents
OBJECTIVE Q Use exponential notation for nth roots. In mathemat
ics we often formulate definitions so that previous rules remain valid. In
Section 6.1 we defined as an exponent in such a way that the rules for
products, quotients, and powers would still be valid. Now we look at expo
nents that are rational numbers of the form ^, where w is a natural number.
For the rules of exponents to remain valid, the product (3^^^)^ = 3^^^ • 3^^^
should be found by adding exponents.
(31/2)2 = 31/2 . 3
= 3I/2 + I/2
= 31
= 3
1/2
However, by definition ( v 3)^ = v 3 • v 3
(V3)^ are equal to 3, we must have
3^/2 = V3.
This suggests the following generalization.
3. Since both (3^^^y and
,l/n
If va is a real number, then
«i/« = V «.
Sexample
^ Evaluatin.
expression.
g Exponentials of the Form a^'"
Evaluate each
(a) 64i'3 = ^
'64 = 4
(b) 1001'^ = ^
/lOO = 10
(c) 256i'4 =
(d) (256)1/4
= V^256 = ■
4
= ^256 is
not a real number because the radicand, —256,
is negative and the index
is even.
(e) (32)1/5 =
^V^^=
2
/i\l/3
\8/
3/i_i
V8~2
OBJECTIVES
Use exponential notation
for nth roots.
Define a""^".
Convert between radicals
and rational exponents.
Use the rules for
exponents with rational
exponents.
Evaluate each exponential.
(a) 8^/3
(b)9
1/2
(c) 811/4
(d) (16)
1/4
(e) 64
1/3
CAUTION
Notice the difference between parts (c) and (d) in Example 1 . The radi
cal in part (c) is the negative fourth root of a positive number, while the
radical in part (d) is the principal fourth root of a negative number^
which is not a real number.
(f)
32
1/5
Work Problem 1 at the Side.
Answers
1. (a) 2 (b) 3 (c) 3
(d) not a real number
(e) 4 (f)
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
490 Chapter 9 Roots, Radicals, and Root Functions
O Evaluate each exponential.
(a) 642«
OBJECTIVE Q Define a™'". We now define a number like 8^'^. For
past rules of exponents to be valid,
82/3 = 8(1/3)2 = (81/3)2.
Since 8'^^ = ^,
82/3 ^ (^)2 = 2^ = 4.
Generalizing from this example, we define a^^^ as follows.
(b) 100
3/2
^mln
Ifm and n are positive integers with m/n in lowest terms, then
provided thaia^'"' is a real number. If a^^" is not a real number, then a
is not a real number.
^mln
(c) 16
3/4
EXAMPLE 2
Evaluating Exponentials of the Form a'"^"
Evaluate each exponential.
(a) 36^/2 =(361/2)3 =63 = 216
(b) 125^/3 =(1251/3)2= 52=25
(C) 45/2 = (45/2) = _(4 1/2)5 = _(2)5 = _32
(d) (27)2/3 = [(27)i/3]2 = (3)2 = 9
Notice how the — sign is used in parts (c) and (d). In part (c), we first evalu
ate the exponential and then find its negative. In part (d), the  sign is part
of the base, —27.
(e) (— 100)^/2 jg not a real number, since (— 100) ^'^ is not a real number.
Work Problem 2 at the Side.
(d) (16)
3/4
Answers
2. (a) 16 (b) 1000 (c)
(d) not a real number
EXAMPLE 3
Evaluating Exponentials with Negative Rational
Exponents
Evaluate each exponential.
(a) 163/4
By the definition of a negative exponent,
1
16
3/4
3/4'
3/4 _ Ca4/i7^3 _ )3
Since 16^/^ = {V\6y = 2
16
3/4
16
1 1
(b)25
3/2
16^^^ 8'
1 1
25'^" (V25)3 5^ 125
Continued on Next Page
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
(c)
/8\
2/3
1
1
1
1
9
127;*
/
' 8 Y^^
/
3/8'
\2
(^]
2
~ T~
4
\
.27;
V
V27.
m
/
fa
[3)
9
We could also
use
the rule

=
here,
as
follows
\aj
Kb.
/
('
V" 
(ir
v2/3
/
3/27Y
(^^Y
9
[n
) 
u.
/
\
h)
V2y
4
Section 9.2 Rational Exponents 49 I
& Evaluate each exponential.
(a) 363/2
CAUTION
When using the rule in Example 3 (c), we take the reciprocal only of the
base, not the exponent. Also, be careful to distinguish between exponen
tial expressions like — 16^^^ \6~^'^, and —\6~^'^.
16
1/4
2, 16
1/4
, and 16^/4= .
2 2
Work Problem 3 at the Side.
(b)32
4/5
We get an alternative definition of a"^'"^ by using the power rule for
exponents a little differently than in the earlier definition. If all indicated
roots are real numbers, then
SO
{a'^y'r
min
If all indicated roots are real numbers, then
We can now evaluate an expression such as 27 ^^^ in two ways:
272/3 = (271/3)2 = 32 = 9
or 272/3 = (272)1/3 = 7291/3 = 9,
In most cases, it is easier to use (a 1/")^.
This rule can also be expressed with radicals as follows.
Radical Form of a""'"
If all indicated roots are real numbers, then
In words, we can raise to the power and then take the root, or take the
root and then raise to the power.
(c)
5/2
For example,
82/3 =
SO
^64 = 4, and 8^/3
q2/3 _ a3/^ _ {^}/q\2
4,
Answers
1 1 243
3. (a)  (b)  (c) 
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
492 Chapter 9 Roots, Radicals, and Root Functions
O Write each exponential as a
radical. Assume that all
variables represent positive
real numbers. Use the
definition that takes the
root first.
(a) 5
2/3
(b) Ak
3/5
(c) {Ir)
4/3
(d) {m^ + n^)
3\l/3
OBJECTIVE Q Convert between radicals and rational exponents.
Using the definition of rational exponents, we can simplify many problems
involving radicals by converting the radicals to numbers with rational expo
nents. After simplifying, we convert the answer back to radical form.
EXAMPLE 4
Converting between Rational Exponents and
Radicals
Write each exponential as a radical. Assume that all variables represent posi
tive real numbers. Use the definition that takes the root first.
(a) 13"^ =
(c) 9m^'^
(e) r
2/3 _
13
1
„2/3 ~
(b) e'* =
(d) 6x^'^
(^)^
(f) {a^ + by^ = Va^ + b^ Note that Va^ + ^^ ^ ^ + b.
In (g)(i), write each radical as an exponential. Simplify. Assume that all
variables represent positive real numbers.
(g)
(i)
10
^z"
10"^ (h) ^^3*
z, since z is positive.
>8/4
Work Problems 4 and 5 at the Side.
& Write each radical as an
exponential and simplify.
Assume that all variables
represent positive real
numbers
To
(a) Vj
(b) i/Tf_
(c)
Answers
4. (a) (^^5)2 (b) A{^ky
(c) {^rY (d) ^nv' + r?
5. (a) j5 (b) 3j3 (c) t
OBJECTIVE Q Use the rules for exponents with rational exponents.
The definition of rational exponents allows us to apply the rules for expo
nents first introduced in Section 6.1.
Rules for Rational Exponents
Let r and s be rational numbers. For all real numbers a and h for which
the indicated expressions exist:
a'
{ay = «'
EXAMPLE 5
{aby = a'b'
a'
y
a'
1
a
Applying Rules for Rational Exponents
Write with only positive exponents. Assume that all variables represent posi
tive real numbers.
(a) 21/221/4=21/2 +
1/4 .
23/4
5^/^ 1
/Vj\ r ^ ^2/37/3 ^ ^5/3 ^ ^
^ * rin ^ ^ ^5/3
Continued on Next Page
Product rule
Quotient rule
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
Section 9.2 Rational Exponents 493
(c)
(d)
(^1/2^2/3)4
y
_ (X'^2)4(^ 2/3)4
y
Power rule
x^y'''
y'
Power rule
= y■2y%l^\
Quotient rule
= ^2ym
fx'y'\
2/3 (^4) 2/3 (_^ 6) 2/3
" (x2)2^3(_^l/3)2/3
[x'y'^'J
^4/3^2/9
Power rule
= j^8/34/3y4(2/9)
Quotient rule
= _^4y38/9
^38/9
T^^*';^;*;^^ ^f
Definition of negative exponent
© Write with only positive
exponents. Assume that all
variables represent positive
real numbers.
(a) 11^/4. 115/4
73/4
(b):^
(c)
^2/3/.J/3^4
{x"'Y
1/3
The same result is obtained if v^e simplify v^ithin the parentheses first,
leading to (x^j~^^^^)~^^^. Then, apply the pov^er rule. (Shovs^ that the result is
the same.)
(e) m^'\m^'^  m^'^)
^3/4 . ^5/4 _ ^3/4 . ^1/4
m
3/4 + 5/4
m
3/4 + 1/4
Distributive property
Product rule
8/4 4/4
= m^ — m
Do not make the common mistake of multiplying exponents in the first step.
Work Problem 6 at the Side.
CAUTION
Use the rules of exponents in problems like those in Example 5. Do not
convert the expressions to radical form.
EXAMPLE 6
Applying Rules for Rational Exponents
RevsArite all radicals as exponentials, and then apply the rules for rational ex
ponents. Leave ansv^ers in exponential form. Assume that all variables rep
resent positive real numbers.
(a)
\3/ 2
= _^2/3 + 1/4
= ^8/12 + 3/12
.11/12
Convert to rational exponents.
Product rule
Write exponents with a common denominator.
(d)
3 1,4 \l/2
a'b
a'b
2t1/5
(e) fl2/3(^7/3 + ^1/3)
O Simplify using the rules for
rational exponents. Assume
that all variables represent
positive real numbers. Leave
answers in exponential form.
(a) Vm^
(b)
^
(c)
\4/a3,
(b)
(c)
,3/2
,2/3
.3/22/3
.5/6
1/4
(zn
l/4\ 1/2
.1/8
Work Problem 7 at the Side.
Answers
6. (a) 112 or 121 ^^j) _ ^j.) 9x^'^
L21/10
(d) —^ (e) a' + a
7. (a) m^i/io (b) p'^^ (c) x'^'^
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
Section 9.3 Simplifying Radical Expressions 499
9.0 Simplifying Radical Expressions
OBJECTIVE Q Use the product rule for radicals. We now develop
rules for multiplying and dividing radicals that have the same index. For
example, is the product of two wthroot radicals equal to the nth root of the
product of the radicands? For example, are v 36 • 4 and V 36 • V 4 equal?
V36 • 4 = Vi44 = 12
36 • \/4 = 6 • 2 = 12
Notice that in both cases the result is the same. This is an example of the
product rule for radicals.
Product Rule for Radicals
If va and vb are real numbers and w is a natural number, then
Va ' V b = V ab.
In words, the product of two radicals is the radical of the product.
We justify the product rule using the rules for rational exponents. Since
Va = a'^''mdVb = b'^\
X/^. V^ = ^i/«. /,!/«= {ab)
OBJECTIVES
Q Use the product rule for
radicals.
Q Use the quotient rule for
radicals.
Q Simplify radicals.
Q Simplify products and
quotients of radicals with
different indexes.
El Use the Pythagorean
formula.
iil Use the distance formula.
Multiply. Assume that all
variables represent positive
real numbers.
(a) Vs
13
\/n _ \n/
ab.
(b) VTo^ • V3^
CAUTION
Use the product rule only when the radicals have the same indexes.
EXAMPLE 1
Using the Product Rule
Multiply. Assume that all variables represent positive real numbers.
(a) Vs • V? = Vs^ = V35
(b) V2 • Vl9 = V2 • 19 = V38
(c) VlT • Vp = Vilp
(d) V? • VUxyz = Vffxyz
Work Problem 1 at the Side.
(c)
5 11
a v z
& Multiply. Assume that all
variables represent positive
real numbers.
(a) V2 • V?
6/oT:2 . a6/^73
(b) V8r" • V2r
EXAMPLE 2
Using tlie Product Rule
Multiply. Assume that all variables represent positive real numbers.
(a) ^ • ^^12 = ^3 • 12 = ^/36
(b) V8j •
(c) ^/lO^
(d) V 2 • V 2 cannot be simplified using the product rule for radicals,
because the indexes (4 and 5) are different.
Work Problem 2 at the Side.
(c) ^,
yx • ^y&cp
(d) V? • ^
Answers
1. (a) \/65 (b) \/^ (c) ^£
2. (a) N^H (b) ^J^16^ (c) ^lly^x^
(d) cannot be simplified using the product rule
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
500 Chapter 9 Roots, Radicals, and Root Functions
& Simplify. Assume that all
variables represent positive
real numbers.
(a)
100
81
OBJECTIVE Q Use the quotient rule for radicals. The quotient rule
for radicals is similar to the product rule.
Quotient Rule for Radicals
If va and vb are real numbers, b ^ 0, and w is a natural number, then
a
n — ^
In words, the radical of a quotient is the quotient of the radicals.
(b)
(c)
n
25
18
125
EXAMPLE 3
Using the Quotient Rule
Simplify. Assume that all variables represent positive real numbers.
(a)
/l6 _ Vl6 _ 4
(b)
(c)
V25 V25 5
3/ 8 3/8 V8
V 125 V125 ^3/1^
2
5
(d)
V216 V^ 6
(e)
V 32 ^^32 2
(f)
i
36
2
5
36
125
\3/ 6
vm
125
5
Work Problem 3 at the Side.
(e)
<* ^/li
v/ 12
V r
m]
Videol
Answers
10
3. (a) — (b) ^
(c)
(d)
(e)
OBJECTIVE Q Simplify radicals. We use the product and quotient
rules to simplify radicals. A radical is simplified if the following four
conditions are met.
Conditions for a Simplified Radical
1. The radicand has no factor raised to a power greater than or equal to
the index.
2. The radicand has no fractions.
3. No denominator has a radical.
4. Exponents in the radicand and the index of the radical have no common
factor (except 1).
EXAMPLE 4
Simplify.
Simplifying Roots of Numbers
(a) V24
Check to see whether 24 is divisible by a perfect square (the square of
a natural number) such as 4, 9, ... . Choose the largest perfect square that
divides into 24. The largest such number is 4. Write 24 as the product of
4 and 6, and then use the product rule.
/24 = V4^ = V4 • V6 = 2V6
Continued on Next Page
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
Section 9.3 Simplifying Radical Expressions 50 1
(b) V108
The number 108 is divisible by the perfect square 36: V 108
If this is not obvious, try factoring 108 into its prime factors.
V36 • 3.
108
= Vl^ •3^3
= 2 • 3 • V3
= 6V3
Product rule
(c) VIO
No perfect square (other than 1) divides into 10, so
simplified further.
10 cannot be
(d) Vl6
Look for the largest perfect cube that divides into 16. The number 8 sat
isfies this condition, so write 16 as 8 • 2 (or factor 16 into prime factors).
i/u> = VS^ = ^ • ^ = 2^
(e) ^/l62
 ^^81 • 2
■3^2
8 1 is a perfect 4th power.
Product rule
O Simplify.
(a) V32
(b) V45
(c) V300
CAUTION
Be careful with which factors belong outside the radical sign and which
belong inside. Note in Example 4(b) how 2 • 3 is written outside
because v 2^ = 2 and V 3^ = 3. The remaining 3 is left inside the radical.
(d) V35
Work Problem 4 at the Side.
EXAMPLE 5
Simplifying Radicals InvolvingVariables
Simplify. Assume that all variables represent positive real numbers.
(a) V\6m^ = VUm^ • m
= Viw
= 4m vm
No absolute value bars are needed around the m in color because of the
assumption that all the variables vQpvQSQnt positive real numbers.
(b) VlOOkY = VlO' • 2 • (kY • k • iq'y Factor.
= lOPq^Vlk
8x^3;^ is the largest perfect cube that
Remove perfect square
factors.
(c) ^^8?^ = V(8xV') Uj')
= IxyVxy^
\V 2
Vxy
divides 8x 7^.
(d) ^3V= V(16/)(2;.)
= ^16?^
16j^ is the largest 4th power that
divides 32y^.
2y'^2y
(e)
'54
(f) V243
Answers
4. (a) 4V2 (b) sVs (c) loVs
(d) cannot be simplified further
(e) SN^ (f) 3V3
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
502 Chapter 9 Roots, Radicals, and Root Functions
& Simplify. Assume that all
variables represent positive
real numbers.
(a) V257
NOTE
From Example 5 we see that if a variable is raised to a power with an
exponent divisible by 2, it is a perfect square. If it is raised to a power with
an exponent divisible by 3, it is a perfect cube. In general, if it is raised to
a power with an exponent divisible by n, it is a perfect nth power.
Work Problem 5 at the Side.
(b) Vllyh
The conditions for a simplified radical given earlier state that an expo
nent in the radicand and the index of the radical should have no common fac
tor (except 1). The next example shows how to simplify radicals with such
common factors.
(c) #
y'x'z'
(d) ^32a'b'
Video
EXAMPLE 6
Simplifying Radicals by Using Smaller Indexes
Simplify. Assume that all variables represent positive real numbers.
(a) V¥
We can write this radical using rational exponents and then write the
exponent in lowest terms. We then express the answer as a radical.
Vs^ = 56/9 = 52/3 = ^ or V25
^^^ = vp (Recall the assumption thatp > 0.)
(b)
aV 2 2/4
vp = p = p
& Simplify. Assume that all
variables represent positive
real numbers.
These examples suggest the following rule.
If w is an integer, n and k are natural numbers, and all indicated roots
exist, then
kn^
^km ^ ^^n
Work Problem 6 at the Side.
(b) ^
OBJECTIVE Q Simplify products and quotients of radicals with dif
ferent indexes. Since the product and quotient rules for radicals apply
only when they have the same index, we multiply and divide radicals with
different indexes by using rational exponents.
O Simplify Vs • ^.
Answers
5. (a) Sp^Vp (b) eyVlix (c) yhz^^yx^
(d) labVlab^
6. (a) ^ (b) ^t
7. '^J^2000
EXAMPLE 7
Multiplying Radicals with Different Indexes
Simplify V? • V2.
Because the different indexes, 2 and 3, have a least common index of 6,
use rational exponents to write each radical as a sixth root.
Therefore,
V~7
^2
■J 112
21/3
y3/6
22/6
'343
Vv • V2 = ^^^•^^ = Viyji.
Product rule
Work Problem 7 at the Side.
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
Section 9.3 Simplifying Radical Expressions 503
OBJECTIVE Q Use the Pythagorean formula. The Pythagorean
formula relates the lengths of the three sides of a right triangle.
Pythagorean Formula
If c is the length of the longest side of a right triangle and a and b are
the lengths of the shorter sides, then
Hypotenuse
c2 = «2 + ^2
® Find the length of the
unknown side in each
triangle.
(a)
The longest side is the hypotenuse and the two shorter sides are the
legs of the triangle. The hypotenuse is the side opposite the right angle.
In Section 10.1 we will see that an equation such as x^ = 7 has two
solutions: v 7 (the principal, or positive, square root of 7) and — v 7. Simi
larly, c^ = 52 has two solutions, ± v 52 = ±2 V 13. In applications we often
choose only the positive square root, as seen in the examples that follow.
EXAMPLE 8
Using the Pythagorean Formula
Use the Pythagorean formula to find the length
of the hypotenuse in the triangle in Figure 6.
To find the length of the hypotenuse c,
let a = 4 and b = 6. Then, use the formula.
.2 =
.2
2 + ^2
4^ + 62
52
52
Let a = A and b = 6.
13
c = V4 • 13
c = V4
c = 2V13
The length of the hypotenuse is 2 v 13.
Choose the principal root.
Factor.
Product rule
Work Problem 8 at the Side.
OBJECTIVE Q Use the distance formula. An important result in
algebra is derived by using the Pythagorean formula. The distance formula
allows us to find the distance between two points in the coordinate plane, or
the length of the line segment joining those two points. Figure 7 shows the
points (3, 4) and (5, 3). The vertical line through (5, 3) and the hori
zontal line through (3, 4) intersect at the point (5, 4). Thus, the point
(5, 4) becomes the vertex of the right angle in a right triangle. By the
Pythagorean formula, the square of the length of the hypotenuse, d, of the
right triangle in Figure 7 is equal to the sum of the squares of the lengths of
the two legs a and b\
(b)
(Hint: Write the Pythagorean
formula els b^ = c^ — a^ here.)
i
(5, 3)
y
a
\
Q.L..:...:....i....:...i....; '
4sn (3, 4).;
(5, '
xy h
Figure 7
Answers
8. (a) 2V65 (b) iVs
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
504 Chapter 9 Roots, Radicals, and Root Functions
(5, 4)
Figure 7 (repeat)
'/
(^2' 3^2)
A
/
_1
(xi,yi) a
(^2' y\)
Figure 8
& Find the distance between
each pair of points.
(a) (2,1) and (5, 3)
The length a is the difference between the jcoordinates of the end
points. Since the xcoordinate of both points is —5, the side is vertical, and
we can find a by finding the difference between the jcoordinates. We sub
tract —4 from 3 to get a positive value for a.
a = 3 (4) = 7
Similarly, we find b by subtracting 5 from 3.
b = 3 (5) = 8
Substituting these values into the formula, we have
J2 = 72 + 82 Let ^ = 7 and Z? = 8 in d^ = a^ ^ b\
d^ = 49 + 64
d^= 113
d = VTT3.
We choose the principal root since distance cannot be negative. Therefore,
the distance between (—5, 3) and (3, —4) is VI 13.
This result can be generalized. Figure 8 shows the two points (x^ y^) and
(^2, 72) Notice that the distance between (x^ y^) and (x^, y^) is given by
and the distance between (x^, y^) and (x^, y^) is given by
^=3^23^1
From the Pythagorean formula,
d^ = a^ + b^
= (X2  x^y + (^2  y^Y'
Choosing the principal square root gives the distance formula.
Distance Formula
The distance between the points (x^ y^) and (X2, J2) i^
d = Vfe  x^y + (y,  ji)'.
imatJoiir TouTry^l
Video
(b) (3, 2) and (0,4)
EXAMPLE 9
Using the Distance Formula
Find the distance between (3, 5) and (6, 4).
When using the distance formula to find the distance between two
points, designating the points as (Xj, y^) and (Xj, >'2) i^ arbitrary. Let us
choose (Xpjj) = (—3, 5) and (x2,y2) = (6, 4).
J = V(x2xi)'+ (yiyiV
= V(6 ( 3))2+ (45)2 x, = 6,y2 = 4,x^ = 3,y^ = 5
= V9'+ (1)2
Hi
Leave in radical form.
Work Problem 9 at the Side.
Answers
9. (a) 5 (b) V45or3V5
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
Section 9.4 Adding and Subtracting Radical Expressions 5 I I
9 #4 Adding and Subtracting Radical Expressions
The examples in the preceding section discussed simplifying radical expres
sions that involve multiplication and division. Now we show how to simplify
radical expressions that involve addition and subtraction.
OBJECTIVE Q Simplify radical expressions involving addition and
subtraction. An expression such as 4 v 2 + 3V2 can be simplified by
using the distributive property.
4V2 + 3V2 = (4 + 3) V2 = 7V2
As another example, 2 V 3 — 5 v 3 = (2 — 5) v 3 = — 3 v 3. This is simi
lar to simplifying 2x + 3x to 5x or 5y — 8j to — 3j.
OBJECTIVES
Simplify radical
expressions involving
addition and subtraction.
Add or subtract to simplify
each radical expression.
(a) 3'\/5 + 7V5
CAUTION
Only radical expressions with the same index and the same radicand
may be combined. Expressions such as 5 V 3 + 2 v 2 or 3 V3 + 2 v 3
cannot be simplified by combining terms.
(b) 2V11  Vll + 3V44
^
EXAMPLE 1
Adding and Subtracting Radicals
Add or subtract to simplify each radical expression.
(a) 3V24 + V54
Begin by simplifying each radical; then use the distributive property to
combine terms.
Product rule
5x
3V24 + V54 = 3\/4 • V6 + V9 • V6
= 3 • 2 V6 + 3V6
= 6V6 + 3 V6
= 9V6
(b) 2V20^  Va5x = 2 V4 • Vsx  V9
= 2 • iVSx  sVSx
= 4 V 5x — 3 V 5x
= VSx, X >
(c) 2V^  4V5
Here the radicals differ and are already simplified, so 2 V 3 — 4 V 5
cannot be simplified further.
Work Problem 1 at the Side.
Combine terms.
Product rule
Combine terms.
(c) 5V123; + 6V753;, y^O
(d) 3 Vs  eVso + 2V200
(e) 9 Vs  4VT0
CAUTION
Do not confuse the product rule with combining like terms. The root of
a sum does not equal the sum of the roots. For example,
V9 + 16 # V9 + Vie, since
V9 + 16 = V25 = 5, but V9 + V16 = 3 + 4 = 7.
Answers
1. (a) loVs (b) vVll
(c) 40\/3^ (d) 4V2
(e) cannot be simplified further
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
5 I 2 Chapter 9 Roots, Radicals, and Root Functions
& Add or subtract to simplify
each radical expression.
Assume that all variables
represent positive real
numbers.
(a) 7^/81 + 3^24
(b) 2^32  7^/162
EXAMPLE 2
Adding and Subtracting Radicals with Greater
Indexes
Factor.
Product rule
Add or subtract to simplify each radical expression. Assume that all vari
ables represent positive real numbers.
(a) 2^16  5^54 = 2^8^  5^27 • 2
= 2^ • V2  5 V27 • V2
= 22V253V2
= 4V2  I5V2
= I1V2
(b) 2^^ + VSxV' = 2^^ + V(8xy)xV
= 2 V x^j + 2xy V x^j
= (2 + 2xy)vx^y
Combine terms.
Factor.
Product rule
Distributive property
CAUTION
Remember to write the index when working with cube roots, fourth
roots, and so on.
Work Problem 2 at the Side.
(c) i/^'  i/6Apq
EXAMPLE 3
Adding and Subtracting Radicals with Fractions
O Add. Assume that all
variables represent positive
real numbers.
+
Answers
2. (a) 21^ (b) 25^
(c) (m'4)^
4j Vs + 9
Perform the indicated operations. Assume that all variables represent posi
tive real numbers.
,75
(b) 10 ^ij,  3
V8
V25 3 V4 • 2
= 7 + 4
Quotient rule
V32
VI6 VI6 • 2
 ^ f ^^^ 1 4 f ^^^
V 4 / V4V^/
Product rule
5V^ ,
\fy
Multiply; ^= 1.
5V3 4
1
Write with a common
 2 ' 2
denominator.
5V3 + 4
2
^5 ^
^/T
/
 10 V 3 ^
Quotient rule
* Vx"
I0V5 3^
x^ x^
lOx^ 3A^
Write with a common
x^ x^
denominator
10x^^5  3^
Work Problem 3 at the Side.
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
Section 9.5 Multiplying and Dividing Radical Expressions 5 I 5
9«5 Multiplying and Dividing Radical Expressions
OBJECTIVE Q Multiply radical expressions. We multiply binomial
expressions involving radicals by using the FOIL (First, Outer, Inner, Last)
method from Section 6.4. For example, we find the product of the binomials
Vs + 3 and Ve + 1 as follows.
First
Outer
Inner Last
(Vs + 3) (V6 +1) =V5V6+V5l + 3V6 + 3l
= V30 + Vs + 3 V6 + 3
This result cannot be simplified further.
TouTf^l
EXAMPLE 1
Multiplying Binomials Involving Radical Expressions
Multiply using FOIL.
F O I L
(a) (7  Vs) (V5 + V2) = vVs + iVl  Vs • Vs  Vs • V2
= 7V5 + 7V2  Vl5  V6
(b) (Vio + Vs) (VTo  Vs)
= Vio • Vio  Vio • Vs + Vio • Vs  Vs • Vs
= 103
= 7
Notice that this is the kind of product that results in the difference of squares:
(x + y)(x — y) = x^ — y^.
Here, x = VlO and y = Vs.
(c) (V7  3)2 = (V7  3) (V7  3)
= V7 • V7  3V7  3V7 + 33
= 7  6V7 + 9
= 16  6V7
(d) (5  ^^3) (5 + a/s) = 5 • 5 + 5^/3  5^  ^^3 • ^^3
= 25  ^^3^
= 25  ^
(e) {V~k +V^){V~kV^) = (V~ky  (V^y
= ky, A:>Oandy>0
OBJECTIVES
Q Multiply radical
expressions.
Q Rationalize denominators
with one radical term.
Q Rationalize denominators
with binomials involving
radicals.
Q Write radical quotients in
lowest terms.
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
5 I 6 Chapter 9 Roots, Radicals, and Root Functions
Multiply using FOIL.
(a) (2 + VS) (1 + Vs)
(b)
(2V3 + V5)(V6 3V5)
NOTE
In Example 1 (c) we could have used the formula for the square of a
binomial,
(x  yy = x^  Ixy + y^,
to get the same result.
(Vv  3)2 = (V7)2  2(V7)(3) + 32
= 7  6V7 + 9
= 16  6V7
IK
Work Problem 1 at the Side.
(c) (4 + Vs) (4  V3)
(d) (V6  Vs)'
(e) (4 + ^) (4  ^)
(f ) (V^ + V2) (V^  V2)
V3
Answers
1. (a) 2 + iVs
(b) 6V2  6V15
(c) 13 (d) 112
15
3015
30
(e) 16
M9 (f)j92,j9>0
OBJECTIVE Q Rationalize denominators with one radical term. As
defined earlier, a simplified radical expression will have no radical in the
denominator. The origin of this agreement no doubt occurred before the
days of highspeed calculation, when computation was a tedious process
performed by hand. To see this, consider the radical expression ^. To find
a decimal approximation by hand, it would be necessary to divide 1 by a
decimal approximation for v 2, such as 1 .414. It would be much easier if the
divisor were a whole number. This can be accomplished by multiplying rm
by 1 in the form
^1
1 V2 V2
V2 V2
Now the computation would require dividing 1.414 by 2 to obtain .707, a
much easier task.
With current technology, either form of this fraction can be approxi
mated with the same number of keystrokes. See Figure 9, which shows how
a calculator gives the same approximation for both forms of the expression.
.7071067312
■J"(2>^2
.7071067312
Figure 9
A common way of "standardizing" the form of a radical expression is to
have the denominator contain no radicals. The process of removing radicals
from a denominator so that the denominator contains only rational numbers
is called rationalizing the denominator.
EXAMPLE 2
Rationalizing Denonninators witli Square Roots
Rationalize each denominator.
3
(a)
V^
Multiply the numerator and denominator by V 7. This is, in effect,
multiplying by 1 .
Continued on Next Page
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
Section 9.5 Multiplying and Dividing Radical Expressions 5 I 7
3 • V?
V? Vi Vi
In the denominator, V? • Vv = Vv • 7 = V49 = 7, so
3 3V7
V7" 7
The denominator is now a rational number.
5V2 sVzVs 5V10
(b)
(c)
6
10
12
Less work is involved if the radical in the denominator is simplified first.
6 6 6 3
V12 V4^ 2V^ V^
Now rationalize the denominator by multiplying the numerator and denomi
nator by V 3.
3 3 • V^ 3V3
V3 V3 • V3
V~3
& Rationalize each
denominator.
(a)
V3
(b)
V3
Work Problem 2 at the Side.
EXAMPLE 3
Rationalizing Denominators in Roots of Fractions
Simplify each radical. Assume that all variables represent positive real
numbers.
