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®f)e StanUarti Series of JHatfjematics
THE
ESSENTIALS OF ALGEBRA
FOB SECONDARY SCHOOLS
BY
ROBERT J. ALEY, Ph.D.
PROFESSOR Off MATHEMATICS, INDIANA UNIVERSITY
AND
DAVID A. ROTHROCK, Ph.D.
ASSOCIATE PROFESSOR OF MATHEMATICS, INDIANA UNIVERSITY
SILVER, BURDETT AND COMPANY
NEW YORK BOSTON CHICAGO
OopTBresHT, 1904, K.^^-UJ
By SILVER, BUEDETT AND COMPANY.
PREFACE.
In the preparation of this book the authors have made
an earnest effort to retain all the essentials of the older
Algebra text-books, and to introduce and properly empha-
size certain newer features which the mathematical studies
of the present demand.
The following are some of the special characteristics of
the book :
1. The Number System. The number system is pre-
sented in the first chapter, and from the arithmetical
system extension is made to the algebraic number system.
In this way the idea of negative number is introduced
and the fundamental operations are explained.
2. Factoring. This subject is treated with particular
fullness, and use is made of the factorial method wherever
applicable in the study of Algebra. At the first reading,
Sections 79, 80, and 86, covering certain details of factor-
ing, may be omitted if thought desirable. The ordinary
student, however, should have no special difficulty in
mastering these sections.
3. The Graph. The work with graphs is made an in-
tegral part of the book. The graphs of simple and
quadratic equations are used freely to aid the pupil's
understanding of the solutions involved. Graphic illus-
trations are given wherever it is thought they will make
the subject clearer.
IV PREFACE.
4. Type Forms. Type forms play an important part in
the study of Algebra. The work of the student is greatly
simplified if he learns early in his course to recognize and
to understand these types. Type forms are extensively
used in multiplication, division, factoring, and equations.
5. Exercises. The exercises have been selected with a
view of clarifying the text and enforcing fundamental
principles. They are numerous, and are difficult enough
to call for effort on the part of the student.
It is believed that the book contains sufficient matter
to furnish a thorough training in the elements of Algebra
and to meet the entrance requirements of American
colleges.
CONTENTS.
CHAPTER
PAGE
I. Introduction 1
II. Definitions 14
III. Addition and Subtraction 20
IV. Multiplication and Division 39
V. Important Identities 70
VI. Factoring . 83
VII. Divisors and Multiples 104
VIII. Fractions 109
IX. Equations in One Variable 125
X. Linear Equations in Two Variables . . . 147
XL Simultaneous Equations 157
XII. Evolution 174
XIII. Theory op Indices 187
XIV. Radicals, Surds, and Imaginaries . . . 191
XV. Quadratic Equations in a Single Variable . 207
XVI. Simultaneous Equations involving Quadratics 229
XVII. Ratio, Variation, and Proportion . . . 255
XVIII. Permutations and Combinations .... 268
XIX. Series 278
THE
ESSENTIALS OF ALGEBRA.
CHAPTER I.
INTRODUCTION.
1. The Integral Number System is that orderly succession
by ones which we first learn by counting. We are familiar
with it in the Arabic numeral form of 1, 2, 3, 4, 5, 6, 7, 8,
9, 10, 11, and so on. The characters 1, 2, 3, 4, etc., are
symbols of number, but we shall hereafter, by the use of a
common figure of speech, speak of them and other number
symbols as number.
2. Elementary Number Notions. We know that 3 + 4 = 7,
because by counting 3 and then 4 more we reach 7. This
may be seen by counting these groups,
All the results of addition are primarily determined by
counting. In practice, a number of simple addition results
are determined by counting, and then these are made a
matter of memory. 3x4= 12, because by counting 3
groups of 4 each we reach 12. This is seen in the follow-
ing arrangement :
The truth of a multiplication table is also established by
counting. The number system shows that 3+4 = 4 + 3;
2 THE ESSENTIALS OF ALGEBRA.
for by counting 3 and then 4 more, we reach the same
result as by counting 4 and then 3 more.
3x4 = 4x3 because 3 groups of 4 each make the same
sum as 4 groups of 3 each.
• • •
• • •
• • •
Illustrations enough have been given to show that the
integral number system is the real basis of the funda-
mental parts of arithmetic.
3. Fractions in the Number System. As long as no exact
measurements are needed, nor accurate divisions attempted,
the integral number system is sufficient. If a stick is
more than 9 inches and less than 10 inches in length, we
can not express its exact length by means of the integral
number system. A similar difficulty arises in attempting
to answer the question 8 -=- 3 = what ? To answer all such
questions, fractions have been devised and made a part
of the number system. The addition of fractions to the
number system made possible many arithmetical opera-
tions which were before impossible. The field of arith-
metic was thus greatly enlarged.
4. Incommensurables in the Number System. When the
necessity for extracting roots arose in the development of
arithmetic, it was found that many roots could not be ex-
actly determined. For example, the square root of 2 lies
between 1 and 2, between 1.4 and 1.5, between 1.41 and
1.42, between 1.414 and 1.415, etc. We may extend this
INTRODUCTION. 6
process of locating the square root of 2 between consecu-
tive numbers of the number system as far as we please,
but we can, never find its exact value. Such numbers as
the square root of 2, and the square and cube roots of other
numbers which can not be exactly found, are called incom-
mensurable numbers or merely incommensurables. Although
such numbers can not be exactly expressed, the number
system now includes them.
5. Numerical Arithmetic Complete. With the number
system so developed as to include integers, fractions, and
incommensurables, ordinary numerical arithmetic is com-
plete. This means that in performing the operations of
ordinary arithmetic no necessity arises for any other kind
of numbers.
6. Literal Arithmetic. In percentage we frequently
represent the base by b, the rate per cent by r, the per-
centage by p, the amount by a, and the difference by d.
When we do this, we can transform the rules for the cases
of percentage into the following forms :
(1) p = b xr.
(2) r =p-t- b.
(3) b=p+r.
(4) a=b +b xr.
(5) d=b — b xr.
The symbols b, r, p, a, and d may be considered as
particular numbers of the number system. When thought
of in this way, they are mere abbreviations of numbers.
Since they may be the abbreviations of any numbers
whatsoever, we may think of the symbols themselves as
4 THE ESSENTIALS OF ALGEBRA.
numbers. When a symbol, such as any of the above, is
thought of in this way, it is called a general number. Such
a number is frequently called a literal number. These
symbols of general or literal numbers may have particular
numerical values assigned to them. In order to find 8^>
of 250 we take form (1), on page 3, and put 250 instead
of b, and .08 instead of r. We then have
p = b x r = 250 x .08 = 20.
7. Substitution. The process of putting a particular num-
ber in the place of a general one is called substitution.
By substitution all the results of general or literal
arithmetic become particular. The solution of a problem
in ordinary arithmetic is a mere matter of substituting
particular numbers for general ones in the proper literal
form, as is illustrated in the percentage problem of
Section 6.
The area of- a rectangle is the product of its length and
width. If we represent area by a, length by I, and width
by w, we at once have the general form
a = l x.w.
If we wish to find the area of a lot 66 feet long and
30 feet wide, we put 66 for I and 30 for w, and we have
a = lxw = m x 30 = 1980.
8. Algebraic Expression. Any combination of literal num-
bers or of literal and arithmetical numbers by means of any
or all of the signs of addition, subtraction, multiplication,
division, involution, and evolution is an algebraic expression.
x + y — z is an algebraic expression and is read x plus
y minus z. The algebraic expression a x b— c-=-d-J-4is.
INTRODUCTION. 5
read a times b minus c divided by d plus 4. The word
function is frequently used instead of expression. The
parts of an algebraic expression separated by either of
the signs + or — are called terms. In ax + by — cyz, there
are three terms ; viz., ax, by, and cyz.
9. Signs used in Algebraic Expressions. The signs +,
— , x, -h, and V are used as in arithmetic. They denote
addition, subtraction, multiplication, division, and root
extraction, respectively. Multiplication is also indicated
by a dot (•), and by writing the characters adjacent to
each other, a xb, a -b, and ab mean exactly the same
thing; viz., a multiplied by b. Between arithmetical
numbers also the signs x and ■ are used to denote multi-
plication. The multiplication of an arithmetical and literal
number or of two literal numbers is denoted by writing
them consecutively. 5 b means 5 x b. ab means axb.
Qab = 6xaxb = 6-a-b. The first form (6 ab') is the
only one in use and is read six ab.
There are five forms of division sign in general use,
6 -s- 2, 2)6, 6:2, &, and 6/2, all meaning exactly the
same thing; viz., 6 divided by 2. In algebra the
fourth and fifth forms are more frequently used than
the others.
Vx, ~s/a, Vb are read square root of x, cube root of a,
and fourth root of b, respectively.
a 2 , b B , at are read a square, b cube, and x to the fourth
power, respectively.
a
2 _
aa ; b z = bbb ; x^ = xxxx.
In the above expressions the 2, 3, and 4 are called
exponents.
6 THE ESSENTIALS OF ALGEBRA.
EXERCISES.
Read the following algebraic expressions :
1. a + b — c + 4. 2. 4 a — 3 b + ab.
3. ab — 4 ac + 3 ad
c
- is read 6 divided by c, or b over c.
c
The last term of (3) should be read minus the quotient b divided
by c, or minus the fraction b over c.
4. 2a + b + 5abc — 46c. 6. 5a 3 — 4aai 2 + Va6.
5. a?-4ax+Vb. 7. 6 aft 4 - ||J + \/«
8. 5aVS+ — -7a*
5 a Vz is read 5 a times the square root of x.
9. z«V + 8--r7V5y. 10. 3a?</y~z-^-+8a 3 l/x~.
y ab
10. Precedence of Signs in Algebraic Expressions. If only
the signs + and — occur in an algebraic expression, the
operations are to be performed in order from left to right.
For example : 8-4 + 3 + 2-5 = 4.
If only the signs x and -s- occur in an algebraic expres-
sion, the operations are to be performed in order from left
to right.
For example : 6-=-3x4-i-2 = 4.
If the signs +, — , x, and -s- occur in an algebraic ex-
pression, the multiplications and divisions are first per-
formed, and then the additions and subtractions.
INTRODUCTION. 7
For example : 4+3x2- 12 + 6x3 + 2x4.
Performing the multiplications and divisions in the above,
we have 4 + 6—6 + 8. Now performing the additions and
subtraction, we get 12, which is the value of the expression.
11. Signs of Aggregation. The signs of aggregation or
grouping are the parentheses ( ), braces { J, brackets [ ],
and vinculum or bar . Each of these signs indicates
th.at the expression within or under it must be treated as
a whole, (be — ad~) -+- b means that the difference between
bo and ad is to be divided by b. The expressions (be —
ad)-i-b, \bc — ad\^-b, [be— ad^-^b, and be — ad-i-b all
mean precisely the same thing. The four signs of aggre-
gation are all called by the general name parentheses. The
different forms are necessary to avoid confusion when one
or more groups are included within another group.
For example: 5 x $12 -=-(7 x 6 + 2 -=-4 h-[2 x 3+ 1])}.
This becomes 5 x {12 -=-(7 x 8 -=- 4 -=- 7)j, which in turn
becomes 5 x {12 -=- 2}, or 5x6 = 30.
Multiplication of a quantity within a parenthesis by any
quantity is indicated by writing the multiplier before
or after the parenthesis. 5(a + b) means 5x(a + b).
(a + 6)5 means (u + i)x5.
12. Coefficient. In the expressions 5 a, Sx, and 7 y, 5, 3
and 7 are the coefficients of a, x, and y, respectively.
In an indicated product any factor or factors may be
considered the named part, then all the other factors con-
stitute the coefficient. Thus, in 8 axy, 8 is the coefficient
of axy, 8 a is the coefficient of xy, and 8 ax is the coefficient
8 THE ESSENTIALS OF ALGEBRA.
of y. In the first case axy is the named part, in the second
xy, and the third y.
When no numerical coefficient precedes a literal expres-
sion, the coefficient 1 is understood.
Read the coefficients of y 2 in the following expressions :
8 if, 5 ay 2 , 17 axy 2 , 11 a 2 bxy 2 .
EXERCISES IN SUBSTITUTION.
Mnd the value of each of the following literal expressions,
in which a = 4, b = 2, c = 3, d = 5, x = 6, y = l. z = 10 :
1. a + b — c + xy.
Substituting the values given to the above letters, this expression
becomes
4 + 2-3 + 6x1 = 6-3 + 6 = 9.
In a similar manner determine the values of the following
expressions :
/- 1
2. a + x — y + 3d. 11. Va + yz + -•
3. ax + by+c. 12. V6» + Vy.
4. x + y + a. 13. 5 a + 6 b — 4 c.
5. ax 2 + by 2 + d. 14. 3 a — 4 b + c.
6 ®+° + z 15- 3a; - 4 2/ + 1 0.
b d 16. (x + y)z + (y + z)x.
7. x + y + z + d. 17 ahc + xyz .
8. x 2 -y 2 + a. 18 aa . _ by + C2 , _ 3 abc
9. z 2 +-2a: + l. 19. ^ + yS + z s _ 3 xyZ-
10. a; 2 + 2 ay + y 2 . 20. y + ex + (d - Va) ■*- 4 6.
INTRODUCTION. 9
13. Use of General Number in Arithmetical Problems. All
the problems of ordinary arithmetic may be made general
by the use of general or literal numbers in place of the
arithmetical numbers involved.
1. If John has 10 cents and Henry 12 cents, they together
have (10 + 12) cents. This becomes general by stating it thus :
if John has a cents and Henry b cents, they together have
(a + 6) cents.
2. If Mary is 10 years old and Susie is x years older, then
Susie is (10 + x) years old.
3. If a merchant sells a bushels of corn at b cents a bushel,
and c bushels of wheat at d cents a bushel, and divides the
money received equally among his e children, each one will
receive \_(ab + cd) -=- e] cents.
4. A man has a cents and b dimes. How many cents
has he ?
5. A man is a years old. His son is \ as old. What is
the combined age of father and son ?
6. I traveled 6 miles at c cents a mile, and d miles at e
cents a mile. How far did I travel and what did it cost
me?
7. A rectangle is x rods long and y rods wide. How many
acres does it contain ?
8. A farm a rods long and b rods wide is sold at c dollars
per acre. Find the amount for which the farm sold.
9. How many hours will be required to travel x miles if
one third the distance be traveled at a miles per hour and the
remaining two thirds at b miles per hour ?
10. A man has x dollars in cash; b men owe him each y
dollars, c men owe him each z dollars. How much money
would he have if his collections were made ?
10
THE ESSENTIALS OP ALGEBRA.
11. A rectangle is a rods long and b rods wide. Find the
length of its diagonal.
12. How many hours will be required for a train running 30
miles per hour to travel x miles, if it makes n stops of b min-
utes each ?
13. Find the cost of excavating a basement a feet long,
b feet wide, and c feet deep at x cents per cubic yard.
14. A man sold from a flock of n sheep; the mth part of
them for p dollars each. He then increased his flock by a sheep
and sold the whole lot at x dollars per head. How much did
he receive for the whole flock ?
14. Opposite Numbers. On the scale of a thermometer,
temperature is marked both ways from 0. On the centi-
grade thermometer, temperatures above freezing
read from up, and temperatures below freezing
read from down. Longitude is measured both
east and west from a fixed or prime meridian.
Latitude is measured both north and south from
the equator. We may consider direction along
a line to the right or to the left. Rotation may
be opposite to that of the hands of a clock, or
it may be clockwise. Numbers which in some
way indicate such opposites are called Opposite
Numbers. The need of opposite numbers becomes
apparent when we try to generalize the opera-
tions of arithmetic, a — b indicates the sub-
traction of b from a.
If a = 10 and b = 9, a - b = 10 - 9 = 1, which
shows that 9 is 1 less than 10.
If a = 10 and b = 10, a — b = 10 — 10 = 0, which shows
that 10 is less than 10, or that 10 is equal to 10.
INTRODUCTION. 11
If a = 10 and 5 = 11, a- 1 = 10 — 11, which, what-
ever it may be, ought to show that 11 is 1 greater than
10.
If a man has $500 and is in debt $400, he is worth
$500 -| 400 = $100.
If a man has $500 and is in debt $500, he is worth
$500-1500 = 0.
If a man has $500 and is in debt $600, he is worth
$ 500 — $ 600. This statement harmonizes with the state-
ments made in the other two cases. What does it mean ?
We may interpret it by saying that he owes $100 more
than he is worth, or that his liabilities exceed his assets by
$100, or that he is worth $100 less than nothing.
15. Negative Number. Such questions as the above are
answered by the extension of the number system so as to
include negative number. We may think of the arithmeti-
cal number system as starting at and extending indefi-
nitely in a horizontal line to the right. It is an easy
matter to think of a similar system extending indefinitely
to the left from 0. These appear as follows :
200-100-60-50-4 3 2101234-50-60-100-200
^__ q->
This extension doubles the scope of the number system.
Integers, fractions, and incommensurables are all included
in the extension to the left.
16. Positive and Negative. That part of the number
system to the right of is called positive. The positive
character of a number is indicated by the use of a + sign
before it. +4, + «, and + x 2 are positive numbers. In
12 THE ESSENTIALS OF ALGEBRA.
practice the + sign is frequently omitted, so that the ab-
sence of a sign before a number shows that it is positive.
That part of the number system to the left of is called
negative. The negative character of a number is indicated
by the use of a — sign before it. — 6 a, — a^, and — 11 a?y
are negative numbers.
The important idea in the two parts of the number sys-
tem is that of oppositeness. When anything is represented
by a positive number, its opposite is represented by a
negative number. If time a.d. is positive, then time B.C.
is negative. If distance to the right is positive, then dis-
tance to the left is negative.
ILLUSTRATIVE EXERCISES.
1. If two points are on the same meridian in latitude + 30°
and — 20°, respectively, how far apart are they ? This means
that one is in north latitude 30°, and the other in south latitude
20°. They are evidently 30° + 20° = 50° apart.
Draw a diagram illustrating this.
2. On a certain day the lowest temperature recorded was
— /5° and the highest + 12°. What was the difference in tem-
perature between the lowest and highest ?
3. A man was born in the year — 31 and died in the year
+ 43. How old was he ?
4. A man travels + 45 miles from A, and his friend travels
— 80 miles from A. How far apart are they ?
5. Two places are in —63° and +87° longitude, respectively.
How far apart are they ?
17. Extension of Meaning of Negative. In the above
exercises and illustrations we have thought of negative
number as beginning at and extending in the opposite
direction from that of positive number, which also begins
+ 12 ml.
— 4 mi.
w
+11 ml.
<-
—15 mi.
INTRODUCTION. 13
at 0. Beginning at zero is not a necessary part of the
meaning. The necessary part is that of oppositeness.
Number starting anywhere is negative if it denotes exten-
sion in the opposite direction to that of positive number.
A man travels east 12 miles, then west 4 miles, then east 11 miles,
then west .15 miles. If we select east as the positive direction, then
west is the negative direction. We may then say that a man travels
-I- 12 miles, then — 4 miles, then + 11 miles, and then — 15 miles ;
as shown in the following diagram.
E
18. Double Use of the Signs + and — . The signs + and
— in algebra retain their arithmetical sense, and denote
addition and subtraction, respectively. When used in
this sense they are called signs of operation. The signs
+ and — are also used to denote the quality of a number,
that is, to indicate that the number belongs to the posi-
tive or negative part of the number system. In this
sense, the signs + and — denote oppositeness, and are
called signs of quality. Smaller signs slightly elevated are
sometimes placed before a number to denote its quality.
For example : + 5, ~5, + a, ~a, +15 b, ~18c.
19. Algebraic Number. When the quality of a number
is considered, the number becomes algebraic. The signs of
algebraic number are + and — . The absolute value of a
number is its value without regard to quality. + a, +b,
+ 11, —7, are algebraic numbers, a, b,' 11, 7, are their
absolute values.
CHAPTER II.
DEFINITIONS.
20. Identity. In the equality 5x — 2x=2>x we have
merely the statement that 5 x diminished by 2 a; becomes 3 x.
No question need be asked concerning the number or value
represented by x. The equality is true, whatever value x
may have. If x = 1, the equality becomes 5x1—2x1
= 3x1. If x = 5, it becomes 5x5 — 2x5 = 3x5. If a;
be any number a, it becomes 5a — 2 a = 3 a.
An equality, true for all values of the letters considered,
is called an identity.
In an identity both sides of the equality may be reduced
to the same form, x 2 + 2 ax + y 2 — 2 ax = x 2 + y 2 is an
identity because the left side becomes x 2 + y 2 by uniting
2 ax and — 2 ax.
21. Equation. The .equality x + 3 = 7 is entirely dif-
ferent from that considered in the preceding Section. If
x be 1, the equality does not exist, for 1 + 3 is not equal
to 7. Neither does it exist if x be 3, for 3 + 3 is not 7.
If x be 4, the equality exists, for then we have 4 + 3 = 7,
a numerical identity. We see that the equality x + 3 = 1
restricts the value of x to 4.
An equality which contains one or more restricted letters,
is called an equation.
U
DEFINITIONS. 15
An identity is usually distinguished from an equation
by having its sign of equality written thus, = , while the
equation retains the sign =. The sign = is read "is
identical with" or "identically equals." The sign = is
read " is equal to " or " equals. "
5a + 3a = 8ais read 5 a plus 3 a is identical with 8 a.
3 x + 4 = 20 is read 3 x plus 4 equals 20.
22. Variables. The letters of an equality which are re-
stricted in value are called variables. —
They are generally, although not necessarily, represented by
the last letters of the alphabet.
In the equality x + y = 5, x and y are variables.
23. Constants. The letters of an equality which are not
restricted and all arithmetical numbers are called constants. .
They are generally, although not necessarily, represented by
the first letters of the alphabet.
In the equality ax + by = c, the constants are a, b, and c.
EXERCISES.
Distinguish between equation and identity in the following
equalities ; also point out the variables and constants.
1. 5x-ix + x = 2x. 9. x 1 -2ax+a i +2ax=a 2 +x 2 .
2. 8y+x—4:y+6x=4:y+7x. 10. 8z — Sz = 12.
3. 3 a; — 5 = 10. 11. ax+by = c.
4. 12a-3a + 4&=9a + 46. 12. 3 x + i x= 5x+ 2x.
5. 4 a + 10 = 18. 13. 7» + a = 12.
6. 2 ax + 5 ax -3 ax = 4 ax. 14. a?x + 5 a?x = a 2 x(8 — 2).
7. 5 y — 2 y = 15. 15. o x + 9 x — 7 x = 28.
8. 4 xy + 6 xy — 2 xy = 8 xy.
16 THE ESSENTIALS OF ALGEBRA.
24. The Root of an Equation. A value of the variable^
which, when substituted for the variable, reduces an equation
to an identity, is called a root of the equation.
The equation x — 5 = 8 asks what number diminished
by5 equals 8. The answer, £ = 13, is the root, for 13 — 5=8
is an identity.
The process of obtaining the value of the root is called
solving an equation.
25. Axioms. In solving an equation certain elementary
facts are taken for granted; that is, their truths are ac-
cepted without proof. Such self-evident truths are called
axioms. The three following are of use to us at present :
(1) Numbers equal to the same number, or to equal num-
bers, are equal to each other.
Ex. 4 + 2 = 6, 3 + 3 = 6 ; hence 4 + 2=3 + 3.
If 3 a - 5 = 10, and 2 b + 4 = 10, then 3a- 5 = 26 + 4.
(2) If equals be added to or subtracted from equals, the
results are equal.
Ex. 3 + 2 = 5; then 3 + 2-2 = 5-2.
If a + b = e, then a + b — b = c — b.
If x + y = 5, then a; + «/ + 4=5 + 4.
(3) If equals be multiplied or divided by equals, the re-
sults are equal.
Ex. tf=3 + f
Multiply by 3, Y- x 3 = 3 x 3 + i x 3, or 10 = 9 + 1.
If 4 x = 20, then — = — , or x = 5.
4 4
DEFINITIONS. 17
26. Solution of Exercises. The algebraic equation can be
used to advantage in the solution of many exercises found
in ordinary arithmetic. The process consists in first ex-
pressing the exercise as an algebraic equation, and then
applying the axioms so as to find the root. This will be
illustrated in the following exercises.
EXERCISES.
1. A and B have $900; A has $100 more than twice what
B has. How much has each ?
Solution.
Let x = B's money.
2x + $100 = A's money.
x + 2 x + $ 100 = both A's and B's money.
$900 = both A's and B's money.
By Axiom (1), x + 2x + 100 = 900.
By Axiom (2), z + 2x = 900-100 = 800 (subtracting 100 from equals).
3 x = 800.
By Axiom (3), x = 266"'|, B's money.
2 x + 100 = 533 J + 100 = 633}, A's money.
2. A and B have $1500 ; A has $300 less than 3 times B's.
How much has each ?
3. What number added to twice itself will make 900 ?
4. The sum of two numbers' is 84; the larger is 11 times the
smaller. What are the numbers ?
5. John has $ 300 more than Henry ; they both have $2100.
How much has each ?
6. A house and lot cost $3700; if the house is worth $700
more than the lot, what is the value of each ?
18 THE ESSENTIALS OF ALGEBRA.
7. Divide $ 720 among three men so that the second shall
have twice as much as the first, and the third 3 times as much
as the first.
8. Divide 440 into three parts so that the second part shall
be 100 more than the first, and the third part as much as the
sum of the first and second parts.
9. A horse and carriage cost $ 288; the carriage cost f as
much as the horse. How much did each cost ?
Solution.
Let x = cost of horse.
$ x = cost of carriage.
x + $ x = cost of both.
$288 = cost of both.
By Axiom (1), x + $ x = 288.
By Axiom (2), 5 x + 4 x = 288 x 5, multiplying by 5.
9 x = 1440.
By Axiom (3), x = 1440 * 9 = 160, cost of horse.
f x = | of 160 = 144, cost of carriage.
10. What number increased by f of itself is 550 ?
11. One third of a number increased by \ of the number
is 455. What is the number ?
12. A's money is f of B's money ; together they have $ 1300.
How much has each ?
13. Four times a number increased by f of the number
is 475. What is the number ?
14. Two numbers added together make 80 ; the greater is 5
more than 4 times the lesser. What are the numbers ?
15. If to my age you add its half and its third and 50 years
more, the sum will be 3 times my age. What is my age ?
16. If to the double of a number you add its half and 42
more, the sum will be 4 times the number, What is the
number ?
DEFINITIONS. 19
17. One number is f of another; their sum is 156. What
are the numbers ?
18. 7 a; -5 a; + 11 a; = 390. Find a;.
19. Divide the number 99 into three parts so that the first
shall be 2 times the second and 3 times the third.
20. A man bought two houses for $ 4400, paying 10 times as
much for one as for the other. What did each cost ?
21. A man. paid $ 700 more for one house than for another ;
the cost of one being | of the cost of the other. What was the
cost of each ?
22. One seventh of a number exceeds ^ of it by 560. What
is the number ?
23. What number added to £ of itself will make 1000 ?
24. What number diminished by -| of itself will make 60 ?
25. What number increased by \ and \ of itself will make
110?
CHAPTER III.
ADDITION AND SUBTRACTION.
27. Arithmetical Addition. In elementary arithmetic, to
add two numbers, 4 and 5 for example, is to find in the
number system a number 9 by the process of counting,
first 4 and then 5 more. The number so found is called
the sum, and the numbers added are called the addends.
The addition of any number of addends furnishes only an
extension of the above process of continuous counting.
If we think of number represented as heretofore upon a
scale, then addition may be represented as follows :
0-
Let us add 5, 3, and 6.
" First, we count 5 from 0, which takes us to A ; then 3
more, which makes 8 and takes us to B ; and finally 6 more,
which makes 14 and takes us to P- In this illustration
is the zero point from which counting proceeds, and P is
the terminal point. Hence, the sum represents the counted
distance of the terminal point from zero. In practical
addition elementary sums are remembered, and thus the
actual counting is avoided.
28. Algebraic Addition. We have already seen that
ordinary algebra contains not only positive number (for-
ward counting), but also negative number (backward
20
ADDITION AND SUBTRACTION. 21
counting). Part or all of the addends may therefore be
negative. Hence, the definition of algebraic addition
must be extended accordingly.
Algebraic addition is the process of fending a number
(surri) in the algebraic number system represented by the
terminal point reached by the successive forward and back-
■ ward countings indicated by the addends.
Thus to add 4, 5, — 6, and 2 is to count 4, then 5 more,
giving 9, then backward 6 to 3, then forward 2 to 5 ; 5 is
the terminal point of the successive countings aud is the
sum of the given addends.
A
o 1 1 1 *— — i i ■
~p
On the diagram the counting is from to A, A to B,
B to G, to P.
To add 4, — 6, — 3, and + 2 is to count 4, then back-
ward 6 to — 2, then backward 3 more to — 5, and then
forward 2 to —3. The terminal point of the successive
countings is — 3, which is therefore the sum of the addends
given. Show this by a diagram.
EXERCISES.
Find as above the sum in each of the following, and illustrate
by a diagram :
1. 4, 5, -3. 6. -8, -9, -4.
2. 6, -7, -3. 7. 4, -8, +6, -5.
3. -4, 5, -1. 8. -6, +9, -10, +8.
4. 8, -9, + 6, -4. 9. -4, -2, +8, +6.
5. -7, -3, -1, +5. 10. 5, +6, -7, -10.
'22 THE ESSENTIALS OF ALGEBRA.
29. Monomials and Like Monomials. Algebraic expressions
consisting of single terms are called monomials.
Monomials containing the same literal parts, each literal
part having the same exponent, are called like monomials.
§x, — 4 ab, 3 xy, are monomials. 4 a, 2 a, and — 3 a are like mono-
mials. To add 4 a, 2 a, and — 3 a, is to count 4 a, and then 2 a to 6 a,
and then backward 3 o to 3 a. 3 a is the terminal point of the suc-
cessive countings, and is therefore the sum.
EXERCISES.
Add the following monomials :
1. 4 6, 5 6, -2 6, —3 6. 5. 11 abc, — 10 abc, — iabc.
2. 3a?b, -6a 2 6, -5a 2 6. 6. 8 a; 2 , -5 a 2 , + 7 x 2 , -Ox 2 .
3. —5xy, 7 xy, —2xy. 7. —2x*y, +ix 2 y, 3?y, —5x 2 y.
4. 6 ax 2 , 3 ax 2 , — 10 ax 2 + ax 2 . 8. 9axy, — 5axy, — 7a,xy, iaxy.
9. 3^/xy, —2y/xy, + 6 y/xy, — Vxi/.
10. 4 a -yfz, + 5 aV^, — 9 a Vz.
30. The Commutative Law. We know that in arithmetic
3 + 4 = 4 + 3. This is known as the Commutative Law,
and means that the addends may be taken in any order.
The law is applicable to any number of addends. I This
( law holds in the extended number system of algebra.A It
is algebraically stated as follows :
a + b = b + a.
The following diagrams show the truth of this law :
B
5 ' 1 •*
a + b
B
r • »P
o a
b +a
ADDITION AND SUBTRACTION. 23
That a — b = — b + a is shown as follows :
.0 a
->-
B
Z]
<- b °
->-
The distance from to P is the same in each case.
31. The Associative Law. We know that in ordinary
arithmetic, iii adding 2, 3, 4, 5, we may, if we wish, first
add 3 and 4, and add their sum to 2, and then add 5.
We may, in fact, associate the addends in any manner we
choose. This is known as the Associative Law. This law
holds in the extended number system of algebra. It is
algebraically stated as follows :
a + b + c = a + (6 + <?).
This law, like the Commutative Law, applies to both
positive and negative numbers.
32. Addition of Monomials. Let us find the value of the
expression v*
3x — 2x + 4x—7x + 8x.
By applying the Commutative Law we may write this
expression in the equivalent form Bx + ix+8x—2x—7x.
Now, by the Associative Law we may add all the positive
numbers into the one sum of 15 a;, and all the negative
numbers into another sum of —9x, and thus write the
expression in the equivalent form 15 x — 9 x, which we at
24 THE ESSENTIALS OF ALGEBRA.
once know to be 6x. All this may be shown in the
following scheme :
3x-2x-t-4:x-lx + 8x
= 3x + 4x + 8x—2x—7x, by Commutative Law.
= 15 x — 9x, by Associative Law.
= 6 x, by adding.
The sum of a positive and negative number is equal to
the difference of their absolute values with the sign of
the greater prefixed ; e.g. 6 — 4 = + 2, while 9 — 13 = — 4.
Rule. To add like monomials, add the positive terms,
then add the negative terms ; to the difference of the absolute
values of the two sums prefix the sign of the greater.
EXERCISES.
1. Add 3a;, — 5x, —9a;, + lla;, —3a;, +7x.
We may for convenience arrange the solution thus :
+ 3a; — 5x
11a; - 9x
7x — 3x
21 x —Ylx
—17 a;
-+- 4a;
It will be noticed that we have arranged in the first column
all the positive numbers, and in the second all the negative
numbers. This is a mere matter of convenience. The stu-
dent should early accustom himself to pick out and unite the
positive numbers mentally, and likewise the negative numbers,
merely writing down the results. Unless the coefficients are
very large, a little practice will enable the student to do this
with accuracy and rapidity.
ADDITION AND SUBTRACTION. 25
2. Add 5 xyz, — 11 xyz, 18 xyz, — 4 xyz, — 13 ai!/«. '
3. Add 15 a-bc, 10 a?bc, -2arbc, - 18 arbc.
4. Add 10 a?, - 5 a; 2 , - 2 x\ 16 a; 2 , - 8 x*.
5. Add | aa;, \ ax, — % ax, — %ax, — f ax.
6. Add 3V2lc, -5V2a;, - 3V2as, -V2a;, 6V2i».
7. Add 4(a + 6), -3(a + &), -4(a + 6), 2 (a + 6).
Algebraic expressions containing like quantities within a paren-
thesis may be added as monomials. Tims, in Exercise 7, the quantity
a + b is common to each addend, hence the sum is — (a + b).
8. Add &(a? + y% -3(a? + f), 7(<* + f), -%(a? + f).
9. Add ^(ax + by + 2), — %(ax + by + 2), (ax+by + 2).
10. Add 1(^-2^- 1(^-2 y), -1(^-22,). ^
11. Add 3(Vx + Vy), -8(VS + V^), 6(Vx + Vy).
-^'(S+S- 1 )-'^- 1 )-^-
33. Addition of Polynomials. ^4w algebraic expression
consisting of more than one term is a polynomial.
If it has two terms, it is called a binomial : and if three,
a trinomial.
If we desire the sum of 4 a + b and 3 a — 4 b, we may indicate that
sum thus : 4a + 6 + 3a — 4 6.
Now, by the Commutative Law, we may write this 4 a + 3 d -f J - 4 J,
which by the Associative Law becomes 7 a — 8 b.
However many polynomials we might have to add, the
process would be an extension of the above. We have
then the following rule for adding polynomials :
Rule. Unite the like terms of the various polynomials
into sums, and connect these by their proper signs to form
the polynomial sum.
26 THE ESSENTIALS OF ALGEBRA.
EXERCISES.
1. Add 2ax + 3by + 5cz, 4 ax — 5 by, 7by — 3cz, 11 ax —
4-by — Q cz.
For convenience we may arrange these polynomials thus :
2 ax + 3 by + 5 cz
4 ax — 5 by
7by — 3cz
11 ax — 4 by — 6cz
17 c!.c 4- by — 4 cz
It should be noticed that in the arrangement like terms have
been placed in columns.
2. Add 5x+3y+4z, —3x—oy+z, 7 x—2y—z, 10x+y—2z.
3. Add 5 ax— 6 by+cz, 3ax+by+3cz, — 4ax+3by—3cz,
3 ax — 3 by + 3 cz.
4. Add 4x? + 3y 2 -4, -3x> + 2y 2 -6, x 2 -y 2 + 7.
5. Add 3ar + 43/ 2 + 3z 2 , - X 2 -4y 2 + 2z 2 , lltf-tf-z 2 .
6. Add 4 xy — 3 yz + 7 zx, 4yz — 7xy, 3yz + 4zx, —4xy +
3zx — 2yz, — xy — zx.
7._AdA 4y/xy + 3 Vyz + y/zx, —2Vyz — 3Vzx, — 2Vxy —
3-\Zyz+~\/zx.
8. Add 3 abc — 4 xyz + 7 Imn, — 4 Imn + 12 xyz, 5 xyz —
4 a&c + 3 ?mra, — 5 abc — 7 Imn.
34. Identity. The sum of the addends is identically
equal to the sum.
Thus, 3x + 4x-5x=2x.
Also, 2a+b+5a— 4b — 6a + 4b = a + b.
Since an identity is true for all values of the letters involved, we
may make use of the identity existing between addends and sum to
ADDITION AND SUBTRACTION. 27
verify our results in addition. If in the first illustration above we
make x = 1, it becomes
3 + 4 - 5 = 2, or 2 = 2.
If in the second illustration we put a = 1 and b = 1, it becomes
2 + 1 + 5-4-6 + 4 = 1 + 1, or 2 = 2.
The use of 1 for each of the letters is convenient, but not necessary.
We may use any value whatever. Thus, in the second illustration
above, we may put a== 3and6 = 5.
The identity then becomes
* 6 + 5 + 15 - 20 - 18 + 20 = 3 + 5, or 8 = 8.
EXERCISES.
Add the following quantities, and verify the results by sub-
stituting particular values for the letters used :
1. 3a+-4 6, 5a-6&, 7a-4c, 56+-llc.
2. ix — 3y, 5x + 3y, — 7x + y, 3x — 2y.
3. 2a?-3b 2 , 5b 2 + ia 2 , 7a 2 -5b 2 , -3a 2 -b 2 .
4. 5x 2 + ±xy + 3y 2 , 2x 2 -5xy + 6y 2 , 3a?-8y 2 , 8xy.
5. — 3ax + kby + c, 5 ax — 6 by — 3c, ax — 3 by +■ 2 c.
6. 12x-3y + z, 6x-4y + 7z, -8x-3y + 3z.
\ rf + if + z 2 , 3x 2 + ±y 2 + 5z 2 , -ix 2 -8y 2 + z 2 .
8. x 2 -3y, 5a+-42/ + 3, x 2 + iy, y 2 -3x.
9. 3V# — 4 V*/, 5V» + 8Vy, — 6Vaj+-Vy.
10. 4a*-56* + 66, 3a*+-2^-46.
Note, a? means Vo; 5 J means Vb.
XL. 5i/a_-6-fyb + c, -3^/a + 4^6-4c,2^-3^6 + 2c,
5^a-3-\/6 + 3c.
12. 3axy + ibyz + 6czx, 2byz-5axy + 6czx, —<Laxy + 2byz
— 7czx, 2 axy — byz — 7 czx.