(a)
18
125
U
V125
V25 • 5
3V2
5V5
3V2 • Vs
5V5 • Vs
3V10
5 • 5
_ 3V10
25
Continued on Next Page
Quotient rule
Factor.
Product rule
Multiply by ^.
Product rule
(c)
48
(d)
16
Answers
, , , 8V3 ^^^ V2T
2. (a) ^ (b) ^
(c) — (d) 2V2
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
5 I 8 Chapter 9 Roots, Radicals, and Root Functions
& Simplify. Assume that all
variables represent positive
real numbers.
(a)
(b)
(c)
200k^
O Simplify.
«^i
(b),7— , n^O
(b)
50m^
p>
SOm^
VsOm'
V?
p'Vp
Vp
p'Vp'
Sm^Vlp
Vp
P'P
Sm^Vlp
Quotient rule
Product rule
Multiply by ^.
Product rule
EXAMPLE 4
Work Problem 3 at the Side.
Rationalizing Denominators with Cube Roots
Use the quotient rule and simplify the numerator and denominator.
Vie"
^^27 3
3
2^
^16 ^ ^
To get a rational denominator, multiply the numerator and denominator by
a number that will result in a perfect cube in the radicand in the denominator.
Since 2 • 4 = 8, a perfect cube, multiply the numerator and denominator
by^.
^
/27 3
16 ~ 2^ ~
3^ 3^
2^ • ^4 2^
3A^
2 • 2
3^
4
«#=
x> 0,
z>0
(C) .7^, j;>0,W^O
3.
(a)
2\/l0
15
(b)
6\/ly
y
(c)
lOk^Vly
y'
4.
(a)
4
(b)
m'^^
(c)
^6yw^
n
w
CAUTION
In problems like the one in Example 4(a), a typical error is to multiply
the numerator and denominator by v 2, forgetting that
V2 • V2 9^ 2.
We need three factors of 2 to get 2^ under the radical. As implied in
Example 4(a),
^2 • V2 • V2 = 2.
Work Problem 4 at the Side.
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
Section 9.5 Multiplying and Dividing Radical Expressions 5 I 9
OBJECTIVE Q Rationalize denominators with binomials involving
radicals. Recall the special product
(x ^ y)(x — y) = x^
y
To rationalize a denominator that contains a binomial expression (one that
contains exactly two terms) involving radicals, such as
1 + Vi
we must use conjugates. The conjugate of 1 + V 2 is 1 — v 2. In general,
X \ y and x — j are conjugates.
Rationalizing a Binomial Denominator
If a radical expression has a sum or difference with square root radicals
in the denominator, rationalize the denominator by multiplying both the
numerator and denominator by the conjugate of the denominator.
For the expression
we rationalize the denominator by multi
1 + Vi
plying both the numerator and denominator by 1 — v 2, the conjugate of the
denominator.
3 3(1  V2)
1 + V2 (1 + V2)(l  Vi)
3(1 V2) ^^^^mVi)
1
= 1  2 = 1
1
(IV2)
= 3(1  V2) or 3 + 3V2
EXAMPLE 5
Rationalizing Binomial Denominators
Rationalize each denominator.
(a) ^
4  V 3
To rationalize the denominator, multiply both the numerator and
denominator by the conjugate of the denominator, 4 + v 3.
5 5(4 + V3)
4  V3 (4  V3) (4 + V^)
_ 5(4 + V3)
163
_ 5(4 + V3)
13
Notice that the numerator is left in factored form. This makes it easier to
determine whether the expression is written in lowest terms.
Continued on Next Page
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
520 Chapter 9 Roots, Radicals, and Root Functions
V2  V3 (V2  V3)(V5  V^) Multiply the numerator and
& Rationalize each
denominator.
4
(a)
VS + 2
(b)
15
V? + V2
(b)
Vs + V3 (Vs + V3)(V5  V3) denominator by V5  V3.
10  \/6  Vis + 3
53
Vio  Ve VTs + 3
(c)
3 ( V5m + Vp)
5m — vp (V 5m — V/?)( V 5w + vp
3 ( V5w + Vp)
5m — p
; 5m i= p, m > 0, p >
Work Problem 5 at the Side.
(c)
V3 + V5
V2  V?
(d)
Vk + Vz
ki=z,k>0,z>0
& Write each quotient in lowest
terms.
(a)
15  5V3
(b)
24  36V7
16
OBJECTIVE Q Write radical quotients in lowest terms.
EXAMPLE 6
Writing Radical Quotients in Lowest Terms
Write each quotient in lowest terms.
6 + 2V5
(a)
Factor the numerator and denominator, then write in lowest terms.
6 + 2V5 _ 2(3 + Vs) _ 3 + Vs
4 ~ 22 ~ 2
Here is an alternative method for writing this expression in lowest terms.
6 + 2V5 _ 6 2V5 _ 3 Vs _ 3 + Vs
4 ~4^4~2^2~ 2
(b)
57
6j
^^ 5y  ly Vl
6y
J (5 iVl)
6j
5  2V2
, 7 > Product rule
Factor the numerator.
Lowest terms
Note that the final fraction cannot be simplified further because there is no
common factor of 2 in the numerator.
Answers
5. (a) 4 (Vs  2) (b) 3 (V7  Vl
 (\/6 + VlT + VlO + \/35)
(c) 1
(d)
2{Vk Vz)
k z
6. (a) 3  V3 (b)
6  9V1
CAUTION
Be careful to factor before writing a quotient in lowest terms.
Work Problem 6 at the Side.
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
Section 9.6 Solving Equations with Radicals 527
9«6 Solving Equations with Radicals
An equation that includes one or more radical expressions with a variable is
called a radical equation. Some examples of radical equations are
Vx
VSx + 12 = 3V2x  1, and ^6 + x = 27.
OBJECTIVE Q Solve radical equations using the power rule. The
equation x = 1 has only one solution. Its solution set is {1}. If we square
both sides of this equation, we get x^ = I. This new equation has two
solutions: — 1 and 1. Notice that the solution of the original equation is also
a solution of the squared equation. However, the squared equation has
another solution, — 1, that is not a solution of the original equation. When
solving equations with radicals, we use this idea of raising both sides to a
power. It is an application of the power rule.
Power Rule for Solving Equations with Radicals
If both sides of an equation are raised to the same power, all solutions of
the original equation are also solutions of the new equation.
OBJECTIVES
Solve radical equations
using the power rule.
Solve radical equations
that require additional
steps.
Solve radical equations
with indexes greater
than 2.
Solve.
(a)
Read the power rule carefully; it does not say that all solutions of the
new equation are solutions of the original equation. They may or may not be.
Solutions that do not satisfy the original equation are called extraneous
solutions; they must be discarded.
CAUTION
When the power rule is used to solve an equation, every solution of the
new equation must be checked in the original equation.
Video
EXAMPLE 1
Using the Power Rule
Solve VSx + 4 = 8.
Use the power rule and square both sides to get
(VSx + 4)' = 8'
3x + 4 = 64
3x = 60 Subtract 4.
X = 20. Divide by 3.
To check, substitute the potential solution in the original equation.
VBx + 4 = 8
VS • 20 + 4 = 8
^=8
Letx = 20.
True
Since 20 satisfies the original equation, the solution set is {20}.
Work Problem 1 at the Side.
(b) VSx + 1 = 4
The solution of the equation in Example 1 can be generalized to give a
method for solving equations with radicals.
Answers
1. (a) {9} (b) {3}
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
528 Chapter 9 Roots, Radicals, and Root Functions
& Solve.
(a) Vx + 4 = 3
Solving an Equation with Radicals
Step 1 Isolate the radical. Make sure that one radical term is alone
on one side of the equation.
Step 2 Apply the power rule. Raise both sides of the equation to a
power that is the same as the index of the radical.
Step 3 Solve. Solve the resulting equation; if it still contains a radical,
repeat Steps 1 and 2.
Step 4 Check all potential solutions in the original equation.
CAUTION
Remember Step 4 or you may get an incorrect solution set.
EXAMPLE 2
(b) Vx 9 3 =
Using the Power Rule
Solve VSx  1 + 3 = 0.
Step 1 To isolate the radical on one side, subtract 3 from each side.
= 3
Vsx
 1
Step 2
Now square both sides.
T)^
= (3)^
(VSx
Step 3
5x
 1
5x
X
= 9
= 10
= 2
Step 4
Check the potential solutio
equation.
n,2,
=
by substituting
it in the original
VSx 1 + 3
V5 • 2  1 + 3
=
? Letx =
2.
3 + 3
=
False
This false result shows that 2 is not a solution of the original equation; it is
exfraneous. The solution set is 0.
Answers
2. (a) (b) {18
NOTE
We could have determined after Step 1 that the equation in Example 2
has no solution because the expression on the left cannot be negative.
Work Problem 2 at the Side.
OBJECTIVE Q Solve radical equations that require additional steps.
The next examples involve finding the square of a binomial. Recall that
(x + yY = x^ + 2xy + y^.
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
Section 9.6 Solving Equations with Radicals 529
EXAMPLE 3
Using the Power Rule; Squaring a Binomial
Solve V4  X = X + 2.
Step 1 The radical is alone on the left side of the equation.
Step 2 Square both sides; on the right, (x + 2)^ = x^ + 2(x)(2) + 2^.
(V4  xY = (x + lY
4x = x2 + 4x + 4
t
' Twice the product of 2 and x
Step 3 The new equation is quadratic, so get on one side.
= x2 + 5x
Subtract 4 and add x.
= x(x + 5) Factor.
x = or x + 5 = Zerofactor property
x= 5
Step 4 Check each potential solution in the original equation.
If X = 0, then
Ifx= 5, then
V4  X = X + 2
V4  X = X + 2
V4  (5) = 5 + 2
V4  = + 2 ?
?
V^ = 2 ?
V9 = 3
?
2 = 2. True
3= 3.
False
The solution set is {0}. The other potential solution, —5, is extraneous.
O Solve.
(a) V 3x  5 = X
(b) X + 1 = V2x  2
CAUTION
When a radical equation requires squaring a binomial as in Example 3,
remember to include the middle term.
(x + 2)2 = x^ + 4x + 4
EXAMPLE 4
Work Problem 3 at the Side.
Using the Power Rule; Squaring a Binomial
Solve Vx^  4x + 9 = x  1.
Square both sides; (x  1)^ = x^  2(x) (1) + P on the right.
(Vx'  4x + 9)2 = (x  1)2
x2  4x + 9 = x2  2x + 1
' Twice the product of x and — 1
— 2x = — 8 Subtract x^ and 9; add 2x.
X = 4 Divide by 2.
Check:
Vx2  4x + 9 =x  1
V42 44 + 9 = 41
3 = 3
The solution set of the original equation is {4}.
? Letx = 4.
True
Work Problem 4 at the Side.
O Solve.
V4x2 + 2x  3 = 2x + 7
Answers
3. (a) {2,3} (b) {1}
4. {2}
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
530 Chapter 9 Roots, Radicals, and Root Functions
& (a) Verify that 15 is not a
solution of the equation in
Example 5.
(b) Solve.
Vx + 1  Vx
& Solve each equation.
(a) i/x^ + 3x + 12
EXAMPLE 5
Using the Power Rule; SquaringTwice
Solve VSx + 6 + VSx + 4 = 2.
Start by isolating one radical on one side of the equation by subtracting
V 3x + 4 from each side. Then square both sides.
VSx + 6 = 2  VSx + 4
(VSx + 6)' = (2  VSx + 4)'
5x + 6 = 4  4V3x + 4 + (3x + 4)
I Twice the product of 2 and V3x + 4
This equation still contains a radical, so square both sides again. Before
doing this, isolate the radical term on the right.
5x + 6 =
8 + 3x  4V3j
c + 4
2x 2 =
4V3x + 4
Subtract 8 and 3x.
X  I =
2V3x + 4
Divide by 2.
ix  ly =
(2V3x + 4)2
Square both sides again
x^  2x + 1 =
(2)2(V3x +
4)^
(abf = a^b^
x2  2x + 1 =
4(3x + 4)
x2  2x + 1 =
12x + 16
Distributive property
x2  14x  15 =
Standard form
(x+l)(x15) =
Factor
X + 1 = or X
15 =
Zerofactor property
X = — 1 or
x= 15
Check each of these potential solutions in the original equation. Only 1
satisfies the equation, so the solution set, { — 1 }, has only one element.
Hi
Work Problem 5 at the Side.
(b) ^2x + 5 +1 =
Answers
5. (a) The final step in the check leads to
16 = 2, which is false.
(b) {8}
6. (a) {4} (b)
OBJECTIVE Q Solve radical equations with indexes greater than 2.
The power rule also works for powers greater than 2.
EXAMPLE 6
Using the Power Rule for a Power Greater than 2
Solve Vx + 5 = ^Ix  6.
Raise both sides to the third power.
(Vx + 5)^ = (V2x  ey
X + 5 = 2x — 6
11 =x
Check this result in the original equation.
Vx+ 5 = ^Ix 6
Vll + 5 = V2 • 11  6
Vl6 = VT6
Subtracts; add 6.
Letx = 11.
True
The solution set is {11}.
Mi
Work Problem 6 at the Side.
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Section 9.7 Complex Numbers 535
9.7 Complex Numbers
As we saw in Section 1.1, the set of real numbers includes many other num
ber sets (the rational numbers, integers, and natural numbers, for example).
In this section a new set of numbers is introduced that includes the set of real
numbers, as well as numbers that are even roots of negative numbers,
like V^.
OBJECTIVE Q Simplify numbers of tfie form v^, wfiere b>0. The
equation x^ + 1 = has no real number solution since any solution must be a
number whose square is —1. In the set of real numbers, all squares are
nonnegative numbers because the product of two positive numbers or two
negative numbers is positive and 0^ = 0. To provide a solution for the equa
tion x^ + 1 = 0, a new number /, the imaginary unit, is defined so that
P= 1.
That is, / is a number whose square is — 1, so / = v — 1. This definition of/
makes it possible to define any square root of a negative number as follows.
Vb
For any positive number b,
OBJECTIVES
Q Simplif y nu mbers of the
form V—b, where b > 0.
Q Recognize subsets of the
complex numbers.
Q Add and subtract
complex numbers.
Q i^ultiply complex
numbers.
El Divide complex numbers.
ill Find powers off.
Write each number as a
product of a real number
and /.
(a) V^
c^ = iVa.
EXAMPLE 1
Simplifying Square Roots of Negative Numbers
Write each number as a product of a real number and /.
(a) V100 = iVm = 10/ (b)  V^ = iV36 = 6/
(c) V^ = /\/2
(d) V^ = V4 • 2 = V^ • V2 = 2i V2
(b) V^
CAUTION
It is easy to mistake v 2/ for v 2/, with the / under the radical. For this
reason, we usually write v 2/ as / v 2, as in the definition of v—b.
Work Problem 1 at the Side.
When finding a product such as v — 4 • V — 9, we cannot use the prod
uct rule for radicals because it applies only to nonnegative radicands. For
this reason, we change v —bio the form / v b before performing any mul
tiplications or divisions. For example.
(d) V^
Answers
1. (a) 4i (b)
9/ (c) i V 7 (d) 4i V 2
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
536 Chapter 9 Roots, Radicals, and Root Functions
O Multiply.
(a) V^ • V^
CAUTION
Using the product rule for radicals before using the definition of"
gives a wrong answer. The preceding example shows that
V^ V^ = 6, but
V4(9) = V36 = 6,
V^ • V^ 56 V4(9).
so
b
(b) V^ • V^
(c) V^\5 ■ Vl
& Divide.
(a) / —
EXAMPLE 2
Multiplying Square Roots of Negative Numbers
Multiply.
(a) V^ • V^ = iV3 • iVl
= PV3^
= (1)V2T Substitute: /2 = 1.
= V2I
(b) V^ • V^ = iVl ■ iVs
= fV2~^
= (l)Vl6
= (1)4
= 4
(c) V^ • Ve = iVs ■ Ve = ? Vso
Hi
Work Problem 2 at the Side.
The methods used to find products also apply to quotients.
(b)
V^
Video
EXAMPLE 3
Dividing Square Roots of Negative Numbers
Divide.
(a)
(b)
V^
v^
/V75 /75
25 = 5
/V32 /32 ,/T ,
— ^ = i\ — = ' v4 = 2/
Vs V 8
Work Problem 3 at the Side.
(c)
V^
10
OBJECTIVE Q Recognize subsets of the complex numbers. With
the imaginary unit / and the real numbers, a new set of numbers can be
formed that includes the real numbers as a subset. The complex numbers are
defined as follows.
Answers
2. (a) 7 (b) 5V2 (c) /V30
3. (a) 4 (b) 3 (c) 2i
Complex Number
If a and b are real numbers, then any number of the form a + bi is
called a complex number.
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
Section 9.7 Complex Numbers 537
In the complex number a + bi, the number a is called the real part and b
is called the imaginary part.'^ When /? = 0, a + /?/ is a real number, so the real
numbers are a subset of the complex numbers. Complex numbers with a = Q
and /? 9^ are called pure imaginary numbers. In spite of their name, these
numbers are very useful in applications, particularly in work with electricity.
The relationships among the sets of numbers are shown in Figure 10.
Complex
numbers
a + bi,
a and b real
Nonreal
complex
numbers
a + bi,
bitO
Irrational
numbers
Real
numbers
a + bi,
b =
/
Integers
<^
y
\
Rational
numbers
y^
<^
^\
Non
integers
^
Figure 10
OBJECTIVE Q Add and subtract complex numbers. The commutative,
associative, and distributive properties for real numbers are also valid for com
plex numbers. Thus, to add complex numbers, we add their real parts and add
their imaginary parts.
EXAMPLE 4
Adding Complex Numbers
Add.
(a) (2 + 3/) + (6 + 40
= (2 + 6) + (3 + 4)i Commutative, associative, and distributive properties
= 8 + 7/
(b) 5 + (9  30 = (5 + 9)  3/
= 14  3i
Work Problem 4 at the Side.
O Add.
(a) (4 + 60 + (3 + 50
(b) (1 + 80 + (9 30
& Subtract.
(a) (7 + 30  (4 + 20
(b) (6   (5  40
EXAMPLE 5
We subtract complex numbers by subtracting their real parts and sub
tracting their imaginary parts.
Subtracting Complex Numbers
Subtract.
(a) (6 + 50  (3 + 20 = (6  3) + (5  2)/
= 3 + 3/
AnimatiooT (b) (7  30  (8  60 = (7  8) + [3  (6)]/
=  1 + 3/
(c) (9 + 40  (9 + 80 = (9 + 9) + (4  8)/
= 04/
= 4/
Work Problem 5 at the Side.
(c) 8  (3  20
*Some texts define bi as the imaginary part of the complex number a + bi.
Answers
4. (a) 1 + 11/ (b) 8 + 5/
5. (a) 3 + / (b) 1 + 3i (c) 5 + 2i
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
538 Chapter 9 Roots, Radicals, and Root Functions
& Multiply.
(a) 6/(4 + 3/)
In Example 5(c), the answer was written as  4/ and then as just 4/.
A complex number written in the form a + bi, like  4/, is in standard
form. In this section, most answers will be given in standard form, but if a
or b is 0, we consider answers such as a or bi to be in standard form.
(b) (6  4/) (2 + 4/)
(c) (3  2/) (3 + 2/)
OBJECTIVE Q Multiply complex numbers. We multiply complex
numbers as we multiply polynomials. Complex numbers of the form a + bi
have the same form as binomials, so we multiply two complex numbers in
standard form by using the FOIL method for multiplying binomials. (Recall
from Section 6.4 that FOIL stands for First, Outer, Inner, Last.)
EXAMPLE 6
Multiplying Complex Numbers
Multiply.
(a) 4/(2 + 30
Use the distributive property.
4/(2 + 3/) = 4/(2) + 4/(3/)
= 8/ + 12/2
= 8/ + 12(1) Substitute: f = \.
= 12 + 8/
(b) (3 + 5/) (4  2/)
Use the FOIL method.
(3 + 5/) (4  2/) = 3(4) + 3 (2/) + 5/(4) + 5/(2/)
First Outer Inner Last
= 12  6/ + 20/  10/2
= 12 + 14/  lO(l) Substitute: f =
= 12 + 14/ + 10
= 22 + 14/
(c) (2 + 3/) (1  5/) = 2(1) + 2(5/) + 3/(1) + 3/(5/) FOIL
= 2 10/ + 3/ 15/2
= 27/ 15(1)
= 2  7/ + 15
= 17  7/
IK
Work Problem 6 at the Side.
Answers
6. (a) 18 + 24/ (b) 28 + 16/ (c) 13
The two complex numbers a + bi and a — bi are called complex conju
gates of each other. The product of a complex number and its conjugate is
always a real number, as shown here.
{a + bi) {a — bi) = a^ — abi + abi — b^i^
= a^ b\l)
(a + bi) (a  bi) = a^ + b^
For example, (3 + 7/) (3  7/) = 3^ + 7^ = 9 + 49 = 58.
OBJECTIVE Q Divide complex numbers. The quotient of two complex
numbers should be a complex number. To write the quotient as a complex
number, we need to eliminate / in the denominator. We use conjugates to
do this.
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
EXAMPLE 7
Dividing Complex Numbers
Find each quotient.
8 + 9/
(a)
^ ^ 5 + 2/
Multiply both the numerator and denominator by the conjugate of the
denominator. The conjugate of 5 + 2/ is 5  2/.
8 + 9/ (8 + 9/) (5  2i)
5 + 2/ (5 + 2/) (5  2i)
40  16/ + 45/  18/'
5  2/ ^
5  2/
1
5^ + 2'
58 + 29/
29
29(2 + /)
29
2 + /
Substitute: f = 1;
combine terms.
Factor the numerator.
Lowest terms
Notice that this is just like rationalizing a denominator. The final result is in
standard form.
1 + /
The conjugate of/ is /. Multiply both the numerator and denominator
by — /.
1 + / (1 + /)(/)
^t
0
i 
e
c
— i —
(1)
(■
1)
/ +
1
Substitute: /2 = 1.
1
1 /
Work Problem 7 at the Side.
Iliil Calculator Tip In Examples 4—7, we showed how complex numbers can
be added, subtracted, multiplied, and divided algebraically. Many current
models of graphing calculators can perform these operations. Figure 1 1
shows how the computations in parts of Examples 4—7 are displayed on a
TI83 Plus or TI84 calculator. Be sure to use parentheses as shown.
<2+3i,> + <6+4i>
S+7i
3+3i
<3+5i><42i,>
22+14i
<S+9iV<5+2i.>
2+i
Figure 11
Section 9.7 Complex Numbers 539
O Find each quotient.
2 + /
(a)
3  /
(b)
8  4/
1  /
(c)
3  2/
(d)
5  /
Answers
7. (a) ^ + ^i (b) 6 + 2i
15 10
(c) — + —i (d) 1  5/
^ M3 13
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
540 Chapter 9 Roots, Radicals, and Root Functions
O Find each power of/.
(a) f'
OBJECTIVE Q Find powers of /_ Because f is defined to be  1, we
can find higher powers of/ as shown in the following examples.
P = //2 = /(!)= /
/4 = 2.2 = (_1)(_1)= 1
/^ = i ' i^ = i ' I = i
f • /4
= /3 . /4 = (/) . 1 =
= /4 . /4 = 1 . 1 = 1
1
As these examples suggest, the powers of/ rotate through the four num
bers /, — 1, — /, and 1. Larger powers of/ can be simplified by using the fact
that i^ = 1. For example,
775
This example suggests a quick method for simplifying larger powers of/.
(b)f
;36
(c) i
;50
>uTry
EXAMPLE 8
Simplifying Powers of i
Find each power of /.
(a) /12 = (/4)3 = 13 = 1
(b) /39 = /36 . p = (,4)9 . p = 19 . (_^) = _i
1 1
(c) /
1
1
(d) /' =
1
To simplify this quotient, muhiply both the numerator and denominator
by —i, the conjugate of/.
i  U0 /
— ?
(1)
— J
T
Work Problem 8 at the Side.
(d) r
Answers
8. (a) i (b) 1 (c) 1 (d) i
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
Quadratic
Equations,
Inequalities,
and Functions
\
Since 1980, the number of multiple births in the United States has
increased 59%, primarily due to greater use of fertility drugs and
greater numbers of births to women over age 40. The number of
higherorder multiple births — that is, births involving triplets or
more — has increased over 400%. One of the most publicized higher
order multiple births occurred November 19, 1997, with the birth of
the McCaughey septuplets in Des Moines, Iowa. All seven premature
babies survived, a first in medical history. {Source: American College
of Obstetricians and Gynecologists; The Gazette, November 19, 2003.)
In Example 6 of Section 10.5, we determine a quadratic func
tion that models the number of higherorder multiple births in the
United States.
TO.T The Square Root Property and
Completing the Square
T0.2 The Quadratic Formula
T0.3 Equations Quadratic in Form
Summary Exercises on Solving
Quadratic Equations
T0.4 Formulas and Further
Applications
T0.5 Graphs of Quadratic Functions
T0.6 More about Parabolas;
Applications
T0.7 Quadratic and Rational
Inequalities
557
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
558 Chapter 10 Quadratic Equations, Inequalities, and Functions
1 0. 1 The Square Root Property and Completing the Square
OBJECTIVES
Learn the square root
property.
Solve quadratic equations
of the form (ox + b)^ = c
by using the square root
property.
Solve quadratic equations
by completing the square.
Solve quadratic equations
with nonreal complex
solutions.
(a) Whichof the following
are quadratic equations?
A. X + 2y =
B. x'
8x + 16 =
C.
D.
2t^  5t = 3
x^ + x^ + 4 =
We introduced quadratic equations in Section 7.4. Recall that a quadratic
equation is defined as follows.
Quadratic Equation
An equation that can be written in the form
ax^ + Ax + c = 0,
where a, b, and c are real numbers, with a ^ 0, is a quadratic equation.
The given form is called standard form.
A quadratic equation is a seconddegree equation, that is, an equation
with a squared term and no terms of higher degree. For example,
4m^ + 4m — 5 = and 3x^ = 4x — 8 Quadratic equations
are quadratic equations, with the first equation in standard form.
«il<
Work Problem 1 at the Side.
In Section 7.4 we used factoring and the zerofactor property to solve
quadratic equations.
(b) Which quadratic equation
identified in part (a) is in
standard form?
& Solve each equation by
factoring.
(a) x^ + 3x + 2 =
ZeroFactor Property
If two numbers have a product of 0, then at least one of the numbers
must be 0. That is, \iab = 0, then a = or b = 0.
We solved a quadratic equation such as 3x^
zerofactor property as follows.
3x2 _ 5^ _ 28 =
(3x + 7) (x  4) =
5x — 28 = using the
3x + 7
3x
7
7
3
or
or
Factor.
4 = Zerofactor property
X = 4 Solve each equation.
The solution set is { 1, 4}.
(b) 3m ^
Sm
(Hint: Remember to write
the equation in standard
form first.)
Work Problem 2 at the Side.
OBJECTIVE Q Learn the square root property. Although factoring is
the simplest way to solve quadratic equations, not every quadratic equation
can be solved easily by factoring. In this section and the next, we develop
other methods of solving quadratic equations based on the following property.
Answers
1. (a) B, C (b) B
2. (a) {2,1} (b)
Square Root Property
If X and k are complex numbers andx^ = k, then
X = vk or X = —vk.
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
Section 10. 1 The Square Root Property and Completing the Square 559
The following steps justify the square root property.
x^  k= Subtract A:,
(x  Vk) (x + Vk) = Factor.
X  vk =0 or X + vk = Zerofactor property
X = vk or X = —vk Solve each equation.
Thus, the solutions of the equation x^ = ^ are x = vk or x = — vk.
& Solve each equation.
(a) m^ = 64
CAUTION
If ^ ^ 0, then using the square root property alv^ays produces two square
roots, one positive and one negative.
EXAMPLE 1
Using the Square Root Property
Solve each equation.
(a) r^ = 5
By the square root property, if r^ = 5, then
r= VS,
r = V 5 or
and the solution set is { V 5, v 5}.
(b) 4x2  48 =
Solve forx^.
4x2  48 =
4x2 = 48
x2= 12
X = Vl2 or X =  Vl2
X = 2V3 or X = 2V3
Check: 4x2  48 =
4(2V3)2  48 = ?
4(12)  48 = ?
48  48 = ?
= Trae
The solution set is {iVd,  iVs}.
1
Add 48.
Divide by 4.
Square root property
Vl2 = V4 • Vs = 2 V3
Original equation
4(2V3)'  48 = ?
4(12)  48 = ?
48  48 = ?
= True
Work Problem 3 at the Side.
(b);;^
(c) 3x2  54 =
NOTE
Recall that solutions such as those in Example 1 are sometimes abbrevi
ated v^ith the symbol ± (read "positive or negative"); w^ith this symbol
the solutions in Example 1 v^ouldbe v^ritten ± V5 and ±2 v 3.
Answers
3. (a) {8,8} (b) {Vv, Vv}
(c) {3\/2, 3\/2}
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
560 Chapter 10 Quadratic Equations, Inequalities, and Functions
O Solve the problem.
An expert marksman can
hold a silver dollar at forehead
level, drop it, draw his gun,
and shoot the coin as it passes
waist level. If the coin falls
about 4 ft, use the formula in
Example 2 to find the time
that elapses between the
dropping of the coin and
the shot.
EXAMPLE 2
Using the Square Root Property in an Application
Galileo Galilei (15641642) developed a formula for freely falling objects
described by
d= I6t\
where d is the distance in feet that an object falls
(disregarding air resistance) in t seconds, regardless
of weight. Galileo dropped objects from the Leaning
Tower of Pisa to develop this formula. If the Leaning
Tower is about 180 ft tall, use Galileo's formula to
determine how long it would take an object dropped
from the tower to fall to the ground. (Source:
Microsoft Encarta Encyclopedia 2002.)
We substitute 180 for d in Galileo's formula.
Mi
■L
«Ja*'
sS
flKUilil.
LJM
Jt
■"/#
iii
"H^i:
h
d= I6t^
180 = I6t^
Vll.25
11.25
or t
t'
Vll.25
LQtd= 180.
Divide by 16.
Square root property
Since time cannot be negative, we discard the negative solution. In applied
problems, we usually prefer approximations to exact values. Using a calculator,
VI 1.25 ~ 3.4 so t ~ 3.4. The object would fall to the ground in about 3.4 sec.
Work Problem 4 at the Side.
OBJECTIVE Q Solve quadratic equations of the form (^ax \ by = c by
using the square root property. To solve more complicated equations
using the square root property, such as
(x  5)2 = 36,
substitute (x — 5)^ for x^ and 36 for k, to get
Check:
X  5 = V36 or
X
 5= V36
X  5 = 6 or
X 5 = 6
X = 11 or
x= 1.
(x  5)2 = 36
Original equation
(11
 5)2 = 36 ?
(1 5)2 = 36
?
62 = 36 ?
(6)2 = 36
?
36 = 36 True
36 = 36
True
Answers
4. .5 sec
EXAMPLE 3
Using the Square Root Property
Solve (2x 3)2= 18.
2x  3 = Vl8 or 2x  3 =  Vl8
2x = 3 + Vl8 or 2x = 3  Vl8
3 + VTs 3  VTs
X = or X =
3 + 3V2
or
Continued on Next Page
Square root property
Adds.
Divide by 2.
vTs = V9 • V2 = 3 V2
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
Section 1 0.1 The Square Root Property and Completing the Square 561
We show the check for the first solution. The check for the second solution
is similar.
Check:
{2x  3)2
3 + 3V2
18
18
(3 + 3V2  3)2 = 18 ?