28 THE ESSENTIALS OF ALGEBRA.
35. Subtraction. Subtraction is the process of finding
one addend, when the other addend and the sum of the two
addends are given.
The given addend is called the subtrahend, the given
sum the minuend, and the addend which is to be found
the remainder or difference.
Referred to the number system, we may say that to
subtract a from b is to find the amount and direction of
counting necessary to pass from a to b, both a and b begin-
ning at the zero point.
This is shown'on our number system diagram as follows :
If P is the terminal point of the subtrahend a and Q
that of the minuend 6, then the distance and direction
from P to Q is the result of subtracting a from b.
36. Cases of Subtraction. A consideration of the defini-
tion of subtraction gives us the following four cases :
(1) If we subtract a positive number from a positive
number, the remainder is the arithmetical difference of the
absolute values, positive or negative, according as the
absolute value of the subtrahend is less or greater than
the absolute value of the minuend.
For example :
7a-(+3«)=4a, 7 a -( + 10a)= - 2>a.
ADDITION AND SUBTRACTION. 29
(2) If we subtract a negative number from a positive
number, the remainder is the positive arithmetical sum of
the absolute values.
For example : hx — ( — 3 x) =8 x.
(3) If we subtract a positive number from a negative
number, the remainder is the negative arithmetical sum of
the absolute values.
For example : — 6 ab — ( + 4 ab) = — 10 ab.
(4) If we subtract a negative number from a negative
number, the remainder is the positive or negative arith-
metical difference of the absolute values according as the
absolute value of the subtrahend is greater or less than
the absolute value of the minuend.
For example :
-1x-(-llx)=±x, _18a!-(-9ar)=-43;.
The student should verify the truth of these cases by
testing them on the number system diagram. ^
EXERCISES.
1. From 19x'y subtract llarfy.
2. From 7 ab subtract — 5ab.
3. From — 9xyz subtract 12 xyz.
4. From — 17 x subtract — 13 x.
5. From —14 a; 2 subtract — 23 a?.
6. From 27 a s y subtract 43 ay
7. From 15 y 2 subtract 15 y 1 .
8. From — 7 abx* subtract — 7 abx*.
9. From 17 ^fxy subtract — 12^/xy.
30 THE ESSENTIALS OF ALGEBRA.
10. From 11 (a + b) subtract 15 (a + b).
11. From — 6(x + y) subtract 9(x + y).
12. From -ll(a 2 + 6 2 ) subtract -7(a 2 + 6 2 ).
13. From 18 (xy + yz + zx) subtract — 7 (xy + yz + zx).
14. From 23(ax + by + c) subtract 40(aas+ by + c).
15. From - 16 (tf + 4 as) subtract — 21 (j/ 2 + 4 oas).
37. Subtraction of Monomials and Polynomials. In the
cases considered in Section 36 it may be noticed that the
subtraction of any number is equivalent to the adding of
an equal opposite number.
Illustrations :
7a-(+4a) = 7a + (— 4a)=3a.
7a-(-4a)s 7 a +( + 4 a)= 11 a.
-7a-(-4a)=-7a + (+4a) = -3a.
-7a-(+4a)=-7« + (-4a)=-lla.
Hence, we have the following rule :
Rule. To subtract one number from another number,
change the sign of the subtrahend and proceed as in addition.
For example : The problem, from 18 ax 2 y take — 5 aa?y is
equivalent to this problem, to 18 ax 2 y add 5 ax 2 y.
In practice the change in sign should always be made men-
tally.
Rule. To subtract one polynomial from another, change
the sign of each term of the subtrahend and proceed as in
addition.
EXERCISES.
1. From 9 x 2 + 3 xy - 11 y 2 + 7
Subtract 7 x 2 — 5 xy — 17 y 2 + 9
2x 2 + 8xy+ 6y 2 ~2
AUDITION ANt) SUBTRACTION. SI
In this we think of the 7 a; 2 as negative, and add it to 9 a; 2 ,
giving 2 a; 2 . The — 5xy is thought of as positive, and added
to 3xy, giving + Sxy, and so on with the other terms of the
subtrahend.
2. From ( da? — lla? + 5y subtract 14 X s — 7 x 2 — 8 y.
3. From 11 a 4 - 15 6 4 - 13 aV subtract 7 a 4 + 10 b 4 - 3 aW.
4. From 4 <n/ + 3 a^ — 11 y 2 subtract 7 xy — 4 a; 2 — 13 y 2 .
5. From 18 ax — 14 % + 11 c subtract 3 aa; + 19 6y — 5 c.
6. From 5 Va; + 8 y/y — 13 a subtract 8 Va; — 5 \A/ + 3 a.
7. From 19 a 3 + 7 a; — 5 subtract 4 aj 3 + 11 a; 2 — 8.
Arranging for subtraction, we have
19a? +7as-5
4a« ! + lla: 2 -8
15 a; 3 - 11 a; 2 + 7 a; + 3
In the minuend there is no term in a; 2 , or we may say there
is the term Oa; 2 . Hence the 11a; 2 is to be subtracted from a; 2 ,
and of course gives a remainder of — 11 x 2 . In the subtrahend
there is no term in x, so there is nothing to subtract from 7 x,
or the remainder is 7 x.
8. From 5a; 4 — 3 a? + 3 a; + 6 subtract 6 a; 4 — 3 x* + x 2 — 5.
9. From 4 x 2 y 2 + 6 xy 3 — 3 v?y + y 4 subtract 3 y 4 + 4 xy s — 6 x 2 y 2 .
10. From 3 (a + b) — 5 (x + y) + 6c subtract 6 (a; + y) —
5 (a + 6) — 5.
11. From 14 Vtf^+13 f— 16 x subtract 15 f+17 y—12 Vxy.
12. From iax + 5y 2 —6xz + 13 subtract 21 + 17 xz — 15 ax
+ 11*/.
13. From 18 j/ 8 - 27 « 3 + 42 y« subtract 18 yz - 36 f + 17 is 3 .
14. From 36 a 3 - 27 a 2 b - 17 a& 2 + b s subtract 13 b s + 17 ab 2
- 37 « 2 Z>.
15. From 13 a; 2 + 17 a;y — 5 subtract 18 y 2 - 17 yz+ 32.
32
THE ESSENTIALS OF ALGEBRA.
38. Identity. Subtraction may be expressed as
Minuend — Subtrahend = Remainder.
This is an identity and will, therefore, be true for any
values of the letters involved. This identical relation
furnishes a convenient method for verifying the results
of subtraction.
For example : 7 ab — 14 a 2 + 11 6 2
4a6-lla 2 + 19S 2
Sab- 3a 2 - 8& 2
If we put a = 1, b = 1, we get
7_14 + 11= + 4
4 -11 + 19 = + 12
3-3- 8 = - 8
which shows our result to be true.
Minuend.
Subtrahend.
Remainder.
EXERCISES.
Perform the following subtractions and verify by putting
the letters each equal to 1 :
1. From 9 a 2 +11 ab - 13 a take 8 a- + 14 ab - 9 a.
2. From 3 x — 4 y + 7 take x — 2y + 8.
3. From 4 a? + 5 y- + 7 ay take 5 a; 2 — 4 xy + 3 y 2 .
4. From 14 ax + 12 6y + 7 take — 3 ax + 2 by — 5.
5. From 4(a: + y) + 7z-4take3(a; + 2/)-53-8.
6. From 16(a; 2 + xy + f) + « 2 — w 2
take 12(ar l + ^+j/ ! )-4« 2 + 3w ! .
7. From 2(6 2 - 4 etc) + ar* + # 2 take 5(6 2 - 4 ac) + 5 jc 2 - 6 y\
8. From aa + by + c take a'a: + b'y + c'.
ADDITION AND SUBTRACTION. 33
9. From aa?+2hxy + by 2 + 2gx + 2fy
take a'x 2 + 2 h'xy + b'y* + 2 g'x + 2f'y+ c'.
10. From x 2 + y 2 + ax + by take 3 x 2 + 3 y 2 + Ix + my.
11. From aVx + b^/y + c take aWx+ b'^/y + c'.
12. From 5 VOF+f + 7-vV + z 2 take 3 Vx 2 + y 2 + 8 Vj^+i 2 .
39. Removal of Parentheses. If we recall the principles
of addition and subtraction already developed, we can by
means of them remove parentheses preceded by the + or
— signs.
For a+(b — c + d)=a + b — c + d,
and a — (b — c + d) = a — b + c — d.
Hence, a •parenthesis preceded by a + sign may be omitted
without any change in the signs of the terms inclosed- A
parenthesis preceded by a — sign may be omitted if the sign
of each term within it is changed.
For example :
ax + (3 ab — 4 cy — 3 az) = ax + 3 ab — 4 cy — 3 az.
ax — (3ab — 4:cy— 3 az) =ax — iab + £cy + 3az.
EXERCISES.
Remove the parentheses in the following and unite like terms:
1. 3a-(2a + 4&-c). 2. 3x-£y-(2x + a-3y).
3. 5a 2 -(6+3a 2 )-4&-(2a 2 -3&).
4. 3 ax — [a 2 — 3ax + xy — (xy — 5ax + a 2 )].
Remove the [ ] first, and we have
3 ax — a 2 + 3 ax — xy + (xy — 5 ax + a 2 ).
Now since the parenthesis is preceded by a + sign it may
be omitted, and we have, when we unite the like terms, ax as
the result.
34 THE ESSENTIALS OF ALGEBRA.
5. 2a-36-[7a-(46-5o-6 + 26)].
6. 3 ax — 5 by — [7 by + 3 ax — (5 ax — 3 by)~\.
7. 11 x — (7 y — 3 x) + 5 y — [3 x — (4 y + 5 a;)].
8. 5 x- \1 a -[3 x -(5 a- 2 x- a)] j.
9. 17 a -12 «- {15 a- 11 a; + (3 a- 2 x)- 5 x -4 a}.
10. 2x-\6y + [6z-(x-y+z)-2x] + 3yl.
In exercise 10 and the following put x = 5, y = 4, and z = 2,
and find the value of the expressions.
11. 5y-3z+[x-y + 2z-(3x + 2y-4:z)].
12. 3z-\2x-[5z-(4 : y-7x-2y + z)-3x'] + 2yl.
13. 2x + 3y-(4y-z + x)-\_5x-(7 y + 3x)~\.
14. xy— (yz — zx — 2xy — 2yz).
15. 3 xy - {Ax 2 -y 2 - [2 xy + 3 x 2 - 5y 2 1- (3 x 2 -iy^l
40. Insertion of Parentheses. Many times it is just as
important to insert parentheses in an algebraical expres-
sion as it is to remove them. Evidently any number of
terms may be inclosed in a parenthesis without change if
the parenthesis is preceded by a + sign. Any number
of terms may be inclosed in a parenthesis preceded by a
— sign, if the sign of every term so inclosed is changed.
For example :
ax + by — cz + ah = ax + (by — cz+ aS),
ax + by — cz + ab = ax — ( — by + cz — ab~).
The number in the parenthesis is thought of as one quan-
tity, and hence may be considered as one term. This gives
an extension in meaning to the word term. In the illus-
trations just given the expressions on the left have four
terms, but those on the right have only two terras.
ADDITION AND SUBTRACTION. 35
EXERCISES.
1. 3a? + 4ab — 3c + d. Inclose the last three terms in a
parenthesis preceded by — .
2. ay + by — cy — xy. Inclose the last three terms in a
parenthesis preceded by — .
In each of the next four exercises inclose all the terms con-
taining x in a parenthesis preceded by — .
3. 5 — 3 x + ax — 4 abx. 4. 7y + 3z + zx — 3x-5yx.
1 > 5. 13 — 5 ax + 11 x — 16 ax + 27 a.
6. 27y-30y + 22xy-13bx + 2lx-30a + 16b.
In the following exercises put all the terms containing a in
a parenthesis preceded by — , and all the terms containing y
in a parenthesis preceded by — .
7. 15 — 3 a + 4 y — 5 ax + 7 xy + 3 ab — 5 by.
Eearranging, we have
15 — 3 a — 5 ax + 3 ab + 4 y + 7 xy — 5 by.
Then 15- (3a + 5ax-3ab) - (-&y- 7xy + 5by).
8. 25 x + 5 y — 3 x + 7 ax — 11 by + ex — 3 ac + 11.
9. b 2 -ab + cy-2ac 2 -4b 2 y + lla-4:y.
10. 3 x + 2 a — 4 y + 5 x — 6 ab + 3 yz — 5 ax + 11 by.
41. Adding and Subtracting with regard to a Named Letter.
7 a + 5 a— 6 a might be written (7 + 5 — 6) a. So if we
had 7 ax + 5 bx — 6 ex, we might write it (7 a + 5 b — &d)x.
In this, a; has been chosen as the element of likeness or
the denomination in the three terms. We have added the
coefficients, but in this case the coefficients are unlike and
we can only indicate the addition. We can subtract 5 ax 2
from 7 bx 2 if we consider x 2 as the element of likeness.
The remainder is (7 b — 5 a) x 2 .
36 THE ESSENTIALS OF ALGEBRA.
EXERCISES.
1. Unite with respect to x, 5 ax + 7 xy — 3 xz.
2. Unite with respect to a, 3 ax + 4 ay — 5 az.
3. Unite with respect to k, &xk + 3 yk — 12 zk.
4. Unite with respect to x and y, 4 ax + 6 ay — 5 &* + 12 cy.
5. Unite with respect to x and y,
3ax + iby-(2ax + 12 by).
6. Unite with respect to x 2 ,
5x> + 16ax 2 -12bx?-x 2 .
7. Unite with respect to x, y, z,
3a; + 4y + 3z— (5 a: — 8y — 7z).
(3-6)a; + (4 + 8)y + (3 + 7)8 = 2a! + 12y + 10sk
8. Unite with respect to x and y,
3x-2y + 7-(2x + 3y + i).
9. Unite with respect to x, y, and z,
— 5x + 8y + 7z + (3x-4 : y + z).
10. Unite with respect to the powers of x,
3a? _5 X 2 -8a^ + 7 a,- 2 -16 x + 20 a; 2 - 21 a; + 6.
11. Unite with respect to the powers of x and y,
ax 2 + by 2 + ex + dy + e — (a'a; 2 + &'y 2 + c'a; + d'y + e').
12. Unite with respect to x 2 , xy, and y 2 ,
4a; 2 -3a3y + 12y 2 -(2ar ! -4ax/ + 8y 2 ) + 16a;y.
13. Unite with respect to powers of k,
12 x + 16 yk - 6 x + 12 «A; + 8 x 2 k 2 + 12 k 2 + k 3 .
14. Unite with respect to x 2 , y 2 , z 2 , and xy,
ax 2 + 2hxy + by 2 + cz 2 - (a'x 2 + 2 h'xy - b'y 2 + c'z 2 ).
ADDITION AND SUBTRACTION. 37
MISCELLANEOUS EXERCISES.
1. A and B have $ 550 ; A has $ 100 more than twice as
much as B. How much has each ?
2. Add llax i +13bx z y+§cxy 2 -5dy i , 4bx 2 y-3aa?+16df,
31bx 2 y — 2<lcxy 2 ! and 10 dy s — 8 cxy 2 + ax 3 .
3. Remove the parentheses and simplify,
13x*-{8y-5x-(3x + 7y)-2yl-(12x + 8y~+2x).
4. Unite with respect to x 2 and y 2 ,
ax 2 — 5 x 2 + 3 y 2 — by 2 + ex 2 — dy 2 .
5. From 5a s -6a 2 &-7a6 2 +ll& 3 take 10 6 3 -3a 3 + 5a6 2
- 12 a 2 b.
6. By means of a diagram show that the sum of 8 and
- 10 is - 2.
7. By means of a diagram show that — 10 taken from 8
leaves 18.
8. By means of a diagram show that 8 taken from —10
leaves — 18.
9. With x = 4 and y = 5, find the value of
8y- \3 x - [4 y + 2x- (6 x -3y -5a;)] \.
10. With the same values of x and y find the value of
{8x-[3y + 2x-(5y-3x)-\\(8x-3y-x).
11. A father's age is 10 years more than 3 times his son's
age ; the sum of their ages is 82 years. Find the age of each.
12. Add, 8 y + 3 xy + 11 x 2 , 7 y - 13 xy - 21 x 2 , -142/ +
10 xy + 12 x 2 , and 2 y — 5 xy — 2 x\
Verify your result by putting y = 2 and x = 1.
13. From 11a — 12 aft— 3 6
take 8a 4- 3ab-10b + ll
Verify your result by putting a = 1 and b = 2.
38 THE ESSENTIALS OF ALGEBRA.
14. Add 8(a + b) — 16 (x - y) + lWax, 5 (x — y) + 2y/ax —
7 (a + b), and — 3 (a + b) + 11 (x — y) — 14Vaa>. (Regard the
quantities in each parenthesis as a single term.)
15; From 8 (a 2 - ft 2 ) — 17 (x + y) — 11 y/x* + f
take 7 (a 2 - b 2 ) -20(x + y)+ 4 Vs 2 + y 2
16. Show by a diagram that the sum of 8, — 6, — 3, + 2,
and — 5 is — 4.
17. Unite with respect to a and x,
5a+3x— 4c a — 7 x — ba + bx + cx — dx.
18. 8 x + 3 y — (x 4 2/) + 2 ?/z. Inclose all the terms after the
first in a parenthesis preceded by a — sign.
19. A farmer exchanged a bushels of wheat at b cents a bushel,
and c bushels of corn at d cents a bushel, for a mowing machine
costing e dollars, and for calves at / dollars each. How many
calves did he get ?
20. John, James, and Henry together have $216. John
has one half as much as James, and Henry has as much as
both John and James. How much has each ?
21. Simplify 8 x 2 + 3 xy - 4 y 2 + 2 xy + 3 y 2 - 7 x 2 + 4 y 2 -
2xy-7f + 5x 2 + 3xy--4:y 2 + 10x 2 .
22. Simplify 17-13 a: + 4a; 2 - f- (3f + 2\x + 5a; 2 - 11).
23. If x = 1, ?/ = 2, and z = 3, find the value of
on/z — \x 2 — Q/ 2 + aiz — (2 z — 3 xy — 4 ?/)] + xz 2 \ .
24. With the same values of x, y, and z, find the value of
3 yz + 5 8 x - [4 y + z 2 — (5 a:?/ +• 4 ?/z) - 8 a:?/z] — 3 ?/ 2 } .
25. Add 3 (a + ?/)- 4(a 2 -&) + 3Vy 2 -b 2
-7(x + y) + 11 (a 2 - b) - 16 VjT^P
4 (x + y)~ 8(a 2 -b)- Wy 2 - b 2
5(x + y)- 3 (a 2 -6)- 9 Vj/ 2 - 6 2
CHAPTER IV.
MULTIPLICATION AND DIVISION.
MULTIPLICATION.
42. Integral Multiplication. To multiply 7 by 8 may be
taken to mean that 7 is to be added 8 times.
7 + 7 + 7 + 7 + 7 + 7 + 7 + 7 = 56.
In such an operation 7 is the multiplicand, 8 is the multi-
plier, and 56 is the product. Similarly, to multiply a by b
is to find the sum of a added b times :
a + a + a + ••• (b times) = ah.
In this, a is the multiplicand, b the multiplier, and ab
(read, a multiplied by b) is the product. It is clear that
this view of multiplication has no meaning when the mul-
tiplier b is fractional or negative. We can not add a one
half times, or three fourths times, or three and a half
times ; neither can we add a negative four times to obtain
the product of a by — 4. All this shows that if we are to
have multiplication of algebraic numbers, we must extend
our definition. No change is made in the meaning of
multiplicand, multiplier, and product.
43. Multiplication Defined. Multiplication is doing to the
multiplicand what has been done to unity to produce the
multiplier.
39
40 THE ESSENTIALS OF ALGEBRA.
For example. To multiply 6 by 5 is to do to 6 what has
been done to unity to produce 5. Unity has been added five
times (1 + 1+1 + 1 + 1 = 5) to make 5, hence we must add 6
five times to get the product 6x5. Or, we may say that
unity has been taken five times to produce 5, hence we must
take 6 five times to produce 6x5. The new definition is seen
to agree with the arithmetical notion of multiplying one integer
by another. To multiply 6 by | is to do to 6 what has been
done to unity to produce |. Unity has been divided into 3
equal parts and 2 of them taken to produce f. Hence to
multiply 6 by | we must divide 6 into 3 equal parts and take
2 of the parts. It is thus seen that the new definition includes
multiplication of fractions. The application of the definition
may be seen by considering the following simple concrete
problem.
A water tank with a capacity of 1000 gallons contains at
the present time 600 gallons. Let us consider the following
four cases :
(1) If it have a supply pipe carrying 10 gallons an hour,
how much water will be poured into the tank in the next
8£ hours ?
Evidently the amount poured in is 10 x 8£. We must do to 10
what has been done to 1 to make 8^. To produce 8J, 1 has been
added eight times, then separated into 2 equal parts and one of the
parts added. Hence we must add 10 eight times, giving 80, then
separate 10 into two equal parts of 5 each and add the 5 to 80, giving
85. Hence the amount poured in is 85 gallons.
(2) With the same rate of flow, how many gallons must be
added to find the amount in the tank 6 hours before the pres-
ent time ?
In this case the 6 hours is negative, as it is the opposite of the
time used in the preceding problem, which we considered positive.
Our problem now is to multiply 10 by — 6. We must do to 10 what
MULTIPLICATION AND DIVISION. 41
has been done to 1 to produce — 6. — 6 is produced by subtracting
1 six times. Hence to multiply 10 by — 6, we must subtract 10 six
times, which gives — 60. Hence we must add — 60 gallons to 600
gallons to get the contents 6 hours before the present time.
(3) A discharge pipe is flowing at the rate of 5 gallons per
hour. How many gallons must be added to give the contents
10 hours from now ?
Since the inflow rate was positive, the outflow rate is negative.
Hence, our problem now is to multiply — 5 by 10. We must do to — 5
what has been done to 1 to produce 10; that is we must add — 5 ten
times, which gives — 50. So - 50 gallons is the amount to be added.
(4) The discharge pipe has been running at the rate of 6
gallons an hour for a number of hours. How many gallons
must be added to give the contents of the tank 8 hours ago ?
Tn this case both the. 6 and 8 are negative, and the problem is to
multiply — 6 by — 8. We must do to — 6 what has been done to 1 to
produce 8; that is, we must subtract —6 eight times, which gives
+ 48. So we must add 48 gallons to get the contents of the tank 8 ^)
hours ago.
The results of the four cases may be arranged thus :
10x8|=+85,
10 x - 6 = - 60,
-5x10= -50,
(-6) x(-8)=+48.
If we should make our reasoning perfectly general by letting
the rate of flow be a and the number of hours be b, then the
four cases would take this form :
axb = +ab,
ax(—b) = —ab,
— axb = —ab,
(-a)x(-b)=+ab.
42 THE ESSENTIALS OF ALGEBRA.
44. Law of Signs in Multiplication. From the definition
and the above considerations, we see that the product is
+ if the multiplier and multiplicand have like signs, and
it is — if they have unlike signs.
In multiplication, like signs in multiplicand and multiplier
give a positive product and unlike signs give a minus product.
45. Continued Products. Products produced by three or
more multiplications are called continued products.
If 6 boys buy 5 oranges each at 3 cents apiece, the total cost of the
oranges is 3 x 5 x 6 cents. If 8 groups of 6 boys each should buy
oranges as above, the total cost of the oranges is 3x5x6x8 cents.
If instead of using arithmetical number we should use algebraic num-
ber, and say that there were d groups, c boys in a group, and each
boy bought b oranges at a cents apiece, then the total cost of the
oranges would beaxbxcxd = abed cents.
46. Factors. As in arithmetic, the numbers multiplied
together are called the factors of the product.
a, b, and c are factors of abc.
5, a, b, and c are factors of 5 abc.
In the latter case, 5 is considered a numerical multiplier
or coefficient, and is usually so designated, instead of being
called a factor.
EXERCISES.
Point out the factors and numerical multipliers in the follow-
ing products :
1. obex. 4. 11 (2 x) (3 y) z.
2. 3abdz. 5. (5 a) (4 a) (3 a).
3. 17 x9(3z) = 51 x9z. 6. (3y)(iz)(2xy).
MULTIPLICATION AND DIVISION. 43
47. Signs of Continued Products.
(-3)x2x3 =-6x3 =-18.
(-3)x(-2)x3 = + 6x3 = + 18.
(- 3) x (- 2).x (- 3)= + 6 x (- 3)= - 18.
If we use algebraic instead of arithmetical number, the above
may be written:
(— a) x 6 x c =(— ab~) x o = — abc.
(— a)x(— J)xc =(+«J)xe = + abc.
(— a)x(— J)x(— <?) = ( + ab) x ( — c) = — abc.
We see that one negative factor produces a negative
product, two negative factors a positive product, and three
negative factors a negative product. Hence,
Products resulting from an even number of negative fac-
tors are + , those resulting from an odd number of negative
factors are — .
EXERCISES.
Give the signs of the following products :
1. (-2 a) x (5a)(-x).
2. (_a)(-&)(-c)(-d>
3. a(-b)(-c}d(-e)f(-g)(-h).
4. x <-(-a)(~b)(f)(-3x).
5. 17(-5)(-3)(-2)(^)(-2)(-3).
48. Commutative Law of Factors. In multiplying to-
gether numbers in arithmetic, the factors may be taken in
any order without changing the product.
5x7=7x5 = 35.
2x3x5=2x5x3=3x2x5=3x5x2
= 5x2x3 = 5x3x2 = 30.
44
THE ESSENTIALS OF ALGEBRA.
If we construct a rectangle of length 5 and width 3, the
area is 5 x 3. If we construct another rectangle of length
3 and width 5, the area is 3 x 5. 3
These two rectangles are readily seen to be equal, for
the second one is merely the first one placed on end.
Hence > 5x3 = 3x5.
The reasoning would be exactly similar if we should use
the general rectangle whose length is a and width b, and
the other general rectangle whose length is b and width a.
In this case our conclusion is
a x b = b x a.
The product is the same whatever the order of the factors.
This is called the Commutative Law of Factors.
This law holds for all algebraic numbers.
a x (— 5) = ( — b~)a = — (ba~)= — («6) = — ab;
or, since —b =(—1)5,
a x (— b) = a x (— 1) x b =(— l)ab = — ah.
EXERCISES.
1. Commute the factors of 3 xy.
Sxy = 3yx = xSy — xy3 = y3x = yx3.
2. Commute the factors of — abc. (—1 is a numerical
multiplier.)
MULTIPLICATION AND DIVISION. 45
3. Commute the factors of ab(x + y). (The quantity in
parenthesis is a factor.)
ab(x + y)=a(x+y)b = (x+y)ab = (x+y)ba = b(x+y)a = ba(x+y).
4. Commute the factors of (a + b)c(x + y).
5. Commute the factors of (a + b)(c + d)(e +/).
49. Associative Law of Factors. If 6 boys buy 4 oranges
each at 3 cents apiece, the total cost is 3 cents x 4 x 6.
We may think of this as each boy paying out 3 cents x 4
and' the 6 boys as paying out (3 cents x 4) x 6 ; or we may
think of total number of oranges bought, 4x6, and then of
the cost of these as 3 cents x (4 x 6) ; or finally we may
think of the six boys as buying 1 orange each, at a cost
(3 cents x 6), and then the total cost of 4 oranges each is
(3 cents x 6) x 4. Hence, we have
3x4x6 = (3 x 4) x 6 = 3 x (4 x 6) = (3 x 6) x 4.
If instead of the arithmetical numbers, 3, 4, and 6, we use
the algebraic numbers a, b, and c, we have
a x b x c = (a x V) x c = a x (6 x c) = (a x c) x b.
The above is an illustration of the Associative Law of
Factors, which may be stated as follows :
The factors of a product may be grouped in any order.
axbxcxd=(ab) x (cd) = a(bx c)d=a(bxcxd).
8x(-3)x(-4)x(-2)={[8x(-3)]}x{(-4)
x (-2)}= etc.
50. Distributive law of Factors. The multiplication of
4 + 5 by 7 may be indicated thus : (4 + 5) x 7. The opera-
tion may be carried out in either of the following ways :
(4 + 5) x 7 = 9 x 7 = 63.
(4 + 5)7 = 4 x 7 + 5 x 7 = 28 + 35 = 63.
46 THE ESSENTIALS OF ALGEBRA.
In the first instance the 4 and 5 have been combined
into 9, and the product of 9 by 7 taken. In the second
instance the 7 has been distributed as a multiplier of the
terms 4 and 5 of the multiplicand, and the sum of the
separate products taken. If these arithmetical numbers
4, 5, and 7 be replaced by the algebraic numbers a, b,
and o, we have
(a + J) xc = axc + bxc = ae + bc.
The fact expressed by the algebraic identity
(« + ?>) x c = ac + be
is known as the Distributive Law. Since the multiplier
and multiplicand are commutative,
V^ c X (a + b) = (a + 6) x c = ac + be.
It may be noticed that the Distributive Law harmonizes,
as it should, with the definition of multiplication. To
multiply a + b by c is to do to a + b what has been done
to 1 to produce c. This would certainly mean that we
are to do the same thing to a and b and add the results.
In distributing factors the law of s.igns must be observed.
(a — b} x 5 c = 5 ac — 5 be.
(_ 2 «2 + }2) x (- 3 c) = 6 a 2 e - 3 b 2 c.
EXERCISES.
Distribute the following factors:
1. (a + b — c) x 2 d.
2. (2a + 3 6-4c)x-(3d).
3. (2 x + y - 3 z)(2 a).
4. (12 a? - 5 f + 13 xy)(i b).
5. (15 xy— 13 yz + 12 xz)(— 2 a&).
MULTIPLICATION AND DIVISION. 47
6. (ab~5b 2 + 3c-a)(5xy).
7. (ax 2 — bx* + cxy)(— 3 m).
8. (4=lx + 5my — 3nz)(2ab).
9. (- 6 a 3 + 5 a 2 6 - 3 a& 2 + 6 3 )(- 3 a?).
10. (a:B 2 + 6a;+c)(-2 2/ 2 2).
51. Index Law of Factors. In a continued product, abed,
any two or more of the factors may become equal. To
indicate the product when two or more factors have
become equal, a convenient notation has been devised. If
b should become equal to a in the above product, we would
have aacd, and it would be written a?cd. The small 2 to the
right of the a and slightly elevated is called the exponent.
If b and c each equal a, we have aaad = a s d. If b, c, and
d each equal a, we have aaaa = a*. It should be noticed
that the exponents 2 , 3 , and i indicate in these cases the
number of times a occurs in the respective products.
a • a = a 2 , read a square.
a • a ■ a = a s , read a cube.
a • a • a ■ a = a 4 , read a fourth power.
a ■ a • a ■ ■ ■ to m as = a m , read a exponent m, or a mth.
It should be noted that a = a 1 .
a h = a ■ a • a ■ a ■ a.
cfix 2 = a ■ a ■ a ■ x • x.
Write out the equivalents of
(1) a 6 x 2 =
(2) 5a s x 2 f =
(3) 1 a • a ■ a • a • x ■ x • x =
(4) aV(3) 3 =
(5) 2-2 .2.2- 2-x-x-x-x-x =
48 THE ESSENTIALS 01? ALGEBRA.
a s xa 2 =a-a-axa-a = a-a-a-a-a = a 5 = a 3+2 .
a m xa* = (a ■ a ■ a ••• to m a's)(a • a ■ a ••• to n a's)
= (a • a • a ••• to (m + n) a's)
= a m+n .
The result a m x a" = a m+n
is the Index Law for positive integral exponents. It is read
a with exponent m multiplied by a with exponent n equals
a with exponent m + n.
a? ■ x 2 = x 3+2 = x 5 .
y* ■ y b = y i+i ' = f-
EXERCISES.
1. X*-X 7 =
2. a s -a*-a s = (a 3 • a 4 ) • a 5 = (a 3 + 4 ) • a 5 = a 3+4+s = a 12 -
3. y 7 -f-y x =
4. a 5 • a 2 • x* • a 9 = (a 5 • a 2 )(a; 4 • a; 9 ) = a 7 • a; 13 -
5. 3 2 -3 3 -2/ 4 .^./.a^ =
6. 2 4 -2 5 -3 7 -3 2 .5 4 -5 7 =
7. a 12 • x 5 • y* ■ a 7 ■ x 12 ■ y 9 =
8. (x + y y-(x + y) 7 -(x + y) 9 =
9. (a + by-(a + by°-(x + yf-(x + yy =
10. 7 3 -7 6 .(y + z) s - (y + z) n - (a-b) 5 ■ (a-6) n =
52. The Multiplication of Any Number of Monomials.
Rules :
(1) Write the product of the numerical coefficients.
(2) Attach the literal factors of the product, observing the
Judex Law for repeated factors.
(3) Prefix the proper sign, determined by the number of
negative factors, + if an even number, — if an odd number.
These rules result from observing the laws already developed. •
MULTIPLICATION AND DIVISION. 49
Multiply together 3 ab, 4 ac, 5 be, - 3 abc.
The indicated result is
3 ab x 4 ac x 5 be x ( — 3 a&c)
= 3 ■ 4 • 5 • ( — 3)a -a-a-6-6-6-c-c-c (by Commutative Law)
= — 180a ■ a ■ a ■ b -b -b • c • c ■ c (by multiplying numerical
factors and observing law of signs) ,
= - 180 a*bW (by Index Law)
The above has been developed by an application of the laws, but
the result is in exact accord with the rules on page 48.
(2a) x (- 3a6)(-2Z>c)=+ 12 a 2 6 2 c.
EXERCISES.
Multiply together the following monomials :
1. 3 ax, 4 aft, — 5 axy. 5.-4 lx\ — 3 l s y 2 , — 8 x*f.
2. 5 a 2 x, 3 b 2 y, 2 aVy*. 6. 2 abc, 3 b 2 <?, - atbc 5 .
3. — 4m 2 w, 2mn s , — 5mW. 7. 3(« + 6) 3 , 42/(a+6) 2 ,5(a+6) 4 .
4. -xyz,5x 2 f, -IxyW. 8. 8ax 2 , 5(a+b) s , -2a»(a+b) 2 .
53. The Multiplication of a Polynomial by a Monomial.
Rules :
(1) Distribute the monomial as a multiplier of each term
of the multiplicand.
(2) Connect the results by the proper algebraic signs as
determined by the law of signs.
By the Distributive Law
(a + 5) X c = ac + be.
b can be any number, as d + e; then (a + b~)c = ac + be
becomes
(a + d + e)c = ac + (d + e)c =ac + dc + ec.
In general, „
(a + b + c + d+ -)m = am + bm + cm + dm H .
50 THE ESSENTIALS QF ALGEBRA.
EXERCISES.
1. (5 x + 3 y) x 2 a = 10 xa + 3 ya = 10 ax + 3 ay.
It is customary to write the letters of a product in alphabetic
order. This can always be done by an application of the Commutative
Law.
2. (3a + 46)x2a = 6o 2 + 8a&.
3. {3a 2 -4tab)xiab =
4. Multiply 3 d*x — 2 ax 2 + 3 xy by 5 axy.
5. Multiply -3l 2 x-\-±lx 2 -5lxy by -2JW.
6. (aa + 3 ay — 5 bey) x 3 axy = 3 a 2 xy + 9 a 2 aw/ 2 — 15 abcxy 2 .
7. Multiply a 3 - a 2 6 + a& 2 - b s by - a 2 6 2 .
8. Multiply 3 a 4 - 4 a?b - 6 a 2 & 2 + 7 aft 3 by a& 8 .
9. Multiply a 2 + & 2 + x 2 + y 2 + z 2 by 6 a6a; 2 .
10. Multiply 5 ab -12 cd -3 fh by -6 6c.
11. Multiply — 4 a; 2 ?/« + 7 xy 2 z — 3 xyz 2 by — 5 xyz.
12. Multiply 3W + 7 6 2 c 2 - 14 c 2 a 2 by 8 aW.
13. Multiply 3(a: + 2/) 2 + 4(ft + &) 2 by 2(x + y)(a + 6).
3(z + 2/) 2 + 4(a + &) 2
2Q + y)(a + &)
6 (a; + 2/) 3 (a + 6) + 8 (x + y)(a + & 3 )
14. Multiply
(x 2 + y 2 f-3x(x 2 + y 2 ) 2 + Ay(x 2 + y^) by 5 ^(^ + y*).
15. Multiply («-6) 4 -3 a:(a-&) 3 +7 y(a-b) 2 by 2 a; 2 !/(a- 6) 2 .
16. Multiply
3(x + yf + 2a 2 (x + y) 2 -3b 2 (x + y) by - a 3 & 2 (a; + y)\
17. Multiply
5(a^ + y) 3 - 3 a 3 (a; 2 + y) 2 - 7 a 4 6(a^ + y) by - 4(a,- 2 + y) 2 .
MULTIPLICATION AND DIVISION. 51
18. Multiply
10(a 2 + 3 by - 6(a 2 4 3 6 2 ) 2 + 12(a 2 + 3 6 2 ) by -5 (a 2 + 3 6 2 ) 3 .
19. Multiply (a + b + c) 3 - 3(a + b + c) 2 by 5 (a + b + c) 4 .
20. Multiply
(ax 1 + bx + cf + 4(aa; 2 + bx + c) 2 by - 5(aa: 2 + 6a; + c).
54. Meaning of x a and a x 0.
(b — c~)a = ab — ae.
If b = c, this becomes
(c — c)a = ac — ac.
But c — e=0 and ac— ac = 0.
Hence, x a = 0.
By the Commutative Law,
Oxa = axO = 0.
55. Multiplication of a Polynomial by a Polynomial. Rule.
To multiply a polynomial by a polynomial multiply each
term of the multiplicand by every term of the multiplier and
take the algebraic sum of the results.
This is a direct consequence of the Distributive Law.
Find the product of (a + J) by (x + y).
Let x + y be replaced by c.
Then (a + 6) x (x + y~) = (a + b) x c = ac + be.
But c = x + y,
and so (a + b~)(ji + y) = ac + be = a (x + y) + b(x + y)
= ax + ay + bx + by.
This product consists of the sum of the products of each
^erm of the multiplicand by each term of the multiplier.
52 THE ESSENTIALS OF ALGEBRA.
If three polynomials are to be multiplied together, two
of them must be multiplied as above and their product
multiplied by the third.
Thus
(a +b~)(x + y)(r + s) = (ax + ay + bx + by)(r + s)
(by multiplying factors one and two together).
(ax + ay + bx + by)(r + s)= arx + ary + brx + bry
+ asx + asy + bsx + bsy.
EXERCISES.