OVlY = 18 ?
18 = 18
. ,3 + 3V2 3  3V2I
The solution set is s r , r r .
Original equation
True
Work Problem 5 at the Side.
& Solve each equation.
(a) (x  3)2 = 25
(b) (3^ + 1)2 = 2
(c) (2r + 3)2 = 8
OBJECTIVE Q Solve quadratic equations by completing the square.
We can use the square root property to solve any quadratic equation by
writing it in the form (x + ky = n. That is, we must write the left side of the
equation as a perfect square trinomial that can be factored as (x + A:) 2, the
square of a binomial, and the right side must be a constant. Rewriting a
quadratic equation in this form is called completing the square.
Recall that the perfect square trinomial
x2 + lOx + 25
can be factored as (x + 5)2. In the trinomial, the coefficient of x (the first
degree term) is 10 and the constant term is 25. Notice that if we take half of
10 and square it, we get the constant term, 25.
Coefficient of x
(10)
Constant
52 = 25
Similarly, in
and in
x2 + 12x + 36,
m — 6m + 9,
■(12)
(6)
36,
= (3)^ = 9.
This relationship is true in general and is the idea behind completing the
square.
Work Problem 6 at the Side.
& Find the constant to be added
to get a perfect square trino
mial. In each case, take half
the coefficient of the first
degree term and square the
result.
(a) x2 + 4x +
(b) ^2  2^ +
(c) m^ + 5m \
id)x'
X +
EXAMPLE 4
Solving a Quadratic Equation by Completing
the Square
Solve x2 + 8x+ 10 = 0.
This quadratic equation cannot be solved easily by factoring, and it is
not in the correct form to solve using the square root property. To solve it by
completing the square, we need a perfect square trinomial on the left side of
the equation. To get this form, we first subtract 10 from each side.
Continued on Next Page
Answers
5. (a) {2,8}
1 + \/2 1  Vl \
3 ' 3 J
3 + 2\/2 3  iVl
(b)
(c)
6. (a) 4 (b) 1 (c) ^ (d) ^
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
562 Chapter 10 Quadratic Equations, Inequalities, and Functions
O Solve n^ + 6n + 4 = 0hy
completing the square.
x2 + 8x + 10 =
^2 + 8x= 10
Original equation
Subtract 10.
We must add a constant to get a perfect square trinomial on the left.
x^ + 8x +
Needs to be a perfect
square trinomial
To find this constant, we apply the ideas preceding this example — we take
half the coefficient of the firstdegree term and square the result.
"2
(8) = 4^ = 16 ^ Desired constant
Now we add 16 to each side of the equation. (Why?)
x2 + 8x+ 16= 10 + 16
Next we factor on the left side and add on the right.
(x + 4)2 = 6
We can now use the square root property.
x + 4=V6 or x + 4
X = — 4 + V 6 or X
Check: x^ + 8x + 10 =
(4 + Vey + 8(4 + V6) + 10 = ?
16  sVe + 6  32 + sVe + 10 = ?
=
= V6
= 4  V6
Original equation
Letx = 4 \ V6.
True
The check of the other solution is similar. Thus, { — 4 + V 6, —4 — v 6} is
the solution set.
Work Problem 7 at the Side.
The procedure from Example 4 can be generalized.
Answers
7. {3 +V5, 3  Vs}
Completing the Square
To solve ax^ + bx + c = (a ^ 0)hy completing the square, use
these steps.
Step 1 Be sure the squared term has coefficient 1. If the coefficient
of the squared term is some other nonzero number a, divide
each side of the equation by a.
Step 2 Write the equation in correct form so that terms with
variables are on one side of the equals sign and the constant
is on the other side.
Step 3 Square half the coefficient of the firstdegree term.
Step 4 Add the square to each side.
Step 5 Factor the perfect square trinomial. One side should now be
a perfect square trinomial. Factor it as the square of a binomial.
Simplify the other side.
Step 6 Solve the equation. Apply the square root property to complete
the solution.
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
Section 10. 1 The Square Root Property and Completing the Square 563
fa"
Video
EXAMPLE 5
Solving a Quadratic Equation with a = 1 by
Completing the Square
Solve ^2 + 5^  1 = 0.
Since the coefficient of the squared term is 1, begin with Step 2.
Step 2 k^ + Sk = \ Add 1 to each side.
Step 3 Take half the coefficient of the firstdegree term and square the result.
(5)
2.
25
4
Step 4
Step 5
Step 6 k +
k +
25
25
k^ + 5k + — =l+ —
k +
4
29
4
29
5
2
5
2
k
k
2
Check that the solution set is
2
5
 +
2
V29
2
5 +
V29
29
4
or
or
or
or
Add the square to each side
of the equation.
Factor on the left;
add on the right.
k +
k +
29 Square root
— property
29
■5 + V29
2
2
k =
5 '
2
2
k =
5  ^
/29
2
5 
V29\
O Solve each equation by
completing the square.
(a) x2 + 2x  10 =
(b) r2 + 3r  1 =
Work Problem 8 at the Side.
EXAMPLE 6
Solving a Quadratic Equation with a ^ 1 by
Completing the Square
Solve 2x2  4x  5 = 0.
First divide each side of the equation by 2 to get 1 as the coefficient of
the squared term.
2x
2 = ^
Step 1
x'2x = ^
2
Step 2
[i*']
'=(1)2 = 1
Step 3
 2x + 1 = 1 + 1
Step 4
(xl)
^ = z
^
Steps
X  1
Continued on Next Page
or X — 1
Step 6
Answers
8. (a) { 1 + Vn,  1  Vii}
f3 +VU 3  Vis
(b)
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
564 Chapter 10 Quadratic Equations, Inequalities, and Functions
& Solve each equation by
completing the square.
(a) 2^2  4r + 1 =
(b) 3z2  6z  2 =
(c) 8x24x2 =
QD Solve each equation.
(a) ^2= 17
{h)(k + 5y
100
(c) 5^2 _ 15;+ 12 =
Answers
9.
[2 + V2 2  V2l
(b){^%^,^3^>^}
^'> {^'^}
10.
(a) {/VTv, /VTv}
(b) {5 + 10/, 5  100
^ ^ ri5 + /Vis 15  iVTs
10
10
1 +
X = 1 +
2
14
or
or
X = 1 
7
14
Addl.
Rationalize denominators.
Since 1 = j. add the two terms in each solution as follows:
1 +
14
2
 +
2 2
14 2 + V14
14
Check that the solution set is
2 _ V14
" 2 2~
2 + Vi4 2
14
IK
Work Problem 9 at the Side.
B J ECTi VE Q Solve quadratic equations with nonreal complex
solutions. In the equation x^ = ^, if ^ < 0, there will be two nonreal
complex solutions.
EXAMPLE 7
Solving Quadratic Equations with Nonreal
Complex Solutions
Solve each equation.
(a) x2= 15
X
X
V^ or X =  V^
I V 15 or X = — / V 15
The solution set is {/ V 15, — / v 15}.
(b) (t + 2y= 16
t + 2 = V^l6 or t + 2 = V^16
t + 2 = 4i or t + 2= 4/
t = 2 + 4i or t = 2  4/
The solution set is {—2 + 4/, —2 — 4/}.
(c) x2 + 2x + 7 =
x^ + 2x= 7
x^ + 2x + 1 = 7 + 1
(x+ 1)2= 6
X + 1 = ± /Ve
X = 1 ± /Ve
Square root property
Square root property
V^ = 4/
Subtract 7.
[^(2)]^ = 1; add 1 to each side.
Factor on the left; add on the right.
Square root property
Subtract 1.
The solution set is { 1 + / v 6,  1  / v 6}.
If
Work Problem 10 at the Side,
NOTE
We will use completing the square in Section 10.6 when we graph qua
dratic equations and in Section 12.2 when we work with circles.
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
Section 10.2 The Quadratic Formula 569
10.2 The Quadratic Formula
The examples in the previous section showed that any quadratic equation can
be solved by completing the square; however, completing the square can be
tedious and time consuming. In this section, we complete the square to solve
the general quadratic equation
ax^ + to + c = 0,
where a, b, and c are complex numbers and a t^ 0. The solution of this
general equation gives a formula for finding the solution of any specific
quadratic equation.
OBJECTIVES
Derive the quadratic
formula.
Solve quadratic equations
using the quadratic
formula.
Use the discriminant to
determine the number
and type of solutions.
OBJECTIVE Q Derive the quadratic formula. To solve ax^ + /?x + c =
by completing the square (assuming a > 0), we follow the steps given in
Section 10.1.
ax^ + to + c
r, b c
x^ + X + 
a a
x^ +

2 \a
2 b b^
X H X H r
a 4a^
c
a
la) 4a^
c b'
 + — 1
a 4a^
Divide by ^. (Step 1)
Subtract f . (Step 2)
(Step 3)
b'
Add — r to each side. (Step 4)
4a
Write the left side as a perfect square, and rearrange the right side.
bV b' c
x + — 1 =^^ + — (Steps)
X +
X +
lay
T
2a J
la
4a^
b^ 4ac
4? 4a^
b^  4ac
4a^
Write with a common denominator.
Add fractions.
X +
la
— 4ac b
—2 — or ^ ^ ^^
4a^ la
b'  4ac Square root
property
4a'
(Step 6)
Since
b^  4ac Vb^  4ac Vb^  4ac
4a'
^4a'
la
the right sides of these equations can be expressed as
X +
?
Vb^  4ac
a
la
b Vb^  4ac
— 1
X
la la
b + Vb^  4ac
or X +
la
or
or
la
X =
VZ)2  4ac
la
b
Vb'
4ac
la
la
b
Vb'
4ac
la
If a < 0, the same two solutions are obtained. The result is the quadratic
formula, which is abbreviated as shown on the next page.
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
570 Chapter 10 Quadratic Equations, Inequalities, and Functions
Identify the values of a, b,
and c. (Hint: If necessary, first
write the equation in standard
form with on the right side.)
Do not actually solve.
(a) 3^2 + 9^ _ 4 =
Quadratic Formula
The solutions ofax^\bx\c = Q{a^Q) are given by
b±Vb^  4ac
X =
2a
CAUTION
In the quadratic formula, the square root is added to or subtracted from
the value of—b BEFORE dividing by la.
(b) 3x2 = 6x + 2
fouTry
& Solve 4x2 llx 3 =
using the quadratic formula.
Answers
1. (a) 3; 9;
4 (b) 3; 6; 2
OBJECTIVE Q Solve quadratic equations using the quadratic formula.
To use the quadratic formula, first write the given equation in standard form
ax^ + /?x + c = 0; then identify the values of a, b, and c and substitute them
into the quadratic formula, as shown in the next examples.
iti
Work Problem 1 at the Side.
EXAMPLE 1
Using the Quadratic Formula (Rational Solutions)
Solve 6x2  5x  4 = q.
Here a, the coefficient of the seconddegree term, is 6, while b, the coef
ficient of the firstdegree term, is 5, and the constant c is 4. Substitute
these values into the quadratic formula.
b±Vb^  4ac
2a
(5)±V(5)24(6)(4)
2(6)
Quadratic formula
a = 6, b = —5, c = —4
5 ± V25 + 96
5 ±
12
/l21
X =
12
5 ± 11
12
This last statement leads to two solutions, one from + and one from — .
5 + 11
12
511
or X
12 12 3 12 12 2
Check each solution in the original equation. The solution set is { — 5, f}.
m
Work Problem 2 at the Side.
We could have used factoring to solve the equation in Example 1 .
6x2  5x  4 =
(3x  4)(2x + 1) = Factor.
3x — 4 = or 2x + 1 = Zerofactor property
3x = 4 or 2x = — 1 Solve each equation.
or
' 1
Same solutions as in Example 1
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
Section 10.2 The Quadratic Formula 571
JuTr
When solving quadratic equations, it is a good idea to try factoring first.
If the equation cannot be factored or if factoring is difficult, then use the
quadratic formula. Later in this section, we will show a way to determine
whether factoring can be used to solve a quadratic equation.
S^EXAMPLE 2 MliiLlJ
:he Qua
idratic Forr
nula (Irrational Solutions)
Solve 4r 2 = 8r  1.
Write the equation in
standarc
I form as 4r^
 8r + 1 = 0.
b ± Vb^
2a
(8) ±A
r =
 4ac
Quadratic formula
2(4)
4(4)(1)
a = 4,b= S,c= I
8 ± V64 
16
8 ±V^
8
8 ±4V3
8
V48 = Vie • V^ = 4\/3
4(2 ± Vs)
4(2)
Factor.
2 ± V3
2
Lowest terms
(2 +
The solution set is s
V3 2 
>
^^l
& Solve each equation using the
quadratic formula.
(a) 6x2 + 4x  1 =
(b) 2F + 19 = 14^
CAUTION
1. Every quadratic equation must be written in standard form
«jc^ + 6jc + c = before we begin to solve it, whether we use
factoring or the quadratic formula.
2. When writing solutions in lowest termSy be sure to factor first;
then divide out the common factor^ as shown in the last two steps
in Example 2.
YDuTf^l
Anij
EXAMPLE 3
Work Problem 3 at the Side.
Using the Quadratic Formula (Nonreal Complex
Solutions)
Solve (9^ + 3)(^ 1)= 8.
To write this equation in standard form, we first multiply and collect all
nonzero terms on the left.
(9^ + 3)(^l)=8
9q^ 6q3 = 8
9q^ — 6q + 5 = Standard form
Continued on Next Page
Answers
3. (a)
10 2
(b)
7 + Vll 7
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
572 Chapter 10 Quadratic Equations, Inequalities, and Functions
I From the equation 9q^ — 6^ + 5 = 0, we identify a = 9,b = —6, and c = 5.
O Solve each equation using the
quadratic formula.
(a) x^ + X + 1 =
_ (6) ± V(6)^  4(9)(5) Substitute in the
^~ ^/ri\ quadratic formula.
2(9)
6 ± V144
18
6 ± 12/
The solution set is
18
6(1 ± 2i)
6(3)
1 ± 2/
1+2/ 1  2/
V'144 = 12/
Factor.
Lowest terms
(b) (z + 2)(z6)= 17
NOTE
We could have written the solutions in Example 3 in the form a + bi,
the standard form for complex numbers, as follows:
1 ±2/
V
Standard form
Mi
Work Problem 4 at the Side.
OBJECTIVE Q Use the discriminant to determine the number and type
of solutions. The solutions of the quadratic equation ax^ + bx + c = sltq
given by
b±Vb^  4ac
Discriminant
2a
If a, b, and c are integers, the type of solutions of a quadratic equation — that
is, rational, irrational, or nonreal complex — is determined by the expression
under the radical sign, b^ — 4ac. Because it distinguishes among the three
types of solutions, b^ — 4ac is called the discriminant. By calculating the dis
criminant before solving a quadratic equation, we can predict whether the so
lutions will be rational numbers, irrational numbers, or nonreal complex
numbers. (This can be useful in an applied problem, for example, where irra
tional or nonreal complex solutions are not acceptable.)
Answers
4. (a)
1 + iV3 1  iV3
Discriminant
The discriminant ofax^ + bx + c = Oisb^ 4ac. If a, b, and c are
integers, then the number and type of solutions are determined as follows.
(b) {2 + i,2 i}
Discriminant
Number and
Type of Solutions
Positive, and the square of an integer
Two rational solutions
Positive, but not the square of an integer
Two irrational solutions
Zero
One rational solution
Negative
Two nonreal complex solutions
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
Section 10.2 The Quadratic Formula 573
Calculating the discriminant can also help you decide whether to solve
a quadratic equation by factoring or by using the quadratic formula. If the
discriminant is a perfect square (including 0), then the equation can be
solved by factoring. Otherwise^ the quadratic formula should be used.
EXAMPLE 4
Using the Discriminant
Find the discriminant. Use it to predict the number and type of solutions for
each equation. Tell whether the equation can be solved by factoring or
whether the quadratic formula should be used.
(a) 6x2 X 15 =
We find the discriminant by evaluating b^ — 4ac.
b^  4ac = (ly 4(6) (15) a = 6,b= l,c= 15
= 1 + 360
= 361
A calculator shows that 361 = 19^, a perfect square. Since a, b, and c are
integers and the discriminant is a perfect square, there will be two rational
solutions and the equation can be solved by factoring.
(b) 3m^  4m = 5
Write the equation in standard form as 3m ^ — 4m — 5 = to find a = 3,
b = —4, andc = — 5.
b^ 4ac = (4)2 4(3) (5)
= 16 + 60
= 76
Because 76 is positive but not the square of an integer and a, b, and c are
integers, the equation will have two irrational solutions and is best solved
using the quadratic formula.
(c) 4x2 + X + 1 =
Since a = 4,b = 1, and c = 1, the discriminant is
I2_4(4)(i)= 15.
Since the discriminant is negative and a, b, and c are integers, this quadratic
equation will have two nonreal complex solutions. The quadratic formula
should be used to solve it.
(d) 4^2 + 9 = i2t
Write the equation as 4^2
c = 9. The discriminant is
4ac
12^ + 9 = to find a = 4, /? =  12, and
= (12)2 4(4) (9)
= 144  144
= 0.
Because the discriminant is 0, the quantity under the radical in the quadratic
formula is 0, and there is only one rational solution. Again, the equation can
be solved by factoring.
Work Problem 5 at the Side.
& Find the discriminant. Use it
to predict the number and
type of solutions for each
equation.
(a) 2x2 + 3x = 4
(b) 2x2 + 3x + 4 =
(c) x2 + 20x + 100 =
(d) 15F+ 1U= 14
(e) Which of the equations in
parts (a)(d) can be solved
by factoring?
Answers
5. (a) 41; two; irrational
(b) —23; two; nonreal complex
(c) 0; one; rational
(d) 961; two; rational (e) (c)and(d)
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
Section 10.3 Equations Quadratic in Form 577
1 • 3 Equations Quadratic in Form
OBJECTIVE Q Solve an equation with fractions by writing it in
quadratic form, A variety of nonquadratic equations can be written in the
form of a quadratic equation and solved by using one of the methods from
Sections 10.1 and 10.2.
EXAMPLE 1
1 1
Solve  +
X X 
Solving an Equation with Fractions That Leads
to a Quadratic Equation
T_
12'
1
Clear fractions by multiplying each term by the least common denomina
tor, 12x(x — 1). (Note that the domain must be restricted to x ^ andx t^ 1.)
12x(x  1) + 12x(x  1)^^ = 12x(x  1)^
12(x
12x
Distributive property
Combine terms.
1 " ^ '' 12
1) + 12x = 7x(x  1)
12 + 12x = 7x2 _ 7^
24x  12 = 7x2 _ 7_^
Combine and rearrange terms so that the quadratic equation is in standard
form. Then factor to solve the resulting equation.
7x2  3ix + 12 =
(7x  3) (x  4) =
7x3 = or x4 =
3
Standard form
Factor.
Zerofactor property
X =
or
X = 4 Solve each equation.
Check by substituting these solutions in the original equation. The solution
set is {f, 4}.
OBJECTIVES
Solve an equation with
fractions by writing it in
quadratic form.
Use quadratic equations
to solve applied problems.
Solve an equation with
radicals by writing it in
quadratic form.
Solve an equation that is
quadratic in form by
substitution.
Solve each equation. Check
your solutions.
5 12
(a)  + — = 2
2
(b)  + 
X X
Work Problem 1 at the Side.
OBJECTIVE Q Use quadratic equations to solve applied problems.
In Sections 2.4 and 8.5 we solved distanceratetime (or motion) problems
that led to linear equations or rational equations. Now we can extend that
work to motion problems that lead to quadratic equations. We continue to
use the sixstep problemsolving method from Section 2.3.
EXAMPLE 2
Solving a Motion Problem
A riverboat for tourists averages 12 mph in still water. It takes the boat 1 hr,
4 min to go 6 mi upstream and return. Find the speed of the current.
Step 1 Read the problem carefully.
Step 2 Assign a variable. Let x = the speed of the current. The current
slows down the boat when it is going upstream, so the rate (or speed)
upstream is the speed of the boat in still water less the speed of the
current, or 12 — x. See Figure 1 on the next page.
Continued on Next Page
4 7
(c) + 9 =
m  I m
Answers
1. (a)
(c)
(b)
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
578 Chapter 10 Quadratic Equations, Inequalities, and Functions
Riverboat traveling
upstream — the current
slows it down.
Figure 1
Similarly, the current speeds up the boat as it travels downstream, so
its speed downstream is 12 + x. Thus,
12 — X = the rate upstream;
12 + X = the rate downstream.
This information can be used to complete a table. We use the dis
tance formula, d = rt, solved for time t, t — ^
fort.
, to write expressions
d
r
t
Upstream
6
12 X
6
12 X
Downstream
6
12 +x
6
12 +x
Times
in hours
Step
3 Write an equation. The
total time, 1 hr an(
1 4 min, can be written as
^^
1 16^
^'+15^ is""
Because the time upstream plus the time downstream equals jf hr.
Time upstream +
Time downstream
= Total time
1
i
1
6
1
6
16
12 X ^
12 + x
15'
Step
4 Solve the equation. Multiply each side by
15(12 x)(12 + x),the
LCD, and solve the resulting quadratic equation.
15(12 + x)6+ 15(12 x)6
= 16(12 x) (12
+ x)
90(12 + x) + 90(12 x)
= 16(144 x2)
1080 + 90x + 1080  90x
= 2304  16x2
Distributive property
2160
= 2304  16x2
Combine terms.
16x2
= 144
x^
= 9
Divide by 16.
X = 3 or
x= 3
Square root property
Step
5 State the answer. The
speed of the current cannot be —3, so the
answer is 3 mph.
Step
6 Check that this value satisfies the original
problem.
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
Section 10.3 Equations Quadratic in Form 579
CAUTION
As shown in Example 2, when a quadratic equation is used to solve an
applied problem, sometimes only one answer satisfies the application.
Always check each answer in the words of the original problem.
Work Problem 2 at the Side.
\u
In Section 8.5 we solved problems about work rates. Recall that a
person's work rate is \ part of the job per hour, where t is the time in hours
required to do the complete job. Thus, the part of the job the person will do
in X hours is\x.
& Solve each problem.
(a) In 4 hr, Kerrie can go 1 5 mi
upriver and come back.
The speed of the current is
5 mph. Complete this table.
d
r
t
Up
Down
EXAMPLE 3
Solving a Work Problem
It takes two carpet layers 4 hr to carpet a room. If each worked alone, one of
them could do the job in 1 hr less time than the other. How long would it take
each carpet layer to complete the job alone?
Step 1 Read the problem again. There will be two answers.
Step 2
Assign a variable. Let x rep
resent the number of hours for
the slower carpet layer to com
plete the job alone. Then the
faster carpet layer could do the
entire job in (x — 1) hours.
The slower person's rate is \,
and the faster person's rate is
7^. Together, they can do the
job in 4 hr. Complete a table
as shown.
Rate
Time Working
Together
Fractional Part
of the Job Done
Slower j^
Worker x
4
1,4,
Faster 1
4
7^<^»
Worker ^ _ ^
Sum is 1
whole job.
Step 3 Write an equation. The sum of the fractional parts done by the
workers should equal 1 (the whole job).
Part done by slower worker
I
4
part done by faster worker
I
4
X 1
1 whole job.
1
(b) Find the speed of the boat
from part (a) in still water.
(c) In 1 1 hr. Ken rows his boat
5 mi upriver and comes
back. The speed of the
current is 3 mph. How fast
does Ken row?
Continued on Next Page
Answers
2. (a) rowl: 15; x 5;
row 2: 15; x+ 5;
15
X  5
15
X + 5
(b) 10 mph (c) 7 mph
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
580 Chapter 10 Quadratic Equations, Inequalities, and Functions
& Solve each problem. Round
answers to the nearest tenth.
(a) Carlos can complete a
certain lab test in 2 hr
less time than Jaime can.
If they can finish the job
together in 2 hr, how long
would it take each of
them working alone?
Rate
Time
Working
Together
Fractional
Part of the
Job Done
Carlos
Jaime
Step 4 Solve the equation from Step 3.
= 1
4 4
 +
X X  1
4 4
x{x !)( + ——
x{x  1) ( 1 ) Multiply by the LCD.
4(x  1) + 4x = x(x  1)
4x
v2
4 + 4x
9x + 4
Distributive property
Distributive property
Standard form
This equation cannot be solved by factoring, so use the quadratic
formula.
9± VSI  16 9
65
+
2
65
5.5 or
(2 = 1,Z? = 9, c = 4
.5 Use a calculator.
Step 5 State the answer. Only the solution 8.5 makes sense in the original
problem. (Why?) Thus, the slower worker can do the job in about
8.5 hr and the faster in about 8.5 — 1 = 7.5 hr.
Step 6 Check that these results satisfy the original problem.
Hi
Work Problem 3 at the Side.
Video
(b) Two chefs are preparing a
banquet. One chef could
prepare the banquet in 2 hr
less time than the other.
Together, they complete
the job in 5 hr. How long
would it take the faster
chef working alone?
Answers
3. (a) Jaime: 5.2 hr; Carlos: 3.2 hr (b) 9.1 hr
OBJECTIVE Q Solve an equation with radicals by writing it in
quadratic form.
EXAMPLE 4
Solving Radical Equations That Lead to Quadratic
Equations
Solve each equation.
(a) k = Vek 8
This equation is not quadratic. However, squaring both sides of the
equation gives a quadratic equation that can be solved by factoring.
k^ = 6k
8 Square both sides.
k^ 6k+S =
Standard form
(k4)(k2) =
Factor
k
4 = or k2 =
Zerofactor property
A = 4 or k = 2
Potential solutions
Recall from Section 9.6 that squaring both sides of a radical equation can
introduce extraneous solutions that do not satisfy the original equation. All po
tential solutions must be checked in the original (not the squared) equation.
Check: If A
k
4
4
4
4, then
V6/t
V6(4)"
16
If A
k
2
2
2, then
4. True
Both solutions check, so the solution set is {2, 4}
Continued on Next Page
V6/t
V6U)"
2 = 2.
True
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
Section 10.3 Equations Quadratic in Form 581
(b) X + Vx = 6
V X = 6 — X
X = 36 — 12x + x^
= x2  13x + 36
= (x  4) (x  9)
x4 = or x9 =
X = 4 or X = 9
Check both potential solutions in the original equation.
If X = 4, then
X + Vx = 6
4 + V4 = 6 ?
6 = 6. True
Isolate the radical on one side.
Square both sides.
Standard form
Factor.
Zerofactor property
Potential solutions
If X = 9, then
X + Vx = 6
9 + V9 = 6
12 = 6.
Only the solution 4 checks, so the solution set is {4}.
False
O Solve each equation. Check
your solutions.
(a) X = VTx  10
Work Problem 4 at the Side.
»i.
EXAMPLE 5
OBJECTIVE Q Solve an equation that is quadratic in form by
substitution. A nonquadratic equation that can be written in the form
au^ \ bu \ c = 0,
for a 7^ and an algebraic expression w, is called quadratic in form.
Solving Equations That Are Quadratic In Form
Solve each equation.
(a) x4  13x2 + 36 =
Because x^ = (x^Y, we can write this equation in quadratic form with
u = x^ and u^ = x^. (Instead ofu, any letter other than x could be used.)
x4  13x2 + 36 =
(x^y  13x2 + 36 =
u^  13w + 36 =
(u4)(u9) =
u 4 = or u9 =
u = 4 or u = 9
To find X, we substitute x^ for u.
(b) 2x = V^ + 1
.4 =
(X^)
2\2
Let u = x^.
Factor.
Zerofactor property
Solve.
Square root property
x^ = 4 or x^ = 9
X = ±2 or X = ±3
The equation x"^ — 13x^ + 36 = 0, a fourthdegree equation, has four solu
tions.* The solution set is { — 3, —2, 2, 3}. Check by substitution.
4x4 + 1 = 5x2
4(xy + 1 = 5x2
4^2 + 1 = 5u
Continued on Next Page
(b)
x4 = (x^y
*In general, an equation in which an /ithdegree polynomial equals has n solutions, although some of
them may be repeated.
Answers
4. (a) {2,5} (b) {1}
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
582 Chapter 10 Quadratic Equations, Inequalities, and Functions
& Solve each equation. Check
your solutions.
(a) m4  10^2 + 9 =
(b) 9k*  37P + 4 =
(c) X*  4x^
4m2  5m
+ 1 =
Standard form
(4m  1) (m 
1) =
Factor
4m 1
= or M
1 =
Zerofactor property
u
1
= 4 «'•
H= 1
Solve.
x'
1
= 4 
x'= 1
Substitute x^ for u.
X
1
= ± or
x= ±1
Square root property
Check that the solution set is { 1, j, j, 1}.
(c) X* = 6x23
First write the equation as
x^  6x2 + 3 =
or
(x^  6x2 + 3 = 0,
which is quadratic in form with m = x2. Substitute uforx^ and m 2 for x"* to get
m2 _ 6h + 3 = 0.
Since this equation cannot be solved by factoring, use the quadratic formula.
6 ± V36  12
a = \,b
6, c = 3
M =
6 ± V24
2
6±2V6
2
2(3 ±V6)
M = 3± V6
x2 = 3 + Ve
V24 = V4 • V6 = 2V6
Factor
Lowest terms
or X
Ve
X = ± V3 + Ve or x= ±V3  V6
The solution set contains four numbers:
{V3 + V6,  V3 + V6, V3  V6,  V3  Ve}
Substitute x^ for u.
Square root property
Answers
5. (a) {3,1,1,3} (b)
1 1
2, ,,2
3 3
(c) {V2 + Vl,  V2 + Vl,
V2  \/2,  V2  \/2}
NOTE
Some students prefer to solve equations
and (b) by factoring directly. For example,
like those in Examples 5(a)
x4  13x2 + 36 =
Example 5(a) equation
(x2  9) (x2  4) =
Factor.
(x + 3)(x
 3) (x + 2) (x  2) = 0.
Factor again.
Using the zerofactor property gives the same solutions obtained in
Example 5(a). Equations that cannot be solved by factoring (as in
Example 5(c)) must be solved by substitution and the quadratic formula.
Hl<
Work Problem 5 at the Side.
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
Section 10.3 Equations Quadratic in Form 583
EXAMPLE 6
Solving Equations That Are Quadratic in Form
Solve each equation.
(a) 2(4m  3)^ + 7(4m  3) + 5 =
Because of the repeated quantity 4m — 3, this equation is quadratic in
form with u = 4m — 3.
2(4«i  3)2 + 7(4m  3) + 5 =
2m2 + 7h + 5 = Let 4ot  3 = u.
(2m
+ 5)(w + 1) = Factor.
2m + 5 =
or
1/ + 1 = Zerofactor property
5
"=2
or
w = 1
4/M  3 = 
or
4m — 3 = — 1 Substitute 4m  3 for u.
4m = 
2
or
4m = 2 Solve for m.
1
1
m = 
or
 = 2
Check that the solution set of the
; original equation is {, j}.
(b) 2fl2/3 11^1/3+ 12 =
=
Leta''^ = m; then a 2/3
= (a^^^y = u^. Substitute into the given equation.
2m2
lli/+ 12 = LQt a'^' = u;a^^' = u\
(2m
3)(u
 4) = Factor.
2m  3 =
or u
— 4 = Zerofactor property
3
" = 2
or
« = 4
 = f
or
ai/3=4 J, = ^1/3
<"'"»' = (f)'
or (c
, 1/3)3 = 43 Q^]jQ gJ^^,[J gj(jg
27
^= 8
or
a = 64
Check that the solution set is {y,
64}.
© Solve each equation. Check
your solutions.
(a) 5(r+ 3)2 + 9(r + 3) = 2
(b) 4m
2/3
3^1/3 + 1
CAUTION
A common error when solving problems like those in Examples 5 and 6
is to stop too soon. Once you have solved for w, remember to substitute
and solve for the values of the original variable.
Work Problem 6 at the Side.
Answers
6. (a) 5^
14
(''>^^''
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
Section 10.4 Formulas and Further Applications 591
1 0.4 Formulas and Further Applications
OBJECTIVE Q Solve formulas for variables involving squares and
square roots. The methods presented earlier in this chapter and the previous
one can be used to solve formulas with squares and square roots.
EXAMPLE 1
Solving forVariables Involving Squares
or Square Roots
s,^I^J^0 Solve each formula for the given variable.
Video
[^
(a) w
kFr
for V
kFr
w
Isolate V on one side.
v^w = kFr
kFr
w
kFr
w
±VkFr Vw ±VkF}
rw
V =
(b) d
w
' — for A
77
W
W
Multiply by v^.
Divide by w.
Square root property
Rationalize the denominator.
4A
77
~ Isolate A on one side.
Square both sides.
^d^ = 4A Multiply by 77.
77^^
= A
Divide by 4.
Work Problem 1 at the Side.
NOTE
In many formulas like v
' kFrw •
in Example 1(a), we choose the
positive value. In our work here, we will include both positive and
negative values.
EXAMPLE 2
Solving for a Squared Variable
Solve s = 2f + kt for t.
Since the equation has terms with t^ and t, write it in standard form
ax^ \ bx \ c = Q, with t as the variable instead of x.
2f + kt
2f + kt s
Continued on Next Page
s
Subtract s.
OBJECTIVES
Solve formulas for vari
ables involving squares
and square roots.
Solve applied problems
using the Pythagorean
formula.
Solve applied problems
using area formulas.
Solve applied problems
using quadratic functions
as models.
O Solve each formula for the
given variable.
(a) A = irr^ for r
(b) s = 30^/ — for a
Answers
±VA7T
1. (a) r =
77
(b) a
900
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
592 Chapter 10 Quadratic Equations, Inequalities, and Functions
' 2f + kt s = Q Standard form
Now use the quadratic formula with a = 2, b = k, and c = —s.
k ± Vk^  4(2) (^)
Solve 2t^  5t + k=0 for t.
Hypotenuse
Leg a
Leg b
Pythagorean Formula
& Solve the problem.
A 13ft ladder is leaning
against a house. The distance
from the bottom of the ladder
to the house is 7 ft less than
the distance from the top of
the ladder to the ground. How
far is the bottom of the ladder
from the house?
Answers
_ 5 
5 + V25 
'■'= 4
3. 5ft
 8^
, t =
V25 
4
 8^
t =
t =
2(2)
k ± Vk^ + 8^
Solve for t.
, . k + \/e + 8^ , k \/e + 8^
The solutions are t = and t = .
HIK
Work Problem 2 at the Side.
OBJECTIVE Q Solve applied problems using the Pythagorean formula.
The Pythagorean formula
illustrated by the figure in the margin, was introduced in Section 9.3 and is
used to solve applications involving right triangles. Such problems often re
quire solving quadratic equations.
EXAMPLE 3
Using the Pythagorean Formula
Two cars left an intersection at the same time, one heading due north, the
other due west. Some time later, they were exactly 100 mi apart. The car
headed north had gone 20 mi farther than the car headed west. How far had
each car traveled?
Step 1
Step 2
Step 3
Read the problem carefully.
Assign a variable. Let x be the distance
traveled by the car headed west. Then
X + 20 is the distance traveled by the car
headed north. See Figure 2. The cars are
100 mi apart, so the hypotenuse of the
right triangle equals 100.
Write an equation. Use the Pythagorean
formula.
North
e
100/
West
90'
x + 20
a + b = c
x2 + (x + 20)2
100^
Step 4 Solve. x^ + x^ + 40x + 400 = 10,000
2x2 + 40;(. _ 9600 =
x2 + 20x  4800 =
(x + 80) (x  60) =
X + 80 = or X  60 =
X = 80 or X = 60
Intersection
Figure 2
Square the binomial.
Standard form
Divide by 2.
Factor.
Zerofactor property
Step 5 State the answer. Since distance cannot be negative, discard the neg
ative solution. The required distances are 60 mi and 60 + 20 = 80 mi.
Step 6 Check. Since 60^ + 80^ = 100^, the answers are correct.
iti
Work Problem 3 at the Side.
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
Section 10.4 Formulas and Further Applications 593
Animatk
OBJECTIVE Q Solve applied problems using area formulas.
EXAMPLE 4
Solving an Area Problem
A rectangular reflecting pool in a park is 20 ft wide and 30 ft long. The park
gardener wants to plant a strip of grass of unift)rm width around the edge of
the pool. She has enough seed to cover 336 ft^. How wide will the strip be?
Step 1 Read the problem carefully.
Step 2 Assign a variable. The pool is shown in Figure 3. If x represents the
unknown width of the grass strip, the width of the large rectangle is
given by 20 + 2x (the width of the pool plus two grass strips), and
the length is given by 30 + 2x.
O Solve the problem.
Suppose the pool in
Example 4 is 20 ft by 40 ft
and there is enough seed to
cover 700 ft^. How wide
should the grass strip be?
Figure 3
Step 3 Write an equation. The area of the large rectangle is given by the
product of its length and width, (30 + 2x) (20 + 2x). The area of
the pool is 30 • 20 = 600 ft^. The area of the large rectangle, minus
the area of the pool, should equal the area of the grass strip. Since the
area of the grass strip is to be 336 ft^, the equation is
Step 4 Solve.
Step 5
area of area of
pool = grass.
I \
 600 = 336.
336
Area of
rectangle —
(30 + 2x) (20 + 2x)
600 + lOOx + 4x2  600
4x2 + lOOx  336
x2 + 25x  84 =
(x + 28) (x  3) =
X = —28 or X = 3
State the answer. The width cannot be
should be 3 ft wide.
Step 6 Check. If x = 3, then the area of the large rectangle is
(30 + 2 • 3) (20 + 2 • 3) = 36 • 26 = 936 ft^. Area of pool and strip
The area of the pool is 30 • 20 = 600 ft^. So, the area of the grass
strip is 936 — 600 = 336 ft^, as required. The answer is correct.
Work Problem 4 at the Side.
Multiply.
Standard form
Divide by 4.
Factor.
Zerofactor property
28 ft, so the grass strip
OBJECTIVE Q Solve applied problems using quadratic functions as
models. Some applied problems can be modeled by quadratic functions,
which can be written in the form
/(x) = ax^ + /?x + c,
for real numbers a, b, and c, with a ?^ 0.
Answers
4. 5ft
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
594 Chapter 10 Quadratic Equations, Inequalities, and Functions
Videoj'
Solve the problem. Vid^n
A ball is propelled
vertically upward from the
ground. Its distance in feet
from the ground at t seconds is
^(0= 16^2 + 64^
At what times will the ball
be 32 ft from the ground?
Use a calculator and round
answers to the nearest tenth.
{Hint: There are two
answers.)
© Use a calculator to evaluate
28.6±V(28.6)'  4(3.37) (17)
2(3.37)
for both solutions. Round to
the nearest tenth. Which
solution is valid for this
problem?
Answers
5. at .6 sec and at 3.4 sec
6. 9.0, .6:9.0
EXAMPLE 5
Solving an Applied Problem Using
a Quadratic Function
If an object is propelled upward from the top of a 144ft building at 1 12 ft
per sec, its position (in feet above the ground) is given by
^(0= 16^2+ ii2t+ 144,
where t is time in seconds after it was propelled. When does it hit the ground?
When the object hits the ground, its distance above the ground is 0. We
must find the value of t that makes s{t) = 0.
16^2 +112^+ 144
It
t =
7 ± V49 + 36 7 ± V85 7 ± 9.2
LQts(t) = 0.
Divide by —16.
Use the quadratic
formula and a calculator.
The solutions are ^ — 8.1 or ^ — — 1.1. Since time cannot be negative, dis
card the negative solution. The object will hit the ground about 8.1 sec after
it is propelled.
[Work Problem 5 at the Side.
EXAMPLE 6
Using a Quadratic Function to Model Company
Bankruptcy Filings
The number of companies filing for bankruptcy was high in the early 1990s
due to an economic recession. The number then declined during the middle
1990s, and in recent years has increased again. The quadratic function
defined by
f(x) = 3.37x2 _ 28.6x + 133
approximates the number of company bankruptcy filings during the years
1990 through 2001, where x is the number of years that have elapsed since
1990. (Source: www.BankruptcyData.com)
(a) Use the model to approximate the number of company bankruptcy filings
in 1995.
For 1995, x = 5, so find/(5).
3.37(5)2 _ 28.6(5) + 133
74.25
/(5)
Letx = 5.
There were about 74 company bankruptcy filings in 1995.
(b) In what year did company bankruptcy filings reach 150?
Find the value of x that makes/(x) = 150.
f(x) = 3.37x2 _ 28.6x + 133 Let/(x) = 150.
150 = 3.37x2 _ 28.6x + 133 Standard form
= 3.37x2 _ 28.6x  17
Now use a = 3.37, b = 28.6, and c =  17 in the quadratic formula.
Work Problem 6 at the Side.
The positive solution is x — 9, so company bankruptcy filings reached 150
in 1990 + 9 = 1999. (Reject the negative solution since the model is not
valid for negative values of x.)
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
Section 10.5 Graphs of Quadratic Functions 599
1 • 5 Graphs of Quadratic Functions
OBJECTIVE Q Graph a quadratic function. Figure 4 gives a graph of
the simplQst quadratic function, defined by j = x^.
X
y
2
4
1
1
1
1
2
4
Figure 4
As mentioned in Section 6.3, this graph is called a parabola. The point
(0, 0), the lowest point on the curve, is the vertex of this parabola. The verti
cal line through the vertex is the axis of the parabola, here x = 0. A parabola
is symmetric about its axis; that is, if the graph were folded along the axis,
the two portions of the curve would coincide. As Figure 4 suggests, x can be
any real number, so the domain of the function defined hyy = x^ is (— °o, ^).
Since y is always nonnegative, the range is [0, oo).
In Section 10.4, we solved applications modeled by quadratic functions.
We now consider graphs of general quadratic functions as defined here.
OBJECTIVES
Graph a quadratic
function.
Graph parabolas with
horizontal and vertical
shifts.
Predict the shape and
direction of a parabola
from the coefficient of x^.
Find a quadratic function
to model data.
Quadratic Function
A function that can be written in the form
f(x) = ax^ \ bx \ c
for real numbers a, b, and c, with a ?^ 0, is a quadratic function.
The graph of any quadratic function is a parabola with a vertical axis.
NOTE
We use the variable y and function notation f{x) interchangeably.
Although we use the letter/ most often to name quadratic functions,
other letters can be used. We use the capital letter F to distinguish
between different parabolas graphed on the same coordinate axes.
Parabolas, which are a type oi conic section (Chapter 12), have many
applications. Cross sections of satellite dishes and automobile headlights
form parabolas, as do the cables that support suspension bridges.
objectiveQ Graph parabolas with horizontal and vertical shifts.
Parabolas need not have their vertices at the origin, as does the graph of
f{x) = x^. For example, to graph a parabola of the form F{x) = x^ + k,
start by selecting sample values of x like those that were used to graph
f(x) = x^. The corresponding values of F(x) in F(x) = x^ + k differ by k
from those of/(x) = x^. For this reason, the graph of F(x) = x^ + k is
shifted, or translated, k units vertically compared with that of/(x) = x^.
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
600 Chapter 10 Quadratic Equations, Inequalities, and Functions
Graph each parabola. Give
the vertex, domain, and
range.
(a) fix) = x2 + 3
(b)/(x) = x2l
Answers
1. (a) y
^3 f{x)=x' + 3
vertex: (0, 3); domain: (—0°, 0°);
range: [3, 0°)
(b) y
f{x)=x'\
vertex: (0, —1); domain: (—0°, 0°);
range: [1, 0°)
EXAMPLE 1
Graphing a Parabola with aVertical Shift
Graph F(x) = x^  2.
This graph has the same shape as that of/(x) = x^, but since k here
is —2, the graph is shifted 2 units down, with vertex (0, —2). Every function
value is 2 less than the corresponding function value of/(x) = x^. Plotting
points on both sides of the vertex gives the graph in Figure 5.
Notice that since the parabola is symmetric about its axis x = 0, the
plotted points are "mirror images" of each other. Since x can be any real
number, the domain is still (—0°, ^)\ the value of j (or F{x)) is always
greater than or equal to —2, so the range is [—2, oo). The graph of/(x) = x^
is shown for comparison.
X
f(x) = X2
F(x) = x2  2
2
4
2
1
1
1
2
1
1
1
2
4
2
F{x)=x^2
Figure 5
Vertical Shift
The graph of F(x) = x^ + A: is a parabola with the same shape as the
graph of/(x) = x^. The parabola is shifted vertically: k units up if ^ > 0,
and I ^1 units down if ^ < 0. The vertex is (0, k).
%\
Work Problem 1 at the Side.
The graph of F(x) = (x — hY is also a parabola with the same shape as
that of/(x) = x^. Because (x  A)^ > for all x, the vertex of F(x) = (x  h)^
is the lowest point on the parabola. The lowest point occurs here when F(x) is
0. To get F(x) equal to 0, let x = /z so the vertex of F(x) = (x — h)^ is (/z, 0).
Based on this, the graph of F(x) = (x — A)^ is shifted h units horizontally com
pared with that of/(x) = x^.
EXAMPLE 2
Graphing a Parabola with a Horizontal Shift
Graph F{x) =
When X
(X  2)2.
= 2, then F(x)
0, giving the vertex (2, 0). The graph of
F{x) = (x — ly has the same shape as that of/(x) = x^ but is shifted 2 units
to the right. Plotting several points on one side of the vertex and using
symmetry about the axis x = 2 to find corresponding points on the other side
of the vertex gives the graph in Figure 6. Again, the domain is (—0°, ^)\ the
range is [0, oo).
Continued on Next Page
nil
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
Section 10.5 Graphs of Quadratic Functions 601
Ludjt
X
F(x) = (X  2)2
4
1
1
2
3
1
4
4
Figure 6
Horizontal Shift
The graph of F(jc) = (jc  hy is a parabola with the same shape as the
graph of/(x) = x^. The parabola is shifted h units horizontally: h units to
the right if A > 0, and /z  units to the left if A < 0. The vertex is (/z, 0).
& Graph each parabola. Give
the vertex, axis, domain, and
range.
(a) fix) = {x 3)2
CAUTION
Errors frequently occur when horizontal shifts are involved. To deter
mine the direction and magnitude of a horizontal shift, find the value
that would cause the expression x — h io equal 0. For example, the
graph of F(x) = (x  5)^ would be shifted 5 units to the right, because
+ 5 would cause x  5 to equal 0. On the other hand, the graph of
F{x) = (x + 5)2 would be shifted 5 units to the left, because —5 would
cause X + 5 to equal 0.
Work Problem 2 at the Side.
A parabola can have both horizontal and vertical shifts.
EXAMPLE 3
Graphing a Parabola with Horizontal and
Vertical Shifts
Graph F(x) = (x + 3)^2.
This graph has the same shape as that of/(x) = x^, but is shifted 3 units
to the left (since x + 3 = if x = 3) and 2 units down (because of the 2).
As shown in Figure 7, the vertex is (—3, —2), with axis x = —3. This func
tion has domain (oo^ oo) and range [2, oo).
X
FW.
5
2
4
1
3
2
2
1
1
2
f{x)=x''
Figure 7
{h)f{x) = {x + lf
Answers
2. (a) y
f(x) = (x3y
vertex: (3, 0); axis: x = 3;
domain: (0°, oo); range: [0, 0°)
(b)
fix) = ix + 2y
vertex: (—2, 0); axis: x = —2;
domain: (00,0°); range: [0, 0°)
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
602 Chapter 10 Quadratic Equations, Inequalities, and Functions
& Graph each parabola. Give
the vertex, axis, domain, and
range.
(a)/(x) = (x + 2)2l
(b)/(x) = (x2)2 + 5
Answers
3. (a) y
m = (x+2yi
vertex: (—2, —1); axis: x = —2;
domain: (0°, oo); range: [ 1, °°)
(b) y
k/
f(x) = (x2y+5
I I I I I I *■ X
vertex: (2, 5); axis: x = 2;
domain: (—00,00); range: [5, 00)
The characteristics of the graph of a parabola of the form F(x)
(x — hy + k are summarized as follows.
Vertex and Axis of a Parabola
The graph ofF(x) = (x — hf \ kissi parabola with the same shape as
the graph of/(x) = x^ with vertex (A, k). The axis is the vertical line x =
h.
Work Problem 3 at the Side.
OBJECTIVE Q Predict the shape and direction of a parabola from the
coefficient of x^. Not all parabolas open up, and not all parabolas have the
same shape as the graph of/(x) = x^
EXAMPLE 4
Graphing a ParabolaThat Opens Down
1
Graph /(x) = ^^^•
This parabola is shown in Figure 8. The coefficient —\ affects the shape
of the graph; the \ makes the parabola wider (since the values of ^x^ increase
more slowly than those of x^), and the negative sign makes the parabola open
down. The graph is not shifted in any direction; the vertex is still (0, 0). Un
like the parabolas graphed in Examples 13, the vertex here has the greatest
function value of any point on the graph. The domain is (oo, oo); the range
is (00, 0].
/(x)=ix2
f ^^
fix) '
2
2
1
1
2
1
1
2
2
2
Figure 8
Some general principles concerning the graph of F(x) = a(x — hy + k
are summarized as follows.
General Principles
1. The graph of the quadratic function defined by
F(x) = a{x  hy + k, a^ 0,
is a parabola with vertex (A, k) and the vertical line x = A as axis.
2. The graph opens up if a is positive and down if a is negative.
3. The graph is wider than that of/(x) = x^ifO<a<l. The graph
is narrower than that of/(x) = x^ if  a  > 1 .
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
Section 10.5 Graphs of Quadratic Functions 603
Work Problems 4 and 5 at the Side.
HI
EXAMPLE 5
Using the General Principles to Graph a Parabola
Graph F(x)= 2(x + 3)2 + 4.
The parabola opens down (because a < 0), and is narrower than the
graph of/(x) = x^, since  — 2 = 2 > 1, causing values of F(x) to decrease
more quickly than those of/(x) = — x^. This parabola has vertex (—3, 4) as
shown in Figure 9. To complete the graph, we plotted the ordered pairs
(4, 2) and, by symmetry, (2, 2). Symmetry can be used to find additional
ordered pairs that satisfy the equation, if desired.
Figure 9
Work Problem 6 at the Side.
OBJECTIVE Q Find a quadratic function to model data.
 ?!?!yyrji=gf> Finding a Quadratic Function to Model the Rise in
Multiple Births
The number of higherorder multiple births in the United States is rising. Let
X represent the number of years since 1970 and;; represent the rate of higher
order multiples born per 100,000 births since 1971. The data are shown in
the following table. Find a quadratic function that models the data.
U.S. HIGHERORDER
MULTIPLE BIRTHS
1 Year
X
y
1971
1
29.1
1976
6
35.0
1981
11
40.0
1986
16
47.0
1991
21
100.0
1996
26
152.6
2001
31
185.6
Source: National Center
for Health Statistics.
A scatter diagram of the ordered pairs (x, y) is shown in Figure 10 on the
next page. The general shape suggested by the scatter diagram indicates that
a parabola should approximate these points, as shown by the dashed curve in
Figure 11. The equation for such a parabola would have a positive coeffi
cient for x^ since the graph opens up.
Continued on Next Page
O Decide whether each
parabola opens up or down.
(a) fix) = ^x'
(b)/(x) =x'+l
(c) f(x) = 2x'
id)f(x) = 3x^ + 2
& Decide whether each
parabola in Problem 4 is
wider or narrower than the
graph of/(x) = x^
O Graph
Ax) = \{x  2)' + 1.
Answers
4. (a) down (b) up (c) down (d) up
5. (a) wider (b) wider (c) narrower
(d) narrower
6.
^(x2)2+l
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
604 Chapter 10 Quadratic Equations, Inequalities, and Functions
O Tell whether a linear or
quadratic function would be a
more appropriate model for
each set of graphed data. If
linear, tell whether the slope
should be positive or
negative. If quadratic, tell
whether the coefficient a of
x^ should be positive or
negative.
(a)
AVERAGE DAILY
EMAIL VOLUME
35
^ 30
o
§ 25
^ 20 
^ 15 
10
12 3 4
Years Since 2000
Source: General Accounting Office.
nfi INCREASES IN WHOLESALE
DRUG PRICES
ID D
y 4
13 5 7 9
Years Since 1990
Source: IMS Health, Retail and
Provider Perspective.
O Using the points (1, 29.1),
(6, 35), and (26, 152.6), find
another quadratic model for
the data on higherorder
multiple births in Example 6.
U.S. HIGHERORDER
MULTIPLE BIRTHS
U.S. HIGHERORDER
MULTIPLE BIRTHS
, 1.
200

•
150

•
Rate

•
50
n
•
•
1
•
1
•
1
1
1
1
200
150 
^ 100
50 
•*
15 25
Years Since 1970
15 25
Years Since 1970
Figure 11
Figure 10
To find a quadratic function of the form
y = ax^ + to + c
that models, or fits, these data, we choose three representative ordered pairs
and use them to write a system of three equations. Using (1, 29.1), (11, 40),
and (21, 100), we substitute the x and jvalues from the ordered pairs into
the quadratic formj = ax^ \ bx \ do get the three equations
a{\Y + b{\) + c = 29.1 or a + b + c = 29.1 (1)
a{l\Y + bill) + c = 40 or 121a + IIZ? + c = 40 (2)
a{llY + b{ll) + c = 100 or 441a + 21/? + c = 100. (3)
We can find the values of a, b, and c by solving this system of three equa
tions in three variables using the methods of Section 5.2. Multiplying equa
tion (1) by  1 and adding the result to equation (2) gives
120a + 10/? = 10.9. (4)
Multiplying equation (2) by  1 and adding the result to equation (3) gives
320a + 10/? = 60. (5)
We can eliminate b from this system of equations in two variables by multi
plying equation (4) by — 1 and adding the result to equation (5) to get
200a = 49.1
a = .2455. Use a calculator.
We substitute .2455 for a in equation (4) or (5) to find that b =  1.856. Sub
stituting the values of a and b into equation (1) gives c = 30.7105. Using
these values of a, /?, and c, our model is defined by
y = .2455x2  1.856x + 30.7105.
Work Problems 7 and 8 at the Side.
NOTE
If we had chosen three different ordered pairs of data in Example 6, a
slightly different model would have resulted.
Answers
7. (a) linear; positive (b) quadratic; positive
H. y = .188x2 .136x + 29.05
lilll CalculatorTip The quadratic regression feature on a graphing calcu
lator can be used to generate a quadratic model that fits given data. See
your owner's manual for details on how to do this.
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
Section 10.6 More about Parabolas; Applications 61 I
1 • 6 More about Parabolas; Applications
OBJECTIVE Q Find the vertex of a vertical parabola. When the
equation of a parabola is given in the form/(x) = ax^ + to + c, we need to
locate the vertex in order to sketch an accurate graph. There are two ways to
do this:
1. Complete the square, as shown in Examples 1 and 2, or
2. Use a formula derived by completing the square, as shown in Example 3.
Video!
EXAMPLE 1
Completing the Square to Find the Vertex
Find the vertex of the graph of/(x) = x^ — 4x + 5.
To find the vertex, we need to write the expression x^ — 4x + 5 in the
form (x  hy + k. We do this by completing the square on x^  4x, as in
Section 10.1. The process is a little different here because we want to keep
/(x) alone on one side of the equation. Instead of adding the appropriate
number to each side, we add and subtract it on the right. This is equivalent
to adding 0.
/(x) = x2  4x + 5
= (x^  4x ) + 5
Group the variable terms.
= iiy = 4
= (x2  4x + 4  4) + 5
= (x2  4x + 4)  4 + 5
fix) = (X  2)2 + 1
The vertex of this parabola is (2, 1).
Add and subtract 4.
Bring —4 outside the parentheses.
Factor; combine terms.
Work Problem 1 at the Side.
OBJECTIVES
Find the vertex of a
vertical parabola.
Graph a quadratic
function.
Use the discriminant
to find the number of
xintercepts of a vertical
parabola.
Use quadratic functions
to solve problems
involving maximum or
minimum value.
Graph horizontal
parabolas.
Find the vertex of the graph
of each quadratic function.
(a) fix)
6x + 7
(b) fix) = x2 + 4x  9
EXAMPLE 2
Completing the Square to Find the Vertex
When a ^ 1
Find the vertex of the graph of/(x) = 3x^ + 6x  1.
We must complete the square on — 3x^ + 6x. Because the x^term has a
coefficient other than 1, we factor that coefficient out of the first two terms
and then proceed as in Example 1 .
fix) = 3x2 + 6x  1
= —3(x^ — 2x) — 1 Factor out 3.
"1
(2)
= (ir = 1
= 3(x2  2x + 1  1)  1 Add and subtract 1.
Bring — 1 outside the parentheses; be sure to multiply it by —3.
= —3(x^ — 2x + 1) + (— 3)(— 1) — 1 Distributive property
= 3(x22x+ 1) + 3  1
/(x) = — 3(x— 1)2 + 2 Factor; combine terms.
The vertex is (1, 2).
Work Problem 2 at the Side.
O Find the vertex of the graph
of each quadratic function.
(a) fix) = 2x2  4x + 1
(b)/(x) = x2 + 2x3
Answers
1. (a) (3, 2)
2. (a) (1,1)
(b) (2,
(b) (2, 
13)
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
612 Chapter 10 Quadratic Equations, Inequalities, and Functions
& Use the formula to find the
vertex of the graph of each
quadratic function.
(a) /(x) = 2x2 + 3x  1
To derive a formula for the vertex of the graph of the quadratic function
defined by/(x) = ax^ + bx + c, complete the square.
f(x) = ax^ + bx + c (a ^ 0) Standard form
Factor a from the first
two terms.
a\ X H X ] + c
a

2V«
2a J 4«^
b b'
a[ x^ \ — X \ r
4a
, b b^
a[ X \ — X \ r ) + a
, b b^
a[ X H X \ r
a 4a^
"''^ih
4a
4ac  b^
+ c
b^
~ 4a^
+ c
fix) = a
b_
2a
h
+
4a
4ac  b^
4a
+ c
Add and subtract — .
Distributive property
Factor; combine terms.
fix) = {xh)^ + k
(b) fix) = 4x2  X + 5
Thus, the vertex {h, k) can be expressed in terms of a, b, and c. It is not nec
essary to remember the expression for k, since it can be found by replacing
X with ^. Using function notation, if j = /(x), then the j value of the vertex
is/(if).
Vertex Formula
The graph of the quadratic function defined by/(x) = ax^ \ bx \ c has
vertex
2a' Ala '"
and the axis of the parabola is the line
b
X =
2a
Answers
3. (a)
3 1
(b)
i 11
8' 16
EXAMPLE 3
Using the Formula to Find the Vertex
Use the vertex formula to find the vertex of the graph of/(x) = x^ — x — 6.
For this function, a = 1, /? =  1, and c = 6. The xcoordinate of the
vertex of the parabola is given by
b (1) 1
2a 2(1) 2'
The jcoordinate is /(^) = /() .
i) = fiY_l_<, = ii6 = ?i
.2/ V2/ 2 4 2 4
/
The vertex is (^f).
«l<
Work Problem 3 at the Side.
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Section 10.6 More about Parabolas; Applications 613
BJ ECTi VE Q Graph a quadratic function.
approach for graphing any quadratic function here.
We give a general
Graphing a Quadratic Function f
Step 1 Determine whether the graph opens up or down. If a > 0,
the parabola opens up; if a < 0, it opens down.
Find the vertex. Use either the vertex formula or completing
the square.
Find any intercepts. To find the xintercepts (if any), solve
f{x) = 0. To find the jintercept, evaluate /(O).
Step 2
Step 3
Step 4
Complete the graph. Plot the points found so far. Find and plot
additional points as needed, using symmetry about the axis.
Y QuTrviy
Video
EXAMPLE 4
Using the Steps to Graph a Quadratic Function
Graph the quadratic function defined by /(x) = x^  x  6.
Step 1 From the equation, a = 1, so the graph of the function opens up.
Step 2 The vertex, (^,  ^), was found in Example 3 by substituting the
values a = \,b = — 1, and c = —6 in the vertex formula.
Step 3 Now find any intercepts. Since the vertex, {\,^), is in quadrant
IV and the graph opens up, there will be two xintercepts. To find
them, let/(x) = and solve the equation.
f{x) = x^ — X — 6
= x2x 6
= (x  3) (x + 2)
x3 = or x + 2 =
X = 3 or X = — 2
The xintercepts are (3, 0) and (—2, 0). Find the jintercept.
Let/(x) = 0.
Factor.
Zerofactor property
/(x) = x^ — X — 6
0^06
Letx = 0.
/(O)
/(O) = 6
The jintercept is (0, —6).
Step 4 Plot the points found so far and additional points as needed using
symmetry about the axis x = \. The graph is shown in Figure 12.
The domain is (— oo, co), and the range is [
25
0.
X
y
2
1
4
6
1
2
25
4
2
4
3
'f{x)=x^x6
O Graph the quadratic function
defined by
fix) = x^  6x + 5.
Give the axis, domain, and
range.
(6,5)
Work Problem 4 at the Side.
5,0)
^(3,4)
f{x) =x^6x + 5
axis: x = 3; domain: (— °o, °o);
range: [4, oo)
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
614 Chapter 10 Quadratic Equations, Inequalities, and Functions
Use the discriminant to
determine the number of
xintercepts of the graph of
each quadratic function.
(a) fix) = 4x2  20x + 25
B J ECTi VE Q Use the discriminant to find the number of
A^intercepts of a vertical parabola. Recall from Section 10.2 that
the expression b^ — 4ac is called the discriminant of the quadratic equation
ax^ + to + c = and that we can use it to determine the number of real so
lutions of a quadratic equation. In a similar way, we can use the discriminant
of a qusidrsitic function to determine the number of xintercepts of its graph.
See Figure 13. If the discriminant is positive, the parabola will have two x
intercepts. If the discriminant is 0, there will be only one xintercept, and it
will be the vertex of the parabola. If the discriminant is negative, the graph
will have no xintercepts.
(b) /(x) = 2x2 + 3x + 5
/\
^
b^  4ac>0
Two xintercepts
T\
IZ
^2 _ 4^^ = Q
One xintercept
Figure 13
A
\J
b^ 4ac<0
No xintercepts
(c) f(x) = 3x2 x + 2
EXAMPLE 5
Using the Discriminant to Determine the Number
of xlntercepts
Use the discriminant to determine the number of xintercepts of the graph of
each quadratic function.
(a) /(x) = 2x2 + 3x  5
The discriminant isb^ — 4ac. Here a = 2,b = 3, and c = —5, so
b^ 4ac = 9 4(2)(5) = 49.
Since the discriminant is positive, the parabola has two xintercepts.
(b)/(x)= 3x21
In this equation, a = — 3, Z? = 0, and c = — 1. The discriminant is
b^4ac = 0 4(3)(l) = 12.
The discriminant is negative, so the graph has no xintercepts.