Find the product of the following polynomials :
1. (3a + b)(a + b) = 3a? + ab + 3ab + W=3a ! + 4 : ab + b 2 .
The distribution may be arranged conveniently as follows :
3a + b
a + b
3 a 2 + ab (the product of (3 a + b) by a)
3ab + b* (the product of (3 a + b) by b)
3 a 2 + 4 ab + b 2 (the total product)
This is called long multiplication, and should only be used
when the distribution can not easily be made in the straight
line form illustrated above.
2. (x + y){x — y) = x i + xy-xy-y 2 = a?-y 2 .
3. (2x + y)(2x-y) = 4x>-y 2 .
4. (ix + 3y)(Ax-3y). 8. (x + 5)(x-2).
5. (ax + by) (ax -by). 9. (y 2 + 4) (y 2 - 3).
• 6. ( x + i)(x + 2). 10. (ax 2 + b)(x + c).
7. (a:-2)(a! + 3). 11. (3x + 5)(ix + 2).
12. (ix s + 3x 2 + 5x-6)(3x s -2x 2 -3x + 2).
MULTIPLICATION AND DIVISION. 53
Multiplicand and multiplier are both arranged according to
the powers of a; ; that is, the exponents of x decrease uniformly
from left to right. The arrangement will be just as good if we
reverse both factors and have the powers of x increase from
left to right. For multiplication we may arrange the work as
follows :
4a?'+3ar ! + 5x - 6
$3?-2a?- 3a; + 2
12 a; 6 + 9 ar>+ 15a: 4 -180? (product of the multiplicand by 3a?)
-8af- 6a; 4 -10af ! +12a? (by -2a: 2 )
-12a; 4 - 9a?-15a: 2 +18a; (by 3 a;)
8a?+ 6a^+10a;-12 (by 2)
12a: 6 + a?— 3a; 4 -29a?+ 3a?+28a;-12 (the total product)
The orderly arrangement according to powers of x merely
insures that the terms of the product will come in an orderly
way, thus making it easier to arrange like terms in columns
ready for adding. When arranged in this way the highest
power in the product appears first, and is the product of the
two highest terms in the factors.
13. (4a 2 -5a 3 + 3a-6)(3-a 2 + 2a).
When these factors are arranged according to powers of a,
we have
(_ 5 a » + 4 o 2 + 3a - 6) (-a 2 + 2 a + 3).
Now multiply as in Exercise 12.
14. (tf-6y + 7tf-12)(4:-2y + tf).
When arranged, this exercise becomes
(y s + 7y 2 -6y-12)(f + 0y 2 -2y + 4 : ).
Observe that in the multiplier the term y 2 does not occur ;
in the arrangement according to y we write y 2 .
54 THE ESSENTIALS OF ALGEBRA.
15. (&«-5 + 66 2 -3& + 2& 3 )(& 2 -26 + 2).
16. (a 3 -7« + 12a 2 + 6)(a 2 -3a-5).
17. (3ar J + 12x 2 -10« + 4)(-^-5 + 8a;).
18. (5 a?y» - 6 xy + 12 a?}? -4) (3 -43?!? + }f *?).
19. (4z 3 -4 + 6z 2 -5z)(> 2 -4z + 3).
20. (3a 2 + 4a 4 + a + 5a 3 -4)(a 4 -a 3 + a 2 -a + l).
21. (2ar i + 3ar ! -4a;-l)(3a; + 4).
Solutions.
(1) (2)
2o?+ 3x 2 - 4x - 1 2+ 3- 4- 1
3x + 4 3+4
6a; 4 + 9 a 8 -12 a 2 - 3x 6+ 9-12- 3
8^ + 12 a; 2 -16a;-4 8 + 12-16-4
6 a 4 + 17 2? + a; 2 -19 a; -4 6 + 17+ 0-19-4
In solution (1) the multiplication is carried out in the usual
way, while in solution (2) merely the coefficients are used.
In the answer on the right the x's should be inserted, beginning
with x*. The method used on the right is called multiplying
by detached coefficients. It is a device which saves time by the
omission of all letters.
In using detached coefficients the following directions should
be observed :
(1) The multiplicand and multiplier must be arranged
according to the same letter.
(2) must be used as the coefficient of every power of the
letter of arrangement which does not occur.
(3) The letter of arrangement is inserted in the product by
beginning at the left with a power equal to the sum of the
highest powers in the multiplicand and multiplier, and decreas-
ing uniformly to the right.
The following exercise illustrates this :
MULTIPLICATION AND DIVISION. 55
22. (3ar i + 4ar J +5a 2 + 7a; + l)x(5ar i -3a;).
3 +0 +4 +5 + 7 +1
5+0-3
15 + +20 + 25 +35 + ~5
-9 - -12 -15 -21 -3
15O& + 0X 1 + llafi + 25x !i + 23as i -10sc s -2lx i -3x
The highest powers of x in the two factors are 5 and 3, so
the product begins with X s .
In the following exercises use detached coefficients.
23. (a 4 -5a 2 + 4a 3 -3 + 2a)(3a 2 -5a + 2).
24. (a 4 - 4 a 3 + 5 a 2 - 2 a + 7) (a 4 + 4 a 3 - 5 a 2 + 2 a - 7).
25. (afy*.— 4 aft/ 2 + 6 aiy — 5) (aft/ 3 + 4 aft/ — 6xy + 5).
26. (3 ar> - 7 ar 5 + 4 a; -5)(2 a: 2 -3 a; + 4).
27. (7 2/ 4 -5 2 / 3 + 3 2 /-4)(4 2 / 3 -8y+l).
28. (a 3 6 3 - 6 ab + 7) (5 aft - 4 a 3 & 3 ).
29. (4 a*b* - 7 - 6 a 2 & 2 + 3 ab)(ab - 5 + a 3 & 3 ).
30. (3a; +4a;-5)(ar ! + a; + l)(a; 2 -3a; + 3).
56. The Identity in Multiplication. The multiplicand
multiplied by the multiplier is identically equal to the
product.
Multiplicand x Multiplier = Product.
(a + b) x c = ac + be.
Let a=.b = o=l.
(l + l)xl = lxl + lxl.
2x1 = 1 + 1.
2 = 2.
56 THE ESSENTIALS OF ALGEBRA.
This furnishes a convenient method of verifying the results
in multiplication. If in Exercise 22 above we put x = 1, we
have
(3+4+5+7+1) X (5-3) = 15+ll+25+23-10-21-3.
20x2=40.
40=40.
This merely verifies the coefficients. In order to verify the
exponents, some value other than 1 would have to be sub-
stituted. In case more than one letter occurs, numerical values
must be given to each.
(x + y)(x-y) = x 2 -y\
Let x = 1 and y = 2.
Then (1 +2)(l-2) = l 2 -2 2 .
(3)(-l)=l-4.
-3= -3.
EXERCISES.
Perform the following multiplications and verify by means
of the identity :
1. (a? + ab)(2a+'3b). 5. (ox 2 -3xy + 2y !! )(2x + 3y).
2. (4 a + 6 6) (2 a - 4 b). 6. (3 x + 4 a) (3 x - 4 a).
3. (7a; 2 + 3a; + l)(a; 2 -a; + l). 7. (5 ab - 7 cd) (5 ab + 7 erf).
4. (5 a6 + 3 cd) (2 aft - 4 cd). 8. (6x i y-ixy 2 +f)(3xy-7y v ).
57. Involution. In multiplication when the factors are
alike, the operation is called involution, and the result a
power. „
1 a • a • a = a 6 .
a 2 • a 2 • a 2 = (a 2 ) 3 = aa aa aa = a 6 .
(a 3 • a 3 • a 3 ) = (a 3 ) 3 = aaa aaa aaa = a 9 .
MULTIPLICATION AND DIVISION. 57
58. Meaning of (a m )".
(a m )" means a m ■ a m • a m ••• to n factors.
Since each of the n factors, a™, contains m a's, there
are in the product mn a's.
But a mn means a ■ a • a • a ■•• to mn factors.
Hence, (a m ~) n = a mn .
■ More generally, (a m V) n = a mn ¥ n .
The exponent n is- distributive as to factors within the
parenthesis. ^ y = ^
The exponent n is not distributive as to terms within
the parenthesis.
O + y) 3 = (x + y} (x + y~)(x + y).
(x + yY is not equal to x 3 + y z .
The difference between the following forms should be
noted: 2 ( «\
a? ! =a 9 . a m = a<» >•
(«3)2 _ a 6_ ( a »)» = a mn -
EXERCISES.
Eemove the ( ) and simplify :
1. (a 2 ) 4 (a 3 ) 2 = a 8 • « e = a". 6. 3* ■ (3 2 ) 3 .
2. (2y(fyY. ?• (a 4 a; 2 ) 4 (aa; 2 ) 3 .
3. ((z + 2/) 2 ) 4 - 8- GW-WPWW-
4. (4 s ) 5 (a^ 4 ) 6 . 9 - O 2 + ^ 2 ) 82 • C* 2 + ff-
5. 2 32 • (2 3 ) 2 = 2 9 • 2 e = 2 15 . 10. (a& 2 c) 4 • (a 2 6c 3 ) 5 .
58
THE ESSENTIALS OF ALGEBKA.
DIVISION.
59. Division Defined. — Division is the process of finding
one number, when, the product of two numbers and one of
them are given. The given product is the dividend, the given
number the divisor, and the required number the quotient.
Division is the inverse of multiplication, the dividend
corresponding to the product, the divisor to the multiplier,
and the quotient to the multiplicand.
Since a x b = ab,
ab -=- b = a.
60. Law of Signs in Division.
From multiplication we have
(+«)(+ I) = + ab.
(+a)(- b) =- ab.
(- «)(+&) = — ab.
(— a)(— b) = + ab.
From the definition of division it follows that
+ ab -f- ( + b) = + a.
— ab -=- ( — b) = + a.
— ab-h(+b) = — a.
+ ab --(-£) = - a.
Like signs in dividend and divisor give a positive quotient,
and unlike signs give a negative quotient.
61. Index Law. We already know that
as" = a m+n -
a" = a m = a"
a 7 " x a" = a"
-,ni+n _s_ «rc n m n m+n~n
Hence,
Suppose m + n = p,
then a" -s- a" = aP~".
More generally, aPb^C -*- a"b'c* = a"-"b"- s c''- t .
MULTIPLICATION AND DIVISION. 59
For the present it is understood that the exponents of
the factors of the dividend are not less than the exponents
of the corresponding factors of the divisor.
The above considerations show that the exponents of the
factors of the divisor are subtracted from the exponents of
the like factors of the dividend in order to obtain the expo-
nents of the factors of the quotient.
Illustrations :
(1) a 7 -^ a* = a 7 ~ 4 = a 3 .
(2) a 10 & 6 -=-a 3 6 5 = a 10 - 3 6 6 ~ 5 =a 7 5.
(3) aWc 8 -=- a s b s c* = abW.
(4) (x + y~) 3 (z + wy -=- O + y~) (s + w) 2 = (x + y)\z + wf.
62. Meaning of and a .
a
We know that a x = 0.
Hence, - = 0.
a
a"xl= a n .
a n
Hence, — = 1.
a n
a"
-= a"
a n
But, by Index Law,
Now by Axiom 1, a = 1.
Any quantity with an exponent is equal to 1.
63. Division of One Monomial by Another Monomial.
Rules :
(1) Divide the numerical coefficients as in arithmetic.
.(2) Attach the literal part determined by the Index Law.
(3) Prefix the proper sign determined by the law of signs.
18 a?yhl -5- 6 xyh* = 3 a?y°z = 3 x*z, since y° = 1.
60 THE ESSENTIALS OF ALGEBRA.
EXERCISES.
Divide :
1. 21 o 5 ^ 2 by 7 aVty.
2. 72 a*xy 2 z n by 24 ay« 10 .
3. 108 x 7 y*z 2 by 36 xYz 2 .
4. 81 a 5 6 8 cd 2 by 27 a 5 b 2 d.
5. 144 a 4 (a: + y)° by 48 a 3 (a; + yf.
6. _63(o-6) 5 (a;-2/) 7 by 21(a-b)\x-yf.
7. 45 aryCiB 2 + y 2 ) 2 by - 5 */(a; 2 + ?/ 2 ) 2 .
8. 3 5 a*xSjz by 3 2 a 2 xY
9. - 7 4 ZV/O - *) 5 by - 7 3 ZV(« - a;) 4 .
10. 3 4 • 5%a - ») s (6 - y) A by 3 2 ■ 5 3 (a - ») 2 (6 - y) 4 .
64. Division of a Polynomial. by a Monomial. From the
Distributive Law of Factors we know that
(a + b + c) xk = ak+blc + eh.
Hence, (a/c + bk + ck~) s-k = a + b + a,
which shows that k is distributed as a divisor to every
term of the dividend.
Rule. Divide each term of the polynomial by the mono-
mial and add the results.
10 a 2 x 3 + 5 ax 2 y — 20 a s x 2 y 2 n , , ,
-— s - ■■?- = 2 ax + y — 4 a 2 y 2 .
5 ax*
EXERCISES.
1. (12 ay - 8 a*xy 5 — 4 a 5 x*if) -=- 4 a 3 f.
2. (9 abcx* - 18 a 2 bV + 27 a s b<?x) h- 9 abx.
3. (30 x*y*z + 25 a?y*z 2 - 35 as>yW) -f- 5 x*tfz.
MULTIPLICATION AND DIVISION. 61
. (x + y) s + 5 (x + y) 2 — a (x -f y) /T> , .
4. * — — ** — I ^ — '-M-i i — ZLJLL . ( Regard x + y as a term. )
(?> + y)
5. [4 (a -by -5 x(a - b) s + 11 xy (a - &) 5 ] -=- (a - &) s .
6. Divide 5 xy (x + y)* — 10 a?y 2 (x + y) s — 15 xy 2 (x + y) s by
5xy(x + y) 2 .
7. Divide 11 a 2 b 2 (x 2 +a 2 ) + 22 a?b (x 2 + a 2 ) 5 - 33 ab^x 2 + a 2 ) 2
by lla&(a: 2 + a 2 ).
8. Divide
7 a?y% (a 2 - 6 2 ) 2 + 21 xfz 2 (a 2 - 6 2 ) 11 by - 7 xyz (a 2 - ft 2 ) 2 .
9. Divide
24 a s b 2 c {ax + by - 36 aW (a» + b) 7 by - 12 (a 2 6 2 c) (ax + b)\
10. Divide
-33(az 2 +&aj+c) 2 +44 a^atf+bx+c) 5 by 11 (aar'+ta+c) 2 .
65. Division of a Polynomial by a Polynomial. The divi-
dend and divisor should be arranged in descending or
ascending powers of some common leading letter. This
gives a quotient arranged with respect to the same letter.
The first term of the quotient is found by dividing the
first term of the dividend by the first term of the divisor.
The process is illustrated in the following solutions :
(1) Divide
x 5 - x* - 11 X s + 16a 2 - 2x - 3 by a? - ±x + 3.
a?-4x+3')x 5 -x i -llx s + 16x i -2x-3(3? + 3a?-2x-l
^ x 6 -4:x i + 3a? =a^> 2 -4 z+3)
3x i -14a? + 16a? - 2 x- 3 = 1st partial div.
3s*-12s 3 + 9 a? = 3x%x*-4x+3')
-2a? + 7x>-2x - 3 = 2d partial div.
-2a? +8x 2 -6x = -2x(a?-4x + 3')
— x 2 + 4x— 3 = 3d partial div.
-aJ+4a;-3=-l(a*-4aH-3)
62 THE ESSENTIALS OF ALGEBRA.
This scheme of division is merely a separation of the
dividend into parts. In the example just solved we have
separated x 6 — x* — llx 3 + 16x 2 — 2x — 3 into these parts:
(a^-4a^+3a^) + (3a^-12a^ + 9a; 2 ) + (-2^ + 8a; 2 -6a;)
+ (-a; 2 + 4a:-3).
Now, regarding the dividend in this separated form, we
have the division thus :
afi-iat + Sa* 3 a; 4 - 12 x 3 + 9a? -2 x 3 + 8 x 2 -6x
x % — 4 a; + 3 a; 2 — 4 a; + 3 a; 2 — 4 a; + 3
- f + + x - S = 3 * + 3 x 2-2x- 1.
a; 2 - 4 x + 3
(2) Divide a^ - 16 by 2 + x.
x + 2)x* - 16(a* -2a; 2 + 4a;-8
a; 4 + 2 x s
-2a£
- '2 x 3 - 4 a?
+ 4 a; 2
+ 4 a; 2 + 8 x
-8a;-
- 8a;-
-16
-16
Divisions such as the above, which terminate without,
any remainder, are called exact.
EXERCISES.
Divide :
1. o 4 -a? ! -9a; 2 + 13a;-12 by a;-3.
2. x i -5x i -3x + l& by x-5.
3. 6ar i + 7ar ! -18a; + 5 by 2a; + 5.
4. 2 a; 4 -90? + 17 a; 2 -14a; by ar*-2a;.
5. 3a 4 -6a 3 + 2a 2 + 14a-21 by o 2 -2a + 3.
MULTIPLICATION AND DIVISION. 63
6. a 4 + 4a 3 & + 6a 2 6 8 + 4a& 3 + & 4 by a? + 2ab + b 9 :
7. x s - a 6 by x 2 - a 2 .
8. & 2 c 6 — a 8 by 7>c 3 + a 4 .
9. m 6 -3m 4 + 3m 2 -l by m 2 -l.
10. a 2 " + 2 a"6" + 6 2 " by a" + 6".
11. o 2n — 6 4 " by a n — 6 2 ".
12. a 3 " — 6 s " by a" — 6".
13. a; 4 -13ar ! + 47a: 2 -31a; + 4 by a; 2 -6a + l.
14. x* - 12 a; 3 + 54a; 2 -108 x + 81 by a 2 -6a; + 9.
15. a"b 3 - 3 « 2 6 2 cd + 3 abc 2 d 2 - c 3 d 3 by ab - cd.
16. 12 aft/ 6 - 17 *y + 10 arty 2 -3 by 4 a; 2 ?/ 2 -3.
17. ar 5 -10a: 4 +40ar , -80a; 2 +80a;-32 by a; 2 - 4 a; + 4.
18. 7 + 15 a; -21 3^ + 18 3^-4 a- 4 by 7 — 6 a; + 4 a; 2 .
19. (a + 6) 2 -5(a + 7>)+4 by (a + 6) - 1.
20. (aj + y)* + 7(ai + y)-18 by (as + y) + 9.
21. (a + xf - (a + x) - 42 by (a + a;) — 7.
22. (m + n) 2 — 7 (m + m) — 44 by (m + n) + 4.
23. (a + b) s - X s by (a + b) — x.
24. (»+ ?/) 3 - (a + 6) 3 by (a; + ?/) - (a + 5).
66. Detached Coefficients^ When the dividend and divisor
are arranged in descending powers of some common letter,
the quotient is also thus arranged. We may then perform
the division by the use of the coefficients only.
(1) Divide x i + x s ~Sx 2 +7x^6 by x*-x + 2.
1_1 + 2)1 + 1-3 + 7- 6(1 + 2-3
1-1 + 2
2-
2-
-5 + 7
-2 + 4
- 8 + 3 -
-3 + 3-
-6
-6
64 THE ESSENTIALS OF ALGEBRA.
Since x i s-x 2 = x i , we know that the quotient begins
with x 2 , and is x 2 + 2 x — 3.
In the use of detached coefficients all powers of the
letters from the highest to the lowest power must be
present in both divisor and dividend. If any powers are
absent, they must be inserted with zero coefficients. _
(2) Divide X s - 8 by x-2.
In the dividend neither x 2 nor x appears. We insert
them, writing the dividend x 3 + 0x 2 + 0x— 8.
l_2)l + + 0-8(l + 2 + 4
1 — 2 x 2 + 2 x + 4, quotient.
2 +
2-4
4-8
4-8
EXERCISES.
Divide, solving by detached coefficients :
1. a; 4 — 5 a; 2 + 4 by a — 1; by a; + 2.
2. a; 4 .— 7 a? + 11 x* + 7 x — 12 by a; — 1; by a; — 4.
3. a; 4 -13a: 2 + 36 by a; 2 + a;-6.
4. a^-18a;y-1752/ 4 by a?-25y*.
5. 2m 4 -17m s m + 31mW-23mm 3 + 12ji 4 by2m-3n.
6. a 4 - 256 by a 2 + 16.
7. 6 8 -729 by 6 2 -9.
8. 2/ 8 -4096 by */ 2 +8.
9. a 8 + a 4 + l by a 4 — a 2 + l.
10. 16a 4 + 4a 2 +l by 4a 2 + 2a + l.
11. x 4 + 4 by a; 2 + 2 a; + 2.
12. 13& + 156 3 -176 2 -3 by 5& 2 -46+3.
MULTIPLICATION AND DIVISION. 65
13. a 6 -b 6 by a 3 + 2a 2 b + 2ab 2 + b s .
14. at + f by tf-tfy + rftf-xtf + y*.
15. 10 a*- 48 a 3 b + 26 a 2 b 2 + 2iab s by -5a 2 + 4a&+3& 2 .
67. Inexact and Continued Division.
(1) Divide x* + 1 by x + 1.
a;+:L> 2 + 10-l
3? + X
— X+l
-x-1
2
In this example there is a remainder of 2, and the
division is inexact. In such examples the division should
continue until the largest exponent of the remainder is
less than the largest exponent of the divisor.
(2) Divide 1 + x 2 by 1 + x. '
l + x)l + x\^-x+2x-2a?
— x + x 2
2*r
j£x 2 + 2x*
-2x*
-2a^-2a^
2x±
This division may end with two terms of the quotient
and the remainder 2x?, or with three terms and the re-
mainder — 2 x 3 , or with four terms and the remainder 2 z 4 .
Evidently the division might be continued to any number
of terms desired. When inexact division takes this form,
it is called continued division.
66 THE ESSENTIALS OF ALGEBRA.
EXERCISES.
Find quotients and remainders :
1. Divide a? — 7 x 2 + 11 x — 7 by x + 3.
2. Divides 4 + 7 a 3 -8 a; + 13 by a; 2 + 3a;-2.
3. Divide a 3 - 11 a 2 + 21 by a + 5.
4. Divide & 4 - 14 b 2 + 11 by b 2 + 9.
5. Divide o¥ + 9 aV - 7 by ax + 3.
In the next five exercises continue the division to four terms :
6: Divide 1 + x by 1 — x.
7. Divide 2 + 3a; -4 a 3 by 2-x.
8. Divide 3 - 6 x + 8 x 2 by 3 + 5 x.
9. Divide 1 - 17a; + 13a; 2 - 8b 3 by 1 - 16a; + bx 2 .
10. Divide 10 - 20 x + 25 a; 2 - 31 X s by 2-5a;.
68. The Identity in Division.
Dividend -=- Divisor = Quotient.
(x 3 -l)-i-(x -l) = x 2 + x + 1.
This is true for all values o# x.
Let a; = 2, and we have
(23-l)-(2-l) = 2^2+*lV
(8
1)^-1
= 4 + 2 + 1,-
7-=-l
= 7,
7
E=7.
In this case, should we make x = 1, we get -=- on the
left of the sign =. -r- is indeterminate. In using the
identity to verify divisions, avoid substitutions that will
produce this form.
MULTIPLICATION AND DIVISION. 67
EXERCISES.
Divide and verify by substituting particular values :
1. 6Z 3 -17Z 2 + 24Z-16by 2Z 2 -3Z + 4.
2. y 5 — 1 by y — 1.
3. a,- 5 + 1 by x + 1.
4. a 3 - 12 a 2 + 48 a - 64 by a 2 - 8 a + 16.
5. 625-500z + 150z 2 -20z 3 + z 4 by 5-z.
6. 625 - z 4 by 25 + z 2 .
7. 2/ 4 -3i/ 2 -154by2/ 2 -14.
8. a 4 6 4 + 4 a 2 & 2 - 117 by a 2 & 2 + 13.
9. a; 6 — y 6 by x* — y 2 .
10. a 4 -4a 3 -34a 2 + 76a + 105 by a-7.
REVIEW EXERCISES.
1. Find the value of 6 a; 2 — \xy + 12 y 2 , when a; = 4, y = — 1.
2. Find the value of X s — 64 ?/ 3 + 8 z 3 — 3 zi/z, when a; = 0,
2/ = 2,z = 5.
3. From 16 .r' - 4 / + 12 z 3 — 14 arty + 3 a;;/ 2 subtract 12 f +
8x" + ixy-- 10 x 2 y + 11 z 3 .
4. Remove parentheses and unite like terms :
16 - { 12 a + [4 b - 3 c] + 8 - [8 a - 3 (4 - 2 6)] J .
5. Remove parentheses and unite like terms :
-3a; 2 +4[a;?/-a;(3a;-42/)-32/(4a; + 22/)]-Ja; 2 + 3(a;2/-!/ 2 )j.
6. Unite terms in x, y, z:
ax + by + cz — 4:(— a'x + b'y — c') + 3 (Ix + my + nz).
7. Simplify x — 4 3/ — [z — y— (x + y — z)], and find value
when x = y=z = 1.
b'8 THE ESSENTIALS OF ALGEBRA.
8. Simplify 4 (a - 5 \b - c\) - [3 b + \2 b - (c- a)}], and
find value when a = 1, 6 = 2, c = 3.
9. Find value of Va." 2 + t/z -f Vy 2 +2 «a; + V« a + a;?/ — ~\Zxyz,
when a; = 1, ?/ = 0, z = 2.
10. Find value of 3 Va * + ? -iVx 2 + f + z 2 - xy-yx- »
a; + ?/ — a
when x = A, y = 5, z = 6.
11. Eemove parentheses and simplify :
x(y-z)+y(z-x) + z(x — y).
12. Eemove parentheses and simplify :
x ' (y — «) + ?/ 2 (2 — «) + a; 2 (» - 2/).
13. Multiply x 2 + x + 1 by x 2 — x + 1. •
14. Multiply x 2 + y 2 + 1 — x — y — xy by x + y + 1.
15. Multiply ar 2 - 4 ?/ 2 by x 2 + xy 2 .
16. Multiply 12a; 4 -3ar ! +10z 2 -5a; + 4 by 3« 3 -a^ + 5a;
-4.
17. Find the value of a? -Ax 2 + 3 x -5, (1) when x = 2;
(2) when a; = — 1 ; (3) when x = 0.
18. Find the remainder after dividing x 3 — 4 : x 2 + 3x — 5
(1) by as-2; (2) by a> + l; (3) by z.
iVote that the remainders are the same as the results found in
Exercise 17.
19. Divide 2 a 5 —3 a 4 6 - 6 a s b + 13 a 2 b 2 - 6 «6 3 by 2 a - 3 6.
20. Divide 3 0^ + 14 a? + 9 a; + 2 by ^+5 a + 1.
21. Divide 2 a 2 + a& - ac - 3 ft 2 - 4 6c - c 2 by 2 a + 3 b + c.
22. Divide x 2 — (a + b)x+ abby x — a.
23. Divide ar 5 - (a + & + c) x 2 + (ab + be + ca) x — abc by x-a.
24. Divide z 3 - (I + m + n) z 2 + (Im + mn-\-ln)z — Zwjri by
? 2 - (Z + n) z + Zvj,
MULTIPLICATION AND DIVISION. 69
25. Multiply together (x n + a") and (x + a).
26. Multiply x n — y" by x n + y n .
27. Multfply 3 a: 2 " + 5 x n + 7 by 2 a" - 4 a; - 3.
28. Divide 3 a; 2 " + 13 a; 2 "" 1 + 15 a?'- 2 + 9 a: 2 — 3 by x n + 3 a;"- 1 .
29. Divide x m+n + y"x m — y m x n — y m+n + x"+y n by x n + y n .
30. Multiply a 3 + ft 3 by a—b, and divide the product by a+b.
31. Multiply 3x f + ix q -2x r by 2 a;* — 3 afl + x r .
32. Divide a 3 " — 6 s " by a"— 6".
33. Multiply a!" +2 + 3 x n+1 -5x n by ai n - z - 2 a;"- 1 + a;".
34. Divide x 5 " — y 5 ' 1 by a; 4 " + arV + a; 2 Y" + x n y in + y 4 ".
35. Divide a 6n —b V2n by a 2 "— 6*".
36. Divide (a + ft) 2 " - x 4r by (a + 6)» + a*.
37. Divide (x 2 + yf" - 1 by (x 2 + y)»- 1.
38. ( VoaT+5) 3 + J/ 3 by Vaa; + 6 + «/.
39. Simplify (a; 2 - xy + y 2 ) (x + y)(x- yf -=- (x* + y*).
40. 5(a 3 + a 2 b + ab 2 + 6 3 ) X 4(a-6) 2 -^[2(a 2 -6 2 ) x lO^+fc 2 )].
CHAPTER V.
IMPORTANT IDENTITIES.
Many expressions in algebra appear in standard or type
forms. When this is the case, multiplications and divi-
sions can be performed mentally by remembering certain
identities.
69. Multiplication Identities.
1. The Product of Two Binomials with a Common Term.
This form is given by
O + a) (* + b) = x 2 + (a + b) x + ab,
or (a + x)(b + x) = ab + (a + b)x + x 2 .
(1) (z + 10)0> + 5) = z 2 + (10 + 5>e + 50
= x 2 + 15z + 50.
To find the product of x + 10 and x + 5 it is only necessary
to see that a is 10 and b is 5 and -make these substitutions.
(2) ( a; _7)(a; + 8)=ar ! +(-7 + 8)a;-56
= x 2 + a; — 56.
In this a is — 7, and b is 8.
(3) (5 + x) (11 -x)= 55 + (11 - 5) x - x 2 .
In this exercise it is necessary to note the signs of the x's.
(4) (3x + y-5)(3x + y + 7) = (3x + yy + 2(3x + y)-35.
In this exercise the common part is 3 x + y.
70
IMPORTANT IDENTITIES. 71
EXAMPLES.
Write out the products of the following :
1. (® + .10)(a>-2). 6. (ax-ll)(ax + 6).
2. (3a: + 6)(3a; + l). 7. (3 xy -5)(3xy-6).
3. (7a-5)(7a + 4). 8. (4 x 2 + 7)(4x 2 -5).
4. (5+2a)(5 + 6o). 9. (3- <Lxy) (3 + 6 xij).
5. (2 2/-6)(2?/ + 7). 10. ( a + 6-6)(a + & + 5).
11. (a + 6 + 7)(a + 6-8).
12. (a; + 2a6-3)(a; + 2a6 + 7).
13. \4-(x + 2y)\\5 + (x + 2y)}.
14. [(a + &) 2 -4x][(a + 6) 2 + 7a;].
15. [3a6-(«-2/) 2 ][5a6 + (a;-2/) 2 ].
2. 2%e Square of a Binomial Sum.
If, in the identity
(x + a~)(x + b) = x 2 + (a + b)x + ab,
we let b = a,
it becomes (2; + a) (z + a) = x 2 + (a + a~)x + aa.
(jr + a) 2 = ^ 2 +2ajr + a 2 .
The square of the sum of two quantities is equal to the
sum of their squares increased by twice their product.
(5z + 4y)2 = (5z) 2 +2(5:K)(4y) + (4 202
s25 x 2 + 40 xy + 16 y 2 .
EXERCISES.
Write out the results in the following :
1. (x + yf. 2. (2x + a) 2 .. 3. (3* + 4 6) 2 .
4. [(x + y) + ay=(x + yy + 2a(x + y)+a 2 .
5. [>+(« + ft)] 2 . 7. [(a-3)+5yf.
6. [3a;+(2a + c)] 2 . 8. [(a + b) + (x + y)J.
72 THE ESSENTIALS OF ALGEBRA.
3. The Square of a Binomial Difference.
If, in the identity
(x + a~) 2 = x 2 + 2ax + a 2 ,
we change a to — a, it becomes
(jr — a) 2 = x 2 — 2 ax + a 2 .
The square of the difference of two quantities is equal to
the sum of their squares diminished by twice their product.
(2 a- y) 2 = (2 a) 2 -2(2 a)y + y 2
= 4 a 2 — 4 ay + y 2 .
EXERCISES.
Write out the results in the following :
1. (a -a) 2 . 5. [(a + 6) - xyf.
2. (3 a-!/) 2 . 6. [(3a + 2/)-a&] 2 .
3. (xy-4b) 2 . 7. \_( x + y)-(a + b)f.
4. (3 a 2 -by) 2 . 8. [(2 x - y) - 3 x'yj.
4. The Product of the Sum and Difference of Two Quan-
tities.
If, in the identity
(x + a)(x + b~) = x 2 + (a + b~)x + ab,
we let b = — a,
it becomes (x + a)(x — a) = x 2 -\- (a — a)x — aa.
(x + a)(x — a)=x 2 — a 2 .
The product of the sum and difference of two quantities
is equal to the difference of their squares.
(1) (7 x - 3 y)(l x + 3 y)= (1 x~) 2 -(3 ^2=49^-9^.
(2) (ax + by-c) (ax + by + c) = (ax + by) 2 - c 2 .
IMPORTANT IDENTITIES. 73
EXERCISES.
Write out the results in the following :
1. (x-y)(x + y). 2. (3a-b)(3a + b).
3. (lx + ab)(4:X — ab).
4. [(a + x)+d] [(a + x)-a] = (a + x) 2 -a 2 =a 2 +2 ax+x'-a?
= 2ax + x 2 .
5. [(as + 2y)-ai][(sB + 2y) + a;].
6. (x 2 + 2x + 16)(x 2 + 2x-16).
7. [(a + 6)-(a! + y)][(o + 6) + (<B + y)].
8. (ax 2 + bx + c)(ax 2 + bx — c).
5. The Square of a Polynomial.
If, in the identity
(x + ay = x? + 2ax + a?
we put a = y + z,
it becomes (x + y + z~) 2 = x 2 + 2 x(y + 2) + (y + s) 2 .
+y + zf = jr 2 +/ 2 + z* + 2 xy + 2 xz + 2/z.
This may easily be extended to include the square of a
polynomial of any number of terms, the result being that
The square of a polynomial equals the sum of the squares
of each term increased by twice the product of each term by
every other term.
(1) (x + y-a) 2 = x 2 + y 2 + a 2 + 2xy + 2x(-a)+2y(-a)
= a? + y 2 + a 2 + 2 xy — 2 ax — 2 ay.
(2) (3x + 2a-±b) 2
= (3xy+(2ay+(-4 : by+2(3x)(2a)+2(3x)(-ib)
+ 2(2a)(-4&)
= 9 x 2 + 4 a 2 + 16 b 2 + 12 ax - 24 bx - 16 ab.
74 THE ESSENTIALS OF ALGEBRA.
EXERCISES.
Write out the results in the following :
1. (a-x-y) 2 . 6. (x-y + a + b) 2 .
2. (3x — y + b) 2 . 7. (m — n—p — q) 2 .
3. (2a + 3b-y) 2 . 8. (2x + y- 3z + af.
4 (3a-5x-2y) 2 . 9. (a? - ab + 2x-3yf.
5. (2 + 3a-4&) 2 . 10. (2x-y 2 + xy-ay.
6. The Product of Three Binomials.
By actual multiplication, it is found that
(r + a)(x + b)(x + c) = x s +(a + b + c)x 2
+ (ab + be + ca) x + abc.
The product is arranged according to the powers of the
common letter x. The coefficient of x 2 is the algebraic
sum of the second terras of the binomials, the coefficient
of x is the algebraic sum of their products in pairs, and the
term free from x is the product of the second terms of the
binomials.
(1) (Hl)(x+2)(x + 3)
= x- + (1 + 2 + 3) a; 2 + (1 x 2 +1 x 3+2 x 3)as+l x2 x3
= X s + 6 x* + 11 x + 6.
(2) (aj + 2)(*-3)(a! + 4)
= sr) + (2-3 + 4y+[2(-3) + 2x4 + (-3)(4)]a!
+ 2(-3)(4)
= x 3 + 3x 2 -10a;-24.
EXERCISES.
Write out the results in the following :
1. (x + c)(x + d)(x + e). 3. (6 + 2)(&-l)(6 + 3).
2. ( a + x){a + y)(a + z). 4. (y -3)(y + 2)(y -1).
IMPORTANT IDENTITIES. 75
5. (m + 5)(m-4)(m-3). 7. (as 2 - S).(a? - 3)(a; 2 + 8).
6. (xy + 2)(xy-l)(xy-l). a (y* -4)(jf + ll)<y_ 7)-
9. (» + y + 5)(a: + y + 3)(as + y + 2).
10. (3as + y-2)(3a! + y-4)(3a; + 3/ + 6).
11. (a* + 6 + 2) (aa; + 6 -f 8) .(a* + 6 — 5).
12. (aa? + bx + 4) (aa^ + 6a; - 2) (ax 2 + bx- 1).
7. 2%e Cube of a Binomial.
If, in the identity
(x + a)(x + b)(x + <?) = X s + (a + b + c~)x*
+ Cab + be + ca)x + abc,
we put b and c each equal to a, it becomes
O + «)(> + a) O + a) = x 3 + (a + a + a> 2
+ (aa + aa + aa)a; + aaa,
or (* + a) 3 Es;r 3 + 3fljr 2 + 3a 2 ,r + a 3
= * 3 + a 3 + 3ajr(* + a).
If, in the above identity, we change a to — a, we have
O - a) 3 = x s - 3 a* 2 + 3 a 2 x - a s
= x s — a 3 — 3 ax(x — a).
The cube of a binomial is the cube of the first term plus
three times the algebraic product of the square of the first
term and the second term, plus three times the algebraic
product of the first term and the square of the second term,
plus the algebraic cube of the second term.
(1) (a+2 20 3 =z 3 +3O) 2 (2 J0+3(a!)(2 2/) 2 +(2 yf
=s'+6 a?y+12 xy*+8 f.
(2) (2 o-36) = (2 a) 3 +3(2 a) 2 (-3 6)+3(2 a)(-3 bf
+ (-3&) 3
= 8 a 3 -36 a 2 6+54 a& 2 -27 W.
76 THE ESSENTIALS OP ALGEBRA.
EXERCISES.
Write out the results in the following :
1. (2 a + b) s . 9. (4 ab - 5 y)\
2. (a -3 b) s . 10. (-+-)*.
3 - (3 * + 4) - ax. [(. + ,)-2j.
4. (2 3-Sy)'. 12 [(a ._ y)+o] ..
5. (3 a; + aft) 3 . 13 [2 (a + ,,) _ 3 a] 3 .
6. (2mn-pq) 3 . 14 . [(«» + &) +c] s .
7. (5a: 2 -l) 3 . 15. [(a + 6) + (* + y)]». _
8- (62/-£) 3 . 16. [(aa;4-6)-(ca; + d)] 3 .
8. The Binomial Theorem.
(a + b)" = a» + /la"" 1 * + fl( f "^ a- V + " ( "~ ^V 2) ""'^
+ +/ia6— 1 + 6"-
The above identity is known as the Binomial Theorem.
A generaljgroof^for it will not be given. For the present
we will limit the exponent n to integral values.
If we multiply both sides of the identity
(a + by= a 3 + 3 a?b + 3 aJ 2 + b s by a + b,
we have (a + 6) 4 = a* + 4 a 3 5 + 6 a 2 6 2 + 4 a& 3 + b\
If in the binomial theorem we make m = 4, we have
(ffl + 5)^ ffl 4 + 4 a 3 6 + ^ a252 + 4^| a&3 + 4.3-2.1 54
= a* + 4 a 3 J + 6 a 2 6 2 + 4 a& 3 + J*,
a result which agrees with that found by multiplying.
IMPORTANT IDENTITIES. 77
If we multiply both sides of the identity
(a + 6>* = a* + 4 a 3 b + 6 a 2 J 2 + 4 a6 3 + b* by a + 4,
we have *
(a + by == a 5 + 5 a 4 6 + 10 a 8 6 2 + 10 a 2 £ 3 + 5 aV + ¥>.
If, in the binomial theorem, we make n = 5, we have
J. • Z 1 • ^ • o 1 • Z • o • 4
1-2-3-4-5
= a 5 + 5 a*b + 10 a 3 5 2 + 10 a 2 6 3 + 5 a6 4 + 6 5 ,
a result which agrees with that found by multiplying.
The identity
(a + b y= a n + na n-i b + <±zl_V 2 5 2 + n(n-l)(n-2) ^^
+ +nab n - 1 + b n
is often called the binomial expansion.
The following Laws should be observed in regard to
the Exponents and Coefficients of the successive terms of
the binomial expansion.
(1) Law of Exponents. The sum of the exponents of
a and b in any term is always n ; the leading letter a ap-
pears in the first term with the exponent n which decreases
by unity in each succeeding term ; the letter 5 appears in
the second term with the exponent 1 which increases by
unity in each succeeding term to the last term b".
(2) Law of Coefficients. If any term be taken, the
coefficient of the next succeeding term is obtained by
multiplying the coefficient of the given term by the ex-
ponent of the leading letter a, and dividing this product
by the number of the given terra in the series.
78 THE ESSENTIALS OF ALGEBRA.
Thus, in
(a + by = a 4 + 4 efib + 6 aW + 4 ab s + J 4 ,
the coefficient 6 in the third term is obtained by taking the
product of 4, the coefficient of the preceding term, by
the exponent of a, 3, giving 4x3, and dividing by 2, the
number of the term 4 a s b in the series. Hence, the coeffi-
cient of the third term is 4 x 3 ■+■ 2 = 6.
Pascal's Triangle. The coefficients of the terms in the
expansion of
(a + 6)1, (a + by, (a + by, (a + by, etc.,
may be arranged in a table forming what has been called
Pascal's Triangle. The arrangement follows.
Coefficients of (a + by are
1
1
Coefficients of (a + by are
1
2
1
Coefficients of (a + b) 3 are
1
3
3
1
Coefficients of (a + by are
1
4
6
4 1
Coefficients of (a + by are
1
5
10
10 5
etc.
Each number appears as the > sum of the number im-
mediately above and the one to the left. Thus, the first
10 is the sum of 6 and 4 ; the second 10 is the sum of
4 and 6 ; the last 5 is the sum of 1 and 4. By this simple
arrangement the binomial coefficients for any power of
a + b may be easily written out, provided we know the
coefficients of the expansion of a + b for a power one
lower. Knowing
1 5 10 10 5 1
IMPORTANT IDENTITIES. 79
to be the coefficients for (« + 6) 5 , the coefficients for the
sixth power of a + b are
1 6 15 20 15 6 1.
(1) (a + 2 b)* = a 4 + 4 a 3 (2 b) + 6 a 2 (2 &) 2 + 4 a(2 6) 3 + (2 ft) 4
ee a 4 + 8 a s b + 24 a 2 b 2 + 32 aft 3 + 16 b*.
(2) (2x- yy = (2 a) 5 + 5(2 a) 4 ( - y) + 10 (2 a;) 3 ( - <,) 2
+ 10(2af(-2/) 3 + 5(2aO(-2/) 4 +(-?/) 5
= 32 k 5 -80 afy + 80 a?f - 40 aft/ 3 + 10 V - */ 5 .
EXERCISES.
1. Extend Pascal's triangle to include the expansion of
(a + b) 10 .
Write out the following expansions :
2. (a + ft) 6 . (Use Pascal's triangle.) / y\ s
3. (x + y) 7 . 9. (3y-2x)\ ' \ V'
4. (m + nf. 10. (2a + 6) 5 . 16. (3 s- 4 y )'.
,(l + *)». ,,(,-3,), £[££*
e. (,+i)- i2. (21 «+«.)• 19 . [3a _ ( ;_^,
7. (2a: + a) 4 . 13. (x 2 -2yf. fx_2j L V
8. (p-q) 8 . 14. (??-yy. ' \B 3/
EXERCISES.
By comparison with types wrrte dowm the following products :
1. (x + 5)(x + 3). 5. (xy + a)(xy-b).
2. (a! + 10)(a!-2). 6. (4 as + 2) (4 a; -5).
3. (x-B)(x + 8). V- (a: 2 + 5)(a; 2 -10).
4. (acc + 3)(a« + 5). 8. (» + a + 3)(x + a- 7).
SO THE ESSENTIALS OP ALGEBRA.
9. (3 + by)(i + by). 14. (3x + y)(3x-y).
10. (x 2 + 3x + 2)(a^ ! +3x-5). 15. (2 x -iy) (2 x + 4 y).
11. (3a + y)(3x-2»). 6 (2^+^ + 8)(2^+^-8).
12. (a + & + 12)(a + &-6). ^ _
f s_l2V -*-2\ "■ (^ + 2/ 2 -4)(^ + 2/ 2 + 4).
13. ^-5+^x ft -J- i8 {ax + hy + c){ a X + h y_ c) .
19. (ar' + l + aOO^ + l-x).
20. (xy + yz + zx — a)(xy + yz + zx + a).
Perform the operations indicated :
21. (x + y+z) 2 . 27. (x" + y n f — 4 x"y n -
22. (3x + 2y + z) 2 . 28. (x + y + z + w) 2 .
23. (as + &i/ + c) 2 . 29. (a; + 3) (a; + 5){x + 6).
24. (a + 3) 2 - 12 x. 30. (x+3)(x-3)(x+5)(x-5).
25. (x + 3 2/-4) 2 +16(x+3?/). 31. (a: + 3 a)(x 2 -3 ax + 9 a 2 ).
26. 3x4-2y+6) 2 -24(3x+2y). 32. (x + 3yf.
33. (a; + yf — 3 xy(x + y).
34. (a 2 + 6 2 + c 2 — a& — 6c — ca)(a + &+c).
35. (x 2 + 4?/ 2 + l — 2 a;?/ — 2y — x)(x + 2y + l).
Show the truth of the following identities :
36. (x + a) 2 — (x — a) 2 = 4 ax.
37. (x ■+- a) 2 — 4 ax = (x — a) 2 .
38. (x 2 + xy + 2/ 2 )0" — 2/) = X s — y s .
39. (x 2 — xy + y 2 )(x + y) = a? + y 3 .
40. (x 3 -)- x 2 y + xy 2 + y s )(x — y) = x* — y*.
41. (x 3 — x 2 y + xy 2 — j/ 3 )(x + y) = x A — y*.
42. (x 2 + xy + ^(a 2 — ay + 2/ 2 ) = x 4 + x 2 y 2 + y 4 .
43. (x 2 + y 2 + z 2 — xy — yz — zx)(x + y + z) = x 3 + y s + z i —3 xyz.
IMPORTANT IDENTITIES. 81
70. Division Identities. From the relation of division
to multiplication, every multiplication identity gives rise to
at least two division identities. The following division
identities are of importance :
1. [je*+(a + b)x + ab] + (x + a) = x + b.
2. (x 2 + 2ax + a*) + (x + a) = x + a.
3. (x 2 -2ax + a-)^(x-a) = x-a.
4. (x 3 -a 2 ) + (x-a) = x + a.
5. (jr s - f) -=- (x -y) = x - + xy +/.
6. (x 3 +/) -h (x +/) = x 2 - xy + f.
7. jr* -/ -=- (x —y) = x 3 + x 2 y + xy 2 +/.
It should be noted that in each of these identities, the
quotient might be the divisor, and the divisor the quotient.
(1) (a; 2 -64) -(a; + 8) = ?
This is an example of type 4. By a comparison with that
type we see at once that the quotient is x — 8.
(2) (x 2 - 11 a; + 30) -=- (x - 6) = ?
This is an example of type 1.
a = — 6 ; ab = 30 ; hence, b = — 5.
The quotient is x — 5.
(3) (27 - a 3 ) -;- (3 - a) = ?
If we notice that 27 = 3 3 , we see that this is of the form of
type 5.
Hence, (27 - u 8 ) -h (3 - a) = 9 + 3 a + a 2 .
EXERCISES.
Perform the following divisions by comparison with
type forms :
1. (a 2 -2a-63) + (a + 7).
3. [(2aj) 4 -& 4 ]^-(4a; z -6 2 ).
82 THE ESSENTIALS OF ALGEBKA.
3. (9 +6 a + a 2 ) -=- (3 + a).
4. (8a 3 + 6 3 )-=-(2a + &)-
5. (81a 2 -25 6 4 )-=-(9a + 5& 2 ).
6. (l-10a + 25a 2 )-^(l-5a).
7. (y* + lly-26) + (y + 13).
8. [(a + 6) 3 -64]-^(a + &-4).
9. [ZIc?-(x-yY]+[3a-(x-y)].
10. [(as + 2/) 2 - 4 a 2 & 2 ] -=- (x + y + 2 a&).
11. (^-M^-S^-^C^-IT).
12. [(a + y) 2 + ll(a + 20-60]H-( a! + 2/-4).
13. [0 + y) 2 - 16 (x + y) (a + b) + 48(a + &) 2 ] -=- [(a; + y)
-4(a + &)].
14. (a; 2 " -9 a;" -112) -(a;" -16).
15. (2/ 2n + 16 2/" + 64)^-(2/° + 8).
16. [(a + 6) 2 - 6 x(a + 6) + 9 a; 2 ] -4- (a + b - 3 a;).
17. (a;* 1 - 25 y 2 ") -=- (x* a + 5 f).
18. [(x + yy n -(a + by-\ + [(x + y)"-(a + b) 2 "]. .
19. [(aa; + b) s + 125 a?] -=- [(aa; + 6) 3 - 5 x (ax + b) + 25 a; 2 ] .
20. [(aar 2 + 6a; + c) 2 — (Ix + m) 2 ] -*- (ax 2 + bx + c + Ix + n).
21. (fn — x 3 *) -i-(y n — x").
22. j/ 4 " — K 4 " -s- G/ n — «")•
23. [(V3 a; + yf - a 2 ] -*- ( V3 x + y - a).
24. [(Vaa; + b) s — 64 y 8 ] -=- (Vax + b — 4 y).
25. [( VaaT+ty) 3 + 125 a 6 ] -r- (VaaT+ty + 5 a 2 ).
CHAPTER VI.
, FACTORING.
71. Products, Factors. Numbers which are multiplied
together to form a product are called factors of that product.
For example, in 5 x a x 6 = 5 aft, 5, a, 6 are factors of the product
5 ab. Only expressions free from divisions and roots will be con-
sidered in factoring. In 7 a (ft + c)(x + y + z), 7 is a numerical factor
or numerical multiplier, a is a monomial factor, b + c is a binomial factor,
andz + y + zisa trinomial factor.
In multiplication, we have the factors given to find the
product; in factoring, the product is given to find the
factors.
72. The Degree and Number of Factors. The degree of an
algebraic monomial is the number of letters composing it.
Thus, a 2 b is'an expression of the third degree, being made up of the
product of a x a x ft. aVy is an expression of the fftk degree in
a, x and y ; it is an expression of the second degree in a, also in x ;
it is of the first degree in y.
The degree of an algebraic polynomial is the highest num-
ber of letters found in any term.
Thus, x 2 + 3 x + 4 is an expression of the second degree, containing
a; x a; in its highest term. ax 2 +bx is an expression of the third degree
in a and x, but is an expression of the second degree in x.
The number of factors of an algebraic expression is not
greater than the degree of the expression.
83
84 THE ESSENTIALS OF ALGEBRA.
EXERCISES.
Determine the degree of the following with regard to all the
letters :
1. x 2 + ax. 6. X s + y 3 + 2 s — 3 xyz.
2. x + y + z. 1.x 2 — a 2 .
3. x 2 + xy + y 2 . 8. a 2 + 2 aa; + a 2 .
4. a? + 3a? + ix + 16. 9. abc + a? +b 2 + a 2 b\
5. ax 2 + bx -f c. 10. a 2 6c + 6 2 ca + c 2 a6.
73. Monomial Factors. Factors of monomial expressions
may be written down by inspection.
Thus, a 2 bx 2 = a • a • b • x • x;
also, 5 x 2 y s = b • x ■ x • y ■ y ■ y.
Here, 5 is not an algebraic factor in the sense of determin-
ing degree ; it is a numerical multiplier.
Monomial factors contained in polynomials may be seen
as the inverse of the distributive law.
{a + b + e)rn = am + bm + cm.
Reverse this identity, and we have the factors of
am + bm + cm, namely m and a + b + c.
Factor :
EXERCISES.
1. 14 a 2 x 4 y.
7.
4 a 2 bc*.
2. — 3 xy 2 z.
8.
4x 2 + 7a; = x(4a; + 7)
3. -| x'yz 2 .
9.
ax 2 + ay.
4. £aW.
10.
3 ax + 6 a¥.
5. -10a 8 6c.
11.
5 a 2 + 10 ab + 5 abc.
6. 12 ay 2 .
3,2.
3xy-&x 2 y 2 + l a?f.
FACTORING. 85
13. Sm + 4 rff 4- 2 mxy. 17. 5 x?yz 4- 30 aVz — 40 xyz 3 .
14. _2fa + 4i«y + 6Jy. 18 - 7ffl! *+l«VH7aV.
19. 3a 2 6cV»+2/— 6a& 2 cVa;— ?/.
15. aay + o^+SaaY. 20l .6 *WS£+&
16. 3 a 2 a*/ — 15 av?y 4- 21 a 3 ^. +10xy 2 zVax— b.
I. TYPES.
A great number of algebraic expressions may be fac-
tored by comparison with some known form of product.
The identities of the preceding chapter are reversible ;
when so written they become Types in factoring.
74. The Type x 2 — a 2 , the Difference of Two Squares. This
expression is recognized as the product of x + a by x — a.
Hence,
jr 2 — o 2 = (x 4- a)(z — a).
The difference of two squares equals the product of the
sum and difference of the square roots of the two numbers.
Thus, a?-16=(a: + 4)(a;-4);
also, (a + S) 2 - 16s (a + S + 4)<>4-&- 4).
EXERCISES.
Factor :
1. a 2 -4Z> 2 . 6. 25ary-36zV.
2. 4 a 2 - 9 ft 2 . 7. (x + yf-a*.
3. a 2 6 2 -c 2 . 8. (z + 3.y) 2 -s 2 .
4. 16a 2 6 2 -25c 2 . 9. (3 a - 2 t/) 2 - (s + af.
5. 4-9ar\ 10. 4(x + 22/) 2 -9(a + &) 2
11. se* - y 4 = (s 2 - f) (o? + f) = (x + y) (x - y) (z 2 + f).
86 THE ESSENTIALS OE ALGEBRA.
In Exercise 11 we have two first-degree factors, x + y and
x — y, also the second-degree factor, x 2 + y 2 . No factors of
x 2 + y 2 can be found . unless radicals be employed. Such a
factor may be called irreducible.
Factor :
12. 4a 2 -16(a:-34) 2 . 14. (x + y)* - 4 a?y 2 .
13. 16-a 4 . 15. (3z-|-4?/ + 5) 2 -9.
16. (-3x + 2y-5y-(x+yf.
17. (Ix + my + nf — 4 (ax + by + cf.
18. (ax + by) 2 — 4 (Iz + mw) 2 .
19. (a"x n ) 2 — (&"2T) 2 . 22. (sk + 3) 2 — 42/ 2 .
20. x 2n y 2n — z 2n w 2 *- 23. (3 a; + a) 2 - 9 6 2 .
21. (w + l) 2 - a 2 . 24. (# + a) 4 — (a + &) 4 .
25. ( a , + 3y) 2 -(3a; + 2 /) 2 .
75. The Type x 2 + (a + b)x + ab. This expression is the
product obtained by multiplying x + a by x + b. Hence,
x 2 + (a + b)x + ab = (x + a)(x + b).
Examples belonging to this type assume the form
x 2 + sx+p,
where p is the algebraic product of two numbers, and s
is their algebraic sum.
If * and p be integers, the factors of p may sometimes
be found by inspection such that their sum shall be s.
Thus, to factor x 2 + 6 x + 8, we must find two factors of
8 whose sum is 6. These are seen to be 4 and 2. Hence,
a? + 6 x + 8 = O + 4) (x + 2) .
To factor a 2 + 10 a — 24, we must find two factors of
— 24 whose sum is 10. These are 12 and — 2. Hence,
a 2 + 10 a - 24 = (a - 2)(a + 12).
FACTORING. 87
It should be observed that if in x 2, + sx + p, p be posi-
tive, the factors of p chosen must be of like signs, and if s be
positive, both factors of p must be positive. If s be negative,
both two factors of p must be negative ; if p be negative, one
factor of p must be positive and the other negative, and the
sign of s shows which is numerically the larger-
EXERCISES.
1. Factor y 2 — 5 y — 24.
Here the factors of — 24 are — 1, 24 ; - 2, 12 ; — 3, 8 ;
— 4, 6 ; and also these numbers with their signs changed. A
pair of factors must be chosen whose sum is — 5; this is seen
to be 3, — 8. Hence,
y 2 -5y-2i = (y + 3)(y-8).
2. Factor {x + 2) 2 - 5 (x + 2) - 14.
The two factors of — 14, whose sum is — 5, are — 7 and 2.
Hence,
(x + 2) 2 - 5 (x + 2) - 14 = (aT+2 + 2) (^+2 - 7)
= (a; + 4) (x — 5).
3. Factor {a? + 3 x) 2 - 8 (x 2 + 3 x) - 20.
The factors of — 20, whose sum is — 8, are — 10 and 2.
Hence,
(a; 2 + 3 a;) 2 - 8 (a» + 3 x) - 20 = (a,- 2 + 3 x + 2) <V+ 3 x - 10)
= \(x + l)(x + 2)}\(x + 5)(x-2)\.
Factor :
4. a; 2 +3a; + 2. 9. x>+ 5 xy + 6 y\
5. a; 2 - 3 a; + 2. 10. v?y 2 -3xy- 10.
6. sc 2 +a;-2. 11. 1 + 3 xy - 10 x>y 2 .
7. a^-a;-2. 12. G«?-5xy + y 2 .
8. a 2 x 2 +5ax + 6. 13. (x + yf+9(x + y) +20.
88 THE ESSENTIALS OF ALGEBRA.
14. (a + 3 ft) 2 - (a + 3 b) -20.
15. (6a; + 32/) 2 +10(6a; + 3?/)+16.
16. (x>+ ff- 6 (x>+ y 2 ) - 27.
17. (2x + 3y + 5)*+9(2x + 3y + 5) + 18.
18. (a; 2 +5a;) 2 +10(a; 2 +5a;)+24.
19. a; 4 - 13 ar>+ 36.
20. (ax + by) 2 + 8 (ax + by) + 7.
21. (ax + by) 2 — (I + m) (ax + by) + Irn.
22. (a; 2 +6a;) 2 +17(a; 2 +6a;) + 72.
23. (x 2 -5x + 4) 2 -(x 2 -5x + ±)-2.
24. a; 2 "— 10 a;* 1 +16.
25. (aa;) 2,l + (Z + m) (aa;)"+ i!m.
76. The Types x 2 + 2ax + a 2 and * 2 - 2 a* + a 2 . These
two trinomials are perfect squares of x + a and x — a,
respectively.
x 2 + 2 ax + a 2 = (x + a) (x + a) = (x + a) 2 .
x 2 -2 ax + a 2 = (x - a)(x - a) = (x - a) 2 .
The sum of the squares of two numbers, increased (or
diminished) by twice the product of the numbers, equals the
square of the sum (or difference) of the two numbers.
EXERCISES.
1. Factor x 2 + 6 x + 9.
Here a; 2 and 9 are the squares of x and 3. and 6 a; is twice
the product of 3 and x ; hence,
x 2 + 6 x + 9 = (x + 3) (a; + 3) = (a; + 3) 2 .
2. Factor 9 x 2 + 6 x + 1.
9 x 2 + 6 x + 1 = (3 a; + 1) (3 x + 1) = (3 x + 1) 2 .
FACTORING. 89
3. Factor 16 a 2 + 40 ab + 25 b 2 .
16 a 2 + 40 ab + 25 6 2 = (4 a) 2 + 2 (4 a) (5 6) + (5 6) 2
e= (4 a + 5 b) (4 a + 5 6) s (4 a + 5 b) 2 .
4. Factor y 2 — 16 #z + 64 z 2 .
y 2 - 16 yz + 64 a 2 = f - 2 (y) (8 z) + (8 z) 2
= (y-8z)(y-8z) = (y-8z) 2 .
5. Factor (a; 2 + 4 a;) 2 — 4 (a: 2 + 4 as) + 4.
In this expression we may consider x 2 + 4 a: as a single quantity.
(a; 2 + 4 a;) 2 - 4 (a; 2 + 4 *) + 4 = (a; 5 + 4a;) 2 - 2(x 2 + 4 a;) (2) + (2) 2
= (a? + 4 x - 2) (a; 2 + 4 a; - 2)
= (a,- 2 + 4 a; - 2) 2 .
Factor :
6. a: 2 -8 a; + 16.
7. 4 a; 2 - 12 x + 9.
8. a 2 a: 2 + 10 aa;?/ + 25 y 2 .
9. 49 a 2 & 2 - 14 ab + 1.
10. 100 - 20 ab + a 2 b 2 .
11. (ax + b) 2 + 2 c (ax + b) + c 2 .
12. (3 x + 4 wV - 6 (3a; + 4 y) 4- 9.
13. 4 a^ + 9 2/ 2 - 12 an/.
14. («a; + by + cf + 8 (ax + by + c) + 16.
15. (a 2 + y 2 ) 2 - 2 (x 2 + f) z 2 + z\
The next three exercises are squares of trinomials. See page 73, 5.
16. x 2 + y 2 + z 2 + 2xy + 2yz + 2 zx.
17. a 2 + b 2 + c 2 - 2 ab + 2 be - 2 ca.
18. ix 2 + 9 y 2 + z 2 + 12 xy + 6 yz + i zx.
90 THE ESSENTIALS OF ALGEBRA.
19. x> a + 12 x n + 36.
20. y 4 " — 14 y 2n + 49.
21. a 2 " + 12 art> 2 " + 36 ft 4 ".
22 (a; + yf 1 — 6 a(a: + y) n + 9 a 2 .
77. The Types jr 3 — / and x s +/. It has been shown by
actual multiplication that
M »- J ?=(,-y ) ( J * + jry+/),
and jr 3 +/* = (* +/)(jr 2 — jr/ +/ 2 ).
EXERCISES.
1. Factor 8 a 3 - 27 6 s .
8 a 3 -27 ft 3 =(2 a) 3 (3 -ft) 8
= (2 a - 3 ft) [(2 a) 2 + (2 a) (3 6) + (3 ft) 2 ],
= (2 a - 3 6) (4 a 2 + 6 aft + 9 ft 2 ).
2. Factor (a,- 2 + 2/ 2 ) 3 - 8 X s f.
(x 2 + ff -8a?tf= (a; 2 + tff - (2 a*,) 3
= [(x* + y*)-2xy][(x* + y*)>+(x* + y*)2xy
+ {2xyf]
= (*-y) 2 [(a? + y*) 2 -\-2xy(x l + y*)+4a~ 1 y 2 ].
3. Factor (a; + 5) 3 + 8 ft 3 .
(x+5) s + 8 ft 3 = (x + 5) 3 + (2 ft) 8
= + 5) + 2 6] |> + 5) 2 - (x + 5)2 6 + (2 ft) 2 ]
= [a; + 5 + 2&]|> + 5) 2 -2&( a; + 5) + 4ft 2 ].
Factor : J
4 - ^-8 6 s - 8. (a; + 2/) 3 - 125 z 3 .
5. 8a 3 -27& 3 . 9. (3a; + 4) 3 + 8^.
6. aV + 6i. 10. (a; 2 + 3a; + 4) 3 -64.
7. a s x* - 27 fz*. ii. (ax+ft^-cV.
12. (3a; + 4 2 /) :! + (2a : + 2/) 3 .
FACTORING. 91
Certain expressions may be transformed into the sum or
difference of two cubes, and the factors then found.
13. Factor a; 6 + y 6 .
In this case we may write
a« + y°=(??y+(yy
= [a? + tf][x*-x>f + y i l.
14. Factor a,- 6 — y 6 .
As in Exercise 13, we may write
= (f?-tf)[(x*y + xhf+{ y y-\
= {x + y){x-y)\_x i +x 2 y i + y A ']
= (x + y)(x- y) (x 2 + xy + f) (x 2 - xy + f).
(See Exercise 42, page 80.)
Of course, the factors of X s — y e could have been obtained by
comparing with the type x 2 — a 2 . Thus,
x e -y s = (x 3 ) 2 -(y s f
= (x + y) (af-xy + y 2 ) (x - y) {x 2 + xy + y 2 ).
Other types of binomials, such as a* — b\ a 5 ± b 5 , and so on,
may appear for factoring, but such special cases will not be
considered at this time.
Factor :
15. 8a 3 -& 6 . 18. a 6 x"+(y + zy.
16. (axy+(by) 6 . 19. (ax + byy-(cz) e .
17. x* -1. 20. 64 a 6 + (6c) 6 .
21. [(x + yyj-l(x-yyj.
22. (a 3 + 3a?b+3ab 2 + b s y + c 6 .
92 THE ESSENTIALS OF ALGEBRA.
78. The Type Ax 2 + Bxy + Cy 2 or Ax 2 + Bx + C. If the
first of these expressions is capable of separation into
factors free of radicals, it must be composed of two
binomials of the form ax + by and Ix + my, where a, b, I,
m are algebraic numbers, i.e., they may be + or — , inte-
gral or fractional.
Multiplying these supposed factors together, we have
(ax + by) (lx + my) = alx 2 + (am + bl) xy + bmy 2 .
Hence, to factor Ax 2 + Bxy + Cy 3, we are to find four
numbers, a, b, I, m, such that
al = A, bm = 0, and am + bl= B.
This method is illustrated by the following examples :
(1) Factor 6 z 2 + 31 xy + 35 y 2 .
The factors of 6 are 6, 1 and 3, 2 ; the factors of 35 are 35,
1, and 7, 5. A good plan is to arrange the letters thus,
x + y ^
x + y
Now attach the factors of 6 and 35 to x, y, respectively, as a
trial arrangement. ' Let us place them thus,
3x + 7y- _
2x+5y J
The square terms appear correctly, 6 x-, 35 y 2 , but the cross
products, 3 x ■ 5 y and 2 x ■ 7 y, do not add so as to give 31 xy.
Hence, our trial arrangement is not correct. Let us try
2x+7y]
3x + 5y}'
This gives 6 x*+(2 x 5+3 x 7)xy+ 35 y 2 , the correct product.
Hence, 6 x~ + 31 xy + 35 y 2 = (2 x + 7 y) (3 x + 5 y).
FACTORING. 93
The case in which y is equal to unity, giving Ax 2 + Bx + C,
requires no special mention when the factors may be found by
inspection as above.
(2) Factor 3 a? + 16 xy + 5 y 2 .
In this example we are to find the arrangements of the
factors of 3 and 5 with the letters x, y in
03 + 2/'
® + y .
such that the sum of the cross products shall be 16 xy.
By trial, the arrangement is seen to be
x + 5y _
3x+ y .
Hence, 3x? + 16xy + 5y 2 = (x + 5y)(3x + y).
(3) Factor 15 x 2 + 58 x + 11.
The factors of 15 are 3, 5 and 1, 15 ; the factors of 11 are 1,
11. Our trial arrangement may be
93 + 11
15a;+ 1
But the cross products 11 x 15+1 x 1 do not give 58. Another
trial may be S r 4- 11 1
533+ 1 J
which gives for the middle term 3 x + 55 x = 58 x.
Hence, 15 x 2 + 58 x + 11 = (3 x + 11) (5 * + 1).
Factor: exercises.
1. 3 a; 2 + 7 xy + 2 y 2 . 4. 12 x 2 + 2 xy - 2 f.
2. 6 x 2 - 5 xy - 6 y 2 . 5. 3 aa^ + (9 + <x)6a3 + 3 b 2 .
3. 21 x 2 + 31 xy + 4 y\ 6. 20 y 2 - 22 yz + 6 z 2 .
7. 24 y 2 - 26 yz - 8 z\
94 THE ESSENTIALS OF ALGEBRA.
79. The Type ax 2 + bx + c. This expression is called
the General Quadratic in a single variable x. For differ-
ent values of a, b, e this expression represents every
quadratic that may be written. Thus, with a = 5, b = 6,
c = 2, we have 5 x 2 + 6 a; + 2 ; with a = 4, b = — 5, «? = — 1,
we have 4 a; 2 — 5 a; — 1, etc.
We shall illustrate the method of factoring the general
quadratic by a few special examples.
(1) Factor 2 ar> + 16 a; -20.
By dividing out 2, we get
2 a? + 16 x - 20
= 2(a; 2 + 8a;-10)
= 2|y+8 a;+16-10-16], adding and subtracting 16.
= 2[(x + 4) 2 — 26], rearranging terms.
26 may be written ( V26) 2 , and we have
2 a? + 16 x - 20
= 2 [(a; + 4) 2 — (V26) 2 ], the difference of two squares.
= 2 [x + 4 + V26] • |> + 4 — V26]; the product of sum
and difference.
Note. Adding 16 is called Completing the Square of the first two
terms. To complete the square of x 2 + 2 mx we must add m 2 ; i.e., we
add the square of half the coefficient of x. To complete the square of
/ 6 \ 2
i 2 + 6 x we add - s3 J . To complete the square of x 2 + kx we add
(§)'
Complete the square :
1. a; 2 + 5 x.
Here (f) 2 must be added.
If we wish the expression to remain unchanged, we must
also subtract (|) 2 .
Factoring. 95
Hence, v? + 5 x = x 1 + 5 x + -\ 5 - — - 2 ^,
= (^ + l) 2 --¥-
2. x*-3x. 7. lla^ + 33a;.
3. a 2 + 7*. 8. 5x* + 1x.
4. a^-8a;. 9. 3^-83;.
5. 3ar ! + 9a; = 3(ar ! + 3a;). 10. 7^-35^
6. 5a? -25a;. 11. 9a^-25a;.
(2) Factor 5 ^ + 15 x -10.
5 x 1 + 15 x — 10
= 5 |y + 3 x - 2], dividing by 5.
s 5 [ar + 3 x + (§) 2 — (f) 2 — 2], completing the square.
= 5 [(a; + f) 2 — | — 2], rearranging.
= 5 [(* + !)■-¥],
= 5 [(as + |) 2 — ( V- 1 /-) 2 ]) difference of two squares.
V17 ]
2 J
^[.+3 + ^
a; -) ^^- I , productof sum and
difference.
2 2
(3) Factor 4 as 2 + 6 a; + 2.
4a? + 6a; + 2
= 4 [a 2 + | x + %], dividing out 4.
= 4 [a; 2 + f ^ + (I) 2 - (f f + $]> completing the square.
= 4[(a; + |) 2 - 1 VJ, rearranging.
= 4 [(as + f ) 2 - (}) 2 ], rearranging.
= 4 [a; -|- 3 + i-] . [a; + | — y, product of sum and difference.
£=40 + 1] -|> + i].
According to the method here illustrated we may factor
the general quadratic.
96 THE ESSENTIALS OP ALGEBRA.
ax 2 + bx + c
a a
= a
= a
^ + -, + l -
h V
2 a)
X 2 a) 4 a 2
r , b , V& 2 -4,
2a la
4 a<A 2 "
2"
dividing out a.
- I, completing square.
combining terms.
difference of two
squares.
~ x b_ _ Vff 2 - 4 ac
* 2a ' 2a
Hence, to factor any quadratic in which a, 5, c have
been replaced by numerical values, we need merely to
replace those letters in the general factors above by the
special values found in the given example.
(4) -Factor 3 x 2 + 8 x + 2.
Here a = 3, 6 = 8, c = 2; hence, we have
3 a? + 8 x + 2
= 3
x +
2x3
+
" , 4 , V40"
x-\
3 6
V64 -4-3^2"
2-3
" , 4 V40'
XA
3 6
' 8_ _ V64 -4-3
2-3 2-3
-1
(5) Factor 5 x 2 — 7 x + 3.
In this case a = 5, 6 = — 7, c = 3; hence,
5z 2 -7a; + 3
= 5
= 5
■ _7 V(_7 }! _4^3-
2-5 2-5
, + ^-M - - 5 - 3 l
2-5 2-5 J
-7 V(-7) 2 -4-5-3 ~
7 , V-ll"
x \-~
10 10
7 , V-ll"
x \-—
10 10
FACTORING.
97
Note. The factors in this case, involving the square root of a
n egati ve numb er are called imaginary. Such numbers as V— 11,
Y — 5' V— a 2 , etc., do not belong to the algebraic number system ex-
plained in Chapter I.
(6) Factor - 3 x 2 + 4 x + 2.
Here a=— 3, & = 4,c = 2; hence,
-3x 2 + ix + 2 = -
* + .-
V4 2 -4(-3)-2 '
2.(-3) + _ 2. (-3) _
4 V4 2 - 4( - 3) ■ 2"
2. (-3)
2. (-3;
3 -6
= -3|>
3 -6
j-lV40][.-| + JV40].
Factor :
1. 5 a; 2 - 25 a; - 15.
2. — 2 a; 2 + 14 x — 10.
3. 3 or' — 5 a; + 9.
4. 7^-35^ + 49.
5. -a^-fll a; -3.
EXERCISES.
6. 11 a; 2 -55 a- + 99.
7. -4 ar -28 a; + 32.
8. G .i' 2 - 30 x + 48.
9. —8 a; 2 + 40 a; -8.
10. 10 a: 2 — 70 x + 20.
II. FACTORS BY REARRANGEMENT AND GROUPING
OF TERMS.
The method of factoring certain algebraic expressions
will often be suggested by a proper rearrangement or
grouping of terms. Two general plans of grouping are
worthy of attention,
98 THE ESSENTIALS OF ALGEBRA.
80. Grouping with Regard to the Descending Powers of
Some Letter.
(1) Factor ix 2 + y 2 + z 2 + 4:xy + 4xz + 2yz.
If this example is not recognized as a perfect square, we
should proceed thus :
4 x 2 +y 2 +z 2 +A xy+A xz+2 yz = i x'+i x(y+z)+y 2 +2 yz+z 2
= ix 2 +Ax{y+z) + {y+zf
= (2x+y+z) 2 .
(2) Factor x 2 + 6 ax - 8 bx + 9 a 2 - 24 ab + 16 b\
Arrange with regard to the letter x, giving
x 2 + 6ax-8bx + 9a 2 -24ab + 16b 2
= x 2 + 2x(3a-4 c b)+9a 2 -24:ab + 16b 2
= a^ + 2a;(3a-4 6) + (3a-4 6) 2
= (» + 3a-46) 2 .
(3) Factor a? + 7 ax + 6 bx + 10 a 2 + 21 ab + 9 b 2 .
Here we may arrange with regard to the letter a.
x 2 + 7ax + 6bx + 10a 2 + 21ab + 9b 2
= 10 a 2 + 7a(x + 3 6) + x 2 +6 bx + 9 6 2
= 10a 2 + 7a(a; + 3 6) + (a! + 3&) 2
= (5 a + a; + 3 6) (2 a + a; + 3 &).
EXERCISES.
Factor :
1. 2a? + 3ax + x 2 + 7ab+3b 2 + 4:bx.
2. 2a^-6 2 + 3y 2 + 6a;-26y + ra!2/.
3. x 2 + iy 2 + 9z 2 -Axy-12yz + 6xz.
4. 4a 2 + 96 2 + 16c 2 + 12a&-16ac-246c.
5. 2x 2 -3y 2 -3z 2 -xy-5xz + 10yz.
FACTORING. 99
81. Grouping with Regard to Some Letter that Enters in
One Degree Only. By forming a product of a number of
factors, one of which contains a letter not found in the others,
we shall see the purpose and application of this method.
Let us multiply together the following :
(x + 3 a + b~)(x + a + m).
The product is
x 2 + 4 ax + 3 a 2 + bx + ab + m(x + 3 a + b~).
It will be noticed that m appears in but three terms,
and that these terms when collected constitute m times the
first factor of the product. Hence, if a literal expression
contain a single letter entering to a single degree only, the
coefficient of that letter contains a factor of the given expres-
sion, if the expression has any factors.
An example will illustrate this method.
Factor 3 x 2 + 8 xy + 3 kx + 5 ky + 5 f.
Here k enters to a single degree.
Arrange with regard to k, and we have
3a^ + 8a;?/ + 5 2/ 2 + (3x + 52/)fc.
If the expression can be factored, 3 x + 5 y must be one of
the factors ; hence, 3 x 2 + 8 xy + 5 y 2 must contain 3 x + 5 y as a
factor.
3x 2 + 8xy + 5y 2 + 3kx + 5ky = 3x 2 + 8xy + 5y 2 + k(3x + 5y)
= (3x + 5y)(x + y)+k(3x + 5y)
= (3x + 5y)(x + y + k).
EXERCISES.
Factor: 1. 4 x> + 4 xy — 35 y 2 + 5 ay — 2 ax.
2. x 2 + zx — iyz — xy—12y 2 . 4. xy + 4 ay + x 2 — ax — 20 a 2 .
3, 4 ac + be -b 2 + 12 a 2 -ab. 5.. 14 x 2 + 7 px- 5 xy +py -y\
100 THE ESSENTIALS OF ALGEBRA.
82. Binomial Factors by Trial. Let us divide a; 2 +6 a; + 5
by x — m in the ordinary way.
x — m)x 2 + 6 x + 5 (a; + 6 + ra, quotient.
(6 + m)x + 5
(6 + m)x — m(6 + rn)
m 2 + 6 m + 5, remainder.