(c) fix) = 9x2 + 6x + 1
Here, a = 9, Z? = 6, and c = 1. The discriminant is
b^ 4ac = 364(9)(l) = 0.
The parabola has only one xintercept (its vertex) because the value of the
discriminant is 0.
Work Problem 5 at the Side.
Answers
5. (a) discriminant is 0; one xintercept
(b) discriminant is — 3 1 ; no xintercepts
(c) discriminant is 25; two xintercepts
OBJECTIVE Q Use quadratic functions to solve problems involving
maximum or minimum value. The vertex of a parabola is either the high
est or the lowest point on the parabola. The jvalue of the vertex gives the
maximum or minimum value of j, while the xvalue tells where that maxi
mum or minimum occurs.
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
Section 10.6 More about Parabolas; Applications 6 I 5
PROBLEMSOLVING HINT
In many applied problems we must find the largest or smallest value of
some quantity. When we can express that quantity as a quadratic func
tion, the value of ^ in the vertex {h, k) gives that optimum value.
© Solve Example 6 if the
farmer has only 100 ft of
fencing.
EXAMPLE 6
Finding the Maximum Area of a Rectangular Region
A farmer has 120 ft of fencing to enclose a rectangular area next to a build
ing. See Figure 14. Find the maximum area he can enclose.
Figure 14
Let X represent the width of the rectangle. Since he has 120 ft of fencing,
X + X + length =120 Sum of the sides is 120 ft.
2x + length = 120 Combine terms,
length = 120  2x. Subtract 2x.
The area A{x) is given by the product of the width and length, so
A{x) = x(120 2x)
= UOx  2x\
To determine the maximum area, find the vertex of the parabola given by
A (x) = 120x  2x^ using the vertex formula. Writing the equation in stan
dard form as A (x) = 2x^ + 120x gives a = 2, b = 120, and c = 0, so
h
b
2a
120
120
30;
2(2) 4
^(30) = 2(30)2 + 120(30) = 2(900) + 3600 = 1800.
The graph is a parabola that opens down, and its vertex is (30, 1800). Thus,
the maximum area will be 1800 ft^. This area will occur if x, the width of the
rectangle, is 30 ft.
CAUTION
Be careful when interpreting the meanings of the coordinates of the
vertex. The first coordinate, x, gives the value for which thQ function
value is a maximum or a minimum. Be sure to read the problem
carefully to determine whether you are asked to find the value of the
independent variable, the function value, or both.
Work Problem 6 at the Side.
Answers
6. The rectangle should be 25 ft by 50 ft with
a maximum area of 1250 ft^.
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
616 Chapter 10 Quadratic Equations, Inequalities, and Functions
& Solve the problem.
A toy rocket is launched
from the ground so that its
distance in feet above the
ground after t seconds is
^(0= 16^2 + 208^.
Find the maximum height it
reaches and the number of
seconds it takes to reach that
height.
EXAMPLE 7
Finding the Maximum Height Attained
by a Projectile
If air resistance is neglected, a projectile on Earth shot straight upward with
an initial velocity of 40 m per sec will be at a height s in meters given by
^(0 = 4.9^2 + 40^^
where t is the number of seconds elapsed after projection. After how many
seconds will it reach its maximum height, and what is this maximum height?
For this function, a = 4.9, b = 40, and c = 0. Use the vertex formula.
b 40
2a
h =
4.1 Use a calculator.
2(4.9)
Thus, the maximum height is attained at 4.1 sec. To find this maximum
height, calculate 5'(4.1).
^(4.1) = 4.9(4.1)2 + 40(4.1) « 81.6 Use a calculator.
The projectile will attain a maximum height of approximately 81.6 m.
Work Problem 7 at the Side.
VideoJ
OGraphx = (y + If  4,
Give the vertex, axis, domain,
and range.
OBJECTIVE Q Graph horizontal parabolas. If x and j are interchanged
in the equation j = ax^ + bx + c, the equation becomes x = ay^ + by + c.
Because of the interchange of the roles of x and j, these parabolas are horizon
tal (with horizontal lines as axes).
Answers
7. 676 ft; 6.5 sec
8. y
x = (3;+l)24
vertex: (—4, —1); axis: J = —1;
domain: [4, oo); range: (00^ oo)
Graph of a Horizontal Parabola
The graph of
x = ay^ \ by \ c or x = a(y — Kf \ h
is a parabola with vertex {h, k) and the horizontal line y = ksis axis. The
graph opens to the right ifa>0 and to the left if a < 0.
EXAMPLE 8
Graphing a Horizontal Parabola
Graph X = (y  2)^  3.
This graph has its vertex at (—3, 2), since the roles of x andj are reversed.
It opens to the right, the positive xdirection, and has the same shape as j = x^.
Plotting a few additional points gives the graph shown in Figure 15. Note that
the graph is symmetric about its axis, y = 2. The domain is [ — 3, oo), and the
range is (— oo, oo).
X
y
3
2
2
3
2
1
1
4
1
Figure 15
Hi
Work Problem 8 at the Side.
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
Section 10.6 More about Parabolas; Applications 6 I 7
When a quadratic equation is given in the form x = ay^ \ by \ c, com
pleting the square on y will allow us to find the vertex.
EXAMPLE 9
Completing the Square to Graph a Horizontal
Parabola
Graph X = 2;;^ + 4y  3. Give the domain and range of the relation.
X = ly^ + 473
= 2(j22j)3
= 2(j22j+ 1  1) 3
= _2(3;2_23;+l) + (2)(l)3
X= 2(7 1)2 1
Factor out 2.
Complete the square;
add and subtract 1 .
Distributive property
Factor; simplify.
Because of the negative coefficient (—2), the graph opens to the left (the
negative xdirection) and is narrower than the graph of 7 = x^. As shown in
Figure 16, the vertex is ( 1, 1). The domain is (00,  1], and the range is
(—00^ 00).
X
y
3
2
3
1
1
Figure 16
Work Problems 9 and 10 at the Side.
In summary, the graphs of parabolas studied in Sections 10.5 and 10.6
fall into the following categories.
GRAPHS OF PARABOLAS
Equation
^^^^^^^^ Graph ^^^^^^^^^^^
y — ax^ \ bx + c
j; = a{xhf + k
\
/ /
Ah, k)
\
7^ ^
(h, k) jhese graphs
<3 > represent functions.
a<0
X = ay^ \ by \ c
x = a(y kf + h
>
(h, k) ^
^(h,k)
^^
These graphs are not
(3 > graphs of functions.
^ > X
a<0
O Find the vertex of each
parabola. Tell whether the
graph opens to the right or to
the left. Give the domain and
range.
(a) X = 272  67 + 5
(b)x
y^ + 2j + 5
CD) Graph X = j^ + 27 + 5.
Give the vertex, axis, domain,
and range.
Answers
9. (a) (—, — ); right; domain:
range: (—00^ 00)
(b) (6, 1); left; domain; (^, 6];
range: (—00^ 00)
10.
 X = };2 + 2>^ + 5
vertex: (6, 1); axis:_y = 1; domain: (—0°, 6];
range: (00^ 00)
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
618 Chapter 10 Quadratic Equations, Inequalities, and Functions
CD (a) Tell whether each of the
following equations has a
vertical or horizontal
parabola as its graph.
A. y= x^ + 20x + 80
B. X = 2j;2 + 6j + 5
C. X + 1 = (j + 2)2
D. fix) = (X  4)2
CAUTION
Only quadratic equations solved for y (whose graphs are vertical
parabolas) are examples of functions. The horizontal parabolas in
Examples 8 and 9 are not graphs of functions, because they do not sat
isfy the vertical line test. Furthermore, the vertex formula given earlier
does not apply to parabolas with horizontal axes.
«i
Work Problem 11 at the Side,
(b) Which of the equations in
part (a) represent functions?
Answers
11. (a) A, D are vertical parabolas; B, C are
horizontal parabolas.
(b) A,D
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Section 10.7 Quadratic and Rational Inequalities 623
10.7 Quadratic and Rational Inequalities
We combine methods of solving linear inequalities and methods of solving
quadratic equations to solve quadratic inequalities.
Quadratic Inequality
A quadratic inequality can be written in the form
ax^ + 6jc + c < or ax^ + 6x + c > 0,
where a, b, and c are real numbers, with a i^ 0.
OBJECTIVES
Solve quadratic inequalities.
Solve polynomial inequali
ties of degree 3 or more.
Solve rational inequalities.
As before, < and > may be replaced with < and >.
OBJECTIVE Q Solve quadratic inequalities. One method for solving
a quadratic inequality is by graphing the related quadratic function.
EXAMPLE 1
Solving Quadratic Inequalities by Graphing
Solve each inequality.
(a) x^  X  12 >
To solve the inequality, we graph the related quadratic function defined
by/(x) = x^ — X— 12. We are particularly interested in the xintercepts,
which are found as in Section 10.6 by letting /(x) = and solving the
quadratic equation
x2x 12 = 0.
(x  4) (x + 3) =
Factor.
= or X + 3 =
Zerofactor property
= 4 or X = 3
Thus, the xintercepts are (4, 0) and (3, 0). The graph, which opens up
since the coefficient of x^ is positive, is shown in Figure 17(a). Notice from
this graph that xvalues less than —3 or greater than 4 result in jvalues
greater than 0. Therefore, the solution set of x^ — x — 12 > 0, written in
interval notation, is (oo, 3) U (4, oo).
/(x)=x2x12 1
xvalues for
which J >
xvalues for
which J >
x12
The graph is above the xaxis for
(00, 3) U (4, oo).
(a)
The graph is below the xaxis for
(3, 4).
(b)
Figure 17
Continued on Next Page
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
624 Chapter 10 Quadratic Equations, Inequalities, and Functions
Use the graph to solve each
quadratic inequality.
(a) x^ + 6x + 8 >
(b)x^
12 <0
Here we want values of j that are less than 0. Referring to Figure 17(b)
on the previous page, we notice from the graph that xvalues between —3
and 4 result in jvalues less than 0. Therefore, the solution set of the
inequality x^  x  12 < 0, written in interval notation, is (3, 4).
NOTE
If the inequalities in Example 1 had used > and <, the solution sets
would have included the xvalues of the intercepts and been written in
interval notation as (—0°, — 3] U [4, oo) for Example 1(a) and [ — 3, 4]
for Example 1(b).
(b) x^ + 6x + 8 <
& Graph/(x) = x^ + 3x  4
and use the graph to solve
each quadratic inequality.
(a) x2 + 3x  4 >
(b) x^ + 3x  4 <
Answers
1. (a) (^, 4) U (2, ^) (b) (4, 2)
2. (a) (^, 4] U [1, ^) (b) [4, 1]
Work Problems 1 and 2 at the Side.
In Example 1, we used graphing to divide the xaxis into intervals. Then
using the graphs in Figure 17, we determined which xvalues resulted in
jvalues that were either greater than or less than 0. Another method for
solving a quadratic inequality uses these basic ideas without actually graph
ing the related quadratic function.
EXAMPLE 2
Solving a Quadratic Inequality UsingTest Numbers
Solve x^ X 12>0.
First solve the quadratic equation x^ — x— 12 = by factoring, as in
Example 1(a).
(x  4) (x + 3) =
x4 = or x + 3 =
X = 4 or X = — 3
The numbers 4 and —3 divide the number line into the three intervals
shown in Figure 18. Be careful to put the lesser number on the left. (No
tice the similarity between Figure 18 and the xaxis with intercepts (—3, 0)
and (4, 0) in Figure 17(a).)
Interval
A
Interval
B
Interval
C
T 3 F 4 T
Figure 18
The numbers 4 and —3 are the only numbers that make the expression
x^ — X — 12 equal to 0. All other numbers make the expression either pos
itive or negative. The sign of the expression can change from positive to
negative or from negative to positive only at a number that makes it 0.
Therefore, if one number in an interval satisfies the inequality, then all the
numbers in that interval will satisfy the inequality.
To see if the numbers in Interval A satisfy the inequality, choose any
number from Interval A in Figure 18 (that is, any number less than —3).
Substitute this test number for x in the original inequality x^ — x — 12 > 0.
If the result is true, then all numbers in Interval A satisfy the inequality.
Continued on Next Page
/(x)=x2 + 3x4
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
Section 10.7 Quadratic and Rational Inequalities 625
We choose —5 from Interval A. Substitute —5 forx.
Original inequality
True
x^ X 12>0
(5)2  (5)  12 >
25 + 5  12 >
18 >0
Because —5 from Interval A satisfies the inequality, all numbers from Inter
val A are solutions.
Try from Interval B. If x = 0, then
0^0 12 >0 ?
12 >0.
The numbers in Interval B are not solutions.
False
Work Problem 3 at the Side.
In Problem 3 at the side, the test number 5 satisfies the inequality, so the
numbers in Interval C are also solutions.
Based on these results (shown by the colored letters in Figure 18), the
solution set includes the numbers in Intervals A and C, as shown on the
graph in Figure 19. The solution set is written in interval notation as
(00, 3) U (4,00).
C I II
■f
H \ \ \ \ h
Figure 19
f
III )
This agrees with the solution set we found by graphing the related quadratic
function in Example 1(a).
& Does the number 5 from
Interval C satisfy
x2  X  12 > 0?
O Solve each inequality, and
graph the solution set.
(a) x^ + X  6 >
In summary, a quadratic inequality is solved by following these steps.
Solving a Quadratic Inequality
Step 1 Write the inequality as an equation and solve it.
Step!
Step 3
Step 4
Use the solutions from Step 1 to determine intervals. Graph
the numbers found in Step 1 on a number line. These numbers
divide the number line into intervals.
Find the intervals that satisfy the inequality. Substitute a test
number from each interval into the original inequality to deter
mine the intervals that satisfy the inequality. All numbers in
those intervals are in the solution set. A graph of the solution set
will usually look like one of these. (Square brackets might be
used instead of parentheses.)
■^
f=
or
^
^
Consider the endpoints separately. The numbers from Step 1
are included in the solution set if the inequality is < or >; they
are not included if it is < or >.
Work Problem 4 at the Side.
(b) 3^2  13m  10 <
Answers
3. yes
4. (a) (^, 3)U (2,^)
C) I I I I ( I >
3 2
(b)
, 5
I l [ l I I I I ] I I
1
3 5 7
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
626 Chapter 10 Quadratic Equations, Inequalities, and Functions
& Solve each inequality.
(a) (3^  2)2 > 2
Ifou Tf^y i\
Special cases of quadratic inequalities may occur, as in the next
example.
EXAMPLE 3
Solving Special Cases
Solve (2t  3)2 > 1. Then solve (2t  3)^ < 1.
Because (2t — 3)^ is never negative, it is always greater than — 1 . Thus, the
solution set for (2^ — 3)^ > — 1 is the set of all real numbers, (—^, ^). In the
same way, there is no solution for (2^ — 3)^ < — 1 and the solution set is 0.
Work Problem 5 at the Side.
(b) (5z + 3)2 < 3
& Solve each inequality, and
graph the solution set.
(a) (x 3)(x + 2)(x+ 1)>0
(b) (k 5)(k+ 1)(^ 3)<0
Answers
5. (a) (^, ^) (b)
6. (a) (2, 1)U (3,00)
I I I (I ^
2101234
(b) (
1]U[3,5]
C I ] I I I [ I ] I
10 1 3 5
OBJECTIVE Q Solve polynomial inequalities of degree 3 or more.
Higherdegree polynomial inequalities that can be factored are solved in the
same way as quadratic inequalities.
EXAMPLE 4
Solving aThirdDegree Polynomial Inequality
Solve (x  1) (x + 2) (x  4) < 0.
This is a cubic (thirddegree) inequality rather than a quadratic inequal
ity, but it can be solved using the method shown in the box by extending the
zerofactor property to more than two factors. Begin by setting the factored
polynomial equal to and solving the equation (Step 1).
(x l)(x + 2)(x4) =
xl=0 or x + 2 = or x4 =
X = 1 or X = — 2 or x = 4
Locate the numbers 2, 1, and 4 on a number line, as in Figure 20, to
determine the Intervals A, B, C, and D (Step 2).
Interval
A
Interval
B
h
Interval
C
Interval
D
F 1 T
Figure 20
Substitute a test number from each interval in the original inequality to
determine which intervals satisfy the inequality (Step 3). It is helpful to
organize this information in a table.
1 Interval
Test Number
Test of Inequality
True or False?
A
3
28 <0
T
B
8<0
F
C
2
8<0
T
D
5
28 <0
F
Verify the information given in the table and graphed in Figure 21. The num
bers in Intervals A and C are in the solution set, which is written as
(00, 2] U [1,4].
The three endpoints are included since the inequality symbol is
(Step 4).
^ — \ h
^^
H h
[ i i ]
H \ h
1
Figure 21
Work Problem 6 at the Side.
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
Section 10.7 Quadratic and Rational Inequalities 627
OBJECTIVE Q Solve rational inequalities. Inequalities that involve
rational expressions, called rational inequalities, are solved similarly using
the following steps.
Solving a Rational Inequality
Step 1 Write the inequality so that is on one side and there is a
single fraction on the other side.
Determine the numbers that make the numerator and
denominator equal to 0.
Divide a number line into intervals. Use the numbers from
Step 2.
Step 2
Step 3
Step 4
Step 5
Find the intervals that satisfy the inequality. Test a number
from each interval by substituting it into the original inequality.
Consider the endpoints separately. Exclude any values that
make the denominator 0.
EXAMPLE 5
Solving a Rational Inequality
Solve
1
Write the inequality so that is on one side (Step 1).
1
1 P
 1 >
 3
3 p  3
I  p + 3
p + 2
Subtract 1 .
> Use p — 3 SiS the common denominator.
> Write the left side as a single fraction;
be careful with signs in the numerator.
> Combine terms in the numerator.
The sign of the rational expression J^_ 3 will change from positive to
negative or negative to positive only at those numbers that make the numera
tor or denominator 0. The number 2 makes the numerator 0, and 3 makes the
denominator (Step 2). These two numbers, 2 and 3, divide a number line
into three intervals. See Figure 22 (Step 3).
Interval
A
Interval
B
Interval
C
T
Figure 22
Testing a number from each interval in the original inequality, ^z^ > 1,
gives the results shown in the table (Step 4).
1 Interval
Test Number Test of Inequality True or False? 1
A
i>l
F
B
2.5
2>1
T
C
4
1 >1
F
Continued on Next Page
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
628 Chapter 10 Quadratic Equations, Inequalities, and Functions
O Solve each inequality, and
graph the solution set.
(a)
X  4
< 3
The solution set of ^z^ > 1 is the interval (2, 3). This interval does not in
clude 3 since it would make the denominator of the original inequality 0; 2 is
not included either since the inequality symbol is > (Step 5). A graph of the
solution set is given in Figure 23.
i — ^
Figure 23
m
Work Problem 7 at the Side.
CAUTION
When solving a rational inequality, any number that makes the denom
inator must be excluded from the solution set.
(b)
z+ 1
>4
k + 2
O Solve < 5, and graph
k — \
the solution set.
Answers
7. (a) (00,4) U
14
— c
3'
14
3
C I I I I ) ( I :>
12 3 4 5 6
(b) 1,
I (D I I
21012
8. (00, 1) u
< I ) [ l h^
3
1 L
4
EXAMPLE 6
Solving a Rational Inequality
Solve
m
m \ 2
Write the inequality so that is on one side (Step 1).
2
m
m + 2
z.
m
 2
+ 2
2{m +
2)
m
m + 2
m —
1lm
 4
m + 2
— m 
 6
m \ 2
Subtract 2.
Use m + 2 as the common denominator.
Write as a single fraction.
Combine terms in the numerator.
The number 6 makes the numerator 0, and 2 makes the denominator
(Step 2). These two numbers determine three intervals (Step 3). Test one
number from each interval (Step 4) to see that the solution set is the interval
(00, 6] U (2, 00).
The number —6 satisfies the original inequality, but —2 cannot be used
as a solution since it makes the denominator (Step 5). A graph of the
solution set is shown in Figure 24.
c I I I
M
H \ h
f
I I I I I D
Figure 24
Work Problem 8 at the Side.
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
Exponential and ^
Logarithmic *'
Functions
\
With the many advances made in electronics over the past
decade, home theater is now a reality. The operating instruc
tions for the Pioneer PDF1009 compact disc player, a component in
one author's system, includes a warning that loud noises can
cause hearing damage, and a list of sound levels, measured
in decibels and shown in the accompanying table, that can be
dangerous under constant exposure.
In Exercise 39 of Section 1 1.5 we examine the meaning of decibel,
which is based on logarithmic functions, one topic covered in this
chapter.
T T .1 Inverse Functions
11.2 Exponential Functions
1 1 .3 Logarithmic Functions
1 1 .4 Properties of Logarithms
1 1 .5 Common and Natural
Logarithms
1 1 .6 Exponential and Logarithmic
Equations; Further Applications
Decibel Level
Example
90
Subway, motorcycle,
truck traffic, lawn
mower
100
Garbage truck, chain
saw, pneumatic drill
120
Rock concert in front of
speakers, thunderclap
140
Gunshot blast, jet plane
180
Rocket launching pad
Source: Deafness Research Foundation.
651
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
652 Chapter I I Exponential and Logarithmic Functions
11.1 Inverse Functions
OBJECTIVES
In this chapter we will study two important types of functions, exponential
Q Decide whether a function
is onetoone and, if it is,
find its inverse.
Q Use the horizontal line
test to determine
whether a function is
onetoone.
Q Find the equation of the
inverse of a function.
Q Graph f "^ from the graph
off.
and logarithmic. These functions are related in a special way: They are
inverses of one another. We begin by discussing inverse functions in general.
M CalculatorTip A calculator with the following keys will be essential
in this chapter.
(¥J (^o^ or Q£2J, (^ or Qn^
We will explain how these keys are used at appropriate places in the chapter.
OBJECTIVE Q Decide whether a function is onetoone and, if it is,
find its inverse. Suppose we define the function
G= {(2, 2), (1,1), (0,0), (1,3), (2, 5)}.
We can form another set of ordered pairs from G by interchanging the x and
yvalues of each pair in G. Call this set F, with
F= {(2, 2), (1,1), (0,0), (3,1), (5, 2)}.
To show that these two sets are related, F is called the inverse of G. For a
function/ to have an inverse,/ must be onetoone.
OnetoOne Function
In a onetoone function, each x value corresponds to only one j value,
and each jvalue corresponds to just one xvalue.
The function shown in Figure 1(a) is not onetoone because the j value
7 corresponds to two xvalues, 2 and 3. That is, the ordered pairs (2, 7) and
(3, 7) both appear in the function. The function in Figure 1(b) is onetoone.
Domain
Range
Domain
Range
Not onetoone
(a)
Onetoone
(b)
Figure 1
The inverse of any onetoone function/is found by interchanging the
components of the ordered pairs of/. The inverse of/ is written/"^. Read
f~^ as "the inverse of/" or "/inverse."
CAUTION
The symbol/" ^(x) does not represent
fix)
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
Section I I.I Inverse Functions
653
The definition of the inverse of a function follows.
Inverse of a Function
The inverse of a onetoone function/, written/"^ is the set of all
ordered pairs of the form (y, x), where (x, y) belongs to/. Since the
inverse is formed by interchanging x and j, the domain of/ becomes
the range of/~^ and the range of/ becomes the domain of/~^.
Find the inverse of each
function that is onetoone.
(a) {(1,2), (2, 4), (3, 3), (4, 5)}
For inverses/ and/ ^ it follows that
f(f^(x)) = x and f'(f(x)) = x.
f
Video
EXAMPLE 1
Finding the Inverses of OnetoOne Functions
Find the inverse of each function that is onetoone.
(a) F = {(2, 1), ( 1, 0), (0, 1), (1, 2), (2, 2)}
Each xvalue in F corresponds to just one jvalue. However, the jvalue
2 corresponds to two xvalues, 1 and 2. Also, the jvalue 1 corresponds to
both —2 and 0. Because some jvalues correspond to more than one xvalue,
F is not onetoone and does not have an inverse.
(b) G = {(3, 1), (0, 2), (2, 3), (4, 0)}
Every xvalue in G corresponds to only one jvalue, and every jvalue
corresponds to only one xvalue, so G is a onetoone function. The inverse
function is found by interchanging the x and j values in each ordered pair.
Gi = {(l,3),(2,0),(3,2),(0,4)}
Notice how the domain and range of G become the range and domain,
respectively, of G~^.
(c) The U.S. Environmental Protection Agency has developed an indicator of
air quality called the Pollutant Standard Index (PSI). If the PSI exceeds 100
on a particular day, that day is classified as unhealthy. The table shows the
number of unhealthy days in Chicago for the years 19912002, based on new
standards set in 1998.
(b) {(0,3), (1,2), (1,3)}
Year
Number of
Unhealthy Days
Year
Number of
Unhealthy Days
1991
24
1997
10
1992
5
1998
12
1993
4
1999
19
1994
13
2000
2
1995
24
2001
22
1996
7
2002
21
Source: U.S. Environmental Protection Agency, Office of Air
Quality Planning and Standards.
Let/ be the function defined in the table, with the years forming the
domain and the numbers of unhealthy days forming the range. Then/ is not
onetoone, because in two different years (1991 and 1995), the number of
unhealthy days was the same, 24.
Work Problem 1 at the Side.
(c)
A Norwegian physiologist
has developed a rule for
predicting running times
based on the time to run
5 km (5K). An example for
one runner is shown here.
(Source: Stephen Seller,
Agder College,
Kristiansand, Norway.)
Distance
Time
1.5K
4:22
3K
9:18
5K
16:00
lOK
33:40
Answers
1. (a) {(2,1), (4, 2), (3, 3), (5, 4)}
(b) not a onetoone function
(c)
Time
Distance
4:22
1.5K
9:18
3K
16:00
5K
33:40
lOK
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
654 Chapter I I Exponential and Logarithmic Functions
& Use the horizontal line test to
determine whether each
graph is the graph of a
onetoone function.
(a)
OBJECTIVE Q Use the horizontal line test to determine whether
a function is onetoone. It may be difficult to decide whether a function
is onetoone just by looking at the equation that defines the function. How
ever, by graphing the function and observing the graph, we can use the
horizontal line test to tell whether the function is onetoone.
Horizontal Line Test
A function is onetoone if every horizontal line intersects the graph of
the function at most once.
The horizontal line test follows from the definition of a onetoone func
tion. Any two points that lie on the same horizontal line have the same
jcoordinate. No two ordered pairs that belong to a onetoone function may
have the same jcoordinate, and therefore no horizontal line will intersect
the graph of a onetoone function more than once.
(b)
EXAMPLE 2
Using the Horizontal Line Test
Use the horizontal line test to determine whether the graphs in Figures 2 and 3
are graphs of onetoone functions.
{xi,y)
N
;
j=/(^X.
V
\
\
\
Figure 2
Because the horizontal line
shown in Figure 2 intersects the graph
in more than one point (actually three
points), the function is not onetoone. i one.
Figure 3
Every horizontal line will inter
sect the graph in Figure 3 in exactly
one point. This function is oneto
Work Problem 2 at the Side.
Answers
2. (a) onetoone (b) not onetoone
OBJECTIVE Q Find the equation of the inverse of a function. By
definition, the inverse of a function is found by interchanging the x and
jvalues of each of its ordered pairs. The equation of the inverse of a func
tion defined by j = f{x) is found in the same way.
Finding the Equation of the Inverse of y = f (x)
For a onetoone function/ defined by an equation j = /(x), find the
defining equation of the inverse as follows.
Step 1 Interchange x andj.
Step 2 Solve for J.
Step 3 Replace y with/" ^ (x).
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
Section 1 1. 1 Inverse Functions
655
EXAMPLE 3
Video
Finding Equations of Inverses
Decide whether each equation defines a onetoone function. If so, find the
equation of the inverse.
(a) fix) = 2x + 5
The graph of j = 2x + 5 is a nonvertical line, so by the horizontal line
test,/is a onetoone function. To find the inverse, let j =/(x) so that
J = 2x + 5
X = 2j + 5
2y = X — 5
X — 5
Interchange x andj. (Step 1)
Solve for J. (Step 2)
& Decide whether each
equation defines a onetoone
function. If so, find the
equation that defines the
inverse.
(a) fix) = 3x  4
f\x)
Replace J with /"I (x). (Step 3)
Thus,/"^ is a linear function. In the function withj = 2x + 5, the value ofj
is found by starting with a value of x, multiplying by 2, and adding 5. The
equation for the inverse has us subtract 5, and then divide by 2. This shows
how an inverse is used to "undo" what a function does to the variable x.
(b)7
+ 2
This equation has a vertical parabola as its graph, so some horizontal
lines will intersect the graph at two points. For example, both x = 3 and
X = — 3 correspond to j = 11. Because of the x^term, there are many pairs
of xvalues that correspond to the same jvalue. This means that the function
defined by j = x^ + 2 is not onetoone and does not have an inverse.
If this is not noticed, following the steps for finding the equation of an
inverse leads to
y
X
2
.Vx
■■y^ + 2
Interchange x and y.
Solve for J.
Square root property
The last step shows that there are two jvalues for each choice of x > 2, so
the given function is not onetoone and cannot have an inverse.
(c) fix) = (X  ly
Refer to Section 6.3 to see from its graph that a cubing function like this
is a onetoone function.
J =
(X  ly
Replace /(x) withj;.
X =
(y  2y
Interchange x and y.
Vx =
V(j  2)^
Take the cube root on each side.
^x =
y2
y =
^x + 2
Solve for j;.
f'M =
^x + 2
Replace y with/~^(x).
(b)/(x) = x3+l
(c) fix) = (X  3)^
Work Problem 3 at the Side.
Answers
3. (a) onetoone function; /" ^ (x)
(b) onetoone function; / \x) = vx — 1
(c) not a onetoone function
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
656 Chapter I I Exponential and Logarithmic Functions
O Use the given graphs to graph
each inverse.
(b)
^
Answers
4. (a)
(b) y
OBJECTIVE Q Graph f~^ from the graph of f. One way to graph the
inverse of a function /whose equation is known is to find some ordered
pairs that belong to/, interchange x andj to get ordered pairs that belong
tof~\ plot those points, and sketch the graph of/"^ through the points.
A simpler way is to select points on the graph of/ and use symmetry to find
corresponding points on the graph of/~^.
For example, suppose the point (a, b) shown in Figure 4 belongs to a
onetoone function/ Then the point (/?, a) belongs tof~\ The line segment
connecting (a, b) and (b, a) is perpendicular to, and cut in half by, the line
y = X. The points (a, b) and (Z?, a) are "mirror images" of each other with
respect ioy = x. For this reason we can find the graph of/~ ^ from the graph of
/by locating the mirror image of each point in/with respect to the line y = x.
EXAMPLE 4
Graphing the Inverse
Graph the inverses of the functions shown in Figure 5.
In Figure 5 the graphs of two functions are shown in blue. Their inverses
are shown in red. In each case, the graph of/~^ is symmetric to the graph of
/ with respect to the line y
X.
Figure 5
Work Problem 4 at the Side.
(C) 3^
•^"^
I I I I I I I
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
Section I 1.2 Exponential Functions 661
W.2i Exponential Functions
OBJECTIVE Q Define exponential functions. In Section 9.2 we
showed how to evaluate 2^ for rational values of x. For example,
2' = 8, 2^ = ^,
21/2 ^ y^^ 2^/4 = V^ = A^^.