It will be noticed that the remainder found on dividing
x 2 + 6 x -f- 5 by x — m is precisely the value that the divi-
dend becomes when x has been replaced by m. Thus, if
in x 2 + 6 x + 5, we put x = m, we get m 2 + 6 m + 5, the
same as the remainder above.
This remainder is, of course, true for any value oi'm.
We make a few special illustrations for various values of m.
(1) What is the remainder on dividing
^ + 6x + 5byx — 4?
Our result above shows the remainder to be the value of
x 2 + 6 x + 5 when x = i; hence, the remainder after division is
4 2 + 6 x 4 + 5 = 45.
Verify by actual division.
(2) Find the remainder after division of
x^ + dx + Sbyx — 5.
Replace x by 5, and the remainder is
5 2 + 6 x 5 + 5 = 60.
Verify by division.
(3) Find the remainder when x 2 + 6 x + 5 is divided by x + 5.
Here, m = — 5 ; hence, the remainder found by dividing by
c5 + 5ls (-5) 2 + 6(-5) + 5 = 0,
FACTORING. 101
The remainder being zero shows that x + 5 is an exact divisor
of x 2 + 6 x + 5.
(4) Find the remainder on dividing x 2 — 8 x + 12 by x — 8.
The remainder is (8) 2 - 8(8) + 12 = 12.
Hence, x — 8 is not an exact divisor.
(5) Find the remainder on dividing x 2 — 8 x + 12 by x — 6.
The remainder is (6) 2 - 8(6) + 12 = 0.
Hence, x — 6 is an exact divisor, i.e., a factor.
In factoring by trial, the number of trials is limited to tlie
number of divisors of the constant term.
(6) To find the factors of a; 3 — 7 x — 6.
The factors of 6 are ± 1, ± 2, ±3, ±6.
When divided by x — 1 the remainder is 1 — 7 — 6 = — 12.
When divided by x + 1 the remainder is— 1+ 7 — 6 = 0.
When divided by x — 2 the remainder is 8 — 14 — 6 = — 12.
When divided by x + 2 the remainder is — 8 + 14 — 6 = 0.
When divided by x — 3 the remainder is 27 — 21 — 6 = 0.
No other divisors need be tried, for we already have the
three {x + 1), (x + 2), and (a; — 3), and an expression of the
third degree can not have more than three factors.
EXERCISES.
Factor each of the following :
1. a^_ ar>_4 x + 4. 6. a? + 2 x 2 - 9 x - 18.
2. & + 2X 2 -4 x -8. 7. a? + 9 a 2 + 26 a; + 24.
3. 3? + 3 a? -13 a; -15. 8. ar 5 + 15 a: 2 - x - 15.
4. a; 3 -6 a; 2 + 11 x- 6. 9. ar 5 + 12 a; 2 + 47 x + 60.
5. x 3 - 10 a; 2 + 31 x - 30. 10. a? - 12 a; 2 + 48 x - 64.
102 THE ESSENTIALS OF ALGEBRA.
EXERCISES IN FACTORING.
1. bx'—b. 13. 27 -64 a; 6 .
2. 10 cm 2 -40 c 3 . 14. 16 x*y 2 - a?.
3. x 2 + xy + xz + yz. 15. a 3 — 2a 2 — a + 2.
4. Zm + mr— Zr — r 2 - 16. m 4 -4roW+4n , l 1 .
5. 2a; 2 + 3a:y-2a;z-32/«. 17. (a + b + c) 2 — (a — 6 - c) 2 .
6. 6-5a + a 2 . 18. 14a 3 - 8a 2 -21 a + 12.
7. a: 4 + aft/ 2 - 72 ?/ 4 . 19. p 2 - 2pq + q 2 - r 2 .
8. m 2 -4m + 96. 20. 25 a 2 - 10 a*/ - 9 z 2 + 2/ 2 -
9. aZ> 2 + 2a 2 6 + a 3 . 21. 9 a 2 - 36 «6 + 36 b 2 .
10. 3a^+4a;+l. 22. a 2 + 29a + 120.
11. 2Z 2 -8(m + l) 2 . 23. a: 4 + 2a^-99.
12. 64a 4 -81y 4 . 24. 24Z 2 + 18Z + 3.
25. a 2 (a;-y)+3a(y-a;) + 2(a!-3f).
1 i 6 ^ 28. aW-2aW + a 4 6 2 c 4 .
81 y* 29. 9^-6^-1 + ^.
27. m 5 + m 3 -m 2 -l. 30. 646 4 -126 2 -l.
REVIEW EXERCISES.
1. Eemove parentheses and simplify
2x — 3y — \5x-[3y + 5x — (Ax + y — 3x — 4?/)]}.
2. Put x = 5 and y = 1 in the above exercise, and find the
value.
3. From the sum of 3x — 8y + 2z and 5y — 7x—3z take
their difference.
4. Multiply out (a 4 + 4) (a 2 + 2) (a 2 - 2).
5. Divide ar 1 " — 3x 2n y + 3x n y 2 — f by a;" — y.
FACTORING. 103
6. Factor ax- + bx i — a — bx.
7. Simplify
(a; + a) 2 - {x - a) 2 - [_{x + of +(x-a)(x-a)- x 2 '].
8. Divide x i + iy* by x 2 — 2 xy + 2 y 2 .
9. Two numbers differ by 17. One third of the smaller is
one greater than % of the larger. What are the numbers ? (Let
x = smaller, x + 17 = larger.)
10. Divide (x 3 + y 3 ) (x 3 — y 3 ) by X s — 2 x'y + 2 xy 2 — y 3 .
11. Factor a; 4 - 10 x 2 + 9.
12. Add with respect to x, a?x + a(b — c)x — y s x + 11 x.
13. Find the value of
x + yz — y \x> — (3 y + xz) (5 x + y — 6) — 2 x - y\,
when x = 5, y = 2, and z = l.
14. Find the value of ~"" ' ~ — =-> when x = — I.
x(l + 3x) — a? *
15. (2a 3 -3a; + 5) 2 =what?
16. Divide a? m+1 — X 2 ™ y n+1 + x m+1 y 2a — y 3 "* 1 by « 2m + 2/ 2n .
17. Multiply a m+3 b + a 2 &"+ 2 by a m ~ 2 b + ab n ~ 2 .
18. Multiply (a" + l)(a"-l)(a 2n + 2a" + l).
19. Verify a (a + 1) (a + 2) (a + 3) s (a 2 + 3 a + l) 2 - 1.
20. From
(x + y + z)(x + y-z) take x 2 - \y 2 - \_2y 2 - (-2xy + z*)]\.
21. Find the sum, difference, product, and quotient of
4x(y — z) 2 ™ -4 and x(y — z) 2m - i .
22. Factor a 3 — 4a 2 + a + 6.
23. Divide 8 a; 3 — j/ 3 + « 3 + 6 xyz by y — z - 2 as.
24. From $(sc - 3y) - K 9 2/ - 2a5 ) take tW^" 9 ^-
25. Divide ^a 3 -^a 2 + i«-| by |a-^.
CHAPTER VII.
DIVISORS AND MULTIPLES.
83. Highest Common Factor.
A Common Factor of two or more numbers is a factor of
each of them.
a is a common factor of ax, a 2 y, and ab s . x — y is a
common factor of x 2 — y 2 and x 3 — y s .
The Highest Common Divisor of two or more numbers is
the product of all their common factors.
a 2 x is the Highest Common Divisor (H.C.D.) of a s xy,
3 a 2 x 2 , and 5 a 2 xz. a 2 is common, and so is x.
In arithmetic, the term Grreatest Common Divisor is fre-
quently used. This term is not applicable in algebra. In
the above example, a 2 may or may not be greater than a.
If a is less than 1, then a 2 is less than a. Hence, in alge-
bra the term Highest Common Divisor is used.
84. Highest Common Divisor of Monomials. Rule. To
the Greatest Common Divisor of the numerical coefficients
affix each letter common to all the monomials, and to the
lowest power it occurs in any one of them.
ZaWfz*, iaWyW, 5a¥j%.
The G. C. D. of 3, 4, and 5 is 1.
The common letters with the proper exponents are
a 2 , 3?, y 3 , s.
Hence, the required H. C. D. is 1 aWyh.
104
DIVISORS AND MULTIPLES. 105
„. , , EXERCISES.
Find the H. CD. of:
1. 4a?y% 8xy s z 2 , 12 abyz 2 .
2. 5a?bc 2 , 10 aWxy, 25 a?b 4 cx.
3. 16aVy*z, 48a 4 6aryz 4 , 36 a 5 b 4 x*y 7 z B .
4. 14 aWmW, 21 a 4 b 4 m 4 , 42a 5 & 2 m 6 » 8 .
5. 22 p s q 4 afiy, Up 4 q 2 xY, G6pVxy.
6. 14(a-&) 2 affy, 18(a-6) 3 a:y, 12(a-6) 2 a;y
7. 15(as-y)V, 21 (as -y)*^, 33 (x-yfz™.
8. 18 (a 2 -& 2 ) 2 *!/, 27(a 2 -b 2 fx 2 y, 36 (a 2 - ft 2 )^ 2 .
85. Highest Common Divisor of Polynomials. The H.C.D.
of polynomials may be found by factoring. If each factor
is considered as a single quantity, the method is the same
as used in finding the H. C. D. of monomials.
' (1) The H.C.D. of a; 2 -8a; + 7, 3? -1, ar' + 3a;-4 is found
asfollows: rf _ 8a! + 7 s (a! _ 7)(a! _ 1))
a; 2 ^l ==(a; + l)(a;-l),
a? + 3x-i = (x + 4) (a; - 1).
It is seen at once that x — 1 is the H. C. D.
(2) Find the H. C. D. of
X s + 5 x 2 — 14 x, x 4 — 8x, x 4 — 4a^ + 4«.
^ + 5 & _ 14 » = x (x + 7)(x - 2),
x 4 -%x =x(x 2 + 2x + ±)(x-2),
<c*-4x* + 4:X* = x 2 (x - 2)(x - 2).
Here we see that x (x — 2) is the H. C. D.
The H. G. D. is sometimes used in reducing fractions to their
lowest terms.
106 THE ESSENTIALS OE ALGEBRA.
EXERCISES.
Find the H. C. D. of :
1. x 2 — y~, 3? — 2 xy + y 2 , x 2 — xy.
2. a 2 ~b 2 , a?-ab 2 , a 2 + 2ab + b 2 .
3. rf — y 2 , tf + y 3 , xP + xPy.
4. x 2 -7x + 12, x 2 + 2x-lh, a^-9.
5. a 2 + 8a + 15, o 2 -2a-35, a 2 + 3a-10.
6. &2-146 + 49, b 2 + b-56, 6 2 -&-42.
7. 3^-12:c + 12, (x - 2f, 3^-12.
8. a 2 b 2 -b\ ab 2 + b 3 , ab-b 2 .
9. x s — xPy 1 , x 2 (xy — y 2 ) 2 .
10. (tf-Sx 2 ) 2 , af-x*-&3?.
11. a? -2 ^-35 x\ afi-25x i .
12. a^ + 2 ay + y 2 , a^ + y 3 + 3xy(x + y).
13. m 2 — 3 to — 70, to 3 -11 to 2 + 10 m.
14. r* — a;?/ + « — yz, xy — y 2 .
15. a 3 -8, a*6 2 -4a 2 & 2 , 4 a 2 -16 a + 16.
86. Lowest Common Multiple.
When two algebraic expressions are so related that the
first is an exact divisor of the second, the second is said
to be a multiple of the first.
6 a 2 bcx is a multiple of 2 ale, because 2 abc is a divisor
of 6 cfibcx.
A Common Multiple of two or more algebraic expressions
is exactly divisible by each of them.
12 a z 3?y z z z is a Common Multiple of 3 a 2 xy and 4 a 2 yz 2 .
The Lowest Common Multiple (L. CMS) of two or more
algebraic expressions is the expression of lowest degree which
is exactly divisible by each of them.
DIVISORS AND MULTIPLES. 107
We use the term Lowest Common Multiple in algebra because we
are concerned about the degree and not about the numerical value.
a s x is the L. C. M. of a 3 and x ; a 4 x is also a multiple, and if a is less
than 1, a 4 x is numerically less than a 3 x.
87. Least Common Multiple of Monomials. Rule. To the
Least Common Multiple of the numerical coefficients affix
every letter found among the monomials, and to the highest-
power it occurs in any of them.
3 ax, 5 a 2 xy, 4 ay*. By the above rule we write 60 a 2 xy*
at once as the L. C. M. of these expressions.
EXERCISES.
Find the L. C. M. of:
1. 2 ax, 3aVy, iabxy. 3. Spqr-, 2ip i q i r, 12 pq^r 2 :
2. 3 a 2 xy, 5 aV, 15 ofuPy. 4. 10 ISnhi, 15 Z 8 mn 3 , 25 l*m 2 n.
5. 7 ab s xy, 14 aWf, 21 ab*a?y.
6. 3(a-b)xy, Q{a-bfa?y, 12 (a - 6) tff.
7. A{x-yfab, 5(a;-2/)W, 10 (x - y)\abf.
8. A0(a 2 -x)Yz, 60(a 2 -a;)V 2 , 120 (a 2 - xfyh 2 .
88. Least Common Multiple of Polynomials. The L. C. M.
of polynomials may be found by factoring. Consider each
factor as a single quantity and proceed exactly as in the
case of monomials.
The L. C. M. of (x - yf, (z 3 - y s ), x*-6xy + 5y 2 is
found as follows :
(x-yf=(x-y)(x-y),
(x 3 - y 3 ) = (x- y~)(x* + xy + «/ 2 ),
(V _ 6 xy + 5 y 2 ) = (x - y) (x - 5 y).
The L. C. M. is (x - y~)\x 2 + xy + y 2 )(x - 5 y).
108 THE ESSENTIALS OP ALGEBRA.
Iii general the L. C. M. should be left in its factored form ;
that is, its factors should not be multiplied together.
The L. C. M. is used to a limited extent in reducing fractions to a
common denominator.
EXERCISES.
Find the L. C. M. of :
1. a 2 -b 2 , a?-2ab + b\ a?-ab. 8. r 2 -5r + 6, r 2 + 5r-24.
2. ^-16, a; 2 -9a!-20. 9. x 3 + 4a: 4 , a) 4 -16a; 6 .
3. p 2 -25, pt+p-30. 10. y 2 - 9 y + 14,/ -4 y- 21.
4. Z 2 — 36, Z 2 -13Z + 42. 11. a 3 + 8& 3 , a 2 -4& 2 .
5. a 3 - 4 ah 2 , a 4 - 2 a 3 &. 12. c 3 - 27 (f, c 2 - cd - 6 d 2 .
6. 6a 2 , a 4 + 3a 2 , a s -3a. 13. 2 -a, 4-x 2 , i + x 2 , 16-a 4 .
7. m 2 + m + l, m 3 -l. 14. 3 + 6, 9 - b 2 , 27 -b 3 .
15. x — 3, x + 3, a? — &x + 9, a 2 + 6a: + 9, a? — 9.
CHAPTER VIII.
FRACTIONS.
89. Algebraic Fraction; Numerator; Denominator; Terms.
An algebraic fraction is an indicated division.
a-i-b, ax 2 -=- by, (a + b) -+- c.
It is usual to write these indicated divisions thus :
a ax 2 a + b ,, , ,, , IN ,
-, — , — — , or a/b, ax 2 /by, (a + S)/c.
- is read a over b, the fraction a over b, or a divided
o
by b. The preferred reading is a over b.
The dividend is called the numerator, the divisor the
denominator, and the two together the terms of the fraction.
In the fraction — , x is the numerator, y the denominator,
y
and x and y the terms of the fraction.
Any expression may be put into a fractional form by
writing it with a denominator 1; a = -, x + y= '& .
Since a fraction is an indicated division, we know that
~ X b = a; for the fraction - may be regarded as the quo-
b o
tient of a -s- b ; but the quotient multiplied by the divisor
equals the dividend. Hence, -xb=a.
109
110 THE ESSENTIALS OF ALGEBRA.
90. The Sign of a Fraction. The sign of a fraction is
placed before the line which separates the numerator and
denominator. is read minus the fraction x over y.
y
Since a fraction is a quotient and the terms are dividend
and divisor, the sign of a fraction is determined precisely
as the sign of the quotient in division is determined.
+ n _ n _ — n _ n _ — n _ _n % + n _ _ n
+ d d' —d d' +d d' —d d
A fraction preceded by a minus sign is equal to the same
fraction preceded by a plus sign, provided either the numera-
tor or denominator be preceded by a negative sign.
Thus, _« = ^L = ^.
b -b b
If the sign of either term of a fraction be changed, the sign
of the fraction is changed.
Let - be a fraction. Change the sign of n, and it be-
comes — —— = — — ; change the sign of d, and it becomes
(t a
— d d
Jf the signs of both terms of a fraction are changed, the
sign of the fraction is unchanged.
If the signs of n and d are both changed, the fraction
— becomes -^-^ = -•
d — d d
From the above it is evident that the value of a fraction
is unaltered by changing the signs of both terms, or by
FRACTIONS. Ill
changing the sign of one term, provided in the latter case
the sign of the fraction is also changed.
n_ — n _ n _ n . ~ n _ ~ n _ n
d — d — d d d — d d
The line separating numerator and denominator acts as
a vinculum on both terms of the fraction.
•; means (x + y) -=- (a + 5) .
a + b
a — b — a + b a a a
c o a—x —a+x x—a
91. Law of Signs. If the terms of a fraction are made
up of factors, the signs of an even number of factors in
either or both terms may be changed without affecting
the sign of the fraction. If the signs of an odd number
of factors in either term are changed, the sign of that term
is changed, and hence the sign of the fraction is also
changed. ffl f_ a;-)(_ y )( s ) = axyz
c c
(a-b)(b -c)(c-a) _ (b - a)(b - c)(c - a)
(* -y)(y -»)(»-*) (*-y)(y -*)(*-*)
(b — a)(c — b)(a — c~)
92. Reduction of Fractions. A fraction is reduced when
its form is changed without changing its value.
Reduction of fractions depends upon the following
principle :
Multiplying or dividing both terms of a fraction by the
same number does not change its value,
n _ n x m
d d x m
112 THE ESSENTIALS OE ALGEBBA.
Proof. Let -=/.
d
Then n =fd, by multiplying both sides by d.
nxm =fdxm, by multiplying both sides by to.
n x to
d x m
=/,
by dividing both sides by d x m.
Hence, - =
a
n x
d x
m , since both — /.
m
Also,
n n-k-m
d d-i-m
Proof. Let
n j.
d~' '
Then
TTonna
n=fd,
n^-m =fd -f- to,
n-t-m _ /.
d^-m
n n-i- m
d d -i-m
Fractions are reduced to higher terms by multiplying
both terms by the same number ; - = -sl_ (by multiplying
both terms by y 2 z). * *
Fractions are reduced to lower terms by dividing both
terms by the»same number ; J 1 = -&— (by dividing both
terms by cpzr). " a
A fraction is in its lowest terms when its terms contain
no common factors.
To reduce a fraction to its lowest terms, divide or cancel
all common factors out of its terms.
FRACTIONS. 113
(1 > a ^wr^vy hy dividin ^ out a2(a+b) -
Selecting and canceling the common factors can gen-
erally be done mentally.
(2) x 2 - 5 x + 6 = (a; - 2) (x - 3) _ x - 3
^ ' x 2 - 4 ~~ (x - 2) [x + 2) ~ x + 2
,„. (a; — a) (x — b) __ (a — x)(b — x) _ 1
^ (a 2 - a 2 )^ 2 - a?) _ (a 2 - x 2 )(b 2 -x 2 )~(a + x)(b + x)
EXERCISES.
Cancel factors common to numerator and denominator in
the following :
a 4 b-x .ax — bx „ a n b n
1.
5aW abx " a'-^"-
3 a 2 x 2 y ax 2 — x 3 8 x 2 + c kc
' 15 <rV ' &ic 2 ' a; 2 -a 2 '
3 17 a&V 6 aa;" | 9 a?-f
' 51 Vex 2 ' ' bx"- 1 ' ' a?-y 2
x*-xY 16 <*-*>
x i — y i x 3 + f
11 ^-9 17 a^ + y 3
' (a + 3) (x - 2) ' (x + yf
a? -25a; 18 (a; + y)(a;-y)(g-a;)
(a; -5) (a; + 5) a; 2 ' (x - z)(y -x)(-x-y)
3a?-27x ' 19 a? + Sx + 2
x
■*-6x 2 + 9 a^-1
s 2 -! 20 a^-5a; + 6
a* — l" ' 3* + a; -12
rfy-xtf, 21 2a^ + a;y-y 2
xHi-xtf (? + y)( x -y)
114 THE ESSENTIALS OF ALGEBKA.
22. ( a + y) 2 - z \ 27. ^ + x y + y 1 .
(x — y + z) 2 X s — y 3
23 ^ ~ ( x + 5 > 2 28 ^ + 6^ + 11^ + 63;
2^+5* ' ' (x + l)Q^ + Sx + 6)'
24 ar*- (a + &)«; + «& 2g x* — 6 ar 5 + 11 a? - 6 a;
a; 2 - (6 + e)a:+ 6c' ' (a; - 3) (x 2 - 3 a; + 2) '
25 3»-7a; + 12 1 + a + b + ab
^-9a; + 20' ' (1 - a 2 ) (1-6 2 )'
26 a; 8 + (a — fr) x — ab a? — b 2 + c 2 +2ca
x 2 -(b-c)x-bc' ' a 2 + b 2 + 2ab-c 2 '
93. Proper and Improper Fractions. A proper fraction is
one whose numerator is of lower degree in a named letter than
the denominator.
x 2 + 2 x 4- 5
« . q 2 — 7 1 * s a P ro P er fraction because its numer-
ator is of second degree in x, while its denominator is of
the third degree.
An improper fraction is one whose numerator is of degree
equal to or greater than the denominator.
x3 + 3 x 2+F>x— 4.
2 is an improper fraction.
An improper fraction may be reduced to an integral
expression and a proper fraction by dividing the numerator
by the denominator.
x i + 2x + l)x s + Sx i + 5x-4(x + l
x 3 + 2x 2 + x
a? + 4 x — 4
a; 2 +2a; + l
2a:-5
■ a g + 3a^ + 5 g-4 v , « . 2x-B
■ — X -f- A. -f-
a? + 2» + l ^ a* + 2f + l"
FRACTIONS. 115
The process is similar to the arithmetical process of
reducing an improper fraction to a mixed number.
EXERCISES.
Eeduce the following to integral or mixed expressions :
, x* + x x* - 4a? + 3af-12x
X. 5. ; .
x ic — 4
n x 2 — 4a; + 1 x 2 + 2 xy + f + z 2
2. • 6. 7 — r-= "
x (x + yf
3 3 a: 2 — 5a; + l X s — y* — 3 xy(x-y)
x+1 " x— y
4 a: 4 — s 3 + a: + 5 4 a; 4 -6 :e 3 + 12g! + 5 |
a^ + ar' + a; ' a; 2 + a: + l
1 - J
9. Prove — — == 1 + a; + a; 2 + ar s 4
1 — a; 1 — a;
10. ; divide as in Exercise 9 to four terms.
1 + x
11. 1 ■ " «
12.
1-a; 2 '
1
l-3x
a "
94. Reduction of Fractions to a Common Denominator.
Since a fraction is not changed in value by multiplying
both of its terms by the same number, we may make the
denominator any number we please by properly selecting
our multiplier. y ana 4 mav ^ e maae *° have the com-
b a
mon denominator bd.
a _ a x d _ad
b b x d bd
c _ c x b __ eb
d d xb bd
116 THE ESSENTIALS OF ALGEBKA.
The method of reducing fractions to a common denomi-
nator is the same as in arithmetic. It may be stated as
follows :
Rule. Find a common denominator, in general the least
common denominator (L. C. D.). Divide it by the denom-
inator of each fraction, and multiply the terms of the frac-
tion by the quotient.
(1) Reduce a and — — to equivalent fractions having
w 6-a; 6 + a: ^
the least common denominator.
The L. C. D. is (6 - as) (6 + as).
L. C. D. -j- 6 — 3; = 6 + x,
L.C.D.-r-6 + as=6-as.
a a(b + x)
6-as~ (6-as)(6 + 3s)'
C _ c(6 — 3!)
b + x~ (b — x)(b + x)
(2) Reduce -^—, -, ~ r-, ; 2 ~ x „ s to equiv-
v ; a 2 -4' (as-2)(as-3)' (a; + 2) (a; + 3) M
alent fractions having the least common denominator.
The L. C. D. is (as - 2) (a; + 2) (as- 3) (x + 3).
L. C. D. -=- (s; 2 - 4) = as 2 - 9,
L. C. D. -=- (x - 2) (s; - 3) = (3; + 2) (3; + 3),
L. C. D. -=- (3! + 2) (3; + 3) = (as - 2) (as - 3).
3s(3! 2 -9) Q?-9x
(as 2 -4) (a: 2 -9) (as - 2) (x + 2) (as - 3) (as + 3)'
4 as (as +2) (a? + 3) = 4 as 8 + 20 3s 2 + 24 x
(a; -2) (3! -3) (as + 2) (as + 3) (3s - 2) (3s + 2) (as - 3) (3s + 3)'
(2 -as) (as -2) (as -3) = -ar^ + Tas 2 - 16as+ 12
(as +2) (as + 3) (as -2) (as -3) (as - 2) (x + 2) (as -3) (as + 3)'
TRACTIONS. 117
EXERCISES.
Eeduce to equivalent fractions having lowest common
denominators :
.1225,1 3 n 4
- 1 - ~ n > k> ^~> tt and - 3 - — r- and — .
a b 3a 3b c x + y x — y
1 1 ._._, 1 . a b c
2. -, -, and 4. -, -, and
x' y' x + y ' x x + 1' x + 2
- a b c j d
5 - r> 7) ~> and
x + 1' x-1' x + 2 x — 2
_ x> + x 3x , 5x
6. , , and
cc 2 -4 x + 2' x-2
x , 3x
and
:B2-6a; + 8 x 2 -9x + 20'
8 1 1 and 1
a; 2 + 5a; + 6' af + ix + S' &n af + 3x + 2'
a b , c
9 - ZT-. 7/ ZT-^i> and
x 2 — jy 2 ' a; 4 — y*' X s — a;?/ 2
io ^ y 2 and ^
xr — y' x' — xy — 2 y* xr + xy — 2 y'
li 1 + ic 3a: 5x , 1 — x
1 — x' 1 + x 2 ' 1 — x 4 ' 1 + x
12. , , 5 , „ , , , „ 6 , „ , and
2 + 5» + 6' a* + 6 a; + 8' ar ! + 7a; + 12
12 3
13. -r :, — -, and
a; 2 — a;?/ + 1/ 2 ' a; 2 + an/ + ^ 2 ' a; 4 + aft/ 2 + ?/ 4
,.. « j 3a ,_ 5 i 6
14. and — -• 15. — and
b n c m 6»+i c »+i ' x n (a+b) 3 a; n - 2 (a+6) 4
95. Addition and Subtraction of Fractions. From division
we know that
a+b+c+d— e — / _ «,^,£,^_«_/
9~~ ~9 9 9 9 9 9
118 THE ESSENTIALS OF ALGEBRA.
If we read this identity from the right, we have the
result of adding a number of fractions. Hence, we con-
clude that ^ b_a + b
c c c
Rule. To add or subtract fractions, reduce them to the
same denominator and then deal with the numerators accord-
ing to' the rules for addition and subtraction, writing the final
result over the common denominator.
1 2 3x ,_ i0
+ ^^-^-^=what?
a — x a-\-x a? — x*
The common denominator is (a — x) (a + x)
1 a + x
a — x a , — or
2 2a-2z
a + x a- — x 2
3x 3x
a 2 — x 2 a 2 — x 2
Thus we have
a + x 2a— 2x 3x _a-\-x + 2a — 2x — 3x 3a — Ax
a 2 — x 2 a 2 — x- a 2 — x 2 a'- — x- ' a 2 — x 2
EXERCISES.
Combine and simplify :
1. x + —?— ■ 5. x 2 + 2x + 5+- 10
x— 1 x—2
2.3 a; ?-. 6. x + a + ^^
x + 1 ^ ^ (x + a)
, K 5a; 2 5 3
3. 5 a; — ■ 7. --I —
x -\- 10 a; a;
4. as + 1-
ar' + Sa; 7a; , 5a; 3a;
+ -
a: + 2 " a;+a a;+a x+a a;+a'
FRACTIONS. 119
3x* + x 5 a; 2 - 3 a; + 4 6 a; 2 + 4 a; + 1
' a? + a 3 X s + a 3 ar* + a 3
10. ^J_ + _l_. 13 . « 6 3a&
*+l w — 1 "' ' a + 6 a — 6 a 2 — 6 2
11 2 5 3je + 5_ 5a; + 2
x 2 + 2 a; 2 - 2 a; 2 - 9 a; 2 - 5 a; + 6
12 3a?+1 5a;-2 1_ lg 2a;+l , 6a; 8a:+l
a; 2 -l (x+iy x-1' ' a?+2x+l a?-l (x-lf
5 . 3
16
(a; - 1) (a; - 2) (x + 4) (x - 1) (a; + 4)
17 _ 2z + x + 2
18.
19.
20.
(a; + 3) (as - 2) (» + 1) {x + 4) (a; 2 - x - 2)
1-1
(a
+ 2/) 2
-3 2
ar*-
- (y + *) 2
1
l
ar*
— 5 a;
+ 6
a; 2 -
-6a: + 8
a;
- + -
y
(x-y)(y- z ) (y-z}( z - x ) -»)(«- y)
96. Multiplication of Fractions.
To prove
a c ac
b X d~bd'
Let
a j.
b~ Jv
and
t-f
Then
But from (1)
«=/A
and from (2)
c=f 2 d.
(1)
(2)
(3)
(4)
120 THE ESSENTIALS OF ALGEBRA.
Now multiplying Equation (3) by (4), member by
member, ~ j.j_,
— =/j/ 2 , by dividing both sides by bd.
bd
But fxj-/i/r
TT a c ac
HenCe ' ft X 5 = M
Rule. The product of two fractions is the product of the
numerators over the product of the denominators.
This covers all possible cases of multiplication involving
fractions, for all integers can be put in fractional form.
a a c ac
b xc= b x i = r
c _a c _ac
d 1 d d
Any number of fractions are multiplied together by
placing the product of all the numerators over the product
of all the denominators.
In all problems in multiplication of fractions free use of
cancellation should be made.
x + y a—b a+b x—y
a 2 —b 2 x — y x + y x 2 + l
In this, if we cancel the common factors, our work
appears thus:
-*-t-ff x a-^T x ^H^ x -a^-y =
1
(^HQ(^-^fT) aj— y x^r-y a^ + 1 a? + l'
TRACTIONS.
121
x
EXERCISES.
Perform the operations indicated, reducing the fractions to
lowest terms : /
1 3 s 10 y
' by 12a?
2 5x* 18 f
' 3y 2 15 a; 3
3 3(x + y) 6(x-yf
' 2(x-y) 9(x + yf
4 a(a?-y*) ^W ^x + y
b(x + y)' i a s x — y'
s 12 aa?y x 146V x f(^+f)
16 bxy 2 6 a?a? x(x + y)
6 . 5a(x + y) 2 x
10.
11.
-*)(H
(!■
\b d)[b'
fdV\
\a)
§b(x-
-y)
13.
14
2fx-yV
3\x+y)'
m-m.
12.
X
.y ■>-
X
y a
+ 1
15.
a v
b d
(1
--^XH
16 (x + 3)(x + 4) x * + <Lx + 3
3? + 5x + & (x + l)(x + 4)'
(Cancel common factors.)
17 a? + 7x + 12 z 2 + 6a: + 5
x^ + Ax + S a? + 9x + 20
18 a?-9x + 20 a?-4:X + 3
' a?-7x + 12 X a?-6x + 5'
20.
x* n -
- 4 a" + 4 w 2 '
-1
tf m + 2y m + l a?" — 4
122 THE ESSENTIALS OF ALGEBRA.
97. Division of Fractions.
To prove
a . c a d
b d b e
Let
a j.
b~ Jv
--f
Then
a . G -fi—f ■ f
~b ' d~f- Jl - Jv
But from (1) a =f-J),
and from (2)
c =f 2 d.
Then
" = fA=fl x L
-xf=4 bymultipl;
a d ad _-a d _f±
c b cb b c f 2
Hence,
a . c a d
b d b o
(1)
(2)
Rule. To divide one fraction by another, invert the divisor
and multiply.
Since all integers can be expressed in fractional form,
this rule suffices for all forms of division involving the
fraction.
/1X x x a x 1 x
(1) --=-a = --=--- = -x- = —
y V 1 y a ay
This is dividing a fraction by an integer.
(2) a + ? = ± + Z = ± x !l = «
y 1 y 1 x i
This is dividing an integer by a fraction.
FRACTIONS. 123
•o\ a? — b 2 a — 6 a 2 — ft 2 _ a; 2 — y 2 , , , w .
W) s- -5 ; = X ~ = (a + b) (x — y).
x + y x 2 -y 2 x + y a — b K yv SJ
In this case the Common factors (a — 6) and (a; 4- y) are can-
celed. The student should constantly be on the alert for
common factors and cancel them as soon as they appear.
1.
EXERCISES.
3 ax . 6a 2 a; a +3b . ab + 3b 2
5 a 2 y 10 a s y 2 a 2 + 5 b ' a s + 5ab
4 a'tE 8 ^_ 8 aa; a; — 4y ___ 3 a; 3 — 5 aft/
15 &Y ' 36y ' 3 a? -5 a:?/ ^2 a; 2 -8 a;?/'
12 a^yV . 4 aa;y 2 a; + y . (x + y) 2
9 a^/z 3 3 aft/« x 2 — y 2 ' (x — y) 2
4 x + y . (a; + ?/) 3 fl x* — y 2 . x + y
(x — y) 2 ' x — y " 3? — f ' x 2 + xy + y 2
9.
10.
x 2 — 9 y 2 a; — 3 y
+ 3 y) 2 : aT+Ty"'
(a; + ft) Q» — &) . a 2 + (a — 6) % — ab
(x — a) (x + b) x 2 — (a — b) x— ab
a? — y 2 . (x — y) 12 a 2 — (b + cf . a-b — c
■ tf-tf " (a? + y 2 )' ' 6 2 -(c+a) 2 ' -a+6-c'
98. The Complex Fraction. A fraction which has a frac-
tional expression for either or both of its terms is called a
complex fraction.
all 1
b x~~ X + ~
-, - — ^, — are complex fractions.
c 11 a+x
d x y y
Sinoe a fraction is an indicated division, a complex
fraction is simplified by performing the division indicated.
124 THE ESSENTIALS OF ALGEBRA.
a
m b _a c __a d _ad
^ ' c~b ' d~ b c~ be
d
11 y x y—x
(2)
x y _xy xy • xy y — x y-\-x
1 , 1 ~ y . x~y+x~ xy xy
x y xy xy xy
_y — x xy _y — x
xy y + x y + x
EXERCISES.
Simplify the following :
* + l 5 . _f_ + _»_. 6-2 «_
a;^ fi 6 + 3
JB+--5
j , __a_ a; ' 1
a,_ a • — "g e' g' + 3g.y+ 2y'
2 _ ^« xr x x 2y
x+a x+a_x-a x ~+y x + 2 y
3. * 7. a; - a X + a . 11. 3+*
1 i « ± a + ; ^-a :
a;
a; + y
4.
a;
1 — 8.
^ +y
x—a x+a
x__y_
y' 2 x 2
12.
2-
a:
1
2-1
a;
I+A + l
CHAPTER IX.
EQUATIONS IN ONE VARIABLE.
99. Identity; Conditional Equation. Distinctions be-
tween an identity and a conditional equation have already
been made in Chapter II. We have had illustrations
of the identity in all the fundamental operations and in
factoring.
O + 2> + 4 = 3? + 2 x + 4,
and (x + 1) (x — 5) = a? — 4 x — 5
are identities ; that is, they are equalities that are true for
all values of x.
A conditional equation restricts the value of some one
letter, the letter so restricted being called the variable^,
The other numbers of an equation are called constants.
(1) x = 5 restricts x to the value 5.
(2) 3 x = 18 restricts x to the value 6.
(3") ax=b restricts x to the value -•
a
(4) a? = 4 # restricts x to the values and 4.
Each of these equations becomes an identity for the
restricted value of the variable.
Thus, (1) becomes 5 = 5 for x = 5,
(2) becomes 3 x 6 = 18 for x = 6,
125
126 THE ESSENTIALS OF ALGEBRA.
7 7
(3) becomes a x - = b for x = -■>
a a
(4) becomes 2 = 4 x for x = 0,
and (4) 2 =4x4 for x = 4.
100. Root of an Equation. A value of the variable for
which the equation becomes an identity, is called a root of the
equation.
A root of an equation is frequently called a solution.
Any root of an equation when substituted for the vari-
able is said to satisfy the equation ; that is, an equation
is satisfied by any value of the variable which reduces it
to an identity.
2 x + 3 = 15 has 6 for a root. Substituting 6 for x,
this equation becomes 2 x 6 + 3 = 15, which is an identity.
6 is the value of x which satisfies the equation 2 x + 3 = 15.
101. Classes of Equations. A rational equation is one in
which the variable is free from radical signs.
Sx 2 — 72 + 11 = is a rational equation.
An irrational equation is one in which the variable is
affected by a radical sign.
5x 2 — QVx = 12 is an irrational equation.
An integral equation is one in which the variable appears
only in the numerators of its terms.
3« 2 + — =- + 3-r = — is an integral equation.
J. i lo 11
A fractional equation is one in which the variable appears
in one or more denominators.
EQUATIONS IN ONE VARIABLE. 127
7 = 3-
x x' z — 4 x + 1
3 x + - = — j — is a fractional equation.
A linear equation is one in which the variable appears to
the first degree only.
3 re + 7 = 11 is a linear equation.
A quadratic equation is one in which the highest power of
the variable is two.
baP + l x — 16 = is a quadratic equation.
A cubic equation is one in which the highest power of the
variable is three.
x 2 — 6 x 2 + 11 x — 6 = is a cubic equation.
It should be noticed that several of these terms may be
applied to the same equation.
Sx 2 — 7 X + 13 = is a rational, integral, and quadratic
equation.
EXERCISES.
Classify the following equations :
1. ax* + bx+c = 0. 5. ar' + lie 2 -! as = £f.
2. ax + b = 0. 6. 3Va = H— x.
3. atf+btf + cx + d^O. 3_5±l^ = io
4. a; 2 + fa; + ! = 0. ' x a? + 1
102. Equivalent Equations. It will be noticed that the
equations (1) % _ 4=Q and
(2) 3z-12 =
have the same root, viz., x = 4.