In more advanced courses it is shown that 2^ exists for all real number values
of X, both rational and irrational. (Later in this chapter, we will see how to
approximate the value of 2^ for irrational x.) The following definition of an
exponential function assumes that a^ exists for all real numbers x.
Exponential Function
For a > 0, a i^ 1, and all real numbers x,
fix) = a
defines the exponential function with base a.
OBJECTIVES
Define exponential
functions.
Graph exponential
functions.
Solve exponential
equations of the form
a^ = a^forx.
Use exponential functions
in applications involving
growth or decay.
NOTE
The two restrictions on a in the definition of an exponential function are
important. The restriction that a must be positive is necessary so that the
function can be defined for all real numbers x. For example, letting a
be negative (a = —2, for instance) and letting x = \ would give the
expression (—2)^^^, which is not real. The other restriction, a ?^ 1, is
necessary because 1 raised to any power is equal to 1, and the function
would then be the linear function defined by/(x) = 1 .
OBJECTIVE Q Graph exponential functions. We can graph an expo
nential function by finding several ordered pairs that belong to the function,
plotting these points, and connecting them with a smooth curve.
m\
Video]
EXAMPLE 1
Graphing an Exponential Function with a > 1
Graph/(x) = 2\
Choose some values of x, and find the corresponding values of/(x).
X 3
2
1
1
2
3
4
f (X) = 2 I
1
4
1
2
1
2
4
8
16
Plotting these points and drawing a smooth curve through them gives the
graph shown in Figure 6 on the next page. This graph is typical of the graphs
of exponential functions of the form F(x) = a^, where a > I. The larger
the value of a, the faster the graph rises. To see this, compare the graph of
F(x) = 5^ with the graph of/(x) = 2^ in Figure 6.
Continued on Next Page
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
662 Chapter I I Exponential and Logarithmic Functions
Graph.
(a) fix) = 10
10
F(x) = 5"
8
fl fj \
6
:fw:i::\
4
liiiiiii
2
j\^m = r\
2
\ 2 \ \ \
Figure 6
By the vertical line test, the graphs in Figure 6 represent functions. As
these graphs suggest, the domain of an exponential function includes all real
numbers. Because y is always positive, the range is (0, oo). Figure 6 also
shows an important characteristic of exponential functions where a> 1: as x
gets larger, y increases at a faster and faster rate.
(b) g{x)
CAUTION
Be sure to plot a sufficient number of points to see how rapidly the
graph rises.
Answers
1. (a)
10
if
5
~\f{x) = 10^
1 1 1 1 1 1
1
1 1 1 1 1 1 •■ ^
r 1
EXAMPLE 2
Graphing an Exponential Function with a < I
Graph g{x)
Again, find some points on the graph.
X 3
2
1
1
2
3
g(x) = ar 8
4
2
1
1
2
1
4
1
8
The graph, shown in Figure 7, is very similar to that of/(x) = 2^ (Figure 6)
with the same domain and range, except that here as x gets larger,
y decreases. This graph is typical of the graph of a function of the form
F{x) = a^, where < a < I.
g(x) = (^V
Figure 7
Work Problem 1 at the Side.
Based on Examples 1 and 2, we make the following generalizations
about the graphs of exponential functions of the formF(x) = a^.
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
Section I 1.2 Exponential Functions 663
Characteristics of the Graph of F(x) = a""
1. The graph contains the point (0, 1).
2. When a > 1, the graph rises from left to right. When < a < 1, the
grwph falls from left to right. In both cases, the graph goes from the
second quadrant to the first.
3. The graph approaches the xaxis, but never touches it. (Recall from
Section 8.4 that such a line is called an asymptote.)
4. The domain is (—0°, 0°), and the range is (0, oo).
& Graph y = 2
4x3
EXAMPLE 3
Graphing a More Complicated Exponential
Function
Graph/(x) = 3^'\
Find some ordered pairs.
Ifx = 0,thenj; = 3'^^^"
51'
Ifx
2,then7 = 32(2)4=30= i.
These ordered pairs, (0, gj) and (2, 1), along with the other ordered pairs
shown in the table, lead to the graph in Figure 8. The graph is similar to the
graph of/(x) = 3^ except that it is shifted to the right and rises more rapidly.
/(x) = 32^4
X
y
1
81
1
1
9
2
1
3
9
Work Problem 2 at the Side.
OBJECTIVE Q Solve exponential equations of the form a^ = a^ for x.
Until this chapter, we have solved only equations that had the variable as
a base, like x^ = 8; all exponents have been constants. An exponential
equation is an equation that has a variable in an exponent, such as
9' = 27.
By the horizontal line test, the exponential function defined by F(x) = a^ is
a onetoone function, so we can use the following property to solve many
exponential equations.
Property for Solving an Exponential Equation
For a> and a ?^ 1, if a^ = ay then x = y.
This property would not necessarily be true ifa=l.
Answers
2.
J
y = 2'^
4
1
3
/
2
/
1
J
1
_ 1
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
664 Chapter I I Exponential and Logarithmic Functions
To solve an exponential equation using this property, follow these steps.
& Solve each equation and
check the solution.
(a) 25^ = 125
Solving an Exponential Equation
Step 1 Each side must have the same base. If the two sides of the
equation do not have the same base, express each as a power of
the same base.
Step 2 Simplify exponents, if necessary, using the rules of exponents.
Step 3 Set exponents equal using the property given in this section.
Step 4 Solve the equation obtained in Step 3.
(b) 4^ = 32
(c) 8P = 27
NOTE
These steps cannot be applied to an exponential equation like
3^= 12
because Step 1 cannot easily be done. A method for solving such
equations is given in Section 11.6.
EXAMPLE 4
Solving an Exponential Equation
Solve the equation 9^ = 27.
We can use the property given in the box if both sides are written with
the same base. Since 9 = 3^ and 27 = 3^,
9^ = 27
(32)^ = 33
32X = 33
2x = 3
3
Write with the same base. (Step 1)
Power rule for exponents (Step 2)
If ^^ = ay, thenx = y. (Step 3)
Solve for X. (Step 4)
Check that the solution set is {f} by substituting \ for x in the original equation.
Work Problem 3 at the Side.
Answers
EXAMPLE 5
Solving Exponential Equations
Solve each equation.
(a) 43^1 = 16^+2
Since 4 = 22 and 16 = 24,
(22)3xl = (2^)^+2
26X2 — 24x+8
6x  2 = 4x + 8
2x= 10
X = 5.
Verify that the solution set is {5}.
Continued on Next Page
Write with the same base.
Power rule for exponents
Set exponents equal.
Subtract 4x; add 2.
Divide by 2.
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
Section I 1.2 Exponential Functions 665
1
'"?
X = — 3 Set exponents equal.
Verify that the solution set is { — 3}.
(c)
216 = 6^
Write with the same base; —^
6
3
x= 2
4X1
Write with the same base.
Power rule for exponents
Set exponents equal.
Check that the solution set is { 2]
O Solve each equation and
check the solution.
(a) 25
x2
125^
« 4' = 
Work Problem 4 at the Side.
OBJECTIVE Q Use exponential functions in applications involving
growth or decay.
EXAMPLE 6
Solving an Application Involving Exponential
Growth
One result of the rapidly increasing world population is an increase of car
bon dioxide in the air, which scientists believe may be contributing to global
warming. Both population and carbon dioxide in the air are increasing
exponentially. This means that the growth rate is continually increasing. The
graph in Figure 9 shows the concentration of carbon dioxide (in parts per
million) in the air.
(c)
9
Carbon Dioxide in the Air
J2 350
O ^
U c^ 330 
^.2= 310 h
J2 2 .
E g a 290
5 270
1750 1800 1850 1900 1950 2000
Year
Source: Sacramento Bee, Monday, September 13, 1993.
Figure 9
Continued on Next Page
Answers
4. (a) {4} (b)
(c) {2}
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
666 Chapter I I Exponential and Logarithmic Functions
& Solve each problem.
(a) Use the function in
Example 6 to approximate
the carbon dioxide concen
tration in 1925.
The data are approximated by the function defined by
/(x) = 278(1.00084)^
where x is the number of years since 1750. Use this function and a calculator
to approximate the concentration of carbon dioxide in parts per million for
each year.
(a) 1900
Since x represents the number of years since 1750, in this case
X = 1900  1750 = 150. Thus, evaluate/(150).
/(150) = 278(1.00084)i5« Letx= 150.
«= 3 1 5 parts per million Use a calculator.
(b) 1950
Usex
= 1950  1750 = 200.
/(200) = 278(1.00084)200
= 329 parts per million
(b) Use the function in
Example 7 to find the
pressure at 8000 m.
EXAMPLE 7
Applying an Exponential Decay Function
The atmospheric pressure (in millibars) at a given altitude x, in meters, can
be approximated by the function defined by
f(x)= 1038(1.000134)^
for values of x between and 10,000. Because the base is greater than 1 and
the coefficient of x in the exponent is negative, the function values decrease
as X increases. This means that as the altitude increases, the atmospheric pres
sure decreases. (Source: Miller, A. and J. Thompson, Elements of Meteorology,
Fourth Edition, Charles E. Merrill Publishing Company, 1993.)
(a) According to this function, what is the pressure at ground level?
At ground level, x = 0, so
/(O) = 1038(1.000134)o= 1038(1) = 1038.
The pressure is 1038 millibars.
(b) What is the pressure at 5000 m?
Use a calculator to find/(5000).
/(5000) = 1038(1.000134)5000^ 531
The pressure is approximately 531 millibars.
IK
Work Problem 5 at the Side.
Answers
5. (a) 322 parts per million
(b) approximately 355 millibars
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Section I 1.3 Logarithmic Functions 669
11 • O Logarithmic Functions
The graph of j = 2^ is the curve shown in blue in Figure 10. Because j = 2^
defines a onetoone function, it has an inverse. Interchanging x andj gives
X = 2y, the inverse of y = 2^.
As we saw in Section 11.1, the graph of the inverse is found by reflecting the
graph of 7 = 2^ about the line y = x. The graph of x = 2^ is shown as a red
curve in Figure 10.
\
8
:J: r
fj = 2 /
6
JIxj
/y=x
4
j
/ ^^^
2
M
or j = log2X
i
4 6 8
Figure 10
OBJECTIVE Q Define a logarithm. We cannot solve the equation x = l^
for the dependent variable y with the methods presented up to now. The follow
ing definition is used to solve x = 2^ for j.
Logarithm
For all positive numbers a, with a^ \, and all positive numbers x,
y — log^x means the same as x — a^.
This key statement should be memorized. The abbreviation log is used for
logarithm. Read log^x as "the logarithm of x to the base a." To remember
the location of the base and the exponent in each form, refer to the following
diagrams.
Exponent
\
Logarithmic form: j = log^x
t
Base
Exponent
\
Exponential form: x = a^
t
Base
In working with logarithmic form and exponential form, remember the
following.
OBJECTIVES
Define a logarithm.
Convert between
exponential and
logarithnriic fornris.
Solve logarithnnic
equations of the form
log^ b = k for o, b, or k.
Define and graph
logarithmic functions.
Use logarithmic functions
in applications of growth
or decay.
Meaning of log^x
A logarithm is an exponent. The expression log^ x represents the exponent
to which the base a must be raised to obtain x.
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
670 Chapter I I Exponential and Logarithmic Functions
Complete the table.
Exponential
Form
Logarithmic ™
Form
2^ = 32
100i'2 = 10
logs 4 = 
''^^ 1296 = '
OBJECTIVE Q Convert between exponential and logarithmic forms.
We can use the definition of logarithm to write exponential statements
in logarithmic form and logarithmic statements in exponential form. The
following table shows several pairs of equivalent statements.
_ Exponentiay^orm
Logarithmic Forrr^
32 = 9
logj 9 = 2
log,/5 25 = 2
105 = 100,000
logjp 100,000 = 5
43 ^L
64
'^^^i4^'
Work Problem 1 at the Side.
Answers
1. log2 32 = 5; logioo 10
2'
^ 4; 64
1
1296
OBJECTIVE B Solve logarithmic equations of the form log^ b = kfor
a, b, or k. A logarithmic equation is an equation with a logarithm in at
least one term. We solve logarithmic equations of the form log^ b = kfor any
of the three variables by first writing the equation in exponential form.
EXAMPLE 1
Solving Logarithmic Equations
Solve each equation.
(a) log X = 2
By the definition of logarithm, log4 x
Solve this exponential equation.
2 is equivalent to x = 4"
The solution set is [j^].
(b) log. ., (3x + 1) = 2
3x + 1 =
Write in exponential form.
3x + 1
Multiply by 4.
Subtract 4.
Divide by 12; lowest terms
J_
4
12x + 4 = 1
12x = 3
1
The solution set is { — 4}.
(c) log^ 3 = 2
x^ = 3 Write in exponential form.
X = V 3 Take thQ principal square root.
Notice that only the principal square root satisfies the equation, since the
base must be a positive number. The solution set is { V 3}.
Continued on Next Page
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
(d) l0g49 V 7 = X
']2x — 1 1/3
1
1
The solution set is {}.
Write in exponential form.
Write with the same base.
Power rule for exponents.
Set exponents equal.
Divide by 2.
Section I 1.3 Logarithmic Functions 671
& Solve each equation.
(a) log3 27 = X
(b) \og,p = 2
Work Problem 2 at the Side.
For any real number b, wq knovs^ that b^ = b and b^ = I. Writing these
tvsAo statements in logarithmic form gives the follov^ing tv^o properties of
logarithms.
(c) log.Y^= 4
Properties of Logarithms
For any positive real number b, v^ith b ^ I,
log. 6=1 and log, 1 = 0.
(d) W 12 = 3
EXAMPLE 2
Using Properties of Logarithms
Use the preceding tv^o properties of logarithms to evaluate each logarithm.
(a) log, 7 = 1
(c) log, 1 =
(b) logV2 V2 =
(d) log 2 1 =
1
Work Problem 3 at the Side.
& Evaluate each logarithm.
2
5
(a) log2/5
OBJECTIVE Q Define and graph logarithmic functions. Now we
define the logarithmic function with base a.
Logarithmic Function
If a and x are positive numbers, with a i= I, then
Gix) = \og^x
defines the logarithmic function with base a.
To graph a logarithmic function, it is helpful to write it in exponential
form first. Then plot selected ordered pairs to determine the graph.
(b) log^TT
(c) log.l
Video
EXAMPLE 3
Graphing a Logarithmic Function
Graph J = logj/2^
By vsAriting J = logy^ x in exponential form as x = {\y, v^e can identify
ordered pairs that satisfy the equation. Here it is easier to choose values for j
and find the corresponding values of x. See the table of ordered pairs on the
next page.
Continued on Next Page
(d) log, 1
Answers
2. (a) {3} (b){25} (c){2} (d) {^^12}
3. (a) 1 (b) 1 (c) (d)
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
672 Chapter I I Exponential and Logarithmic Functions
O Graph.
(a) y = log3 X
X
y
1
4
2
1
2
1
1
2
1
4
2
3
1 [
1 :Hl 1 i
\ 2
! il
....\^ = iogi^x
i iO
: 1
L.m..2...!...3....:....4....
5*
i....i2
^
Figure 11
Plotting these points (be careful to get the values of x and y in the right
order) and connecting them with a smooth curve gives the graph in Figure
11. This graph is typical of logarithmic functions with < a < 1. The graph
of X = 2y in Figure 10, which is equivalent toy = log2 x, is typical of graphs
of logarithmic functions with base a> I.
Work Problem 4 at the Side.
(h) y = log.,..x
Based on the graphs of the functions defined hy y = log2 x in Figure 10
and 7 = logj/2 ^ i^ Figure 1 1, we make the following generalizations about
the graphs of logarithmic functions of the form G(x) = log^ x.
Characteristics of the Graph of G(x) = log^x
1. The graph contains the point (1,0).
2. When a> 1, the graph rises from left to right, from the fourth quad
rant to the first. When < a < 1, the graph/a//^* from left to right,
from the first quadrant to the fourth.
3. The graph approaches the jaxis, but never touches it. (The jaxis is
an asymptote.)
4. The domain is (0, oo), and the range is (— oo, oo).
Compare these characteristics to the analogous ones for exponential
functions in Section 11.2.
Answers
4. (a) y
OBJECTIVE Q Use logarithmic functions In applications of growth or
decay. Logarithmic functions, like exponential functions, can be applied
to growth or decay of realworld phenomena.
EXAMPLE 4
Solving an Application of a Logarithmic Function
The function defined by
/(x) = 27+ 1.105 log^,(x+ 1)
approximates the barometric pressure in inches of mercury at a distance
of X miles from the eye of a typical hurricane. (Source: Miller, A.
and R. Anthes, Meteorology, Fifth Edition, Charles E. Merrill Publishing
Company, 1985.)
Continued on Next Page
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
(a) Approximate the pressure 9 mi from the eye of the hurricane.
Letx = 9, and find/(9).
/(9) = 27+ 1.105 log,o (9+1)
Letjc = 9.
= 27+ 1.105 logjo 10
Add inside parentheses.
= 27+ 1.105(1)
logio 10 = 1
= 28.105
Add.
The pressure 9 mi from the eye of the hurricane is 28.105 in.
Section I 1.3 Logarithmic Functions 673
& Solve the problem.
A population of mites
in a laboratory is growing
according to the function
defined by
P(0 = 801og^,(^+ 10),
where t is the number of days
after a study is begun.
(a) Find the number of mites at
the beginning of the study.
(b) Approximate the pressure 99 mi from the eye of the hurricane.
/(99) = 27 + 1.105 log^o(99 + 1) Letx = 99.
= 27+1.105 logjQ 100 Add inside parentheses.
= 27 + 1.105(2) logiolOO = 2
= 29.21
The pressure 99 mi from the eye of the hurricane is 29.21 in.
Work Problem 5 at the Side.
(b) Find the number present
after 90 days.
(c) Find the number present
after 990 days.
Answers
5. (a) 80 (b) 160 (c) 240
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Section I 1.4 Properties of Logarithms 679
11 .4 Properties of Logarithms
Logarithms have been used as an aid to numerical calculation for several
hundred years. Today the widespread use of calculators has made the use
of logarithms for calculation obsolete. However, logarithms are still very
important in applications and in further work in mathematics.
OBJECTIVE Q Use the product rule for logarithms. One way in which
logarithms simplify problems is by changing a problem of multiplication
into one of addition. We know that log2 4 = 2, log2 8 = 3, and log2 32 = 5.
Since 2 + 3 = 5,
log^ 32
log2(4 • 8)
log2 4 + log2 8
log2 4 + log^ 8.
OBJECTIVES
Use the product rule for
logarithms.
Use the quotient rule for
logarithms.
Use the power rule for
logarithms.
Use properties to write
alternative forms of
logarithmic expressions.
This is true in general.
Product Rule for Logarithms
If X, J, and b are positive real numbers, where b i^ \, then
In words, the logarithm of a product is the sum of the logarithms of the
factors.
NOTE
The word statement of the product rule can be restated by replacing
"logarithm" with "exponent." The rule then becomes the familiar rule
for multiplying exponential expressions: The exponent of a product is
equal to the sum of the exponents of the factors.
auT(
To prove this rule, let m = log^ x and n = log^ j, and recall that
logft ^ — ^ means b"^ = x.
log^ J = n means b"" = y.
Now consider the product xy.
xy = b"^ ' /?"
xy = Z?"^+"
log^ xy = m \ n
log^^7 = log^^ + log^j
Substitute.
Product rule for exponents
Convert to logarithmic form.
Substitute.
The last statement is the result we wished to prove.
EXAMPLE 1
Using the Product Rule
Use the product rule to rewrite each expression. Assume x > 0.
(a) log,(6 • 9)
By the product rule,
log^Ce • 9) = logj 6 + log^ 9.
(b) log, 8 + log, 12 = log,(8 • 12) = log, 96
Continued on Next Page
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
680 Chapter I I Exponential and Logarithmic Functions
Use the product rule to
rewrite each expression.
(a) log,(5 • 8)
(c) log 3 (3x) = log3 3 + log3 X
= 1 + l0g3 X
(d) log.x^ = log.Cx ' X ' x)
log3 3 = 1
log 4 X + log 4 X + log 4 X Product rule
3 1og4X
Work Problem 1 at the Side.
(b) log4 3 + log4 7
OBJECTIVE Q Use the quotient rule for logarithms. The rule for
division is similar to the rule for multiplication.
(c) log. 8^, k>0
(d) log. m^, m>
Quotient Rule for Logarithms
If X, 7, and b are positive real numbers, where b ^ I, then
logft = logi,x  log^j.
In words, the logarithm of a quotient is the difference between the loga
rithm of the numerator and the logarithm of the denominator.
The proof of this rule is very similar to the proof of the product rule.
& Use the quotient rule to
rewrite each expression.
(a) logy
9
(b) log3;7log3^,
p>0, q>0
BEXAMPLE 21
Using the Qi
rule to rewrite
jotient Rule
Use the
quotient
sach logarithm.
(a) log
7
'9~~
= log
4 7  log4 9
(b) log,
6
 log
, 6
5X = logs,
A'
X >
(C) l0g3
27
5
= log3 27  log3 5
= 3
 log3 5
log3 27 = 3
(C) l0g4Y^
CAUTION
There is no property of logarithms to rewrite the logarithm of a sum
or difference. For example, we cannot write log^(x + y) in terms of
log^xandlog^j.Also,
X log^x
y logz^j
Answers
1. (a) log, 5 + log, 8 (b) log4 21
(c) 1 + logg k (d) 2 log5 m
2. (a) log, 9 log, 4 (b) log3
(c) log4 32
Work Problem 2 at the Side.
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Section I 1.4 Properties of Logarithms 681
OBJECTIVE Q Use the power rule for logarithms. The next rule
gives a method for evaluating powers and roots such as
2^~\ (V2)3/4, (.032)5/8, and ^/u.
This rule makes it possible to find approximations for numbers that could
not be evaluated before. By the product rule for logarithms,
log3 23 = log3(222)
= log^ 2 + log^ 2 + log^ 2
Also,
= 3 log^ 2.
log,74 = W(7777)
2 ' "^fe2V
= log2 7 + log^ 7 + log2 7 + log2 7
= 4 log^ 7.
Furthermore, we saw in Example 1(d) that log4 x^ = 3 log4 x. These examples
suggest the following rule.
Power Rule for Logarithms
If X and b are positive real numbers, where b i^ \, and if r is any real
number, then
log^A:'^ = rlog^x.
In words, the logarithm of a number to a power equals the exponent
times the logarithm of the number.
As examples of this result,
log^ m^ = 5 log^ m and log3 S'* = 4 log3 5.
To prove the power rule, let
log^x = m.
b^ = X Convert to exponential form.
{b^y = x^ Raise to the power r.
j^mr — ^r Power rule for exponents
log^ x^ = mr Convert to logarithmic form,
log^ x^ = rm Commutative property
log^ x'' = r log^ X m = log^ x
This is the statement to be proved.
As a special case of the power rule, let r = ^, so
log^Vx = logf^x^'P = log^x.
For example, using this result, with x > 0,
r 1 / — 4
log^Vx = log^x^/^ = log^x and log^Vx^ = log^x^^^ = log^x.
Another special case is
logz, = logz^x ^ = log^x.
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
682 Chapter I I Exponential and Logarithmic Functions
& Use the power rule to rewrite
each logarithm. Assume
a>0,b>0,x>0,ai^ 1,
and/? ¥^ 1.
(a) WS^
(b)log x^
(c) log,V8
(d) log2^
NOTE
For a review of rational exponents, refer to Section 9.2.
EXAMPLE 3
Using the Power Rule
Use the power rule to rewrite each logarithm. Assume b> 0,x> 0, and b i= I.
(a) logj 42 = 2 logj 4
(b) log^x^ = 51og^x
(c) log.V?
When using the power rule with logarithms of expressions involving
radicals, begin by rewriting the radical expression with a rational exponent.
log,V7 = log, 71/2 V~x
A/2
flog. 7
Power rule
(d) log^Vx^
logjX
2/5
= ^ l0g2 X
^^ = x2/5
Power rule
Work Problem 3 at the Side.
Two special properties involving both exponential and logarithmic
expressions come directly from the fact that logarithmic and exponential
functions are inverses of each other.
O Find the value of each H ;jOs
logarithmic expression. AudioJ
(a) log^, 103
(b) log, 8
(c) 5 ^"^5 3
Answers
3. (a) 21og3 5 (b) 41og^x
(c) ^log, 8 (d) ^
4. (a) 3 (b) 3 (c) 3
Special Properties
lfb>Omdb ^ l,then
Ijiog^x = x, x>0 and log^ 6^ = x.
To prove the first statement, let
y = log^ X.
b^ = X Convert to exponential form.
ly log^ ^ = X Replace y with log^ x.
The proof of the second statement is similar.
"^^^^1 — f^£J^^5X^ Using the Special Properties
Video Find the value of each logarithmic expression.
(a) logj 5^
Since log^ b^ = x,
(b) log3 9 = log3 32 = 2
(c) 4'°84io = 10
log,54 = 4.
AH
Work Problem 4 at the Side.
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
Section I 1.4 Properties of Logarithms 683
Here is a summary of the properties of logarithms.
Properties of Logarithms
If X, y, and b are positive real numbers, where b i^ I, and r is any real
number, then
Product Rule
Quotient Rule
Power Rule
Special Properties
log^xj; = log^x + log^j;
log^ = log^x log^j
OBJECTIVE Q Use properties to write alternative forms of logarithimic
expressions. Applying the properties of logarithms is important for solving
equations with logarithms and in calculus.
EXAMPLE 5
Writing Logarithms in Alternative Forms
Use the properties of logarithms to rewrite each expression. Assume all vari
ables represent positive real numbers.
(a) log4 4x^ = log4 4 + log4 x^
= 1 + 3 log4 X
Product rule
log4 4 = 1; power rule
(b) log7
1 m
 — 10^7
2 ^ n
1/2
Power rule
1
{logjjn — logy w) Quotient rule
(c) log5
be
2
log5 be
Quotient rule
Power rule
Product rule
Distributive property
= 2 log^ a — log^ be
= 2 log^ a  (log^ b + log^ e)
= 2 log^ a  log^ b  log^ e
Notice the careful use of parentheses in the third step. Since we are subtract
ing the logarithm of a product and rewriting it as a sum of two terms, we must
place parentheses around the sum.
(d) 4 log, m — log, n = log, m^ — log, n Power rule
logi
m
Quotient rule
Continued on Next Page
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
684 Chapter I I Exponential and Logarithmic Functions
& Use the properties of
logarithms to rewrite each
expression. Assume all
variables represent positive
real numbers.
(a) log. 36m ^
(e) log^(x + 1) + log^(2x 1)  log^x
= log,(x + 1) + log,(2x  1)  log, x2/3
logz
log^
(x + l)(2xl)
.2/3
2x^ + X 1
Z2J3
Power rule
Product and quotient rules
(f) logg (2p + 3r) cannot be rewritten using the properties of logarithms.
There is no property of logarithms to rewrite the logarithm of a sum.
Work Problem 5 at the Side.
(b) log2V9z
(c) log.
8r^
m
,m > \,q i^ \
(d) 21og^x + 31og^7,a i 1
(e) log4(3x + j)
Answers
5. (a) 2 + 5 logg m (b) log2 3 + log2z
(c) log^8 + 21og^rlog^(m 1)
(d) log^x^y^ (e) cannot be rewritten
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Section I 1.5 Common and Natural Logarithms 687
f
Video
11.5 Common and Natural Logarithms
1^1
As mentioned earlier, logarithms are important in many applications of
mathematics to everyday problems, particularly in biology, engineering, eco
nomics, and social science. In this section we find numerical approximations
for logarithms. Traditionally, base 10 logarithms were used most often
because our number system is base 10. Logarithms to base 10 are called
common logarithms, and log^^ x is abbreviated as simply log x, where the
base is understood to be 10.
OBJECTIVE Q Evaluate common logarithms using a calculator. We
use calculators to evaluate common logarithms. In the next example we give
the results of evaluating some common logarithms using a calculator with a
OQ^ key. (This may be a second function key on some calculators.) For
simple scientific calculators, just enter the number, then press the Clog^ key.
For graphing calculators, these steps are reversed. We will give all approxi
mations for logarithms to four decimal places.
EXAMPLE 1
Evaluating Common Logarithms
Evaluate each logarithm using a calculator.
(a) log 327.1  2.5147 (b) log 437,000 « 5.6405
(c) log .0615 « 1.2111
Notice that log .0615 ~ —1.2111, a negative result. The common loga
rithm of a number between and 1 is always negative because the logarithm
is the exponent on 10 that produces the number. For example,
101.2111 « 0615.
If the exponent (the logarithm) were positive, the result would be greater
than 1 because 10^ = 1. See Figure 12.
Figure 12
OBJECTIVES
Evaluate common
logarithms using
a calculator.
Use common logarithms
in applications.
Evaluate natural
logarithms using
a calculator.
Use natural logarithms
in applications.
O Evaluate each logarithm to
four decimal places using a
calculator.
(a) log 41,600
(b) log 43.5
(c) log .442
Work Problem 1 at the Side.
OBJECTIVE Q Use common logarithms in applications. In chemistry,
pH is a measure of the acidity or alkalinity of a solution; water, for example,
has pH 7. In general, acids have pH numbers less than 7, and alkaline solutions
have pH values greater than 7. The pH of a solution is defined as
pH=log[H30+],
where [1130^] is the hydronium ion concentration in moles per liter. It is
customary to round pH values to the nearest tenth.
Answers
1. (a) 4.6191
(b) 1.6385 (c) .3546
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
688 Chapter I I Exponential and Logarithmic Functions
& Solve the problem.
Find the pH of water with
a hydronium ion concentra
tion of 1.2 X 10Mfthis
water had been taken from a
wetland, is the wetland a rich
fen, a poor fen, or a bog?
EXAMPLE 2
Using pH in an Application
® Find the hydronium ion
concentrations of solutions
with the following pH values.
(a) 4.6
Wetlands are classified as bogs, fens, marshes, and swamps. These classi
fications are based on pH values. A pH value between 6.0 and 7.5, such
as that of Summerby Swamp in Michigan's Hiawatha National Forest,
indicates that the wetland is a "rich fen." When the pH is between 4.0 and 6.0,
the wetland is a "poor fen," and if the pH falls to 3.0 or less, it is a "bog."
{Source: Mohlenbrock, R., "Summerby Swamp, Michigan," Natural History,
March 1994.)
Suppose that the hydronium ion concentration of a sample of water from a
wetland is 6.3 X 10~^. How would this wetland be classified?
Use the definition of pH.
pH = log(6.3 X 103)
= (log 6.3 + log 103)
= [.7993  3(1)]
= .7993 + 3
2.2
Since the pH is less than 3.0, the wetland is a bog.
Work Problem 2 at the Side.
Product rule
Use a calculator.
(b) 7.5
EXAMPLE 3
Finding Hydronium Ion Concentration
Find the hydronium ion concentration of drinking water with pH 6.5.
pH = log[H30+]
6.5 = log[H30+] LetpH = 6.5.
log[H30+] = 6.5 Multiply by 1.
Solve for [H3O+] by writing the equation in exponential form, remembering
that the base is 10.
[H3O+] = 1065
[H3O+] « 3.2 X 10"^ Use a calculator.
m
Work Problem 3 at the Side.
Answers
2. 2.9; bog
3. (a) 2.5 X 105 (b) 3.2 X 10"
OBJECTIVE Q Evaluate natural logarithms using a calculator. The
most important logarithms used in applications are natural logarithms,
which have as base the number e. The number e is a fundamental number in
our universe. For this reason e, like tt, is called a universal constant. The
letter e is used to honor Leonhard Euler, who published extensive results on
the number in 1748. Since it is an irrational number, its decimal expansion
never terminates and never repeats.