128 THE ESSENTIALS OF ALGEBRA.
The equations
(3) 5 x 2 - 10 x = and
(4) 5 x 2 - 10 x + 6 = 6
are satisfied when x = and when x = 2. Such equations
as (1), (2), and (3), (4), having the same roots, are called
equivalent equations.
Equivalent equations are those having precisely the same
roots.
Determine which of the following equations are equiva-
lent:
(1) x — 1 = 0, whose root is 1.
(2) 3 x — 6 = 0, whose root is 2.
(3) 2 x — 2 = 0, whose root is 1. »
(4) x 2 — x = 0, whose roots are and 1.
(5) 2 x — 4 = 0, whose root is 2.
(6) x 2 — 1 = 0, whose roots are 1 and — 1.
(7) Zx— 6 + 5 = 5, whose root is 2.
(8) x 2 — x — 4 = — 4, whose roots are and 1.
(9) - — = 0, whose root is 2.
v J 4 2
(10) — — - + 1 = 1, whose roots are and 1.
(11) x 2 + 3 = 4, whose roots are 1 and — 1.
103. Solution of Linear Equations. To solve an equation
is to find all of its roots.
The solution of an equation consists in deriving one or
more equivalent equations, the last of which is the value
of the variable.
EQUATIONS IN ONE VARIABLE. 129
Thus, to solve
(1) 3z + 4-2 = 6 + 2:z,
we bring 2 a; to the first member of the equation and
(4 — 2) to the second member. This is done by subtract-
ing these quantities from both members and gives us
(2) 3z-2a ; + 4-2-(4-2) = 6 + 2 : z-2a;-(4-2),
which becomes, by omitting the terms that destroy each
other,
(3) 3z-2a;=6-4 + 2.
This process is called transposition. It is effected by
changing the sign of a term when it is moved from one
member of an equation to the other.
We next unite the terms in each member of Equation (3),
which makes
(4) x = 4.
This process is called combining terms. It is effected by
addition. The last equation expresses the value of the
variable x. Hence, 4 is the root or solution of the given
equation.
It should be noticed' that Equations (1), (2), (3), and
(4) are equivalent equations, each having the root 4.
If 4 is put for x in these equations, they become
■(4-2),
(1)
3x4 + 4-2=6 + 2x4,
(2)
3x4-2x4 + 4-2-(4-2)
=6+2x4-2x4
(3)
3x4-2x4 = 6-4 + 2,
(4)
4 = 4.
Each of the above is an identity.
130 THE ESSENTIALS OF ALGEBRA.
As another illustration let us solve
(1) ix + 2-l = 2x-8.
Transposing 2 and — 7 to the second member and 2 x
to the first member, we have
(2) 4:x-2x = -2 + l-8.
Combining, we have
(3) 2 x = -3.
Dividing both terms of the equation by 2, we have
(4) *=-§•
The root or solution of Equation (1) is — |. Equations
(1), (2), (3), and (4) are equivalent equations, each being
satisfied by the root — | .
Putting x — — §, they become
(1) 4(-|) + 2-7 = 2(-|)-8,
or -6 + 2-7=-3-8.
(2) 4(-f)-2(-f)=-2 + 7-8,
or -6 + 3 = -2 + 7-8.
(3) 2(-f)=-3.
(4) -!=-!•
As a further illustration let us solve
(1) £^ + 6 = 1-^=1 + 13.
In order to get rid of the fractions in this equation, we
multiply both members of the equation by the Lowest Com-
mon Multiple of the denominators. Multiply both sides by
30, the L. C. M. of 2, 5, and 3 ; the result is
(2) 45a;-75 + 180=6a;-20:r ; + 10 + 390.
EQUATIONS IN ONE VARIABLE. 131
This process is called clearing effractions. It is always
brought about by multiplying both members of the equa-
tion by the L. C. M. of the denominators.
(3) 45 z- 6 z + 20 z= 75-180+10+390, by transposing.
(4) 59 z=295, by combining.
(5) x=5, by dividing.
5 is the root or solution of (1).
Equations (1), (2), (3), (4), and (5) are equivalent
equations.
Show that each is satisfied by the root 5.
104. Rule for Solving a linear Equation.
(1) Clear of fractions.
(2) By transposition bring all the terms containing the
variable to one member of the equation and all the constant
terms to the other member.
(3) Combine the terms of each member of the equation by
addition.
(4) Divide both members of the equation by the coefficient
of the variable.
Steps (1) and (4) depend upon the axiom that multi-
plying or dividing equals by equals gives equals. Steps
(2) and (3) depend upon the axiom that increasing or
diminishing equals by equals gives equals.
EXERCISES.
Solve the following equations :
1. 3x-5 = 19. 4. 7»-12 + 3 = 5a; + 16-5.
2. 7^-11 = 24. 5. 12-4a; + 2 = 13-7a! + 10.
3. 3a:-5 = a; + 13. 6. 15-14a;-7 = 17-16a;-6.
132 THE ESSENTIALS OF ALGEBRA.
7. 7a;-ll + 4a;-7 = 3a;-8.
8. ll-5a; + 17-3a; = 18-lla; + 23.
9. •5x-16-6x-6 = US-7x-4:X-7.
10. 2x-22 + 7x + U = 6x-8 + 4:X + ±2-5x.
11. 10 (x — 3) = 8 (x — 2). (Remove the parentheses first.)
12. ll(4a;-5)=7(6a;-5).
13. 3(a;-2)+2(2a;-3) = 3(a;-4).
14 §_x±l = x + 12 2Q 2a; — 1 Sx— o = 6a- + 2
2 ~~ 3 ' 3 2 ~ 4 "
5*-4 = 2x + 3. 3^-3 g = 4 «
3 2 5 2 6
16 . l±zl + 1 = lSL±5. a. 2 s-i^ + 5 = 2* + 3.
2 3 5 3 3
17 8a; + 5 _ 2 = 3a + 4 a; + l _ 2 a; — 5 a; = 1
5 5 ' ' 8 9 ~*~7 '
18. 5 + * = 6_25_l. 24. ^l + 2a: - 6 = l + ^.
2 3 3 6 5 2
19 ^_E = 5+1 25 !B + 1 2a; + 1 I l- 3 -'- 9
' 2 3 4 2' ' 9 7 + 8
26 . 2x-l 3a ._5^+3 = 2^_
3 7 5
27. 3 (a; - 2) - 2 (a; - 5) + 2 a; - 20 = 17.
28. aa; + bx = a 2 + 2 a& + 6 2 .
(a + 6)a;=(a 2 + 2a6 + 6 2 ).
„ _ a 2 + 2 ab + b 2
a + b
x = a + b.
EQUATIONS IN ONE VARIABLE. 133
29. ax + a 2 = bx + ft 2 . 34. ex — 3 x = c 2 — 9.
30. ox — bx — a 3 — & 3 . 35. mx — 5 nx = 3 m 2 — 75 w 2 .
31. <m + &a: = a 3 + 6 s . 36. - — - = 6 2 — a 2 .
a 6
32. a?x — abx + b 2 x = a s + b\
317 ^ & or
33. a. 2 a; + & s = a 3 -& 2 a;-a&a. ' a+b a -b~
38. (>-3)(> + 5)-7 = (a: + 4)(a;-8).
39. (2a;-5) 2 + 4 = (a;-6)(4a;-3).
40. 14-(2-a:) 2 = 5-(a: + 3)(a;-2).
EXERCISES.
1. One half of A's money is f 35 more than B's. They
together have $ 280. How much has each ?
Solution.
Let x = B's money.
x + $ 35 = one half of A's money.
2 x + 70 = A's money.
X + 2 x + 70 = 280.
x + 2 * = 280 - 70.
3 x = 210.
a: = 70 = B's money.
2 x + 70 = 210 = A's money.
Verification. One half of $210 is $105, which is $35 more
than $70. Also, $70 + $210 = $280.
2. A has $ 10 more than 3 times as much as B, and they
together have $ 250. How much has each ?
3. Find two numbers whose sum is 81, such that one may
exceed 6 times the other by 4.
134 THE ESSENTIALS OF ALGEBRA.
4. Divide 114 into three parts such that the first may exceed
the second by 15, and the third the first by 21.
5. Divide $ 176 among A, B, and C, so that B may have
$ 16 less than A, and $ 8 more than C.
6. Divide 440 into three parts such that the second is double
the first increased by 10, and the third is the sum of the first
and second.
7. What two numbers have a sum of 861 and a difference
of 221 ?
8. Find a number that exceeds 31 by the same amount that
£ of the number exceeds 1.
Solution.
Let" x = the number.
x — 31 = the excess of the number over 31.
- — 1 = the excess of \ of the number over 1.
6
By the conditions of the problem, these are the same ; hence,
x -
-31
= 5-1.
6
6x —
186:
= x- 6.
Qx
— X
= 186-
5 x
= 180.
X
= 36.
Verification.
36-
-31
= 5.
&of 36
- 1
= 5.
9. What number increased by \ of itself and 80 is 30 more
than double itself ?
10. Eight times the difference between the third and fourth
parts of a certain number is 40 less than the number. What
is the number ?
EQUATIONS IN ONE VARIABLE. 135
11. If 10 be subtracted from a number, \ the remainder
+ 40 is 30 less than the number. What is the number ?
. 12. Find two consecutive numbers such that \ of one plus \
of the other is 44.
Suggestion. Let x = one number, and x + 1 = the other number.
13. Find two consecutive numbers such that \ their sum is
34 less than the larger one.
14. Find three conseciitive numbers such that \ the first
+ \ the second + \ the third is 88.
15. In 10 years John will be twice as old as Henry was
10 years ago. John is 9 years older than Henry. Find their
ages now.
Solution.
Let x = Henry's age.
x + 9 = John's age.
x + 9 + 10 = John's age 10 years hence.
x — 10 = Henry's age 10 years ago.
z + 9 + 10 = 2(>- 10).
x + 9 + 10 = 2 x - 20.
x - 2 x = - 9 - 10 - 20.
- x = - 39.
a: = 39.
x + 9 = 48.
16. A man's age plus that of his wife's is 95 years ; 40 years
ago he was twice as old as she was then. What are their ages
now?
17. Eight years ago a father was 9 times as old as his son
was at that time ; in 37 years the father will be 1\ times as old
as the son is at that time. What are their ages now ?
18. A man left £ his estate to his son, \ to a nephew, \ to a
niece, and the remainder, amounting to $2600, to his wife.
What .was the value of his estate ?
136 THE ESSENTIALS OF ALGEBRA.
19. A house is sold for $2280. This is a gain of 14%.
What did the house cost ?
Solution.
Let x = the cost of the house.
^ x = gain.
x + tf e x = 2280.
100 x + Ux = 228000.
lUx = 228000.
x = 2000.
20. A horse sold at a loss of 7% brought $111.60. What
did the horse cost ?
21. A man invests ^ his capital at 4% and the remainder
at 5 % ■ His income is $ 2800. What is his capital ?
22. What number must be added to each of the terms of the
fraction i^ to make it -§ -| ?
23. What number must be subtracted from both terms of
the fraction ^ to make it -§- ?
24. Divide $ 5600 into two parts such that the income from
one part at 3% may be equal to the income of the other part
at 4%.
25. Divide f 760 among A, B, C, and D so that A and B
together shall receive $ 150, A and C together $ 190, and A and
D together, f 580.
26. $ 7.20 is changed into 36 coins. Each coin is either a
dime or a quarter. How many of each are there ?
27. A bill of f 10.20 is paid in an equal number of dimes,
quarters, and half dollars. How many of each are used ?
28. A man bought sheep at $ 4 a head, calves at $ 9, and
cows at $35. He bought twice as many calves as cows, and
twice as many sheep as calves. The cost of all the stock was
$ 690. How many head of each did he buy ?
EQUATIONS IN ONE VARIABLE. 137
29. Find three consecutive numbers such that the sum of
the quotient of the first divided by 10, the second by 11, and
the third by 61, is 25.
30. Find three numbers such that the second is a times the
first, the third b times the second, and their sum c.
31. One half of A's money is equal to B's, and five eighths
of B's is equal to C's ; together they have $ 1450. How much
has each ?
32. A man walks out at the rate of 4 miles an hour, and rides
back at the rate of 10 miles an hour. How far can he go out if
he must make the round trip in 7 hours ?
33. A man sold 12 acres more than \ of his farm, and had
2 acres less than ^ of it left. How many acres had he ?
34. A train leaves a station at 8 a.m. and runs 30 miles an
hour. At 11 a.m. another train leaves in the same direction
running 45 miles an hour. When and where will it overtake
the first train ?
35. A and B are two towns 120 miles apart. A messenger
starts from A to B at 7 a.m. and travels 10 miles an hour. At
8 a.m. another messenger starts from B to A and travels
12 miles an hour. When and where will they meet ?
36. A man in traveling from New York to Buffalo, goes \ as
far by boat as by train and ^ as far by carriage as by boat.
If the distance to Buffalo from New York be 490 miles, how
far does he travel in each conveyance ?
105. The Linear Type. Every linear equation in a single
variable may be reduced to the type form
ax + b = 0.
In this form a and b represent any positive or negative
numbers whatever".
138 THE ESSENTIALS OF ALGEBRA.
For example, 3 x -7 + 5 ' * '~ X = 2 x -7 + '—-.
Clearing of fractions,
12 x - 28 + 10 x - 2 = 8 x - 28 + 3 x.
Transposing all terms to the first member,
12s + 10a;-8a;-3x-28-2 + 28 = 0.
Collecting, 11 x - 2 = 0.
This is in the type form.
Comparing it with ax + b = 0, we see that a = 11 and 6 = — 2.
The solution of the type form ax + b =
b
is x =
a
Hence, the solution of the above example is x = fa-
Special roots of ax + b = 0.
If 6 = 0, then the solution of ax + b = becomes
*=-° = 0.
a
If a = and b is not 0, then the solution of ax + b =
becomes }
We have here a new form whose value we must in-
vestigate. _l_ J,
~T = + b >
i T = + 106 '
±|= + ioos,
= + 100000000 b.
.00000001
EQUATIONS IN ONE VARIABLE. 139
It appears that, as we decrease the denominator, the
value of the fraction increases. When the denominator
of the fraction is very small, the value of the fraction is
very large. When the denominator becomes 0, the value
of the fraction is large beyond measure. We express this
fact by saying that the value of the fraction is infinity.
The symbol for infinity is co.
I
Any number divided by is equal to co.
3 15 a 1000 J
- = oo, — = oo, - = oo, = co, etc.
If a = and 5 = 0, the solution of ax + b = becomes
- is the symbol of indeterminateness. - may have
any value.
(1) * = l =a;+ l.
x — 1
If in the above we put x = 1, it becomes
If in this we put x = 5, it becomes
« = 5 + 5,or£ = 10.
5 — 5 U
140 THE ESSENTIALS OF ALGEBRA.
106. Equations of Second or Higher Degree which depend
upon the Linear Type.
If we have the equation
x 2 - 5 x + 6 = 0,
we may by factoring write it in the form
(ir-3)(a;-2)=0.
We know that if one factor of a product is 0, the
product is 0. The product (x — 3) (x — 2) may be by
either factor being 0. If x — 3 = 0, then the product is 0,
or if x — 2 = 0, then the product is ; that is, the product
is if x = 3, or x = 2.
This is called equating the factors to 0.
A root is a value of the variable which satisfies the
equation. Hence, in the above equation, 3 is a root
because it satisfies the equation. .2 is likewise a root
because it also satisfies the equation. Therefore, the equa-
tion a; 2 --5a; + 6 = lias the two roots x = 3 and x = 2.
An equation of higher degree than the first may be
solved by the linear type, provided, after all the terms
have been brought to one member, it may be factored into
linear factors. Each factor equated to will give one root.
Hence, the number of roots is equal to the degree of the
squation.
EXERCISES.
1. x 2 - 5 x- 24 = 0.
By factoring, this is written (x — 8)(x + 3)= 0.
Hence, x — 8 = 0ora; = 8,
also a; + 3 = 0ora; = — 3.
The two roots are 8 and — 3.
EQUATIONS IN ONE VARIABLE. 141
2. a; 2 -3a;-40 = 0. 13. 16z 2 -25 = 0.
3. ar* + 6a; + 8 = 0. 14. ar 1 - 7 a; + 10 = 0.
4. a; 2 + 10 a; + 16 = 0. 15. a? 1 + 3 a; -10 = 0.
5. a; 2 -5 a: -14 = 0. 16. 3^ + 8 a; + 15 = 0.
6. a; 2 -16a; + 48 = 0. 17. 4a; 2 -12 x + 9 = 0.
7. a^ + 4=4a;. 18. x 3 — b 2 + 2 ax + a? = 0.
8. 0^-1 = 0. 19. a^-12a; + 35 = 0.
9. a^-25 = 0. 20. a; 2 -21 a; + 20 = 0.
10. a; 2 + ll=36. 21. a^ + 28a; + 75 = 0.
11. a; 2 -16 = 0. 22. a; 2 -7a; = 98.
12. a?-{a-bf=0. 23. a^+7a;-98 = 0.
24. 3a; 2 + lla;-4 = 0.
25. 12(a; + l)- 3(0^-1)+^- 1 = 0.
• 26. (ar ! -a5) 2 -22(ar ! -a;)+40 = 0.
27. (a; 2 + 3a:) 2 -8(a; 2 + 3a;)-20 = 0.
28. a^ + 5a^ + 6a; = 0.
29. a? - 12 a? + 27 a: = 0. 35. « 2 - a 2 -2ab- & 2 = 0.
30. (a;+4) 2 +(2a;-5) 2 =73. 36. z 4 - 13z 2 + 36 = 0.
31. (2a;-5) 2 -(2a;+10) 2 =24. 37. (x + |) 2 - (2 x-\J = 0.
32. (5a: + 4) 2 -(3a;-8) 2 =0. 38. (a? + 2 a;) 2 - (ar* - 4 a;) 2 = 0.
33. y 2_^ = 0. 39. 4a; 4 -8ar i -5ar ! = 0.
34. aV - (6 + e) 2 = 0. 40. a^-3sc 2 + 3a;-l = 0.
107. Fractional Equations. Certain fractional equations
may be reduced to the linear form or to the form dis-
cussed in the last section.
142 THE ESSENTIALS OF ALGEBRA.
Fractional equations are made integral by clearing of
fractions.
The common multiple used in clearing of fractions will
contain the variable. It may give an integral equation
which is not equivalent to the given fractional' equation.
(1) x - - = 0.
x
Clearing of fractions by multiplying by x, we have
a? - 1 = 0,
or O-l)O+l) = 0.
Whence, x = 1 and x = — 1.
These roots both satisfy the given equation.
When x = 1, x becomes 1 — -=1 — 1 = 0.
x 1
When x = — 1, x becomes — 1 = — 1 + 1 = 0.
x — 1
If in clearing of fractions we multiply by x 2 , the
resulting equation is
x s — x — 0,
or x(x — l)(a; + l) = 0.
Whence, x = 0, x = 1, and a; = — 1.
We now have three roots, two of. which, 1 and — 1,
satisfy the given equation, while the other one, 0, does not
satisfy it ; for, when x = 0, x becomes — -, which
is not equal to 0.
The root which is here introduced by clearing of
fractions is called an extraneous root.
The root occurs because we multiplied by a multiple
higher than the L. C. M.
EQUATIONS IN ONE VARIABLE. 143
In integral equations any multiple whatever of the denorn^
inators may be used in clearing of fractions, but in frac-
tional equations the L. C. M. should always be used.
(2) 5 + HL+l 4.
2T— 1
5 re 2 -5+ a; + 1 = 4 a; 2 - 4.
5a^-4a; 2 + a;-5 + l+4 = 0.
x 2 + x = 0.
x(x + 1) = 0.
x = and x = — 1.
When x = 0,
5 + ^±4 = 4 becomes 5 + £±4 = 4, or 5 - 1 =4.
x 2 — 1 — 1
When x= — 1,
5 + 4^ = 4 becomes 5 + ~ 1 + 1 = 4, or 5 + - = 4.
#* — 1 1 — 1
x= — 1 does not satisfy the equation and is therefore an
extraneous root.
The root — 1 occurs in this solution because the fraction
x -4- 1
was not reduced to its lowest terms. By reducing
it to its lowest term the equation becomes
5+^- = 4.
#—1
5x — 5 + l=4a;— 4.
5a;-4a;— 5+1 + 4 = 0.
No extraneous root now appears.
Before beginning the solution of a fractional equation, all
fractions should be reduced to their lowest terms. The safe
144 THE ESSENTIALS OF ALGEBRA.
plan in all fractional equations is to test every root, retain
only those roots that satisfy the equations, and reject all others
as extraneous.
EXERCISES.
1. ?-?=0. 3. ? + ^_ = o.
3 x
X XT
x—3 x—2 x+2
6. The quotient of a number divided by 7 increased by "the
quotient of 63 divided by the number is 6. What is the number?
7. A number is increased by 82 and the sum divided by the
number; the quotient is ^ of 1 more than the number. What
is the number ?
8. The sum of the squares of two consecutive numbers is 85.
What are the numbers?
Solution.
Let x = one of the numbers,
x + 1 = the other.
x 2 + (x + l) 2 = 85.
x* + x" + 2 x + 1 = 85.
2 x 2 + 2 x + 1 - 85 = 0.
2 x 2 + 2 x - 84 = 0.
x 2 + x - 42 = 0, by dividing by 2.
(x + 7)(x - 6) = 0.
x = — 7 or x = 6.
The numbers are 6 and 7 or — 7 and — 6.
9. The sum of the squares of two consecutive numbers is
41. What are the numbers ?
10. Two numbers differ by 5, and their squares differ by 105.
What are the numbers ?
11. Three times the product of two consecutive numbers
lacks 92 of being twice the sum of their squares. What are
the numbers ?
EQUATIONS IN ONE VARIABLE. 145
12. The area of a square field is doubled by increasing its
length 12 rods and its width 5 rods. What is the length of
one side of the field ?
Solution.
Let x = one side of the field.
2x 2 = 0+ 12)0 + 5).
2 x 2 = x 2 + 17 x + 60.
2 x 2 - x 2 - 17 x - 60 = 0.
X 2 _ 17 x _ 60 = 0.
0-20)0 + 3) = 0.
a = 20 and x = — 3.
Both of these roots satisfy the equation, but only 20 can be used
in this problem, as it would not be possible, to have a field one side of
which is — 3 rods in length.
13. The denominator of a fraction is 3 more than its numera-
tor. If 7 is added to each of its terms, the value of the fraction
is increased by T \. Find the fraction.
14. A can do a piece of work in 10 days, B in 8 days, and C
in 6 days. In how many days can they all do it working
together ?
Solution.
Let x = the time required.
- = part done in one day.
x
— = part done in one day by A.
- = part done in one day by B.
8
- = part done in one day by C.
— -| \- - = part done in one day by A, B, and C.
■1 + 1 + 1 = 1.
10 8 6 x
12 x + 15 x + 20 x = 120.
47 x = 120.
x = 2|f, the number of days required.
146 THE ESSENTIALS OF ALGEBRA.
15. A cistern has two pipes; one will fill it in 8 hours, and
the other in 12 hours. If both are open, how long will the
cistern be in filling ?
16. A cistern has three pipes ; one will fill it in 12 hours, one
in 10 hours, and the other will empty it in 15 hours. If all
three are open, how long will the cistern be in filling ?
17. A number added to 22 times its reciprocal makes 13.
Find the number.
- is called the Reciprocal of x.
x
18. A can do a piece of work in a days, B can do it in b
days. In how many days can they together do the work ?
19. The area of a square field is doubled by increasing its
length a rods and its width b rods. What is the length of one
side of the field ?
20. A fraction whose numerator is 3 less than its denom-
inator added to its reciprocal gives 2^. Find the fraction.
CHAPTER X.
LINEAR EQUATIONS IN TWO VARIABLES.
108. Roots of a Linear Equation in Two Variables. The
type form of the linear equation in x and y is
ax + by + c = 0.
a, b, and c may be any positive or negative numbers
whatever.
The equation 2 a; + 5 y — 10 = is a special form of
ax + by + c = ; in which a = 2, & = 5, and e = — 10.
If in 2 x + 5 y — 10 = 0, we transpose 5 y — 10 and di-
vide by 2, we have
J 10-5y
* = — 2
The value of a; depends upon y. It has owe, awe? only
one, value for each value of y.
If y = 0, x = 5. If y = - 1, a; = Jjf.
y = 1, x = f . y = - 2, a; = 10.
y = 2, x = 0. y = - 3, x = - 2 ^.
y = 3, x = - f . y = - 4, x = 15.
y = 4, a; = — 5. y = — 5, x — *£-.
y = 5, x = — -y-. y = — 6, a; = 20.
W
148 THE ESSENTIALS OF ALGEBRA.
The equation 2x + 5 y — 10 = is satisfied by x =5
and y = 0, for these values reduce the equation to
2x5 + 5x0- 10 = 0.
x = | and y = 1 also satisfy the equation, for they reduce
lfct0 '2x| + 5xl-10 = 0.
Therefore, x = 5, y = and a; = |, ^ = 1 are roots of the
equation 2 x + 5 «/ — 10 = 0.
In the set of values of x and y above, every y&lue of x
and the corresponding value of y constitute a root. The
number of such sets of values that are possible is evi-
dently unlimited. Besidesthe roots (5, 0), (&, 1), (0, 2),
(-!, 3), (-5, 4), (--«-, 5), (-V-, -1), (10, -2), and
(■Sg 5 -, — 3), any number more could be worked out at
pleasure. A root of an equation in two variables may
be written (m, n); m is the value of x, and n is the
corresponding value of y, the two together constituting
a root.
109. Graph of the Linear Equation. The coordinate axes,
or lines of reference, are two lines perpendicular to each
other.
The axis of abscissas, or a;-line, is the horizontal line
X'OX.
The axis of ordinates, or y-line,
is the vertical line Y'OY.
Abscissas, or ^-distances, are al-
O ' x ways measured parallel to the a;-line.
They are positive when measured to
the right of the ^-line and negative
Y' ' when measured to the left of it.
Fig. i. Ordinates, or y-distances, are always
LINEAR EQUATIONS IN TWO VARIABLES.
149
measured parallel to the y-line. They are positive when
measured above the a;-line and negative when measured
below it.
The coordinates of a point are its x and y distances.
The x distance is the abscissa, and the y distance the
ordinate. The coordinates of a point completely deter-
mine it with respect to the lines of reference.
A point is designated by its coordinates written (m, n).
This means that m is the abscissa and n the ordinate of
the point.
Y
P,<-4.8>
X'
N
P,(-S,-4)
P,(2,3)
M
X
P 4 (0,-3)
Y'
Fig. 2.
The point P l or (2, 3) is located by measuring from
to M, a distance of 2, and from if to P v parallel to OY,
a distance of 3. The point P 2 or (-4, 2) is found by
measuring from to N, a distance of 4, and then from
N to P 2 , parallel to OF, a distance of 2. The abscissa in
this case is measured to the left of OF because it is nega-
150
THE ESSENTIALS OF ALGEBRA.
tive. The location of the points P 3 or (— 5 — 4) and P t
or (6, — 2) is also shown on Figure 2.
■K-
M-
<-L-
Y
Fig. 3.
EXERCISES.
In Figure 3 the side of each small square is a unit or one.
1. Write the coordinates of each lettered point. Thus,
A is (2, 2), if is (-6, -1).
2. Locate on Figure 3 the following points : (5, 1), (— 2, 1),
(3, - 4), (- 3, - 2), (6, 1), (- 1, - 4), (-3, 2), (0, 3), (- 3, 0),
(0,0), (11,3), (2|,2|),(-3|,i).
On Figure 4 are located the points (10, — 2), (7J, — 1),
(5, 0), (21 1), (0, 2), (- 2J, 3), (- 5, 4), (7J, 5).
LINEAR EQUATIONS IN TWO VARIABLES.
151
It will be seen that these points are in a straight line.
The points located on Figure 4 are some of the roots of the
equation 2x+ 5y — 10 = 0, worked out in Section 108.
i
I
-■'-
-%
,5J_
(-!
,4)
~V
-ax
.3)
(0,
V
p
X'
,.
(5,0
) '
x
V
as.
-1)
no
-3)
•
t
Fig. 4.
2%e graph of an equation is the line upon which are found
all the points indicated by its roots.
The line MN is the graph of the equation !
2z + 5«/-10 = 0.
The coordinates of every point upon this line satisfy
the equation 2a; + 5«/-10 = 0. The point P is (2, 1^).
(2, 1£) is a root of 2 x + 5 y — 10 = 0, for when x = 2
and t/ = 1£, the equation becomes 2 x 2 + 5 x \\ — 10 = 0,
which is an identity.
The graph of x — 2 y = 4.
a; = 4 when y = 0.
% = Q when y = 1.
a; = 8 when y = 2.
Here a; = 4 + 2 y.
x= 2 when y = — 1.
x = when ^ =; — 2.
a = — 2 when y == — 3.
152
THE ESSENTIALS OF ALGEBEA.
The points represented by the roots above worked out
are (4, 0), (6, 1), (8, 2), (2, -1), (0, -2), and (-2,-3)
and are shown on Figure 5.
The points located in Figure 5 lie upon the straight line
PQ, which is the graph of x — 2 y = 4.
The graph of x + y + 5 = Q. Here x = — (y + 5).
x = — 5 when y = 0. x = when y = — 5.
x= — Q when y = 1. a;=l when ^= — 6.
a; = — 2 when y = — 3.
The points represented by the above roots are (—5, 0),
(-6, 1), (-2, -3), (0, -5), and (1, -6).
Locating these points upon Figure 6, we find that they
all lie upon the straight line RS, which is the graph
of x + y + 5 = 0. This line passes through the points
LINEAR EQUATIONS IN TWO VARIABLES.
153
A(-4, - 1), 5(-3, -2), and C(-l, -4). Verify that
these are roots of the equation.
Fig. 6.
EXERCISES.
Draw the graphs of the following equations :
1. 3x — 2 y = 6. 3. 2x — 5i/ = 10. 5. a; + ?/ = 4.
2. ix — y = S. 4. x— y = 4. 6. 2 a; — 3 ?/ = 0.
TAe graph of every linear equation in two variables is a
straight line.
Since two points are sufficient to locate a straight line,
we need but two roots of an equation to draw its graph.
154
THE ESSENTIALS OF ALGEBRA.
The graph of 3 x + 5 y = 15.
Since x = when y = 3, one root is (0, 3) ;
and since x = 5 when y = 0, another root is (5, 0).
Fig. 7.
■ Locate these two points upon Figure 7, and draw through
them the straight line AB. This line is the graph of
3 x + 5 y = 15.
In general, the most convenient pair of roots for deter-
mining the graph of an equation is found hy making x=0
and solving for y, and then hy making y = and solving
for x. These two roots give the points in which the line
cuts the coordinate axes. ■
EXERCISES.
By the above method make the graphs of the following
equations :
1. 3x- 2 y = 6.
2. 4 x — y = 8.
3. 2y-5x = 10.
4. 7x — y = 7.
5. x + 7 y = 7.
6. 3 x + 4 y = 12.
LINEAR EQUATIONS IN TWO VARIABLES. 155
110. Graphs of Two Linear Equations upon the Same
Diagram.
Graphs of x — y = 6 and 2 x + y = 9.
Fig. 8.
The graphs of these lines are AB and CD, respectively.
They intersect at the point P (5, — 1). This point P lies
on both lines, and its coordinates constitute a root of each
equation. By putting x = 5 and y= — 1, x— y=6 becomes
5 — (— 1)=6, and 2x + y = 9 becomes 2x5 — 1 = 9. This
verifies that (5, — 1) is a root of each equation.
Since two straight lines can intersect in but one point,
a pair of linear equations can have but one common root.
156
THE ESSENTIALS OF ALGEBRA.
1.
2.
EXERCISES.
By means of their graphs find the common root of each of
the following pairs of equations :
2x-y=l, ^ (x-2y = l,
x + y = 5. \x + y = — 5.
x + y = 4, (2x + 4y = 6,
o.
x — 2y = l. [x + y=l.
■2x + y = 3, 6 p-y=l,
x — 2 ?/ = 4. [2xi-2y = 9.
7. Graphs of x + y = 1 and 2 x + 2 y = 9.
Fig.
The graphs of these equations are shown in Figure 9. They
are parallel lines, and so do not intersect. The two equations
have no common, finite root. It should be noticed that the
coefficients of x and ?/ in the second equation are just double
the coefficients of x and y in the first equation.
8.
f x-3y=6,
[3x-9y = 9.
9.
f x + 2y = 6,
2x-
y = 2.
CHAPTER XI.
SIMULTANEOUS EQUATIONS.
111. Definitions. Equations in two or more variables,
having the same solutions, are called equivalent when any
one of the equations may be changed to the exact form of the
others.
Thus, x + y = 5
and 3 x + 3 y = 15
are equivalent equations; each has a root (1, 4), and the
first may be changed to 'the second by multiplying both
members by 3.
Equations not equivalent, but having the same solutions are
called simultaneous equations.
Thus, x + y = 6
and Bx + y=12
are simultaneous, having the common solution x=S, y = 3.
These two equations are not equivalent, since the first
can not be changed to the form of the second.
Two equations in two variables form a set of simultane-
ous equations; three equations in three variables form a
set of simultaneous equations, etc.
Elimination. To solve a set of simultaneous equa-
te must so operate upon and combine the given
157
112
tions, we must so
158 ' THE ESSENTIALS OF ALGEBRA.
equations as to produce a single equation containing a
single variable. The processes of obtaining such a single
equation are called elimination.
In the operations of elimination the following principles
are to be carefully noted :
(1) For any expression in an equation an identical
expression may be substituted.
(2) When both members of an integral equation are
multiplied by an integral expression containing the vari-
able, the resulting equation is not equivalent to the
original equation.
For example : x + 3 y = 4.
Multiplying both sides by x — y, we have
(a; - y) (x + 3 y) = 4 (x -y),
an equation which not only has all the roots of £ + 32/ = 4,
but also all the roots of x — y = 0.
(3) All the axioms heretofore given can be used in the
processes of elimination, with the single exception noted
above.
113. Elimination by Substitution. This method will be
understood by noting the following solution :
1(2) 5^-2^=1.
By transposing the 2y in Equation (1), we have
(3) z=5-2y.
Substituting this value of x in Equation (2),
(4) 5(5-2*,)-2*,= l. ,
by transposing.
by collecting.
by dividing.
SIMULTANEOUS EQUATIONS. 159
Removing the parentheses,
(5) 25-10y-2y = l.
(6 ) -10y-2y = l-25,
(7) -12y = -24,
(8) y=2,
Substituting this value of y in (3),
(9) a;=5-2(2).
(10) x=l.
The root is (1, 2).
EXERCISES.
Solve the following equations by the method of substitution :
1.
f x + y = 4,
[3a; — 4 «/ = 5.
f2a; + 3y = ll,
2 ' J2x+ y = 9.
a; + 4?/ = 10,
3a;- y = 12.
f2a!-7y = -5,
L3a; + 2y = 7.
3.
2a; + 32/ = 2,
a; + 3y = l|.
9.
f3a;-22/ = 5,
\q x + y = 20.
2" 5 '
6 15 3
10.
5 x + 6 y = 16,
-8a; + 32/ = -13.
-2 a; + 2y= 10,
+ 3a; + 32/ = -15.
Construct graphs for Exercises 4, 5, 6, and 7.
fa:-f;*/ = -5,
fas + $y=17.
6.
160 THE ESSENTIALS OF ALGEBRA.
114. Elimination by Comparison. The following problem
will illustrate the method :
(1) 5z + 3y = 19,
1(2) 4x- jf = b.
By transposing 3 y and dividing by 5, we get from (1)
/,x 19- 3 v
(3) « = — j-*.
By transposing — y and dividing by 4, we get from (2)
(4) *=6±*.
Equating the two values of x given by (3) and (4),
(5) 19-3y ^5 + y
v 7 5 4 "
(6) 76 — 12 y = 25 + 5 y, by clearing of fractions.
(7) — 12 y — 5 y = 25 — 76, by transposing.
(8) — 17 y = - 51, by collecting.
(9) y = S .
Substituting this value of y in (3),
(10) T = 19-3x3 = 10 = 2|
5 5
The root is (2, 3).
EXERCISES.
Solve by the method of comparison :
1.
2x-Sy = 7,
\2x + y = 3.
ix + y = 3,
[2x + 3y = 4 : .
2 ( 5x + y = 3, 5 f3a;-4y = 4,
L 2 x — 3 ?/ = — 4. : ' \2x + 5y = 10.
SIMULTANEOUS EQUATIONS.
161
5.
6. -I
' 5 x — 2 y = 4,
Zx+ly = 2l.
[x— y = a,
[x + 2y = b.
■5x-iy = 20,
3x + 2y = 12.
8.
9.
10.
-4x-2y = l,
— x + 5 y = 6.
x + 2y = 4,
5x+ y = 3.
5x+ y = m,
3x + Ay = im.
Construct graphs for Exercises 3, 5, 7, and 9.
115. Elimination by Addition or Subtraction. The two
problems here solved will illustrate the method.
Solve
r(l) 2z + 3*/ = 5,
1(2) 7z-2«/ = 5.
We first make the coefficients of the y's have the same
absolute value. This is done by multiplying Equation (1)
by 2 and Equation (2) by 3, thus giving us
(3) 4x + 6t/ = 10,
(4) 21 x - 6 y = 15.
Adding Equations (3) and (4),
(5) 25 a; =25.
(6) x=l.
Substituting this value of x in (1),
(7) 2xl + 3y = 5.
(8) 3^ = 5-2 = 3,
(9) y = l.
The root is (1, 1).
162 THE ESSENTIALS OF ALGEBRA.
j(l) 2z + 7#=38.
2. bolve • n . _ , 01
1(2) 3z + 4# = 31.
We can make the coefficients of the #'s alike by multiply-
ing Equation (1) by 3 and Equation (2) by 2, thus giving
(3) 6x + 21y=lU,
(4) 6 x + 8 y = 62.
Subtracting Equation (4) from (3),
(5) 13 #=52.
(6) # = 4.
Substituting this value of y in Equation (1),
(7) 2z+7-4 = 38.
(8) 2a;=38-28 = 10.
(9) x = b.
The root is (5, 4).
The method of elimination generally used is that of
addition or subtraction. The method by comparison is
merely a disguised form of eliminating by subtraction.
The particular method to be used must be determined by
a careful inspection of the problem.
116. Some Illustrative Examples.
a) ;+!-«,
(2) |-|=2.
Here there is no need of clearing of fractions. By addi-
tion, y is eliminated, and
SIMULTANEOUS EQUATIONS. 163
(3)
3a; „
(4) 3 a; =24.
(5) a; =8.
Substituting 8 for x in (1), we have
(6) J + | = 4.
(7) | = 4-2 = 2.
(8) y=Q.
The root is (8, 4).
2.
K J x y 24
(2)^-I = a
* 2/ 24
In problems of this form never clear of fractions
(3) — - - = — , by multiplying (2
if
(4) — = 24, by adding (1) ai
(5) 39 x =13x24.
(6) x = 8.
Substituting 8 for x in (1),
(7) M = 17.
S ' 8 y 24
m 2_17_3_1_
(8) ^~24 8~3
(9) y = 6.
The root is (8, 6).
164
THE ESSENTIALS OF ALGEBRA.
j(l) x-3tf=10,
1(2) 3z + 5y = 2.
In this form, in which one equation has the variable with
a coefficient 1, use the method of substitution.
From (1),
(3) a; =3 # + 10.
Substituting the value of x in (2),
(4) 3(8y + 10)+5y = 2.
(5) 9y + 30 + 5y=2.
(6) 14y= — 28, transposing and combining.
(7) y=-2.
(8) x= -6 + 10 = 4, from (3).
The root is (4, - 2).
4.
(1)
l + 2z j
8y
X
-
y
(2)
y+i-.
X
_5
X
In examples of this form it is best to clear of fractions.
(3) l + 2z + 3y=3z,
(4)
— x+ Sy = —
(5)
x + y = 5,
(6)
4y = 4,
CO
y=l.
(8)
aj = 4,
by clearing (1) of fractions,
by transposing and collecting.
by clearing (2) of fractions.
by adding (4) and (5).
by substituting 1 for y in (4).
The root is (4, 1).
SIMULTANEOUS EQUATIONS.
165
EXERCISES.
Solve the following simultaneous equations :
2. ]
x + y = 5,
[2x + 3y = 12.
(2x-5y = 12,
{ {3x + 5y = 8.
Ax + 8y = 13,
2x + 3y=7%.
I
ix-
■#y = i,
ii* + iy = 4.
4*-| = 6,
-1,
13
: 5'
x
4 ' 3
5_4_
a; j/~
x y
7. ■!
f ax + by = c,
(. Ix + my =p.
, + -=5,
3as-- = 7.
3 + * = 7,
a; y
a; y
10.
11.
12.
13.
14.
15.
16.
2 - 3 - = ll,
2/
a;
3 A
x y
= 22.
a; + - = 5,
2/
3a;_? = 3.
2/
m
teH — = a,
y
Jcx-
n
y
b.
17. 1
a 6
la; ?/
a; + y = m + n,
m + x n
n + y m
3x-7y = 0,
ax — by = 0,
mx + ny = q.
( mx -{-ny = m 2 ,
I nx + my = n 2 .
166 THE ESSENTIALS OE ALGEBttA.
20.
3y + 7ce + l 2,y-3a; + 8 _ g
5 3
5, y _7a; + 10 3y + 2a; + 6 _g
3 5
Construct graphs for Exercises 2 and 5.
EXERCISES.
l. The sum of two numbers is 32, and one number is 3
times the other. What are the numbers ?
Solution.
Let x and y be the numbers.
2
by the conditions of the problem.
Then
(1)
s + 3f = 32,
and
(2)
z = 3y
(3)
3 y + ?/ = 32.
(4)
42, = 32.
(5)
(6)
y = 8.
x = 24.
2. Eight apples and 5 oranges cost 31 cents, and 5 apples
and 10 oranges cost 40 cents. What is the cost of 1 apple and
of 1 orange ?
3. Three bushels of wheat cost 15 cents more than 5
bushels of corn, and 2 bushels of wheat and 1 bushel of corn
together cost $ 2.05. What is the price per bushel of each ?
4. A fraction is equal to f. If both of its terms are
increased by 12, the value is then |. Find the fraction.
(Let x — numerator, y = denominator, - = fraction.')
y
5. Find a fraction such that if 1 is added to the numerator
it becomes £, and if 5 is added to the denominator it becomes \.
SIMULTANEOUS EQUATIONS. 167
6. The sum of two numbers is 75. The larger contains the
smaller 5 times, with a remainder of 3. Find the number.
7. There are two numbers ; 3 times the first is 8 more than
the second, and their difference is 42. Find them.
8. A man spent $ 225 for sheep at $3.50 a head and calves
at $ 10 a head. H^e bought 42 head in all. How many of each
did he buy ?
9. Ten years ago a father was 5 times as old as his son.
Twenty years hence he will be twice as old. What are the
present ages of each ?
10. A said to B, " Give me $ 60, and I shall have twice as
much as you." B said to A, "Give me $90, and I shall have
as much as you." How much had each ?
11. Find two numbers such that \ the first and \ the second
is 36, and \ the first and \ the second is 13.
12. There are two numbers such that if each is increased by
5, the sums are in the ratio 5 and 11, and if each number be
decreased by 15, the remainders are in the ratio 1 and 7. Find
the numbers.
13. A farmer has two horses and an $18 saddle. If the
saddle is put on the cheaper horse, the horse and saddle are
worth \ of the better horse. The better horse and saddle lack
$ 12 of being worth twice as much as the cheaper horse. What
is the value of each horse ?
14. If the greater of two sums be multiplied by 5 and the
lesser by 7, the sum of the products is 140. If the greater be
divided by 7 and the lesser by 5, the difference of the quotients
is 0. Find the numbers. •
15. There are two numbers which differ by 11. One sixth
of the larger is 1 more than \ of the smaller. Find the
numbers-.
168 THE ESSENTIALS OF ALGEBRA.
16. A number consists of two digits whose sum is 13 ; if 27
be added to the number, the order of the digits is changed.
Find the number.
Solution.
Let x = units' digit.
y = tens' digit.
10 y + x = the number.
Then
(1)
x + y = 13,
and
(2)
(3)
(4)
10
y + x + 27 = 10 x + y,
9 y - 9 x = - 27.
y -x = -Z.
by conditions of problem,
(6)
2y = 10,
by adding (1) and (4).
(6)
y = 5.
'
(7)
z= 8,
from (1)
10 ;/ + £ = 58, the number.
17. If to a certain number of two digits the tens' digit be
added, the sum is 80. If the units' digit be subtracted, the
remainder is 70. Find the number.
18. A number is composed of two digits whose sum is 13.
If their order is inverted, the new number is 4 less than double
the original number. Find the number.
19. A sum of money was divided equally among a certain
number of people. If there had been 3 persons more, the share
of each would have been $ 2 less ; but if there had been 2 per-
sons fewer, the share of each would have been $ 2 more. How
many persons were there, and what was the share of each ?
20. A lost f of his money and then borrowed £ of B's money,
when he had f 12. At first A had f as much as B. Find
how much each had at first.
21. The sum of a number of two digits and the number
formed by reversing the order of the digits is 110. The differ-
ence of the digits is 8. Find the number.
SIMULTANEOUS EQUATIONS. 169
22. A man has a certain number of silver dollars and quar-
ters. He notices that if his dollars were quarters and his
quarters dollars he would have $ 22.50 more than he now has.
He also notices that if his dollars were dimes and his quarters
half dollars, he would have $ 1 more than he now has. How
much money has he ?
23. In a certain school \ of the number of boys is equal to
\ of the number of girls ; twice the whole number of pupils in
the school is 100 more than 3 times the number of girls.
How many pupils in the school ?
• 24. A and B are 45 miles apart. If they walk in the same
direction, A overtakes B in 45 hours. If they walk toward
each other, they meet in 5 hours. Find their rates of walking.
25. If the first of two numbers be divided by 12 and the
second by 15, the sum of the quotients is 12; if the first be
divided by 4 and the second by 3, the difference of the
quotients is 12. What are the numbers ?
26. Find two numbers such that the sum of their reciprocals
is T \\, and the difference of their reciprocals is T |^.
27. If the base of a rectangle be increased by 6 feet and
the altitude by 4 feet, the" area is increased by 216 square feet.
If the base be decreased by 4 feet, and the altitude increased
by 4 feet, the rectangle becomes a square. Find the base and
altitude of the rectangle.
28. If B loans A $ 500, A will then have 3 times as much
money as B has left ; but if A loans B § 200, B will have twice
as much money as A has left. How much money has each ?
29. A sum of money on interest amounted to $ 824 in 9 months
and to $ 840 in 15 months. Find the principal and the rate.
30. If the greater of two numbers be divided by the less,
the quotient is 1, with a remainder of 8 ; if 4 times the less be
divided by the greater, the quotient is 2 with a remainder of
22. What are the numbers ?
170 THE ESSENTIALS OF ALGEBRA.
31. Sixty workmen, consisting of men and boys, did a piece
of work in 5 days and received for it $ 430. The men were
paid $ 1.75 a day, and the boys 80 cents a day. How many
men and how many boys were there ?
32. Find a fraction such that when the numerator is trebled
and the denominator decreased by 4 the value becomes 3, and
when the denominator is trebled and the numerator increased
by 4 the value becomes £.
33. In 10 years I will be 5 times as old as my son was 5
years ago, and 2 years ago I was twice as old as my son will
be 4 years hence. Find my age and that of my son.
34. The lengths of two ropes are as 4 : 5, and when 20 feet
is cut from each rope the remainders are as 3:4. Find the
lengths of the ropes.
35. A river flows 3 miles an hour; a boat going down the
river passes a certain point in 12 seconds and in going up it
takes 18 seconds. Find the speed of the boat in still water
and the length of the boat.
117. System of linear Equations with Three or More
Variables. We have seen that in order to solve linear
equations with two variables, we must have a set of two
independent equations; in like manner, when we have
three variables, we must have a set of three independent
equations ; when four variables, we must have a set of
four equations, etc.
The method of solving. a problem in three variables
will be understood by noting the following solutions :
(1) x-2y + z=l,
l. (2) 3z + */-z = 4,
(3) 2z+?/ + 2 = 12.
By looking at this problem we see that the s's can be
easily eliminated.
SIMULTANEOUS EQUATIONS. 171
04)
4 x - y = 5,
by adding (1)
and (2).
(5)
5 a; + 2 «/ = 16,
by adding (2)
and (3).
(6)
8a:-2# = 10,
by multiplying ■
(4) by 2.
CO
13 a; = 26,
by adding (5)
and (6).
(8)
a? = 2.
(9)
2/ = 3,
by
substituting 2 for
x in (4).
(10)
2=5,
by substituting 2 for a;, 3 for
V in (!)•
The roof
; is (2, 3, 5).
[(1)
\x— 3y + 2a =
:3,
2. j (2)
6a; + 2>y + 3z =
= 7,
1(3) 2a:-6# + 52 = 4.
By adding (1) and (2), we eliminate y, and have
10 a; +52 = 10,
or (4) 2x + z = 2. ■
Multiplying (1) by 2 and subtracting (3), we have
(5) 6 x - z = 2.
Adding (4) and (5),
(6) 8 a; = 4.
(7) x = \.
Substituting x = J in (4),
(8) 1+2=2.
(9) * = 1.
Substituting a; = J and 2 = 1 in (1),
(10) 2-3^+2 = 3.
(11) -3y = -l.
(12) y = f
The root is (l J, 1).
172
THE ESSENTIALS OF ALGEBRA.
EXERCISES.
3.
4.
X + Z = 11,
y + Z=6,
2a + </ = 25.
2a; + 3« = 54,
7?/ + 5z = 106,
3a; + 52/ = 76.
2a — 5 */ + 4 z = 7,
3x — 2y+ z = 5,
5a: + 3j/ — 5z = 2.
+ ^ + - = 2,
2 10
8
= 9,
la; + 2/ + 3z = 0.
1 + ^ + 1 = 9,
x y z
1-1 + 1 = 3,
a; y 2
1 + 1-1 = 1.
a; ?/ z
6.
9.
7.
x y z
x y z
5 - + 1 — 3 - = l.
x y z
^ + ^ = r,
m n
x , z
m I
n I
P , 9
— + - = s,
9 r
': . ?= s .
z'x
X
3~
X
2"
* , .V_£_i
6 4 3 °
(Do not clear of fractions in Exercise 5.)
x + y + z + w — 10,
2» — 2/ + 3« + w = 13,
z+3?/ — 2z — w = — 3,
x — 2y + 3z-2w = -2.
10.
SIMULTANEOUS EQUATIONS. 173
11. Three men have together $ 750 ; £ of A's and £ of C's
is equal to \ of B's; twice A's is $ 150 more than both B's
and C's. How much money has each ?
12. The sum of three numbers taken two and two are 68,
94, and 62, respectively. Find the numbers.
13. There are three numbers such that the first with \ the
second is 68 ; | the first with § the third is 73 ; and the second
with | the third is 90. Find the numbers.
14. A number consists of three digits whose sum is 17. The
hundreds' digit is 3 times the tens' digit. If the order of the
digits be reversed, the number is increased by 297. Find the
number.
15. A and B can do a piece of work in 6 days, B and C in
7^ days, and C and A in 10 days. In how many days can each
do the work separately ?
16. A cistern has three pipes A, B, and C. If A and B fill
while C empties, the cistern will be filled in 60 minutes. If A
and C fill while B empties, the cistern will be filled in 24 minutes.
If B and C fill while A empties, the cistern will be filled in 120
minutes. In what time could each pipe fill it alone ?
17. There are three numbers whose sum is 113; £ the sec-
ond is 2 more than | the third ; f of the first lacks 3 of being
\ the second. Find the numbers.
18. Separate the number 180 into three parts, such that the
second divided by the first equals 2, the third divided by the
second equals 3, and the first divided by the third equals \.
19. A, B, and C have certain sums of money. If A gives
B $100, they will have the same amount; if C gives A $200,
he will have as much left as A and B together then have ; if
B's money were doubled and A's increased by $ 100, they
would then have together as much as C. What sum has each ?
CHAPTER XII.
EVOLUTION.
118. Definitions. Square Root. One of the two equal
factors of a number is called its square root.
25 = 5 x 5, hence 5 is a square root of 25.
Cube Root. One of the three equal factors of a number
is called its cube root.
64 = 4 x 4 x 4, hence 4 is a cube root of 64.
nth Root. One of the n equal factors of a number is
called its nth Root.
a n = a x a x a x «••■ to n factors, hence a is an nth.
root of a".
Evolution. The process of finding roots is called evolu-
tion. It is the inverse of involution.
From the definition we see that the square root of a 4 is
ai = a 2 , the cube root of a 12 is a* = a 4 , and the wth root
m
of a m is an.
The Radical Sign. When the sign V is placed before a
number, a root is to be extracted. The number written over
the radical sign is called the index, and denotes what root
is to be extracted. Thus, Vl6 = 4, V125 = 5, Vl6 =' 2,
V32 = 2, Va" = a ; the indices 2, 3, 4, 5, n, denote, respec-
tively, the 2d, ,3d, 4th, 5th, and nth roots of the number
before which the radical sign (-^) is placed. The square
root sign (\/) is usually written V-
174
EVOLUTION. 175
The Radicand. The number whose root is to he ex-
tracted is called the radicand. Thus, in V25, Vl25, 25
and 125 are radicands.
The root of an expression of two or more terms is
denoted by the radical sign in connection with the vinculum
or parenthesis.
Va 2 + 2 ab + b 2 and V(« 2 + 2ab + b 2 ) each denote the
square root of a 2 + 2 ab + b 2 .
119. Even and Odd Boots. An even root has an even
index, an odd root an odd index.
V3, V8, V5, are even roots. V8, V32, Va 14 , are odd
roots.
An even number of equal positive or negative factors
multiplied together gives a positive product. Hence, only
positive numbers can have real even roots. An even root
of a negative number is called an imaginary number. All
other numbers are real. All numbers are either real or
imaginary.
5, V7, V21, a, b, are real numbers. V— 1, V— 5,
V— 16, V— a 2 , are imaginary numbers.
120. The Law of Signs. (1) An even root of a positive
number is either + or — .
V81=±9, for( + 9)x( + 9) = 81
and (-9)x(-9)=81.
Va 2 = ± a.
(2) A negative number has no real even roots.
(3) An odd root of a negative number is — .
^/327=_3, for (-3)(-3)(-3) = -27. V-S2=-2.
176
THE ESSENTIALS OP ALGEBRA.
EXERCISES.
Extract the indicated roots :
1. V25T 2 =±5a.
2. y/aWc=±a 2 bVc.
3. V16 a'W.
4. V169 afyV.
5. Vvmxhf-
6. ■\/-8a 3 6 s c 9 = -2a6c 3 -
7. Va*-4 a &+4 fc 2 =± (a-2 b).
8. Va 2 + 6aa; + 9a^.
9. v'-125a; l y.
11. V625 aVc 2 .
12. ^/1000 a W.
13. V40Q aVc 18 .
. / l44 a'y
' V fo + y) 2 '
15. ^/-343a 9 6'V.
16
81a;V
'' *256 aW
17. </l6 aW.
18. V- 32 a 25 ^ 5 .
10
19. ^eWc* 1 .
20. -^-aVc 56 .
SQUARE ROOT.
121. The Square Root of a Polynomial. The square root
of a polynomial is found by the reversal of the method
used in squaring a polynomial.
{A + B) 2 = A 2 + 2 AB + B 2 is the general type form of
the square of the sum o'f any two quantities, and is the
type form used in the reversal process.
By comparing any perfect square, whose root consists of
two terms, with this type, its root may be easily determined.
02+102; + 25.
This may be written x 2 + 2 x • 5 + 5 2 .
It is at once seen that A = x and B -■
square root of x 2 + 10 x + 25 is x + 5.
= 5. Hence, the
EVOLUTION. 177
The same method may often be applied to polynomials
whose roots have three terms.
* 2 + y 2 + 9 + 2 xy + 6 x + y.
Arrange this in type form, according to x.
x 2 + 2x(y + S) + y 2 + 6y + 9.
This may be further arranged
z 2 + 2:z<j, + 3) + G/ + 3) 2 ,
and the square root is at once seen to be x + y + 3.
EXERCISES.
Find the square root of the following by comparing with
the type form :
1. ar ! + 16a; + 64.
2. 9 x 2 + 24 x + 16. (Write (3 x) 2 + 2(3 x)l + 16.)
3. a^-lSz + Sl. 5. 16z 2 + 56r!/ + 492/ 2 .
4. a 2 -10 a*/ + 25/. 6. aW + 2 abxy + b 2 y\
7. x 2 + 2xy + y 2 + z 2 + 2zx + 2yz.
8. a? + 9y 2 + z 2 + 6xy + 2zx + 6yz.
9. a 2 -66c + 6 2 + 9c 2 -2a6 + 6ca.
10. 4 a; 2 + y 2 + 9 z 2 — 4 xy — 6 yz + 12 zx.
11. a 2 + 6 2 + c 2 + d 2 + 2a6 + 26c + 2ca + 2ad + 2&d + 2cd
12. x 2 -6xy + 9y 2 + 16 + 8x-24 : y.
122. Formal Extraction of Square Boot. When the root
oan not be easily determined by inspection, we reverse
the type form. A * + 2AB + B* \A + B
A*
2A + B
2AB + B 2
2AB + B 2
178 THE ESSENTIALS OF ALGEBRA.
The first term of the root A is the square root of A 2 .
The second term, B, is contained in 2 AB, and is found
from it by dividing by 2 A. 2 A is called the trial divisor.
(2 A + B) is the complete divisor. "When this is multi-
plied by B, the result is 2 AB + B 2 , which is the part of
the square remaining after A 2 is subtracted.
(1) Find the square root of 36 x 2 — 144 xy + 144 y 2 .
36 x 2 - 144 xy + 144 y 2 \6x-12y
36 x 2
\2x-\2y
- 144 xy + 144 y 2
- 144 xy + 144 y 2
Here the square root of 36 a? is 6 x. 12 x is the trial
divisor, which divided into — 144 xy gives — 12 y as the
next term of the root. The complete divisor is 12 a;— 12 y,
which multiplied by — 12 y gives — 144 xy + 144 y 2 , the
remaining part of the square after 36 # 2 is subtracted.
6x— 12 y is the required square root. This method is
easily extended, as the following example will show :
(2) Find the square root oi x i +6x 3 + x 2 —2ix + 16.
x i + 62?+x 2 -24x+l6 \x 2 + 3x-4:
x*
2x 2 +'Sx
2x 2 + 6x-4:
6 X s + x 2
6 x 3 + 9 x 2
-8^-24 x+ 16
-8^-24^+16
Note that the polynomial is arranged according to
powers of x.
The square root of x* is x 2 . The trial divisor is 2 x 2 ,
which divided into dx 3 gives 3 a;, the next term of the
EVOLUTION. 179
root. The complete divisor is 2a? + 3x. When this is
multiplied by Bx it gives 6x 3 +9z 2 , which leaves, when
subtracted from the remaining part of the square,
— 8 x 2 — 24 x + 16. (x 2 + Bx) is now regarded as the first
part of the root, giving a trial divisor of 2 x 2 + 6 x. This
gives — 4 as the next term of the root. The complete
divisor is 2 x 2 + 6 x — 4. When this is multiplied by — 4
it gives — 8 x 2 — 24 x + 16, which is the part of the square
remaining. The root is x 2 + 3 x — 4.
123. Rule for Extracting the Square Root:
(1) Arrange the terms with respect to the powers of some
letter.
(2) Extract the square root of the first term, place its root
as the first term of the root sought, and take its square from
the given polynomial.
(3) Double the root already found for a trial divisor,
divide the first term of the remainder by the trial divisor,
placing the quotient as the next term of the root, and also
annexing it to the trial divisor to form the complete divisor.
(4) Multiply' the complete divisor by the last term of the
root, and take the product from the first remainder.
(5) Continue this process until the other terms of the root
are found.
EXERCISES.
Extract the square root of the following :
1. x* + 14 x 2 + 4 a? + 20 x + 25.
2. a^-4^ + 10a^-12a; + 9.
3. 4 x* +25 x 2 + 12 ^+240; + 16.
4. a; 6 + 1 + 12 x* + 42 x* + 6 x + 38 a? + 21 x>.
5. 4 x* + 5 x 2 y 2 + y* + 4 a?y + 2 xf.
180
THE ESSENTIALS OF ALGEBRA.
6. 9 x 4 + 37 x>f + 4 y* - 30 x>y - 20 xy 3 .
7. l-2x-6y + x 2 + 6xy + 9f.
8. 4a 2 -12a^/ + 16Kz + 92/ 2 + 16z 2 -242/2.
9. 4 + x 6 + 4z 3 -12a;-6:B 4 + 9ar ! .
10 g + 2£ + 4|p + ^ + *.
124. Inexact Square Boots. The following example will
sufficiently show the method.
Extract the square root of 1 + 3 x to four terms.
l + 3a: |l + |z-| 3? + lJx*
1
2 + f*
2+3z- j
Sx
3x + %a?
_ 9 ~2
— if X
i^-¥^+ii^
2 + 3a;-|a; 2 + f|^
w^+w^-hi* 6
EXERCISES.
Extract the square root of the following to four terms:
1. 1 + x. 5. 1 + 535 + a! 2 .
2. 1+4 a; + 8^.
3. 1 — x.
4. l + a + z 2 .
6. l-4a; + 3ar ! .
7. l-Sa + Sar' + a?.
8. l+s + ar' + ar' + aA
125. Square Root of Arithmetical Number. The following
principles may be easily verified by the student ;
EVOLUTION.
181
(1) The square of a number of one digit consists of one or
two digits.
(2) The square of a number of two digits consists of three
or four digits.
(3) The square of a number of three digits consists of five
or six digits ; and so on.
From these principles we can at once tell how many
digits in the square root of a given number. The square
root of 390625 will consist of three digits.
The number of digits in the root may be indicated by
separating the given number into groups of two figures
each, beginning at the right. The left group may contain
either one or two digits. 39 06 25 and 1 93 21.
Now solve by using the same method as in algebraic
problems.
Extract the square root of 390625.
39 06 25 | 600 + 20 + 5
36 00 00
3 06 25
2 44 00
1200, trial divisor
20
1220, complete divisor
1240, trial divisor
5
1245, complete divisor
It is usual to put the work in the following shorter form :
62 25
62 25
39 06 25
36
625
122
3 06
2 44
1245
62 25
62 25
182 THE ESSENTIALS OF ALGEBRA.
EXERCISES.
Find: 1. Vl 04976. 2. V278784. 3. V57121.
4. V61504. 5. V235225. 6. V1092025.
126. Roots of Decimals. Group both ways from the deci-
mal point, and solve exactly as in whole numbers.
36 74.87 63 64. This shows the root to be made of two
whole number places and three decimal places.
Find:
EXERCISES.
1. V18.3184.
3. V. 133225.
4. V.00300304.
5. V1.110916.
2. V8.6436.
6. V2.
If the result of Exercise 6 is desired to three decimals, we may
write it V2.000000, and then proceed as in the above exercises.
7. VS to three decimals. 8. Vll to three decimals.
127. Cube Root of Polynomials. The type form is
(A + Bf = /»* + 3 A 2 B + 3 AB 2 + B s .
All polynomials in this form may have their cube roots
written by inspection. x 3 + 6 x 2 y + 12 xy 2 + 8 y 3 may be
written, x 3 + 3 x 2 (2 y) + 3 x(2 yj- + (2 y) 3 .
The cube root is readily seen to be x + 2 y.
By exactly reversing the type form we can extract the
cube root of any polynomial which is a perfect cube.
^ 3 + 3 A 2 B + 3 AB 2 + B 3 \A + B
A 3
3A* + 3AB + B*
3 A 2 B + 3 AB 2 + B 3
3 A 2 B + 8 AB 2 + B 3
The term B of the root is found in 3 A 2 B and is obtained
by dividing 3 A 2 B by 3 A 2 . This shows 3 A 2 as the trial
divisor, and 3 A 2 + 3 A B + B 2 as the complete divisor.
EVOLUTION.
183
Extract cube root of
27z 6 -108z 5 + 198z*-
208:c 3 + 132 a; 2 -48 x + 8.
o
M
H
p
O
02
CM
GO
+
+
H
8
■*
CO
|
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«?.
1
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T»
CM
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+
GO
©
CM
1i
GO
OJ
T-H
+
00
o
CM CM
GO
33
1— I
+
•fc
-b
+
CO
o
CO
CO
+
+
s
«
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CO
1
1
1
1
CM
CM
CO
CO
T— 1
T-H
+
+
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CO
s
«
■*
■*
-*
Tfl
1» T»
lO
o
CO
• I— I
03
Ma
§ ^ a
'C 1 ^ o
£ + °
43 ' +
*€ %
t- s to
CM jo CO
II ^ I
« CO ■*
€ + !
CO » ^
CO CM o
:>
r-
'S
|
13
*?,
<B
cS
H
-t=
CD
°£
t—
. — i
+3
CM
&
c?r
11
s
h
in
O
CO
CM
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+
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8
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1
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°u
t-
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CM
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to
II
+ +
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<N
S~\S~\
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8
Tt< "*
CO CO
^ ^<
CO CO
184 THE ESSENTIALS OF ALGEBRA.
128. Rule for Extraction of Cube Root.
(1) Arrange with respect to some letter.
(2) Extract the cube root of the first term for the first
term of the root, and subtract the cube from the polynomial.
(3) Use three times the square of the root found for a
trial divisor, and by dividing the remainder by this divisor
get the second term of the root.
(4) Add to trial divisor three times the product of the first
part of the root and the part of } the root last found and the
square of the root last found.
(5) Subtract the product of the complete divisor and the
part of the root last found from the remainder of the
polynomial.
(6) Repeat this process until the root has been completely
determined.
EXERCISES.
Extract the cube root of :
1. sb 3 + 6 3^ + 12 a; + 8.
2. af i + 3tf + 6x i + 7x s + 6a? + 3x + l.
3. a; 6 - 3 «?y + 6 x*y 2 - 7 x*y* + 6 x>y* - 3 xy 5 + y e .
4. a 3 + 3a 2 (& + l) + 3a(& + l) 2 +& 3 + 3& 2 + 3&+l.
5. « 3 + 3 a 2 b + 6 a 2 + 12 ab + 3 ab 2 + b 3 + 12 a + 12 b + 6 b 2 + 8.
6. 27 X s - 64 J/ 3 -108 a?y + 144 xy 2 .
7. X s - 9 3^ + 42 ib 4 - 117 0?+ 210 ^-225 x + 125.
8. 8 a 6 6 6 - 36 a 5 6 5 + 78 a 4 6 4 - 99 a 3 6 3 + 78 a 2 b 2 - 36 ab + 8.
129. Extraction of Cube Root of Arithmetical If umbers. The
following principles may be verified as in square root:
(1) The cube of a number of one digit consists of one, two,
or three digits.
EVOLUTION. 185
(2) The cube of a number of two digits consists of four,
five, or six digits.
(3) The cube of a number of three digits consists of seven,
eight, or nine digits ; and so on.
From these principles we can at once tell how many
digits in the cube root of a given number.
If the given number is separated into groups of three
figures each, each group will correspond to a digit of the
root.
For example : 95 256 152 263. There are four digits in
the cube root of this number. The group at the left may
contain one, two, or three digits. 3 365 791 and 871 625.
An example will illustrate the method of solution and
show that the same plan is followed as in the extraction
of the cube root of polynomials.
Extract the cube root of 262144.
a 3 + 3 a % + 3 atf + J3 = 262 144 | 60 + 4 = 64
a? = 60 3 = 216 000
(3 a% + Sab+b 2 )b= 46 144
3 a 2 = 3 x 60 2 = 10800
+ 3 ab = 3x60x4= 720
. + IP = 4 2 = 16
(3 o?b + Sab + b 2 )b = 11536 x 4 = 46 144
The above shows the similarity to the general type.
In practice the solution should appear as follows :
262 144 [64
6 3 = 216
Trial divisor 6 2 x 300 = 10800
6 x 4 x 30 = 720
4 2 = 16
Complete divisor 11536
46144
46144
186 THE ESSENTIALS OF ALGEBRA.
EXERCISES.
Extract the cube root of each of the following numbers :
1. 4913. 3. 753571. 5. 2628072..
2. 300763. 4. 614125. 6. 1:728.
In Exercise 6, group both ways from the decimal point. If neces-
sary, annex ciphers to fill the last group. If the root is not exact, by
annexing ciphers the result may be carried to any desired number of
decimal places.
7. 130323.843. 9. .0081 to three decimals.
8. 3 to three decimals. 10. 2.05 to three decimals.
CHAPTER XIII.
THEORY OF INDICES.
130. The Index Laws. The following laws for integral
exponents have already been proved :
1. a m -x a" = a m+n .
2. a m -7- a" = a m ~ n , when m is greater than n.
3. Qa m y = a mn .
4. (ax6) m = a m x b m .
5. f-r =c
m
6. -\/a m = a n , when m is divisible by n.
The laws of algebra should be general. Let us assume
that the above index laws hold true for all values of m
and n, and find consistent meanings for certain special
s.
forms, viz., a , a~ n , at.
Any number affected by an exponent is called the base with
respect to that exponent.
In a n , a is the base. In (x + y) m , x + y is the base.
131. The Form a°. The second index law is
a m -=- a" = a m ~ n .
Let m = n, and this becomes
a" -=- a" = a .
,,-But' a n + a n =l.
Hence, a = a" ■+- a" = 1.
187
188 THE ESSENTIALS OF ALGEBRA.
Any quantity with an exponent zero is equal to unity.
2»=1, for 2 + 2=1; 10° = ±^=1;
t x + y) (x + yy ' \5+bJ
132. The Form ar". The first index law states that
a m x a n = a m+n .
Let us assume that this holds for all values of m and n.
Let m = —n, and we have
ar n x a n = ar" +n = a = 1.
Hence, a - " = — , hy division.
a n
a~ n means the reciprocal of a n -
3" 5 = 4; 10-i = i-; <* + *)- = ■
3 5 10' v * y (as + y)"
oar* = 4; «« _ "2/" m = — ; 3a- = - •
a^ ay a"
A factor may be removed from the numerator to the denomi-
nator, or vice versa, if at the same time the sign of its exponent
be changed.
5 2 x~ s y 2 _ o 2 z 2 y 2 _ 5 2 x 6 4 z 2 a; _1 w -2 _ ,
6"*s-V ~ 6-Vic 4 _ x 2 y~ 2 w 2 -' e c '
p
133. The Form a". We know that (a m ) K = a mn -
p v m
Put m = — and n = q. Then (a m ) =(»)' = «' = a p -
? p
Now extract the <?th root of both sides of (a 9 ) ? = a p , and
p . P
we have ai = ^Ja p ; that is, ai means the qth root of the pth
power of a.
The numerator of a fractional exponent means the power
of the base, and the denominator the root to be extracted.
THEORY OF INDICES. 189
To extract the qth root of a", divide the exponent p by q.
p v
The qth root of a*"is at, or ai =-Va p .
a^ x cfi = a* + * = a 1 .
a? x a 1 = (a^) 2 -
(a i ) 2 =a 1 .
a* _ Va.
a ¥ = V« 3 . a* = Va 5 .
We find that by assuming the generality of the index
laws we have consistent meanings for zero, negative, and
fractional exponents. We will hereafter use the six index
laws of page 187 for all values of m and n.
l.
EXERCISES.
3a~ 3 3 x 2
5 x~' 2 y 5 a s y
x 2 • X s
VrfY.-Z/tf = x* -.y* _y l + i _y\
Vy-^X-s/aP y~i ■ x% x%~% x%
Simplify the following by making all exponents + after
combining like numbers:
A s/aVb 3*af^M
4, _ . y_ -Z .
4 Va V& 3^ x~$y ~*~z^
5» x6^x 10 3 r<o , K , no
5-- —7 Tl 773 a V ax v f5x±l.
5-ix6+JxlO^ 8. — x^^
VoV x -v^&V
VaWy 9 - (o^c*) + (a-*»*c-»).
10. (^^-"c-") • (aVc) • (a*&M)°.
190 THE ESSENTIALS OF ALGEBRA.
Kemove the parentheses and simplify :
11. S[(a 2 ) 2 ]" 2 ]i. 12. ([Kax + by^ff.
13. (x + y-J = x 2 + 2xy-* + y~ 2 = x 2 + ^ + 1-
14. (a?-y-y. 15. {a?-y- 2 f.
It is to be noted that while (c&bty = ab, (a^ + b*) 2 does not
equal a + b. An exponent is distributive to the factors, but not
to the terms of an expression.
(a* + bfy = a + 2 a^ + &.
(a 2 + 6 2 )*=Va 2 + & :! .
Perform the operations indicated in the following ;
16. (x + y 1 )^ - y~').
17. (a; 2 + x + 1 + ar 1 + a;- 2 ) (a; — ar 1 ).
18. (tf-y-^ + fx-y-i).
19. (x 2 - 2 x + aP - 2 x-^x 2 - x + x°).
20. (cr 2 + 2a~ 1 & 1 + 6- 2 )(a- 1 + fe" 1 ).
21. (a VcM) 2 -T- (a* 6* c*a;^) 3 .
22. (a"i& W*) 6 -=- (a' ft'c-'ar 2 )- 1 .
23. (2 a- 3 &- 2 <f * + 3 a-^- 1 ^) -s- (6 a-^-V 1 ).
24. (9a5^$ — 16a*&*)-s-(3aj*yi + 4a&*).
25. (a;" 2 + a;-'?/" 1 + y- 2 )(x~ 2 - arty- 1 + y~ 2 ).
CHAPTER XIV.
RADICALS, SURDS, AND IMAGINARIES.
134. Definitions.
Radicals. A root indicated by means of a radical sign
is called a radical.
As noted in Chapter XII the quantity under the radical
sign is called the radicand.
VT, V 5, Va 3 , V*; 3 + y 3 are radicals, and 7, 5, a 3 , x 3 + y s
are the radicands.
These radicals may also be expressed in equivalent
expressions by means of fractional exponents. Thus,
VT = 7*, V5 = 5 J , .V? = a*, Vx 3 + y 3 = (x 3 + «/ 3 )l
The laws of algebra apply to radicals, since radical signs
may be replaced by exponents. All the laws of exponents
hold for radicals.
Thus, VaJ = Va Vfi,
for Vab = (ab~)n = a n b>> = Va V&.
This law may be extended to any number of factors.
■y/abc = Va V6 Ve.
Commensurable Numbers. A commensurable number
is one whose value may be expressed as a fraction with
integral terms.
Thus, 49 = 4JL = 3g., is a commensurable number.
191
192
THE ESSENTIALS OF ALGEBRA.
Incommensurable Numbers. A number which can not be
expressed as a fraction with integral terms is called an
incommensurable.
Thus, V2 = 1.4142 ••• is incommensurable.
Surds. A surd is an incommensurable root of a com-
mensurable number.
Thus, V2 is a surd, for it is an incommensurable root of
a commensurable number. 2 is commensurable, but V2
is 1.4142-.-, an incommensurable number.
\ 1 +V2 is not a surd in the sense of the above defini-
tion, for 1 + V2 = 2.4142 ••• is itself an incommensurable
number.
Examples of Surds : \ r 3, -y/2, -\/5, V7, Va. The latter
expression, Va, is a surd if a be not a perfect square.
A surd is always a radical, but a radical is not always
a surd.
V5, -Va~\ V2l, Vl6, y/¥f are radicals, but Vl6, Vtff
are not surds.
135. Surds Expressed Graphically. Many surds may be
expressed graphically. In doing this, use is made of the
Pythagorean proposition. In a right triangle the square
of the hypotenuse in equal to the sum of the squares of the
two legs.
If AB = BC=\,
then AC= Vl 2 + l 2 = V2.
RADICALS, SURDS, AND IMAGINARIES. 193
If DE = 2 and EF = 1,
then DF = V2 2 + l a = VS.
If .X" is a cube whose volume is 5,
then KL = V&.
EXERCISES.
Represent graphically
1. VlO. 2. V3. 3. Vl3. 4. VM. 5. VQ.
136. Surd Forms. Radicals whose radicands are alge-
braic numbers are generally considered surds unless the
radicand is the wth power of an algebraic number, n being
the index of the root.
Va + b, V(« + by, Va + b + xy are surds. y
V(a — x) s , Va*^ 8 , Va^ + 2 xy + y 2 are radicals, but not
surds.
These latter expressions are frequently spoken of as
being in the surd form.
137. Irrational Numbers. An expression involving a
surd or surds is an irrational.
5 ■+- V6, 3 -)- V2 — V5 are irrationals.
194 THE ESSENTIALS OF ALGEBRA.
138. Kinds of Surds. The order of a surd is denoted by
the index of the root. V5, -s/i, Vx are surds of second,
third, and nth orders respectively. Surds of the second
order are generally called quadratic surds, those of the
third order cubic surds, etc.
Surds are of the same order when they have the same
root index. •\ / 'l6, -y/x, -y/a + b are surds of the same
order.