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
Section I 1.5 Common and Natural Logarithms 689
V&uTr^l
Video
The first few digits of the decimal value of e are 2.718281828. A cal
culator key Ce^ or the two keys Qnv^ and ( j^ are used to approximate
powers of e. For example, a calculator gives
e^ « 7.389056099, e^ « 20.08553692, and e^ « 1.8221188.
Logarithms to base e are called natural logarithms because they occur in
biology and the social sciences in natural situations that involve growth or
decay. The base e logarithm of x is written In x (read "el en x"). A graph of
7 = In X, the equation that defines the natural logarithmic function, is given
in Figure 13.
Figure 13
A calculator key labeled Qn^ is used to evaluate natural logarithms.
If your calculator has an C^ key, but not a key labeled Qnx^ , find natural
logarithms by entering the number, pressing the C"^ key, and then press
ing the C^ key. This works because y = e^ defines the inverse function of
y = In X (or 7 = log x).
EXAMPLE 4
Finding Natural Logarithms
Find each logarithm to four decimal places.
(a) In .5841 « .5377
As with common logarithms, a number between and 1 has a negative
natural logarithm.
J (b) In 192.7 5.2611
(c) In 10.84 « 2.3832
Work Problem 4 at the Side.
O Find each logarithm to four
decimal places.
(a) In .01
(b) In 27
OBJECTIVE Q Use natural logarithms in applications. A common
application of natural logarithmic functions is to express growth or decay of
a quantity, as in the next example.
(c) In 529
An^matioii
Video
EXAMPLE 5
Applying Natural Logarithms
The altitude in meters that corresponds to an atmospheric pressure of
X millibars is given by the logarithmic function defined by
f(x) = 51,600  7457 In X.
(Source: Miller, A. and J. Thompson, Elements of Meteorology, Fourth Edition,
Charles E. Merrill Publishing Company, 1993.) Use this function to find the
altitude when atmospheric pressure is 400 millibars.
Letx = 400 and substitute in the expression for/(x).
/(400) = 51,600  7457 In 400
— 6900 (to the nearest hundred)
Atmospheric pressure is 400 millibars at approximately 6900 m.
Answers
4. (a) 4.6052 (b) 3.2958 (c) 6.2710
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
690 Chapter I I Exponential and Logarithmic Functions
Use the logarithmic function
in Example 5 to approximate
the altitude at 700 millibars
of pressure.
lull Calculator Tip In Example 5, the final answer was obtained using
a calculator without rounding the intermediate values. In general, it is
best to wait until the final step to round the answer; otherwise, a build
up of roundoff error may cause the final answer to have an incorrect
final decimal place digit.
Hi
Work Problem 5 at the Side.
Answers
5. approximately 2700 m
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
Section I 1.6 Exponential and Logarithmic Equations; Further Applications 695
11 .0 Exponential and Logarithmic Equations; Further Applications
As mentioned earlier, exponential and logarithmic functions are important in
many applications of mathematics. Using these functions in applications
requires solving exponential and logarithmic equations. Some simple equa
tions were solved in Sections 11.2 and 11.3. More general methods for solving
these equations depend on the following properties.
Properties for Solving Exponential and Logarithmic Equations
For all real numbers b > 0,b i^ 1, and any real numbers x and j:
1. If X = 7, then b' = b^.
2. IfZ?^ = by.thmx = y.
3. If X = 7, and x > 0, j > 0, then log^ x = log^ y.
4. If X > 0, 7 > 0, and log^ x = log^ y, then x = y.
We used Property 2 to solve exponential equations in Section 11.2.
OBJECTIVE Q Solve equations involving variables in the exponents.
The first examples illustrate a general method for solving exponential
equations using Property 3.
OBJECTIVES
Solve equations involving
variables in the exponents.
Solve equations involving
logarithms.
Solve applications of
compound interest.
Solve applications involv
ing base e exponential
growth and decay.
Use the changeofbase
rule.
O Solve each equation and give
the decimal approximation to
three places.
(a) 2^ = 9
EXAMPLE 1
Solve 3^ = 12.
Solving an Exponential Equation
3^= 12
log 3^ = log 12
xlog 3 = log 12
log 12
log 3
X =
Property 3
Power rule
Divide by log 3.
This quotient is the exact solution. To find a decimal approximation for the
solution, use a calculator.
X « 2.262
The solution set is {2.262}. Check that 32262
12.
(b) 10^
CAUTION
log 12
Be careful: — is not equal to log 4. Note that log 4
log 12
logs
log 3
2.262.
.6021, but
Work Problem 1 at the Side.
Answers
1. (a) {3.170} (b) {.602}
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
696 Chapter I I Exponential and Logarithmic Functions
Solve e
.ou
.38.
When an exponential equation has e as the base, it is appropriate to use
base e logarithms.
EXAMPLE 2
Solving an Exponential Equation with Base e
Solve eoo^^ = 40.
Take base e logarithms on both sides.
In eC03x = In 40
.003x In e = In 40 Power rule
.003x = In 40 In e = In gi = 1
In 40
.003
X « 1230
The solution set is {1230}. Check that e003(i230)
Divide by .003.
Use a calculator.
40.
Work Problem 2 at the Side.
O Solve log3(x+ 1)5 = 3.
Give the exact solution.
General Method for Solving an Exponential Equation
Take logarithms to the same base on both sides and then use the power
rule of logarithms or the special property log^ h^ = x. (See Examples 1
and 2.)
As a special case, if both sides can be written as exponentials with the
same base, do so, and set the exponents equal. (See Section 11.2.)
OBJECTIVE Q Solve equations involving logarithms. The properties
of logarithms from Section 11.4 are useful here, as is using the definition of
a logarithm to change the equation to exponential form.
EXAMPLE 3
Solving a Logarithmic Equation
Solve log2(x + 5)^ = 4. Give the exact solution.
(x + sy = 2^
Convert to exponential form.
(x + 5)3 = 16
x + 5 = ^^16
Take the cube root on each side.
X = 5 + ^^16
Add 5.
x= 5 + 2^
Simplify the radical.
Verify that the solution satisfies
the equation, so the solution set is
{5 + 2^^}.
Answers
2. {96.8}
3. {1 + WZ}
CAUTION
Recall that the domain of j = log^ x is (0, oo). For this reason, it is always
necessary to check that the solution of an equation with logarithms
yields only logarithms of positive numbers in the original equation.
Work Problem 3 at the Side.
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Section I 1.6 Exponential and Logarithmic Equations; Further Applications 697
EXAMPLE 4
Solving a Logarithmic Equation
Solve log2(x + 1) — log2 X = log2 7.
l0g2(x + 1)  l0g2X = l0g2 7
X + 1
log:
X
 ^^^2 1
yUULlCllL lUlC
X + 1
X
= 1
Property 4
x+ 1
= lx
Multiply by x.
1
6
= X
Subtract x; divide by 6.
Check this solution by substituting in the original equation. Here, both
X + 1 and X must 1
solution set is {^}.
X + 1 and X must be positive. If x = , this condition is satisfied, so the
O Solve
logg(2x + 5) + logg 3 = logg 33.
Work Problem 4 at the Side.
auTr
AnImatlcH
EXAMPLE 5
Solving a Logarithmic Equation
Solve logx + log(x — 21) = 2.
For this equation, write the left side as a single logarithm. Then write in
exponential form and solve the equation.
logx + log(x — 21) = 2
logx(x — 21) = 2
x(x21)= 10^
x2  21x = 100
x^  21x  100 =
(x  25) (x + 4) =
x25 = or x + 4
X = 25 or X
Product rule
log X = logjQ x; Write in
exponential form.
Distributive property; multiply.
Standard form
Factor.
Zerofactor property
Solve each equation.
The value —4 must be rejected as a solution since it leads to the logarithm of
at least one negative number in the original equation.
log(4) + log(4  21) = 2 The left side is undefined.
The only solution, therefore, is 25, and the solution set is {25}.
& Solve
log3 2x — log3(3x +15)
CAUTION
Do not reject a potential solution just because it is nonpositive. Reject
any value that leads to the logarithm of a nonpositive number.
Work Problem 5 at the Side.
Answers
4. {3}
5. {1}
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698 Chapter I I Exponential and Logarithmic Functions
In summary, we use the following steps to solve a logarithmic equation.
& Find the value of $2000
deposited at 5% compounded
annually for 10 yr.
rAudioJ
Solving a Logarithmic Equation
Step 1 Transform the equation so that a single logarithm appears
on one side. Use the product rule or quotient rule of loga
rithms to do this.
Step 2 (a) Use Property 4. If log^ x = log^y, then x = y. (See
Example 4.)
(b) Write the equation in exponential form. If log^ x = k,
then X = b^. (See Examples 3 and 5.)
OBJECTIVE Q Solve applications of compound interest. So far in this
book, problems involving applications of interest have been limited to simple
interest using the formula / = prt. In most cases, interest paid or charged is
compound interest (interest paid on both principal and interest). The formula
for compound interest is an important application of exponential functions.
Compound Interest Formula (for a Finite Number of Periods)
If a principal ofP dollars is deposited at an annual rate of interest r com
pounded (paid) n times per year, the account will contain
A = P[l +
dollars after t years. (In this formula, r is expressed as a decimal.)
EXAMPLE 6
Solving a Compound Interest Problem for 4
How much money will there be in an account at the end of 5 yr if $1000 is
deposited at 6% compounded quarterly? (Assume no withdrawals are made.)
Because interest is compounded quarterly, n = 4. The other values given in
the problem are P = 1000, r = .06 (because 6% = .06), and ^ = 5. Substitute
into the compound interest formula to get the value of ^.
A = 1000 1 + 
M
4
45
A = 1000(1.015)20
Now use the (^ key on a calculator, and round the answer to the nearest cent.
A = 1346.86
The account will contain $1346.86. (The actual amount of interest earned is
$1346.86  $1000 = $346.86. Why?)
Work Problem 6 at the Side.
Answers
6. about $3257.79
EXAMPLE 7
Solving a Compound Interest Problem for t
Suppose inflation is averaging 3% per year. How many years will it take for
prices to double?
We want to find the number of years ^ for $1 to grow to $2 at a rate of
3% per year. In the compound interest formula, we let ^ = 2,P = 1, r = .03,
and n = I.
Continued on Next Page
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
Section I 1.6 Exponential and Logarithmic Equations; Further Applications 699
1
2 = (i.osy
log 2 = log(1.03y
log 2 = Hog 1.03
log 2
t =
t
log 1.03
23.45
Substitute in the compound
interest formula.
Simplify.
Property 3
Power rule
Divide by log 1.03.
Use a calculator.
Prices will double in about 23 yr. (This is called the doubling time of the
money.) To check, verify that 1.03^^ "^^ ~ 2.
Work Problem 7 at the Side.
& Find the number of years it
w^ill take for $500 to increase
to $750 in an account paying
4% interest compounded
semiannually.
Interest can be compounded annually, semiannually, quarterly, daily, and
so on. The number of compounding periods can get larger and larger. If the
value ofn is allov^ed to approach infinity, v^e have an example of continuous
compounding. Hov^ever, the compound interest formula above cannot be
used for continuous compounding since there is no finite value for n. The
formula for continuous compounding is an example of exponential grov^th
involving the number e.
Continuous Compound Interest Formula
If a principal ofP dollars is deposited at an annual rate of interest r
compounded continuously for t years, the final amount on deposit is
A = Pe'K
0(a) How much will $2500
grow to at 4% interest
compounded continuously
for3yr?
EXAMPLE 8
Solving a Continuous Interest Problem
(a) In Example 6, we found that $1000 invested for 5 yr at 6% interest com
pounded quarterly would grow to $1346.86. How much would this same
investment grow to if compounded continuously?
A = Pe^^ Continuous compound interest formula
A = lOOOe^^^)^ LetP = 1000, r = .06, and ^ = 5.
A ~ 1349.86 Use a calculator; round to two decimal places.
The account will grow to $1349.86.
(b) How long would it take for the initial investment amount to double?
We must find the value of t that will cause ^ to be 2 ($ 1 000) = $2000.
A = Pe''
2000 = lOOOe 06^
2 = e^^'
In 2 = .06^
In 2
t
Let^ = 2P = 2000.
Divide by 1000.
Take natural logarithms; In ^^ ^
k.
.06
^« 11.55
It would take about 1 1.55 yr for the original investment to double.
Divide by .06.
Use a calculator.
Work Problem 8 at the Side.
(b) How long would it take
for the initial investment in
part (a) to double?
Answers
7. about 10.24 yr
8. (a) $2818.74 (b) about 17.33 yr
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
700 Chapter I I Exponential and Logarithmic Functions
& Radioactive strontium decays
according to the function
defined by
where t is time in years.
(a) If an initial sample contains
Jq = 12 g of radioactive
strontium, how many grams
will be present after 35 yr?
(b) What is the halflife of
radioactive strontium?
B J ECTi VE Q Solve applications involving base e exponential
growth and decay. You may have heard of the carbon14 dating process
used to determine the age of fossils. The method used is based on a base e
exponential decay function.
EXAMPLE 9
Solving an Exponential Decay Application
Carbon14 is a radioactive form of carbon that is found in all living plants and
animals. After a plant or animal dies, the radioactive carbon14 disintegrates
according to the function defined by
where t is time in years, y is the amount of the sample at time t, and j^ is the
initial amount present at ^ = 0.
(a) If an initial sample contains Jq = 10 g of carbon14, how many grams
will be present after 3000 yr?
Letj^Q =10 and t = 3000 in the formula, and use a calculator.
y = i0e"^^^^2^(^^^^^ « 6.96 g
(b) How long would it take for the initial sample to decay to half of its original
amount? (This is called the halflife.)
Let J = ^(10) = 5, and solve for t.
5 = lOe
.00012U
1
2 = '
_ ^.000121?
In = .00012U
2
In^
^ .000121
t  5728
The halflife is just over 5700 yr.
Substitute.
Divide by 10.
Take natural logarithms; In e^ = A:.
Divide by .000121.
Use a calculator.
IK
Work Problem 9 at the Side.
Video
OBJECTIVE Q Use the changeofbase rule. In Section 11.5 we used
a calculator to approximate the values of common logarithms (base 10) or
natural logarithms (base e). However, some applications involve logarithms
to other bases. For example, for the years 19801996, the percentage of
women who had a baby in the last year and returned to work is given by
f(x) = 38.83 + 4.208 log^x,
for year X. (Source: U.S. Bureau of the Census.) To use this function, we need
to find a base 2 logarithm. The following rule is used to convert logarithms
from one base to another.
Answers
9. (a) 5.20 g (b) 29 yr
ChangeofBase Rule
If (2 > 0, a 9^ 1, Z? > 0, Z? 9^ 1, andx > 0, then
log«x =
log/, a
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Section I 1.6 Exponential and Logarithmic Equations; Further Applications 701
NOTE
Any positive number other than 1 can be used for base b in the change
ofbase rule, but usually the only practical bases are e and 10 because
calculators give logarithms only for these two bases.
CO) (a) Findlog3 17 using
common logarithms.
To derive the changeofbase rule, let log^ x = m.
log X = m
log^(a^) = log^x
m log^ a = log^ X
(log^x)(log^a) = log^x
log^x
Change to exponential form.
Property 3
Power rule
Substitute for m.
log^x
Divide by log, a.
log^a
The last step gives the changeofbase rule.
EXAMPLE 10
Using the ChangeofBase Rule
Find log 5 12.
Use common logarithms and the changeofbase rule.
log 12
logs 12
logs
1.5440 Use a calculator.
(b) Find log3 17 using natural
logarithms.
NOTE
Either common or natural logarithms can be used when applying the
changeofbase rule. Verify that the same value is found in Example 10
if natural logarithms are used.
Work Problem 10 at the Side.
EXAMPLE 11
Using the ChangeofBase Rule in an Application
Use natural logarithms in the changeofbase rule and the function defined by
f(x) = 38.83 + 4.208 log^x
(given earlier) to find the percent of women who returned to work after
having a baby in 1995. In the function, x =
Substitute 1995  1980 = 15 forx.
/(15) = 38.83 + 4.208 log^ 15
'In 15
= 38.83 + 4.208
In 2
 55.3%
This is very close to the actual value of 55%.
represents 1980.
Changeofbase rule
Use a calculator.
Work Problem 11 at the Side.
CD In Example 11, what percent
of women returned to work
after having a baby in 1990?
Answers
10. (a) 2.5789 (b) 2.5789
11. 52.8%; This is very close to the actual value
of 53%.
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
Nonlinear
Functions, Conic
Sections, and
Nonlinear Systems
\
In this chapter, we study a group of curves known as conic sections.
One conic section, the ellipse, has a special reflecting property
responsible for "whispering galleries." In a whispering gallery, a person
whispering at a certain point in the room can be heard clearly at another
point across the room.
The Old House Chamber of the U.S. Capitol, now called Statuary
Hall, is a whispering gallery. History has it that John Quincy Adams,
whose desk was positioned at exactly the right point beneath the
ellipsoidal ceiling, often pretended to sleep there as he listened to
political opponents whispering strategies across the room. {Source:
We, the People, The Story of the United States Capitol, 1991.)
In Section 12.2, we investigate ellipses.
12.T Additional Graphs of Functions;
Composition
T2.2 The Circle and the Ellipse
T2.3 The Hyperbola and Other
Functions Defined by Radicals
T2.4 Nonlinear Systems of
Equations
T2.5 SecondDegree Inequalities
and Systems of Inequalities
721
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
722 Chapter 12 Nonlinear Functions, Conic Sections, and Nonlinear Systems
1 2 • 1 Additional Graplis of Functions; Composition
OBJECTIVES
Recognize the graphs of
the elementary functions
defined by  x  , ^, and V x,
and graph their translations.
Recognize and graph step
functions.
Find the composition of
functions.
In earlier chapters we introduced the function defined by/(x) = x^, sometimes
called the squaring function. This is one of the most important elementary
functions in algebra.
OBJECTIVE Q Recognize the graphs of the elementary functions
defined by \xl \, and \/x, and graph their translations. Another one of
the elementary fiiinctions, defined by f{x) = x, is called the absolute value
function. Its graph, along with a table of selected ordered pairs, is shown in
Figure 1. Its domain is (— oo, oo), and its range is [0, oo).
X
y
±1
1
±2
2
±3
3
nx) = \x\
Figure 1
The reciprocal function, defined by f{x) = \^ was introduced in
Section 8.4. Its graph is shown in Figure 2, along with a table of selected
ordered pairs. Notice that x can never equal for this function, and as a
result, as x gets closer and closer to 0, the graph approaches either co or — oo.
Also, \ can never equal 0, and as x approaches oo or — oo, ^ approaches 0.
The axes are called asymptotes for the function. (Asymptotes are studied in
more detail in college algebra courses.) For the reciprocal function, the
domain and the range are both (oo, 0) U (0, oo).
X
1
y
X
y
1
3
3
1
3
3
1
2
2
1
2
2
1
1
1
1
2
1
2
1
2
3
1
3
3
1
3
fix):
Figure 2
The square root function, defined by f{x) = V x, was introduced in
Section 9.1. Its graph is shown in Figure 3 on the next page. Notice that
since we restrict function values to be real numbers, x cannot take on nega
tive values. Thus, the domain of the square root function is [0, oo). Because
the principal square root is always nonnegative, the range is also [0, oo).
A table of values is shown along with the graph.
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
Section 12. 1 Additional Graphs of Functions; Composition 723
X
y
1
1
4
2
\f(x)= '^■
; i ; ; \2
^.i#rf
i....(0,0).
; L., 4... .;....
Figure 3
Just as the graph of/(x) = x^ can be shifted, or translated, as we saw in
Section 10.5, so can the graphs of these other elementary functions.
' Graph f{x) = Vx + 4.
Give the domain and range.
EXAMPLE 1
Applying a Horizontal Shift
Graph /(x) = \x — 2\.
The graph of j = (x — 2)^ is obtained by shifting the graph of j = x^
two units to the right. In a similar manner, the graph of /(x) =  x — 2  is
found by shifting the graph of y = \x\ two units to the right, as shown in
Figure 4. The table of ordered pairs accompanying the graph supports this,
as can be seen by comparing it to the table with Figure 1. The domain of this
function is (oo, oo), and its range is [0, oo).
X
y
2
1
1
2
3
1
4
2
fix)= \x2\
Figure 4
Work Problem 1 at the Side.
EXAMPLE 2
Graph /(x)
1
Applying aVertical Shift
+ 3.
The graph of this function is found by shifting the graph of y = ^ three
units up. See Figure 5 on the next page. The domain is (—00, 0) U (0, 00), and
the range is (00, 3) U (3, 00).
Continued on Next Page
■fix) =^x + 4
[4, ^); [0, ^)
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724 Chapter 12 Nonlinear Functions, Conic Sections, and Nonlinear Systems
& Graph/(x) =   2.
Give the domain and range.
X
y
1
3
6
1
2
5
1
4
2
3.5
X
y
1
3
1
2
1
1
2
2
2.5
/(*) = J + 3
IK
Figure 5
Work Problem 2 at the Side.
& Graph/(x) = x + 2 + 1.
Give the domain and range.
O Find each of the following.
(a) 1181 (b) 18.71
(c) 151 (d) 16.91
(e)
^]
EXAMPLE 3
Applying Both Horizontal and Vertical Shifts
Graph/(x) = Vx + 1  4.
The graph of j = (x + 1)^
4 is obtained by shifting the graph of j = x
one unit to the left and four units down. Following this pattern here, we shift
the graph ofy = Vx one unit to the left and four units down to get the graph
of/(x) = Vx +14. See Figure 6. The domain is [ 1, oo), and the range
is [4,00).
X
y
1
4
3
3
2
Figure 6
Work Problem J at the Side.
VtouTrylt'
2 —
Vide<
Answers
2. y
(00, 0) U (0, ^); (00, 2) U (2, ^)
3. 3;
f(x) = I X + 2 I + 1
V?^
I I I I I
2
(—00 ooV n 00 ")
4. (a) 18 (b) 8 (c) 5 (d) 7 (e)
OBJECTIVE Q Recognize and graph step functions. The greatest
integer function, usually written/(x) = [x]], is defined as follows:
jc]] denotes the largest integer that is less than or equal to x.
For example,
181 = 8, 17.451 = 7, [771 = 3, [11 =1, [2.61 = 3,
and so on.
IK
Work Problem 4 at the Side.
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Section 12. 1 Additional Graphs of Functions; Composition 725
EXAMPLE 4
Graph/(x) = M
ForW,
Graphing the Greatest Integer Function
if 1 <x<0, then [xl = 1;
if
0<x< 1,
then
if
1 < X < 2,
then
M = 0;
and so on. Thus, the graph, as shown in
Figure 7, consists of a series of horizontal
line segments. In each one, the left end
point is included and the right endpoint is
excluded. These segments continue infi
nitely following this pattern to the left and
right. Since x can take any real number
value, the domain is (—00, 00). The range is
the set of integers {. . . , —4, —3, —2, —1,
0, 1, 2, 3, 4, ... }. The appearance of the
graph is the reason that this function is
called a step function.
Figure 7
The graph of a step function also may be shifted. For example, the graph
of h (x) = [x  2]] is the same as the graph of /(x) = [x]] shifted two units to
the right. Similarly, the graph of g(x) = [x]] + 2 is the graph of/(x) shifted
two units up.
Work Problem 5 at the Side.
M = 1,

r5^
; :/(x) = [[xi[^
]■!
M
I H :
■■ : H :
: H I
;5432 r^
^ H '
i H 3
: H ^ 4
^ H i : 5
] 2 3 4 5 : *
EXAMPLE 5
Applying a Greatest Integer Function
An overnight delivery service charges $25 for a package weighing up to 2 lb.
For each additional pound or fraction of a pound there is an additional
charge of $3. Let D(x) represent the cost to send a package weighing
X pounds. Graph D(x) for x in the interval (0, 6].
For X in the interval (0, 2], y = 25.
For X in the interval (2, 3], 7 = 25 + 3 = 28.
For X in the interval (3, 4], j = 28 + 3 = 31, and so on.
The graph, which is that of a step function, is shown in Figure 8.
40
30
Q 20
10
F^^
HH
FH
12 3 4 5
Pounds
Figure 8
Work Problem 6 at the Side.
& Graph/(x) = [x + IJ. Give the
domain and range.
3^
& Assume that the post office
charges SO^ per oz (or fraction
of an ounce) to mail a letter to
Europe. Graph the ordered
pairs (ounces, cost) for x in the
interval (0,4].
Answers
5. y
4
2
;rt
1 2
);{.
2,1,0, 1,2,
3.20
^ 2.40
d 1.60
.80
M
e^
e^
H
12 3 4
Ounces
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
726 Chapter 12 Nonlinear Functions, Conic Sections, and Nonlinear Systems
OBJECTIVE Q Find the composition of functions. The diagram in
Figure 9 shows a function /that assigns to each element x of set X some
element J of set Y. Suppose that a function g takes each element of set 7 and
assigns a value z of set Z. Using both /and g, then, an element x in X is
assigned to an element z in Z. The result of this process is a new function h,
which takes an element x in X and assigns an element z in Z.
Figure 9
This function h is called the composition of functions g and/ written g o/ and
is defined as follows.
Composition of Functions
If/ and g are functions, then the composite function, or composition,
of g and/ is defined by
(g^f)(x)=glf(x)]
for all X in the domain of/ such that/(x) is in the domain of g.
Read g o/as "g of/."
As a reallife example of how composite functions occur, consider the
following retail situation.
A $40 pair of blue
jeans is on sale for
25% off. If you
purchase the jeans
before noon, the
retailer offers an
additional 10% off
What is the final
sale price of the
blue jeans?
You might be tempted to say that the jeans are 35% off and calculate
$40(.35) = $14, giving a final sale price of $40  $14 = $26 for the jeans.
This is not correct. To find the final sale price, we must first find the price
after taking 25%) off, and then take an additional 10%) off that price.
$40 (.25) = $10, giving a sale price of $40  $10 = $30.
$30(.10) = $3, giving di final sale price of $30  $3 = $27.
This is the idea behind composition of functions.
Take 25% off
original price.
Take additional
10% off
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Section 12. 1 Additional Graphs of Functions; Composition 727
As another example, suppose an oil well off the California coast is leak
ing, with the leak spreading oil in a circular layer over the surface. See
Figure 10.
Figure 10
At any time t, in minutes, after the beginning of the leak, the radius of
the circular oil slick is given by r{t) = 5t feet. Since A (r) = irr^ gives the
area of a circle of radius r, the area can be expressed as a function of time by
substituting 5t for r in^(r) = irr^io get
A{r) = irr^
^[r(0] = 77(50^ = 2577^2
The function ^[r (^J is a composite function of the functions A and r.
^^^^^^^^^ Evaluating a Composite Function
Let/(x) = x2 and g (x) = X + 3. Find (/ o g) (4).
(/°g)(4)=/[g(4)]
Definition
= /(4 + 3)
Use the rule forg(x); g(4) = 4 + 3.
= /(7)
Add.
= 72
Use the rule for/(x);/(7) = 71
= 49
Notice in Example 6 that if we reverse the order of the functions, the
composition of g and/ is defined by g[/(x)]. Once again, letting x = 4,
we have
(g°/)(4) = g[/(4)] Definition
= g(42) Use the rule for/(x);/(4) = 4^.
= g(16) Square 4.
= 16 + 3 Use the rule for g(x); g(16) = 16 + 3.
= 19.
Here we see that (/ o g) (4) i^ (g o/) (4) because 49 ?^ 19. In general,
{f^g)ix)*{gof){x).
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728 Chapter 12 Nonlinear Functions, Conic Sections, and Nonlinear Systems
O Let/(x) = 3x + 6 and
g(x) = x^. Find each of the
following.
(a) (fog) (2)
(b)(go/)(2)
(c) (f°g)(x)
(d)(go/)(x)
EXAMPLE 7
Finding Composite Functions
Let/(x) = 4x — 1 andg(x) = x^ + 5. Find each of the following.
(a) (fog) (2)
(f°g)(2)=f[g(2)]
= f(2' + 5)
= /(9)
= 4(9)  1
= 35
gix)=x^ + 5
fix) = 4x  1
(b)(/°g)(x)
Here, use g (x) as the input for the function/
if°g)(x)=f[g(x)]
= 4 (g(x))  1 Use the rule for/(x);/(x) = 4x  1.
= 4(a:^ + 5)1 gix)=x^ + 5
= 4x^ + 20 — 1 Distributive property
= 4x^ + 19 Combine terms.
(c) Find (/ o g) (2) again, this time using the rule obtained in part (b).
(/ ° g) (x) = 4x^ + 19 From part (b)
(/°g)(2) = 4(2)2 +19 Letx = 2.
= 4(4) + 19
= 16+19
= 35
The result, 35, is the same as the result in part (a).
(d)(go/)(x)
Here, use/(x) as the input for the function g.
(g^f)(x) = g[f{x)]
= [f(x)f + 5 Use the rule for g(x); g(x) = x^ + 5.
= (4x ly + 5 f(x) = 4x 1
= 16x2  8x + 1 + 5 (xyy=x^2xy^y^
= 16x2 — 8x + 6 Combine terms.
Compare this result to that in part (b). Again, (/ ° g) (x) ?^ (g o/) (x).
Work Problem 7 at the Side.
Video
Answers
7. (a) 30 (b) 1728 (c) 3x^ + 6
(d) (3x + 6)3
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Section 12.2 The Circle and the Ellipse 733
imation?
1 2 • 2 The Circle and the Ellipse
When an infinite cone is intersected by a plane, the resulting figure is called a
conic section. The parabola is one example of a conic section; circles, ellipses,
and hyperbolas may also result. See Figure 11.
Ellipse
Parabola
Figure 11
Hyperbola
OBJECTIVE Q Find an equation of a circle given the center and radius.
A circle is the set of all points in a plane that lie a fixed distance from a fixed
point. The fixed point is called the center, and the fixed distance is called the
radius. We use the distance formula from Section 9.3 to find an equation of
a circle.
EXAMPLE 1
Finding an Equation of a Circle and Graphing It
Find an equation of the circle with radius 3 and center at (0, 0), and graph it.
If the point (x, y) is on the circle, the distance from (x, y) to the center (0, 0)
is 3. By the distance formula,
V(X2  x^y + (72  yiV = d Distance formula
Vo
^x  0)' + (j  0)' = 3
;^2 _^ ^2 _ g Square both sides.
An equation of this circle isx^ \ y^ = 9. The graph is shown in Figure 12.
'
1
;
3 1 : 1 1
"""''^iN^ :
CenterV ;
: f : :
^ I MO
^: : 1 :
^^'\yOc,y)
_>< \
x2+/=9
Figure 12
Work Problem 1 at the Side.
»li
OBJECTIVES
Find the equation of a
circle given the center
and radius.
Determine the center
and radius of a circle
given its equation.
Recognize the equation
of an ellipse.
Graph ellipses.
Find an equation of the circle
with radius 4 and center
(0, 0). Sketch its graph.
Answers
1. x2+/= 16
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734 Chapter 12 Nonlinear Functions, Conic Sections, and Nonlinear Systems
A circle may not be centered at the origin, as seen in the next example.
& (a) Find an equation of the
circle with center at
(3, —2) and radius 4.
Graph the circle.
(b) Use the centerradius
form to determine the
center and radius of
(x  Sy + (y + If = 9,
and then graph the circle.
Answers
2. (a) (x  3)2 + (j + 2)2 = 16
(b) center at (5, —2); radius 3
y
(x  sy + (y + 2y = 9
EXAMPLE 2
Finding an Equation of a Circle and Graphing It
Find an equation of the circle with center at (4, 3) and radius 5, and graph it.
Use the distance formula again.
■1
Videoj
Vo
4)2 + [3, _ (_3)]2
(x  4)2 + (y + 3)2
5
25
Square both sides.
To graph the circle, plot the center (4, 3), then move 5 units right, left, up,
and down from the center. Draw a smooth curve through these four points,
sketching one quarter of the circle at a time. The graph of this circle is shown
in Figure 13.
(x4)2+(j + 3)2 = 25
Figure 13
Examples 1 and 2 suggest the form of an equation of a circle with radius
r and center at (A, k). If (x, y) is a point on the circle, then the distance from
the center (A, k) to the point (x, y) is r. By the distance formula,
V(x  hy + (y  kY = r.
Squaring both sides gives us the following centerradius form of the equation
of a circle.
Equation of a Circle (CenterRadius Form)
An equation of a circle of radius r with center at (A, k) is
(x  hf + (j  kf = r\
EXAMPLE 3
Using the Center^Radius Form of the Equation
of a Circle
Find an equation of the circle with center at ( 1, 2) and radius 4.
Use the centerradius form, with h = —l,k=2, and r = 4.
(x  hf + iy kf
[x(l)r + (y2f
(x + 1)2 + (y If
u
42
16
Work Problem 2 at the Side.
(5,2)
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Section 12.2 The Circle and the Ellipse 735
OBJECTIVE Q Determine the center and radius of a circle given its
equation. In the equation found in Example 2, multiplying out (x  4)^
and (y + 3)^ and then combining like terms gives
(x  4)2 + (j + 3)2 = 25
x^  8x + 16 + y^ + 6y + 9 = 25
x^ + y^  Sx + 6y = 0.
This general form suggests that an equation with both x^ and j^terms with
equal coefficients may represent a circle. The next example shows how to tell,
by completing the square. This procedure was introduced in Section 10.1.
& Find the center and radius
of the circle with equation
x^ +y^ 6x+8y 11 = 0.
aulr
EXAMPLE 4
Completing the Square to Find the Center
and Radius
Graph x^ + y^ + 2x + 6y  15 = 0.
Since the equation has x^ and j^terms with equal coefficients, its
graph might be that of a circle. To find the center and radius, complete the
squares onx andj.
x^ + y^ + 2x + 6y = 15
(x^ + 2x ) + (j^ + 6j ) = 15
1
(2)
1
1
(6)
Get the constant on the
right.
Rewrite in anticipation of
completing the square.
Square half the
coefficient of each
middle term.
(x^ + 2x + 1) + (y^ + 6j + 9)
(x + ly + (y + 3)2
[x(l)f + [y(3)r
The last equation shows that the graph is a circle with center at ( 1, 3) and
radius 5. The graph is shown in Figure 14.
15 + 1 + 9
25
52
Complete the squares on
both X and J.
Factor on the left; add on
the right.
Centerradius form
x^ + y^ + 2x + 6y15 =
Figure 14
NOTE
If the procedure of Example 4 leads to an equation of the form
(x  hf + (7  ky = 0, then the graph is the single point (A, k). If the
constant on the right side is negative, then the equation has no graph.
Work Problem 3 at the Side.
Answers
3. center at (3, —4); radius 6
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736 Chapter 12 Nonlinear Functions, Conic Sections, and Nonlinear Systems
OBJECTIVE Q Recognize the equation of an ellipse. An ellipse is the
set of all points in a plane the sum of whose distances from two fixed points
is constant. These fixed points are called foci (singular: yoct/^*). Figure 15
shows an ellipse whose foci are (c, 0) and (— c, 0), with xintercepts {a, 0) and
(a, 0) andjintercepts (0, b) and (0, b). It can be shown that c^ = a^  b^
for an ellipse of this type. The origin is the center of the ellipse.
Figure 15
From the preceding definition, it can be shown by the distance formula
that an ellipse has the following equation.
Equation of an Ellipse
The ellipse whose xintercepts are (a, 0) and (—a, 0) and whose jintercepts
are (0, b) and (0, —b) has an equation of the form
a b
NOTE
A circle is a special case of an ellipse, where a^
p
H!fr
IF
1
J
I
When a ray of light or sound emanating from one
focus of an ellipse bounces off the ellipse, it passes
through the other focus. See the figure. As mentioned
in the chapter introduction, this reflecting property is
responsible for whispering galleries. John Quincy
Adams was able to listen in on his opponents' con
versations because his desk was positioned at one of
the foci beneath the ellipsoidal ceiling and his oppo
nents were located across the room at the other focus.
The paths of Earth and other planets around the
sun are approximately ellipses; the sun is at one focus and a point in space is
at the other. The orbits of communication satellites and other space vehicles
are also elliptical.
Reflecting property
of an ellipse
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Section 12.2 The Circle and the Ellipse 737
Elliptical bicycle gears are designed to respond to the legs' natural
strengths and weaknesses. At the top and bottom of the powerstroke, where
the legs have the least leverage, the gear offers little resistance, but as the
gear rotates, the resistance increases. This allows the legs to apply more
power where it is most naturally available. See Figure 16.
O Graph each ellipse.
Figure 16
OBJECTIVE Q Graph ellipses. To graph an ellipse centered at the
origin, we plot the four intercepts and then sketch the ellipse through those
points.
EXAMPLE 5
Graphing Ellipses
Graph each ellipse.
1
Here, a^ = 49, so a = 7, and the
xintercepts for this ellipse are (7, 0) and
(—7, 0). Similarly, b^ = 36, so b = 6, and
the jintercepts are (0, 6) and (0, 6).
Plotting the intercepts and sketching the
ellipse through them gives the graph in
Figure 17.
49 36
Figure 17
(b)
+
1
y
36 121
The xintercepts for this ellipse are (6, 0) and (6, 0), and the jintercepts
are (0, 11) and (0,  1 1). Join these intercepts with the smooth curve of an
ellipse. The graph has been sketched in Figure 18.
^ + ^ = 1
36 121
Figure 18
Work Problem 4 at the Side.
Answers
4. (a)
(b)
I I I I I I " X I I I I I I
64 49
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738 Chapter 12 Nonlinear Functions, Conic Sections, and Nonlinear Systems
& Graph
(x + 4y , (y  1)
16
+
36
As with the graphs of parabolas and circles, the graph of an ellipse may
be shifted horizontally and vertically, as in the next example.
EXAMPLE 6
Graph
(x  ly
Graphing an Ellipse Shifted Horizontally
and Vertically
(y + 3)'
+ — = 1.
25 49
Just as (x  ly and {y + 3)^ would indicate that the center of a circle
would be (2, —3), so it is with this ellipse. Figure 19 shows that the graph
goes through the four points (2, 4), (7, —3), (2, —10), and (—3, —3). The
xvalues of these points are found by adding ±a = ±5 to 2, and the j values
come from adding ±b = ±7 to —3.
(xif _^{l + yf ^^
Figure 19
Work Problem 5 at the Side.
Answers
(4, 7)
NOTE
The graphs in this section are not graphs of functions. The only conic
section whose graph is a function is the vertical parabola with equation
/(x) = ax^ + to + c.
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Section 12.3 The Hyperbola and Other Functions Defined by Radicals 743
12.3 The Hyperbola and Other Functions Defined by Radicals
OBJECTIVE Q Recognize the equation of a hyperbola. A hyperbola
is the set of all points in a plane such that the absolute value of the difference
of the distances from two fixed points (calledybc/) is constant. Figure 20
shows a hyperbola. Using the distance formula and the definition above, we
can show that this hyperbola has equation
16
2^
12
1.
OBJECTIVES
Recognize the equation
of a hyperbola.
Graph hyperbolas
by using asymptotes.
Identify conic sections
by their equations.
Graph certain square
root functions.
Figure 20
To graph hyperbolas centered at the origin, we need to find their inter
cepts. For the hyperbola in Figure 20, we proceed as follows.
jIntercepts
jcIntercepts
Let J = 0.
16 12 ~ ^
— = 1
16
x^= 16
X = ±4
Letj = 0.
Multiply by 16.
The xintercepts are (4, 0) and
(4,0).
Letx = 0.
0^
16
12
12
= 1
= 1
Letx = 0.
J2 = 12 Multiply by 12.
Because there are no real solutions
to j;2 =  12, the graph has no
yintercepts.
The graph of
1 in Figure 20 has no yintercepts. On the
_ 2^
16 12
other hand, the hyperbola in Figure 2 1 has no xintercepts. Its equation is
= 1, with jintercepts (0, 5) and (0, —5).
25 y
Ky
3
25
/'"X
Figure 21
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
744 Chapter 12 Nonlinear Functions, Conic Sections, and Nonlinear Systems
Equations of Hyperbolas
A hyperbola with xintercepts (a, 0) and (—a, 0) has an equation
of the form
a' b' '
and a hyperbola with jintercepts (0, b) and (0, —b) has an equation
of the form
b' a'
OBJECTIVE Q Graph hyperbolas by using asymptotes. The two
branches of the graph of a hyperbola approach a pair of intersecting straight
lines, which are its asymptotes. (See Figure 22 on the next page.) The
asymptotes are useful for sketching the graph of the hyperbola.
m
Asymptotes of Hyperbolas
The extended diagonals of the rectangle with vertices (corners) at the
points (a, b), (—a, b), (—a, —b), and (a, —b) are the asymptotes of the
hyperbolas
7=1 and
^= 1.
This rectangle is called the fundamental rectangle. Using the methods of
Chapter 4, we could show that the equations of these asymptotes are
y =
and
•^ a
To graph hyperbolas, follow these steps.
Graphing a Hyperbola
Step 1 Find the intercepts. Locate the intercepts at (a, 0) and (a, 0)
if the x^term has a positive coefficient, or at (0, b) and (0, b)
if the j^term has a positive coefficient.
Step 2 Find the fundamental rectangle. Locate the vertices of
the fundamental rectangle at {a, b), {a, b), {a, /?), and
(a, b).
Step 3 Sketch the asymptotes. The extended diagonals of the rectangle
are the asymptotes of the hyperbola, and they have equations
J = ± x.
Step 4 Draw the graph. Sketch each branch of the hyperbola
through an intercept and approaching (but not touching) the
asymptotes.
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
Video
Section 12.3 The Hyperbola and Other Functions Defined by Radicals 745
EXAMPLE 1
Graph
Step 1
Step!
16
25
Graphing a Horizontal Hyperbola
1.
Here a = A and b = 5. The xintercepts are (4, 0) and (—4, 0).
The four points (4, 5), (4, 5), (4, 5), and (4, 5) are the
vertices of the fundamental rectangle, as shown in Figure 22.
Steps 3 The equations of the asymptotes are j = ±  x. The hyperbola
and 4 approaches these lines as x and j get larger in absolute value.
CAUTION
When sketching the graph of a hyperbola, be sure that the branches do
not touch the asymptotes.
Work Problem 1 at the Side.
EXAMPLE 2
Graphing aVertical Hyperbola
y^ x"
0,aph^=l.
This hyperbola hasjintercepts (0, 7) and (0, 7). The asymptotes are the
extended diagonals of the rectangle with vertices at (4, 7), (4, 7), (4, 7),
and (4, —7). Their equations are j = ± x. See Figure 23.
(_4,7)y^f^V(4,7)
(4,7)ih
49 16
= 1
Figure 23
it (4, 7)
Work Problem 2 at the Side.
Graph
3^
25
®°"""'5rii = ''
15 9 3
3 9 15
Answers
1.
1: ^_
81 64
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746 Chapter 12 Nonlinear Functions, Conic Sections, and Nonlinear Systems
OBJECTIVE Q Identify conic sections by their equations. Rewriting a
seconddegree equation in one of the forms given for ellipses, hyperbolas, cir
cles, or parabolas makes it possible to identify the graph of the equation.
SUMMARY OF CONIC SECTIONS
Equation
y — ax^ + bx + c
or
y = a{x — hy + k
X = ay^ + by \ c
or
X = a(y — kY + h
{xhY +
{y  kY = r^
a' b'
= 1
y_
b'
= 1
Graph
(h,k)
Parabola
Circle
(a, 0)
(0,b)
(a,0)
^(0,b)
Ellipse
Hyperbola
Description
It opens up if a > 0,
down if (2 < 0.
The vertex is (h, k).
It opens to the right
if^ > 0, to the
left if ^<0.
The vertex is (h, k).
The center is (/z, k),
and the radius
is r.
The xintercepts
are (a, 0) and
(^,0).
The j^intercepts
are (0, b) and
(0, b).
The xintercepts
are (a, 0) and
(a,0).
The asymptotes
are found from
(a, b), {a, —b),
{—a, —b), and
(a,b).
The jintercepts
are (0, b) and
(0, by
The asymptotes
are found from
(a, b), (a, b),
(—a, —b), and
(a,b).
Identification
It has anx^term.
y is not squared.
It has aj^term.
X is not squared.
x^ andj^^terms
have the
same positive
coefficient.
x^ andj^terms
have different
positive
coefficients.
x^ has a positive
coefficient.
y^ has a negative
coefficient.
y^ has a positive
coefficient.
x^ has a negative
coefficient.
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
Section 12.3 The Hyperbola and Other Functions Defined by Radicals 747
EXAMPLE 3
Identifying the Graphs of Equations
Identify the graph of each equation.
(a) 9x2 = 108 + 12^2
Both variables are squared, so the graph is either an elHpse or a hyperbola.
(This situation also occurs for a circle, which is a special case of the ellipse.)
To see which one it is, rewrite the equation so that the x^ andj^terms are on
one side of the equation and 1 is on the other.
9x2 _ I2j;2
108
12 9
Subtract \2y^.
Divide by 108.
Because of the minus sign, the graph of this equation is a hyperbola.
(b) x^ = J  3
Only one of the two variables, x, is squared, so this is the vertical
parabola J = x^ + 3.
(c) x^ = 9y^
Write the variable terms on the same side of the equation.
x2+3;2 = 9 Add;;^.
The graph of this equation is a circle with center at the origin and radius 3.
Work Problem 3 at the Side.
& Identify the graph of each
equation.
(a) 3x2 = 21  4j2
(b) 6x2 = 100 + 2y^
OBJECTIVE Q Graph certain square root functions. Recall from
Section 4.5 that no vertical line will intersect the graph of a function in more
than one point. Thus, horizontal parabolas and all circles, ellipses, and
hyperbolas are examples of graphs that do not satisfy the conditions of a
function. However, by considering only a part of the graph of each of these
we have the graph of a function, as seen in Figure 24.
(c) 3x2 = 21 Ay
rx
(a)
(b)
In parts (a)(d) of Figure 24, the top portion of a conic section is shown
(parabola, circle, ellipse, and hyperbola, respectively). In part (e), the top
two portions of a hyperbola are shown. In each case, the graph is that of a
function since the graph satisfies the conditions of the vertical line test.
In Sections 9.1 and 12.1 we observed the square root function defined by
/(x) = vx. To find equations for the types of graphs shown in Figure 24,
we extend its definition.
(d) 3x2 = 21  33;2
[5:
iudu
Square Root Function
For an algebraic expression u, with i/ > 0, a function of the form
fix) = \4
is called a square root function.
Answers
3. (a) ellipse
(d) circle
(b) hyperbola (c) parabola
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748 Chapter 1 2 Nonlinear Functions, Conic Sections, and Nonlinear Systems
O Graph f(x) = V36  x\
Give the domain and range.
& Graph
3
Give the domain and range.
EXAMPLE 4
Graphing a Semicircle
Graph /(x) = v 25  x^. Give the domain and range.
Replace/(x) with j and square both sides to get the equation
25  x^ or x^ + y2 = 25.
y
This is the graph of a circle with center at (0, 0) and radius 5. Since/(x), or
7, represents a principal square root in the original equation, /(x) must be
nonnegative. This restricts the graph to the upper half of the circle, as shown
in Figure 25. Use the graph and the vertical line test to verify that it is indeed
a function. The domain is [—5, 5], and the range is [0, 5].
fix) = hSx^
Figure 25
Work Problem 4 at the Side.
I I I I 1 1 •■ ^
fix) = V36  x^
[6, 6]; [0,6]
Xi^
3 VI ^
[2, 2]; [3,0]
EXAMPLE 5
Graphing a Portion of an Ellipse
Graph
y
Give the domain and range.
6 V 16
Square both sides to get an equation whose form is known.
36
x' y'
— + —
16 36
1
1
X
Ye
Add
16'
This is the equation of an ellipse with xintercepts (4, 0) and (4, 0) and
jintercepts (0, 6) and (0, —6). Since f equals a negative square root in the
original equation, y must be nonpositive, restricting the graph to the lower
half of the ellipse, as shown in Figure 26. Verify that this is the graph of a
function, using the vertical line test. The domain is [—4, 4], and the range
is [6,0].
Figure 26
Work Problem 5 at the Side.
Copyrigfit © 2005 Pearson Education, Inc., publisfiing as Pearson AddisonWesley
Section 12.4 Nonlinear Systems of Equations 753
1 2 • 4 Nonlinear Systems of Equations
An equation in which some terms have more than one variable or a variable
of degree 2 or greater is called a nonlinear equation. A nonlinear system
of equations includes at least one nonlinear equation.
When solving a nonlinear system, it helps to visualize the types of
graphs of the equations of the system to determine the possible number of
points of intersection. For example, if a system includes two equations where
the graph of one is a parabola and the graph of the other is a line, then there
may be 0, 1, or 2 points of intersection, as illustrated in Figure 27.
y^ iy
No points of intersection
One point of intersection
Figure 27
Two points of intersection
OBJECTIVES
Solve a nonlinear system
by substitution.
Use the elimination
method to solve a system
with two seconddegree
equations.
Solve a system that
requires a combination of
methods.
OBJECTIVE Q Solve a nonlinear system by substitution. We solve
nonlinear systems by the substitution method, the elimination method, or a
combination of the two. The substitution method (Section 5.1) is usually
appropriate when one of the equations is linear.
EXAMPLE 1
Solving a Nonlinear System by Substitution
Solve the system.
x^ + y^ = 9
2x
■y
(1)
(2)
The graph of (1) is a circle and the graph of (2) is a line. Visualizing the
possible ways the graphs could intersect indicates that there may be 0, 1, or
2 points of intersection. See Figure 28. First solve the linear equation for one
of the two variables, and then substitute the resulting expression into the
nonlinear equation to obtain an equation in one variable.
Substitute 2x
2x  J = 3
J = 2x — 3
3 for J in equation (1).
x^ + j2 = 9
x^ + (2x  3)2 = 9
x^ + 4x2  12x + 9 = 9
5x2  i2x =
x(5x  12) =
12
5
or X
(2)
(3)
(1)
Subtract 9; combine terms.
Factor; GCF is x.
Zerofactor property
Let X = in equation (3) to getj = 3. If x = y, then y = \. The solution
set of the system is {(0, —3), (y, )}. The graph in Figure 29 on the next
page confirms the two points of intersection.
Continued on Next Page
No points of intersection
One point of intersection
Two points of intersection
Figure 28
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754 Chapter 12 Nonlinear Functions, Conic Sections, and Nonlinear Systems
Solve each system.
(a) x2 + j;2 = 10
X = J + 2
(b) x2  2j2 =
y + X = 6
& Solve each system.
(a) xy = 8
X + y = 6
(b) xy + 10
4x + 9 J
Answers
1. (a) {(3,1), (1,3)}
(b) {(4, 2), (20, 14)}
2. (a) {(4, 2), (2, 4)}
^9 _20
v2' 9
(b) (5,2),
Circle:
x^ +3^2 = 9
Figure 29
— II
EXAMPLE 2
Work Problem 1 at the Side.
Solving a Nonlinear System by Substitution
Solve the system.
6x  y = 5 (1)
xy = 4 (2)
The graph of (1) is a line. We have not specifically mentioned equations like
(2); however, it can be shown by plotting points that its graph is a hyperbola.
Visualizing a line and a hyperbola indicates that there may be 0, 1, or 2 points
of intersection. Since neither equation has a squared term, we can solve either
equation for one of the variables and then substitute the result into the other
equation. Solving xy
4
y
24
y
24
= 4 for X gives X
y = 5
y = 5
y^ = 5y
=y^ + 5y  24
= (73)(7 + 8)
= 3 or 7= 8
. Substitute  for x in equation (1).
Letx
4
y
Multiply.
Multiply by J, J ¥^ 0.
Standard form
Factor.
Zerofactor property
We substitute these results into x = ^ to obtain the corresponding values of x.
If 7 = 3, then x
4
3'
If 7 = 8, thenx
The solution set of the system is {(f, 3), (^, 8)}. See Figure 30.
Hyperbola: xy = 4
Line: 6x y = 5
Figure 30
Work Problem 2 at the Side.
Section 12.4 Nonlinear Systems of Equations 755
OBJECTIVE Q Use the elimination method to solve a system with
two seconddegree equations. The elimination method (Section 5.1) is
often used when both equations are second degree.
EXAMPLE 3
Solving a Nonlinear System by Elimination
Solve the system.
2x2  ^^2
(1)
y = 6 (2)
The graph of (1) is a circle, while the graph of (2) is a hyperbola. By
analyzing the possibilities we conclude that there may be 0, 1, 2, 3, or
4 points of intersection. Adding the two equations will eliminate y, leaving
an equation that can be solved for x.
x^+y^= 9
2x^y^= 6
3x2 = 3
x= 1
X"
or
x= 1
Divide by 3.
Square root property
Each value of x gives corresponding values fory when substituted into one
of the original equations. Using equation (1) is easier since the coefficients
of the x2 andj^terms are 1.
If X = 1, then
12 + j;2 = 9
72 = 8
J = Vs or y =  Vs
y = 2 V 2 or y = — 2 v 2.
The solution set is
{(1,2V2), (1, 2V2), (1,2V2), (1, 2V2)}.
Figure 3 1 shows the four points of intersection.
& Solve each system.
(a) x2 + y^
X2/
41
9
Ifx= l,theii
(1)2+^2 = 9
y2 = 8
(b) x2 + 3y^ = 40
y = Vs or y = Vs
4x2 y2 = 4
y = 2V2 or j= 2V2.
Figure 31
Work Problem 3 at the Side.
Answers
3. (a) {(5,4),(5, 4),(5,4),(5, 4)}
(b) {(2,2\/3),(2, 2\/3),.
(2,2V3), (2, 2\/3)}
756 Chapter 12 Nonlinear Functions, Conic Sections, and Nonlinear Systems
OBJECTIVE Q Solve a system that requires a combination of methods.
Solving a system of seconddegree equations may require a combination
of methods.
EXAMPLE 4
Solving a Nonlinear System by a Combination
of Methods
Solve the system.
x^ + 2xy — y^ = 1
y
(1)
(2)
While we have not graphed equations like (1), its graph is a hyperbola.
The graph of (2) is also a hyperbola. Two hyperbolas may have 0, 1,2,
3, or 4 points of intersection. We use the elimination method here in
combination with the substitution method. We begin by eliminating the
squared terms by multiplying each side of equation (2) by  1 and then
adding the result to equation (1).
x^ + 2xy
Next, we solve Ixy
Ixy = 4
4 for J. (Either variable would do.)
Ixy = 4
y = \ (3)
Now, we substitute y =
this with equation (2).
into one of the original equations. It is easier to do
j2 = 3
(2)
!J
■^3
 4 = 3x2
Multiply by x'^,x +
4 =
Subtract Sx^.
f 1) =
Factor.
X^
(x2  4) (x2
x2  4 = or x^ + 1
X = — /
x^ = 4 or x^ = — 1
X = 2 or X = — 2 X = / or
Substituting these four values of x into equation (3) gives the corre
sponding values for J.
If X = 2, then J = 1. If x = /, then j = —li.
Ifx = —2, then J = — 1. Ifx = — /, thenj = li.
Note that if we substitute the xvalues we found into equation (1) or (2)
instead of into equation (3), we get extraneous solutions. It is always wise to
check all solutions in both of the given equations. There are four ordered
pairs in the solution set, two with real values and two with nonreal complex
values. The solution set is
{(2,1), (2,
Continued on Next Page
1), a, 20, (?, 2/)}.
Section 12.4 Nonlinear Systems of Equations 757
The graph of the system, shown in Figure 32, shows only the two real
intersection points because the graph is in the real number plane. The two
ordered pairs with nonreal complex components are solutions of the sys
tem, but do not appear on the graph.
O Solve each system.
(a) x^ + xy + j2 = 3
Figure 32
Work Problem 4 at the Side.
NOTE
In the examples of this section, we analyzed the possible number of
points of intersection of the graphs in each system. However, in
Examples 2 and 4, we worked with equations whose graphs had not
been studied. Keep in mind that it is not absolutely essential to visu
alize the number of points of intersection in order to solve the system.
Furthermore, as in Example 4, there are sometimes nonreal complex
solutions to nonlinear systems that do not appear as points of inter
section in the real plane. Visualizing the geometry of the graphs is
only an aid to solving these systems.
(b) x^ + Ixy  ly^ =
2x^ + 4j2 = 16
Answers
4. (a) {(1,2), (1,2), (2,1), (2,
(b) {(0,2),(0, 2), (2/V2,0),
i2iV2.,0)}
1)}
Section 12.5 SecondDegree Inequalities and Systems of Inequalities 763
±2.5 SecondDegree Inequalities and Systems of Inequalities
B J ECTi VE Q Graph seconddegree inequalities. The linear in
equality 3x + 23; < 5 is graphed by first graphing the boundary line
3x + 23; = 5. A seconddegree inequality is an inequality with at least one
variable of degree 2 and no variable with degree greater than 2. An example is
x^ + j^ < 36. Such inequalities are graphed in the same way. The boundary of
36 is the graph of the equation x^ + y^
36, a circle
the inequality x^ + y^
with radius 6 and center at the origin, as shown in Figure 33.
The graph of the inequality x^ + J^ — 36 will include either the points out
side the circle or the points inside the circle, as well as the boundary. We decide
which region to shade by substituting any test point not on the circle, such as
(0, 0), into the original inequality. Since 0^ + 0^<36isa true statement, the
original inequality includes the points inside the circle, the shaded region in
Figure 33, and the boundary.
OBJECTIVES
Graph seconddegree
inequalities.
Graph the solution set of
a system of inequalities.
x^+y^<36
Figure 33
Graphing a SecondDegree Inequality
Graph J < 2(x  4)^  3.
The boundary, y = — 2 (x — 4)^ — 3, is a parabola that opens down with
vertex at (4, —3). Using (0, 0) as a test point gives
0< 2(04)2 3 ?
0<323 ?
0< 35. False
Because the final inequality is a false statement, the points in the region con
taining (0, 0) do not satisfy the inequality. Figure 34 shows the final graph. The
parabola is drawn as a dashed curve since the points of the parabola itself do not
satisfy the inequality, and the region inside (or below) the parabola is shaded.
3
;
:
3.<2(*4)^3
: K4,3):
^
:,f ....;.... \.:
1 \
1 I
: ; 1 i 1 :
1 \
: r \ :
Figure 34
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764 Chapter 12 Nonlinear Functions, Conic Sections, and Nonlinear Systems
Graph J > (x + 1)^  5.
® Graphx2 + 4j2>36.
Answers
1.
(1,5)
f
x^ + 4y^ > 36
NOTE
Since the substitution is easy, the origin is the test point of choice unless
the graph actually passes through (0, 0).
Work Problem 1 at the Side.
EXAMPLE 2
Graphing a SecondDegree Inequality
Graph I6y^ < 144 + 9x\
First rewrite the inequality as follows.
I6y^  9x^
X
144
1
Subtract 9x
Divide by 144.
This form shows that the boundary is the hyperbola given by
9 16
Since the graph is a vertical hyperbola, the desired region will be either the
region between the branches or the regions above the top branch and below
the bottom branch. Choose (0, 0) as a test point. Substituting into the origi
nal inequality leads to < 144, a true statement, so the region between the
branches containing (0, 0) is shaded, as shown in Figure 35.
Figure 35
Work Problem 2 at the Side.
OBJECTIVE B Graph the solution set of a system of inequalities. If
two or more inequalities are considered at the same time, we have a system
of inequalities. To find the solution set of the system, we find the intersec
tion of the graphs (solution sets) of the inequalities in the system.
EXAMPLE 3
Graphing a System of Two Inequalities
Graph the solution set of the system.
2x + 3y> 6
x^ + y^ < 16
Begin by graphing the solution set of 2x + 3y > 6. The boundary line is
the graph of 2x + Sj = 6 and is a dashed line because of the symbol >.
The test point (0, 0) leads to a false statement in the inequality 2x + 3y > 6,
Continued on Next Page
Copyright ©2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
Section 12.5 SecondDegree Inequalities and Systems of Inequalities 765
SO shade the region above the line, as shown in Figure 36. The graph of
x^ + y^ < 16 is the interior of a dashed circle centered at the origin with
radius 4. This is shown in Figure 37.
"
^
k
^ ^x:
2x + 3y > 6
i 3^^ ^
Figure 36
/ x^+y^< 16 \
Figure 37
Finally, to show the graph of the solution set of the system, determine
the intersection of the graphs of the two inequalities. The overlapping region
in Figure 38 is the solution set.
& Graph the solution set of the
system.
X +7 < 3
;<
^2 ^ „2
2x + 3y > 6
3^1
x^+/< 16
Figure 38
EXAMPLE 4
Work Problem 3 at the Side.
Graphing a Linear System with Three Inequalities
Graph the solution set of the system.
X + y<l
7 < 2x + 3
y^2
Graph each inequality separately, on
the same axes. The graph of x + y < 1
consists of all points below the dashed line
X + y = 1. The graph of j < 2x + 3 is
the region that lies below the solid line
y = 2x + 3. Finally, the graph ofy > 2 is
the region above the solid horizontal line
y = —2. The graph of the system, the inter
section of these three graphs, is the triangu
lar region enclosed by the three boundary
lines in Figure 39, including two of its
boundaries.
3
;
x + y< \K
fy <2x + 3
\
y^l/
\
\
/
2
: :'\„':
/
: J ^^
/
Figure 39
Work Problem 4 at the Side.
Graph the solution set of the
system.
3x  47 > 12
X + 3 J > 6
7<2
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766 Chapter 1 2 Nonlinear Functions, Conic Sections, and Nonlinear Systems
& Graph the solution set of the
system.
y>x^ + 1
x' y' ,
— + — > 1
9 4
7^5
EXAMPLE 5
Graphing a System with Three Inequalities
Graph the solution set of the system.
7
 2x + 1
2x^ + y^> 4
y<4
The graph of j = x^  2x + 1 is a parabola with vertex at (1, 0).
Those points above (or in the interior of) the parabola satisfy the condition
y > x^ — 2x + 1. Thus, points on the parabola or in the interior are in the
solution set of j > x^ — 2x + 1.
The graph of the equation 2x^ + j^ = 4 is an ellipse. We draw it as a
dashed curve. To satisfy the inequality 2x^ + j^ > 4, a point must lie out
side the ellipse.
The graph of j < 4 includes all points below the dashed line y = 4.
Finally, the graph of the system is the shaded region in Figure 40 that lies
outside the ellipse, inside or on the boundary of the parabola, and below the
line J = 4.
y
f
\
i#.j
i \ \ \ \ \
,^^::
Figure 40
— W
Work Problem 5 at the Side.
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