A monomial surd consists of a single surd term.
A binomial surd consists of two surd terms, or a surd
and a rational term. V5 + V2, 5 + VlO are binomial
surds.
A trinomial surd consists of three terms, two of which
at least are surds. 3+V2 — ^5 and VB + -v/2 — 4 V5
are tFinomial surds.
A mixed surd consists of a rational factor and a surd
factor. 5VB, 4a/2, ay/x + y are mixed surds.
A surd is in its simplest form when the root index is
the smallest possible and the radicand the simplest possi-
ble integral expression.
V27 = V9x3 = V9x V3 = 3V3 (simplest form).
V36 = V6 (simplest form).
* a * a* V a 2 a
Similar surds are those which, when reduced to their
simplest form, have the same surd factor.
5V4, 8V4 are similar surds.
In this chapter only quadratic surds will be considered.
139. Transformation of Quadratic Surds. A rational
quantity may be put in quadratic surd form by squaring
RADICALS, SURDS, AND IMAGINARIES.
195
it and indicating its square root. Thus, 5 = VW= V25.
This is readily extended to the case of reducing' a mixed
surd to an entire surd.
5V3 = V25x3 = V75; 4 Vz = Vl6x~2 = V32.
A quadratic surd is reduced to its simplest form by
factoring the radicand and removing to the left of the
radical one of each pair of equal factors.
(1) V72 = V2 2 -b 2 -2 = 6V2.
(2) Simplify V75+V243-VT08-V27.
V75 = V52x~3=5V8.
V248 = VWx~S = 9 V3.
VIM = VWxTz = 6 V3.
V27=VJ?x~3 = 3V8.
The expression now is
5 V3 + 9 V3 - 6 V3 - 3 V3 = 5 VS.
EXERCISES.
Simplify the following surd expressions :
1. 2V3-4V3 + 6V3.
2. VI2+V27-V75.
3. V50-V32 + 2V18-5V8.
4. V300+V108-VM3.
5. 3VSy-2V«y + 13V«/.
6. V25 a 3 6 3 - V81 ab + VlHW.
7. V3 (x + y) 2 - 5 V27 (a? + 2xy + y 2 )+ V12 (x + y)(x + y).
8. VaV(> + zf- V9aV(2/ + z)+ 3 V16 (y + «) 3 -
196
THE ESSENTIALS OF ALGEBRA.
^125 + V245 - V320 + V405 - V720.
10. V8ar ! -24^+18a;-V2!« 3 - 12^+18*.
11. V8 m s - 16 m?n + 8 mw 2 - V2 m 3 + 4 m 2 m + 2 mm B .
12. (a;-y)V3+V3^f6^T37-(a! + y)VI08.
13. V4 cc 8 + 4 afy + V4 a;?/ 2 + 4 y 8 .
14. 5V9 a 2 b + 27 a 2 + 7 ay/25 & + 75.
15. V48 xy 2 + yy/75 x + V3 x (x - 9 y) 2 .
16. -VWa 3 - V3 a 3 + 27 a& 2 - 18 a 2 & + V27atf.
17. V5(a-6) 2 -V20a 2 +40a6+20 6 2 +V20a 2 -40a6+206 2 .
18. a/99-V176 + V539+4V275.
19. VB2-
-3V117 + 5V1573.
20. (a + 6)V(a-&) 2 (a; + 2/) + (a-6)V(a; + 2/)(a + 6) 2
(3 a - 4 &)V(3a + 4&) 2 + 2/).
140. Product of Quadratic Surds. Products of surds obey
the following law : r- ,-r ,—r
(1) V27xV32 = 3V3x4V2
= 12V3xV2
= 12V6.
(2) V3(a + 5)xV2(a-J) = V6(a 2 -62).
(3) VTM-5- V48=5V6^-4V3 = 4V| = |V2.
EXERCISES.
Multiply the following surds :
1. V20xV80. 3. 5 V3 x V48 x V75 x V15.
2. V32 X V200 X V50. 4. 3VaKxVT6AxV48«^;
RADICALS, SURDS, AND IMAGINARIES. 197
5. V(x — yf x V(x — yf.
6. V60 a?f x -ViMxy x V36 aty 4 .
7. V3 (a - b) 3 x V2 (a - b) x V6 (a - bf.
8. V5 (a: — yf x V20(a; + y) i! x V3 (x - yf(x> +xy + y 2 ).
Simplify :
V5, aa: 2 x V72 g 2 (x + y ) a x V2 (* — y)
V25 oV X V32a(a; — y)\x + y)
10. (V32 + V48)h-(V2 + V3).
141. Multiplication of Polynomial Surd Expressions. Such
expressions as
(a + Vb + Vc) x (a - V6 + Vc)
are multiplied together in the same way as integral
.expressions, the extended law for multiplication of radi-
cals being observed.
(1)
a + Vb +
Vc
a — V5 +
Ve
a 2 + a V6 +
ay/e
— aVb
-b-Vbc
+
a^Jc
+ Vbc + o
a 2 +2 a^/c —b +c
(2) (l + V3)x(V2-V5) = V2 + V6-V5-VI3.
EXERCISES.
Multiply the following :
1. (1+V3)x(l-V3). 5. (3V2-4)(3V2 + 4).
2. (2-V5)x(2 + V5). 6. (6V3-2)(3V3-1).
3. (V2+V3)(V2-V3). 7. (V2 + V5) 2 .
4. (2V3-1)(2V3 + 1). 8. (V2 + V3-1) 2 .
198 THE ESSENTIALS OF ALGEBRA.
9. (2V3 + l)(V3 + 2).
10. (V5+v3 + 4)(V5-V3-4).
11. (2V3-3V5 + 1)(5V3 + 2V5-1).
12. ftV7-iVB)(iV7 + iV6).
13. (V3 + 1)(V9-V3 + 1).
14. (Va + Vb — Vc)(Va— Vb + Vc).
15. (2VS-3-\4-5Vi)(2VS + 3Vy-5Vz).
16 V5-V7 y 2V5 + 2V7
2V3-4V5 3V3 + 6V5
142. Conjugate Binomial Snrds.
Two quadratic binomial surds differing only in the sign
of a surd term are called conjugate surds.
Va + V6 and Va — V6, or x + Vy and x — Vy,
are conjugate binomial surds.
( Va + V6) x ( Va - Vb) = Va 2 _ VP = a - i.
(# + Vy) x (a; — Vt/) = x 2 — V«/ 2 = x 2 — y.
The product of two conjugate binomial surds is rational.
143. Rationalizing Factors. When the product of two
surd expressions is rational, one expression is said to be the
rationalizing factor of the other.
For example : Va — V6 is the rationalizing factor of
Va+V6, because (Va + V&)(Va — V3) = a — b, a rational
result.
x + s/y is the rationalizing factor of x — V#, because
(a; -_VV) O + Vy) = a; 2 - y.
V5 is the rationalizing factor of V5, because V5 x V5 = 5.
RADICALS, SURDS, AND IMAGINARIES. 199
V5
Suppose the value of —= -= is required. It will
V5— V3
evidently be very much simpler if we first rationalize the
denominator.
V5 V5(v5 + V3) 5+VI5 5+V15
V5-V8 (V5-Vi5)(V5 + V3) 5-3 2
VT-V2 (VT-V2)V2 VI4-2
Again,
V2 ~ V2V2
EXERCISES.
Rationalize the denominators of the following fractions :
! V|. 5 5 9 2±V3.
V2 4-2V3 V3-1
2 1±V2. 6 _^_. 10. 3+Vg -
V3 5-2V6 3-V5
3 V2+V3 ? 3V6 X1 4+2V3
2V2 4-3V7 ' V5-V3
2 „ 1+V2 12 3-4VB
1+V2 1-V2 2V3-V2
13 5 + 8V5 17 l+SVs-l
2VS-V3 l-3Vas-l
^ 4 3 + 2V2 + V3 , 18 V^ + Vy
2V5 + 3V7 (x + y) + ^/2xy
3 + 4 Vo" 19 _ a; - Va;y + y
Va + V6 ^/x — Vy
2 + 3Va + V& , 20 Vffl-2& + V2a-&
2Va + 4V& Va-26- V2a-6
200 THE ESSENTIALS OF ALGEBRA.
144. Rationalizing a Trinomial Surd. A trinomial surd
expression may be rationalized by two operations.
(1) Rationalize Va+Vi+Ve. First multiply by
Va + y/b — Ve. This gives a + b — c + 2 Vab. Now
multiply by a + b — c — 2 Va5. This gives (a + b — c) 2 — 4 ab,
a rational expression. Hence the rationalizing factor of
Va + Vb + Vc is (Va +Vb— Vc) (a + & — c — 2 Vol).
(2) To rationalize v'2 + VB + V5, we multiply by
( V2 + V3 - V5)(2 +3-5-2 V6). This will give a
product 24.
EXERCISES.
Find the rationalizing factors of the following :
1. 1+V2+V3. 5. V10-V2+V3.
2. V3+V5 + V7. 6. Va + Vfc + c.
3. V3+V2 + V5. 7. 1-2V2 + 3V3.
4. V5-V2 + 1. 8. 2Va-V2& + 3Vc.
145. Rational Numbers and Surds.
TheoremI. If Vz and Vy be surds, then V» can not
equal a + y/y, where a is rational.
For, assuming Vx =a + Vy, and squaring, we have
x=a? + y + 2a Vy,
or a: - ffi2 --y = Vy.
2a y
But the left member of this supposed equality is rational,
and therefore can not equal the surd Vy.
Hence, Vx =£ a + Vy.
The sign =£ is read, is not equal.
RADICALS, SURDS, AND IMAGINARIES. 201
Theorem II. If a + Vx = b + V«/, where a and b are
rational, and if ~Vx and ~\/y are surds, then a = b and x = y.
The proof of this theorem is similar to that of Theorem I.
We have 1 ,— ,-
a — b +Vz = ~vy.
Squaring, (a — S) 2 + 2 (a — b} Vx + x = y.
Transposing, — x + y — (a — J) 2 = 2 (a — V) Vx.
Here, a rational number, — x + y — (a — J) 2 , is equal to
2 (a — 5) times a surd. But this can not be true, except in
the case a = b ; but when a = b the original equality says
that x = y.
146. Square Root of a Binomial Surd. The square root of
certain binomial surds may be extracted.
(1) Find the square root of 5 + 2 Vo.
Let V5 + 2 Vti = Vx + Vy.
Then 5 + 2 V6 = x + y + 2 Vxy, by squaring.
x + y = 5, xy = 6, by Theorem II.
The question now is to find two numbers whose sum is
5, and whose product is 6. These are seen to be 2 and 3.
Hence, V5 + 2 V6 = V2 + V3.
(2) Find the square root of 8 — 2 Vl5.
As in (1), let V8 - 2 Vl5 =Vx-Vy~.
Then 8-2 VlB = x + y - 2 Vay.
x + y = 8, a;y = 15,
then a; = 5, ?/ = 3.
Hence, V8-2V15 = V5 - VS.
202
THE ESSENTIALS OF ALGEBRA.
EXERCISES.
Extract the square root of the following :
1. 10 + 2V2I. 7. 12-V80.
2. 12-2V35. 8. 13+2V42.
3. 11 + 2V24.
4. 7- V40 = 7-2ViO.
.5. 16 + 2V55.
6. o + 6-2V«6.
9. 49 + 12V5.
10. y — 2 Vy — 1.
11. 87-12V42.
12. a + b + c+2Vac + bc.
147. Imaginaries. An imaginary number has already
been defined as the even root of a negative number.
We shall have occasion to use only the square root of
negative numbers.
V^5, V^IO, V^16 = 4V^T, V^,
are imaginaries.
It is to be noted that every square root of a negative
number may be expressed as a real number multiplied by
the square root of — 1. Thus,
V^16 = -
V^o? = -
/16 V-1 = 4V
^a? V — 1 =a V-
-1; V=6 = V5V-1;
■1; V— a = Va V— 1.
The V — 1 is usually denoted by i, and is called the
imaginary unit.
148. Some Properties of / = V^l.
(1) ." = V=1.
(2) i» = (V=l)»=-l.
(3) i» = (V^l)» = t(i») = -i.
RADICALS, SURDS, AMD IMAGINARIES. 203
(4)
i* = (fy* = l.
(5)
v> =i(fi~)={.
(6)
t» = i(i6) = ia=_]
(7)
{i = i (i e ) = — i.
(8)
t * = (#)> = !.
(9)
i 9 = i(i 8 ) = i ; etc.
This table shows that the powers of i repeat the values
V — 1, — 1, — V — 1, 1, in cycles of four.
149. Operations with Imaginaries. All the operations
possible with surds are also possible with imaginaries.
The properties of i must be observed.
(1) V-36+V^81+V^100
= V86(- 1) + V81(- 1) + V100(- 1)
= 6 V^~I + 9V^1 + lOV^T
= 6i + 9i+10i
= 25i.
(2) V-48 + V^75+V^243
= Vl6(- 3) + V25(- 3) + V81(- 3)
= 4iV3 + 5zV3 + 9iV3
= 18 i VS.
(3) y^Tg x V^7 = V5 i x VT *
= V5VT«
= V35i 2
= -V35.
204
THE ESSENTIALS OF ALGEBRA.
(4) (V-3+V-2)(V^3-V^2)_
= (i V3 + »V2) (»VB - iV2)
= i 2 ( V3 + V2) (V 3 - V2)
= i 2 (3 - 2)
= - 1(3 - 2)
= -1.
(5) (3 + zV2) 2 =9 + 6iV2 + i 2 -2
= 9 + 6V2i-2
= 1 + 6V2L
,«n 5+V3z
(6) 5~Vl7
To rationalize and make real the denominator, multiply
both terms of the fraction by 5 + V3 i.
(5 + VSi) (5 + V3Q ^ 25 + 10V3i - 3 _ 22 + 10V3 i
(5-V8t)(5+V30 25 -3J 2 25 -(-3)
_ 22 + 10V3z
28
_ 11 + 5V3^
14
EXERCISES.
1. V :r l8+V-128-V :
(3i + V2i)x5V2i = ?
(4-V3i) 2 =?
(V^3 + V2) (V^2+3)=?
5V3
50 = ?
5
7.
■V-
-2V3i
3 + 2V3i
V3-V-2
(Make denominator real and
rational.)
8. V-75-=-V^25 = ?
RADICALS, SURDS, AND IMAGINARIES.
9. (a;V— x + y\/— y) (a;V— x — y^J—y) = ?
a + bi _ 9
10
12.
a — bi
(- 1 -
+ 2i
V-58! 3
2i
2
_?
205
150. Graphical Representation of Imaginaries. Complex
Numbers. We are accustomed to represent real numbers
upon a straight line, the positive numbers in one direction
and the negative numbers in the opposite direction.
4 3 2 10 1
3 4
Let
0JL = 1,
OA' = - 1,
lxi 2 =-l.
B
-A.
Hence, multiplying 1 by i 2 turns it so as to reverse it in
direction. OA is turned about to the position OA' when
it is multiplied by i 2 ; that is, OA is turned through a half
circle. From this we conclude that, when OA is multi-
plied by i, it will be turned half as far ; that is, to the
position OB. If OA is multiplied by i 3 , it will be turned
to the position OB'. It is customary to regard BB' as
the line of imaginaries.
A number like 3 + 2 i is called a complex number. The
type form of such a number is
a + bi.
206
THE ESSENTIALS OF ALGEBRA.
The following diagram shows how complex numbers are
graphically represented.
(,P S 1 -3+S(
A-
-•
(P,)3+8.«
(IJ-3-1
(IS 1-2 1
B
EXERCISES.
Locate on a diagram the following complex numbers :
1- i + i. 4. -3 + i.
2. -3-3t.
3. 2 -5s.
5. 5-fi.
6- i + ii.
CHAPTER XV.
QUADRATIC EQUATIONS IN A SINGLE VARIABLE.
151. Definition. An equation in which the highest power
of the. variable is two is called a quadratic equation.
Thus, 3z 2 + 4a; + 5 = 0, a^-5ar + l = 0, ax 2 + b = 0,
are quadratics.
152. Type. Every quadratic in a single variable may
be reduced to the type ax 2 + bx + c = 0, where a may be
any quantity except 0, b may be any quantity, and c may
be any quantity.
3 a; 2 — 4x + 5 = 0isa special form of the type, in which
a = 3, b = — 4, and c = 5.
EXERCISES.
Eeduce the following equations to the form of the type
ax 2 + bx + c = 0, and indicate the particular values of a, b, and
c in each case :
1. 13a 2 = 7aj-5. . at-ll 3x-2
'72'
2. 6x = %-x>. g-4_ 7a?-3
(Clear of fractions.) 5 - — j= — q
3 2^7 = 2-^. 6 s+i-5-S.
a; a;
207
208 THE ESSENTIALS OF ALGEBRA.
7 . t^ = 2x + 3 -. 9. 3 J*+± = 2x.
x + 1 4 x
8. *±| = . + 8. 10. **±l- 2 -*±*.
k + 2 as + 1 * + 2
153. The Pure Quadratic. If 6 = 0, the general quad-
ratic ax 2 + bx + c = becomes ax 2 + c = 0, an equation
sometimes called a pure quadratic.
The solutions of ax 1 + c = are easily found.
' Transposing <?, arK 2 = — c.
Dividing by a, a?— —
c
a
Extracting the square root, x = ± -v
If a and c have like signs, the roots ±\j are both
* a
imaginary ; if a and c have unlike signs, the roots are both
real. . — -
Thus, 3 x 2 + 5 = has the roots x=± \/—jr- , which are
o
imaginary; 3a; 2 — 5 = has the roots x= ±\/-, which are
real. *>
7%e roots are rea? and rational if a and c have unlike
signs, and - is a perfect square ; the roots are irrational
a
when - is not a perfect square,
a
Thus,
3 x 2 — 5 = has the roots x = ± a/- , which are irrational ;
o
V27
— = ± 3, which are rational.
QUADKATIC EQUATIONS IN A SINGLE VAKIABLE. 209
EXERCISES.
Keduce the following equations to the form of the type:
ax 2 +c = 0.
1 ^,11_q 4 a; — 5 3x
2.
7 3
(Clear o£ fractions.)
«-3 = _3_
5 x + 3'
7.
3 x
ix + 5
3a; + 4_
5 x-
-3
5* + 3
3x-
-4
4 a; — 5
X
_ x , 2 x 2 — 5 A
3. = + -71 = 0. x ix + 5
7 14 a;
cc-3 5-a; o ? , 4_5.
4- = ^ ■ 5 ^ 7, ~~ 7
a; + 5 * + 3 3
a; a;
3 a: 2 - 7 = x 10 3a;-7 = a;
7a; 11 ' "a; 3as'+7'
EXERCISES.
Solve the following equations :
1. Ix 2 -112 = 0.
First Solution.
Transposing, 7 x 2 = 112.
Dividing by 7, x 2 = 16.
Extracting the square root, a: = ± 4, rational roots.
Second Solution.
The general pure quadratic
ax 2 + c =
c
has the solutions i = ±i
' a
Comparing 7 z 2 - 112 = with this general equation,
a = 7,c = -112.
Hence, x= ±^ -(~ 112 ) = ± y"g = ± V 16 = ±4.
210 THE ESSENTIALS OF ALGEBRA.
x-3_ J2_. , aj . + 3« zl 6 = 3
2 ' 6 ~z + 3 + 2 2
x-3 = rf-Gx + 49. JS*,. 3a; + 30
2 ^5 ' 5
5. 3 a; 2 -fa,- 2 = 21. 9. _^ + 5 + ^±i = f + 16a£
4. (as + 3) (x -3) = 2 a; 2 -45.
la; 2 -
5 a;
a + 4 - 3a; ' 10. - ax 2 + 25 & = 0.
154. Solution of the General Quadratic, ax 2 + bx + c = 0.
Dividing through by a, we have
x* + -x + -=0.
a a
Transposing, x 2 + -x =
TO
Adding — - (the square of half the coefficient of x) to
both sides, we have
, , b , b 2 b 2 c
x -\ — x -\ =
a 4 a 2 4 a 2 a
J 2 - 4 ac
4 a 2
This is called completing the square, because it always
makes the first member of the equation a square.
Extracting the square root of both sides,
, b , Vb 2 - 4 ac
2a
Transposing, x =
_ -&±
2a
6 Vfi>
! — 4ae
2a ■
2a
- 6 ± V6 2 -
- 4ac
QUADRATIC EQUATIONS IN A SINGLE VARIABLE. 211
This result shows that the quadratic ax 2 + bx + c — has
two roots ; namely,
-6 + V6 2 -4ac
2a
and
x = -
■ b-Vb 2 -i
ac
2a
This is the solution of the general quadratic and may
be taken as the type solution for all quadratics.
Solve 3 x 2 + 8^ + 5 = 0.
First Solution.
Dividing by 3, x 2 + | x + 1- = 0.
Transposing, x 2 + f x — — J.
Completing the square,
<r2_i_j$.™i l_g._J_fi._l._l
Extracting the square root,
x= ~~5 ± S
= — 1 and — |^.
Verifying the Solution : Put x — — 1 in the equation
3 x 2 + 8 x + 5 = and
3(_l)2 + 8(-l) + 5 = 3-8 + 5 = 0.
Put x = — f and
3(-!) 2 + 8(-f) + 5=¥-¥ + 5-¥-¥ + ¥ = <>-
Replacing the variable of an equation by a root reduces
the equation to an identity. This process is called verify-
ing the solution.
212 THE ESSENTIALS OP ALGEBRA.
Second Solution.
Comparing 3a?+8x +5= with ax 2 + bx + c = 0,
a = 3, 6 = 8, and e = 5.
Let us substitute these values in the type solution, viz.,
and we have
- b ± V6 2 -
-4ac
2ffl
-8+V64-
-60
6
-8±VI
6
-8±2
1 an
6
It will be noticed that the substitution in the type solu-
tion gives results identical with those obtained by carrying
out all the steps of dividing, transposing, completing the
square, extracting the root, and transposing.
EXERCISES.
Solve the following equations by both processes in the order
of the above illustration and verify your results :
1. 2 a? -5 a; = 3. io. 11 a? ^ 39 x + 20.
2. 4a3 2 +7ar=l5. u. 6^ + 7^-20 = 0.
3. 3»a; 2 = 19a;+14. 12 6^-47 a; + 77 = 0.
13. f a 2 + 4a; + 10 = 0.
4. 5 a 2 -12 = 4a;.
5. 6 3^ + 35-15 = 0.
6. a? = |+18.
7. 5 a; + 77 = 12 x\ "' * X ~12 = "
8. 15 sc 2 - 8 = 37 a:. 16. l+£a; = -3a; 2 .
9. 7a^ + 7=50a;. 17. a; 2 + 1 = - 2.9 a;.
QUADRATIC EQUATIONS IN A SINGLE VARIABLE. 213
18.
19.
1 2i 3
x-2 x + 2 5
20. ?-l «
a; ar
21 ± -3
21. 2 + a;- — = 0.
X
5a; + 20 20-5a;
« » x ~ 3 |
x x+S_2
a; — 3 a;
a; + 3 x 3
23. x 2 + 2 bx + c = i
0.
24. ar* — (m + n)x -f
• mti = o.
25. (a — 6)a; 2 — bx =
-a.
26. (a 2 -6> 2 -(a !
' + b*)x + ab = 0.
1 1
a?-3
27.
b — x b + x b' — x 1
28. Va; 2 — 5 x =
29. V^ 2 — 8a; + 5 =
Va^ — 5 x
-10
Va; 2 -8a; + 5
30. V«a^ + ex =
Vaa; 2 + ca;
155. The Double Root.
Solve 2a; 2 -20a; + 50 = Q.
Comparing with the type ax 2 + bx ■+ c = 0, we have
a = 2, 6 = -20, c=50.
20±V400-400
Hence, a; = ■
4
= 20 ± V0
4
= 5±0.
214 TflE ESSENTIALS OF ALGEBRA.
In this case the roots are 5 + and 5 — ; that is, 5 in
each case. Hence, the equation has two equal roots. In
the above example 5 is said to be a double root. The
quantity under the radical, b 2 — 4 ac, is 400 — 400 or 0.
Hence, the condition for a double root, or two coincident
roots > is 12 a n
b l — 4 ac = 0,
where a, b, c are the coefficients in the general quadratic
ax 2 + bx + c = 0.
EXERCISES.
Solve the following equations, noting those which have double
(coincident) roots :
1. z 2 -4a: + 4 = 0. 6. 3?-\x + ^ = 0.
2. 4^ + 403 + 1 = 0. 7. 5a^-4x-l = 0.
3. 3^-23-1 = 0. 8. 25x 2 + 30a; + 9 = 0.
4. 4^-12 0! + 9 = 0. 9. ^ t x i + ^x + l=0.
5. 3a? 2 + 4a; + l=0. 10. -3 a; 2 + 4 x + f = 0.
156. Irrational Roots.
Solve 3 a; 2 -9a: + 2 = 0.
Comparing with the type
ax 2 + bx + c = 0,
we have a = 3, b = — 9, ,c = 2.
Hence, , = 9±V81-24
6
9±V57
QUADRATIC EQUATIONS IN A SINGLE VARIABLE. 215
Here it will be noticed that the roots are irrational.
The quantity under the radical, 6 2 — 4 ac, is 81 — 24 or 56,
and is not a perfect square. The roots are conjugate surds.
EXERCISES.
Solve the following equations :
1. ix 2 + x-l = 0. _ as 2 ix 1 _ n
2. 3a; 2 -7x + 3 = 0. 6 (x-l){x-2) = l.
3. 2z 2 + lla;-7 = 0. * 7 - 8a 2 -21 = 20».
a 2 , 1 3
4. 5x i -Sx-'3 = 0. ' x-2 3sc-l x-3
9 . 20,-^-^ = ^-251.
2 5 2 *
10. 9 mW — w 2 = 6 m 3 n s x.
157. Complex Boots.
Solve 5z 2 -7a; + 3=0.
Comparing with the type
ax 2 + bz + c=0,
we have a = 5, b = — 7, c = 3.
Hence,
7±V49-60
10
7±V-11
10
7±vTI.*
10
r
In this case the roots are imaginary. The quantity
under the radical, 5 2 — 4 ae, is 49 — 60 or — 11, which is
negative and therefore its square root is imaginary. The
roots in this case are conjugate complex numbers.
216 THE ESSENTIALS OE ALGEBKA.
EXERCISES.
Solve the following equations :
1. x 2 -5x + 8 = 0. 5. - 3x 2 + 13 a; = 20.
2. 2a; 2 + 9a; + ll = 0. 6. x 2 + .5x + .3 = 0.
3. 3a,' 2 -10a; + 9 = 0. 7. - .3^+ .8x- .6= 0.
4. 7a 2 -lla; + 8 = 0. 8. f^-f a + £ = 0.
158. The Discriminant. The solution of the general
quadratic ax' 2 + bx + c = is
_ — b ± y/b 2 — iac
*~ 2a
From the examples just given we see that the character
of the roots is determined by the quantity under the radi-
cal. This quantity, 6 2 — 4 ac, is called the discriminant of
the quadratic.
159. Some Conclusions.
(1) When b 2 — 4 ac = a square, the roots are real, rational,
and different.
(2) When b 2 — 4 ac = 0, the roots are real, rational, and
equal.
(3) When b 2 — 4 ac = a positive number not a square, the
roots are real and conjugate surds.
(4) When b 2 — iac = a negative number, the roots are
conjugate, complex numbers.
(1) 4a; 2 - 7 x + 3 = 0.
The discriminant b 2 — 4 ac = 49 — 48 = 1.
Hence, the roots of this equation are real, rational, and
different.
QUADRATIC EQUATIONS IN A SINGLE VARIABLE. 217
(2) 2 z 2 - 4z + 2 = 0.
The discriminant b 2 — 4 ac = 16 — 16 = 0.
Hence, the roots of this equation are real, rational, and
equal.
(3) 5 x* + 8 x- 2 = 0.
The discriminant b 2 — 4 ac = 64 + 40 = 104.
Since 104 is not a square, the roots are conjugate surds.
(4) 7z 2 -5a;+l = 0.
The discriminant b 2 - 4 ac = 25 - 28 = - 3.
Since the square root of — 3 is imaginary, the roots are
conjugate complex numbers.
EXERCISES.
By means of the discriminant tell what kind of roots belong
to each of the following equations :
1. x 2 — 5a; — 9 = 0. 6. az 2 + 2 ax - (a - 4) = 0.
2. 7ar ! + 3a;-l = 0. 7. 11 x 2 - 3x + ^ = 0.
3. 5a; 2 + 9a; + ll = 0. 8. ar ! -3a; + £ = 0.
4. 6a; 2 — 7a; + 4 = 0. 9. 4 a; 2 + 13 a; + 11 = 0.
5. 9^ — 13a; + 4 = 0. 10. aa? + 5x — l = 0.
EXERCISES.
Solve the following quadratics by comparison with the solu-
tions of the type :
1. 3a^ + 5a;-3 = 0. 7. 5 a; 2 - 500 = 0.
2. 3a^-6as + l = 0. 8. -6a; 2 + 8a; = 0.
3. -2a; 2 +8a; + 6 = 0. 9. 9a^-25a; + 50 = 0.
4 . 6a^ + a;-2 = 0. 10. 3a? + 21 x -5 = 0.
5. 24a; 2 + 14a:-5 = 0. 11. 7a 2 - 4a: + 3 = 0.
6. a; 2 -.55 a; -.065 = 0. 12. -Ja: 2 -5a; + f = 0.
218
THE ESSENTIALS OF ALGEBRA.
13.
-%xr -
■ 6x-if- = 0.
14. X 2 -
■| = 0.
15. lla^-|a; + I i r = 0.
16. ^f-5!S + 8 = 0.
The general
160. Graphical Solution of the Quadratic,
quadratic ax* + bx + c =
is equivalent to the two equations
y = x 2 ,
ay + bx + c = 0.
For if in the second of these, we put the value of y from
the first, we get the quadratic ax 2 + bx + c = 0. We know
that y + bx + c = has a straight line for its graph. (Ses
page 153.)
Let us see what the graph of y = x 2 is.
y can not be negative, because a square can not be
negative.
Solving for x, we have
x = ±y/y.
y=Q, x = 0.
y = %
y=3,
y = 4,
etc.
If
x = + 1 and — 1 .
a; =.+ 1.414 and -1.414.
a; = + 1.732 and -1.732.
x = + 2 and — 2.
etc.
(3,9)
Representing these points with refer-
(2J&6J4) ence to the coordinate axes and draw-
(2,4) ing a smooth curve through them,
we have the curve of the adjacent
figure, which is the graph of y = x 2 -
This curve is a ^-parabola and is
the same for every quadratic.
(%2M>
CU) _ x
QUADRATIC EQUATIONS IN A SINGLE VARIABLE. 219
The line ax + by + c = can only be specifically repre-
sented, when we give particular values to a, b, and c.
If the quadratic becomes particular, then the line is
specific, and we can readily draw its graph.
Determine by means of graphs the roots of
a? - x - 2 = 0.
This equation is equivalent to
y = x\
,y-x-2 = 0.
In this a = 1, b = — 1, and c = — 2.
In the equation y — x — 2 = 0,
when y = 0, x = — 2, and when a; = 0, # = 2.
Hence, the x-intercept is — 2 and the ^-intercept is 2.
Now, drawing the graph upon the same diagram that
contains the y-parabola, we get the result
shown in the adjacent figure.
It is seen that the line cuts the y-pa-
rabola at P and Q. The x of P is OM,
which is — 1, and the x of Q is ON, which
is 2.
The roots of the quadratic x 2 — x — 2
= are — 1 and 2.
Hence, we see that the intersections of
the line ax + by + c = and the t/-parabola
y = x 2 have for their abscissas the roots of the quadratic
ax* + bx + o = 0.
161. Graphical Solution of the Pure Quadratic. If the
quadratic is ax* + e = 0, the .two equations to. which it is
equivalent are
rr
+ c=0.
220
THE ESSENTIALS OF ALGEBRA.
Here again we have the same y-parabola. The line ay +
e = is parallel to the z-axis, and so the abscissas of the
two points of intersection will be equal
in value but opposite in sign.
Determine by means of graphs the
roots of 3:c 2_ 12 = o.
The equivalent equations are
f y = * 2 >
[3 y -12=0.
The graphs show the two roots to be
+ 2 and - 2.
162. Graphical Solution in Case of Equal Roots.
Construct graphs showing the roots of
x 2 - 4 a; + 4 = 0.
The equivalent equations are
y = x 2 ,
.y-4x + 4 = 0.
In this case the line just touches the
parabola at P, whose abscissa is 2.
The quadratic x 2 — 4 z + 4 = has equal
roots, each of them being 2.
163. Graphical Representation in Case of Imaginary Roots.
Construct graphs showing that x 2 — 2 x + 5= has imagi-
nary roots.
The equivalent equations are
,, _~2
y-2x+5 = 0.
QUADRATIC EQUATIONS IN A SINGLE VARIABLE. 221
This equation has imaginary roots. The line
y- 2x + 5 =
does not touch the parabola y = a; 2 .
164. Certain Conclusions.
(1) In the case of real different roots,
the line ay + bx + c = cuts the parabola
y = x 2 in two places.
(2) In the case of real 'equal roots, the
line ay + bx + c = touches the parabola
y=x 2 , but does not cut it.
(3) In the case of imaginary roots, the line ay + bx + c =
is entirely outside the parabola y = x % .
EXERCISES.
Using the parabola in the book, or a similar one of your
own construction, solve by means of graphs the following
equations :
1. 2a?+x-l = 0.
Notk. The line in this case is2y + x — 1 = 0.
x = 0, y = \.
y = 0, x=l.
x and y intercepts are 1 and J.
Merely lay a ruler across the parabola so as to
make these intercepts, and note the abscissas of
the points of intersection with the parabola.
2. a! 2 -3a; + 2 = 0. 6. 5o^-6a;-8 = 0.
3. 4a 2 +4a; + l = 0. 7. a^-4a; + 4==0.
4. 5z 2 -8a; + 3 = 0. 8. 2a; 2 -5a; + 3 = 0.
5. 10 tf-3 x -4 = 0. 9. 2a 2 -4a; + 3 = 0,
222 THE ESSENTIALS OF ALGEBRA.
165. Relations among the Roots and Coefficients of a
Quadratic. The roots of ax 2 +bx + c = are
_6 + V6 2 -
- 4ae
■2a
-J-Vft 2 -
- 4 ac
0-
la
Here we use a (alpha) and /3 (beta) to represent the
two roots.
Adding these two roots we have
, a — b+Vb 2 — 4ac , — b— V6 2 — 4ae
« + £= ^ + o
2a 2a
-25 5
2 a a
Multiplying together these two roots, we have
„ = f - b + Vfi 2 - 4ac V - S - Vft 2 - 4 ac\
_ b 2 —(b 2 — 4 at?) _ 4 ac _ c
4 a 2 4 a 2 a
If the coefficient of a? in the general quadratic
ax 2 + bx + c =
be made unity by dividing by a, the equation takes the
form a r
a a
The sum of the roots a + /3 = , and the product of
c ' a •
the roots aft = —
a
Hence, if a quadratic be written so that the coefficient of
x 2 is unity, the coefficient of x is the negative of the sum
of the roots, and the constant term is the product of the roots.
This has already been seen in factoring, page 86.
QUADRATIC EQUATIONS IN A SINGLE VAttlABLE. 223
The roots of the equation 25 x 2 — 15 x + 2 = are -| and £.
When this equation is divided by 25, it becomes
The sum of the roots is £ + \ = f , and their product is ^.
This sum and product are, respectively, the negative coeffi-
cient of x and the constant term.
166. Formation of Equations with Given Roots. If the
roots of an equation are f and 1, what is the equation ?
Here a =
and /3 = 1.
^ H 3 8 a
«/3=(f)l = f = '-
Hence, the equation is
a^-fz-f f=0,
or 3^-5^ + 2 = 0.
In general, z 2 — (« + /3> + «/3 = is the quadratic
equation whose roots are a and /3.
EXERCISES.
Make equations which have the following roots :
6. 3-iV2, 3 + «V2.
7. 3-AV5, 3 + ^VS".
8. a + hi, a — 6i.
9. 31 + 5 Vmi, 3 Z — 5Vmi.
10. V5 + fV3, VS-fVs.
1.
q 2 . K A . 2
2.
1+V5, 1-V5.
3.
3 + 5i, 3 — 5i.
4.
2-V7 2 + Vf
3 ' 3
S.
5, -V-
224 THE ESSENTIALS OF ALGEBRA.
167. Generalized Quadratic. aX 2 + bX+c=0.
The above is a quadratic in X. Its roots are
X= ~ 6 ± ^ - 4 ag .
2a
X may be any algebraic expression. Whenever an
equation may be arranged like the above type, it is said
to be of the quadratic form.
(1) 3z*+ 7^ + 4=0.
Put s? = X, and the equation becomes
3Z 2 +7Z+4 = 0,
. , v -7±V49-48
whence X= ■
6
= — 1 and — |>
But
X=x 2 ,
hence,
x 2 = — 1 and — £,
and
x = ± i and ± ^ V3 i.
The roots of
3x* + 7z 2 +4 =
are
±i, ±|V3z.
(2) O 2 -5) 2 -70z 2 -5)+12 = 0.
Put x 2 - 5 =
X, and the equation becomes
X 2 -7X+12 = 0.
(X-4)(X-8) = 0.
X= 4 and 3.
Hence,
x 2 — 5 = 4 and 3.
x 2 = 9 and 8.
■
x=±2 and ±2V2.
QUADRATIC EQUATIONS IN A SINGLE VARIABLE. 225
(3) 2a? + 5^/x 2 -5x+'6 = 10x + Q.
Transposing, 2 x 2 — 10 x + 5 V:e 2 — 5x + '6 — 6 = 0.
By adding 6 and subtracting 6, we have
2 z 2 - 1 a; + 6 + 5 Vz 2 -5z + 3-6-6 = 0,
or, 2(a; 2 -5a: + 3) + 5Va; 2 -5a; + 3-12=.0.
Put Vx 2 — 5 a; 4- 3 = X, and the equation becomes
2X 2 + 5X-12 = 0, or (2J-.8)(Z + 4) = 0.
X=-4and*.
Since -X"= Va£ — 5x + 3, we have
Va; 2 — 5a; + 3 = — 4 and |-.
a; 2 -5a: + 3 = 16 and f.
ai 2 -5a;-13 = 0anda; 2 -5a ; + | = 0.
a;:
5±V25 + 52
2
5±V77
2
a;:
•p.
5±V25-3
2
5±VTT and
5±V22
2
5±V22
and
The roots are
EXERCISES.
Solve the following equations :
1. (» 2 + 3a;) + 3Va; 2 + 3a;-4 = 0.
2. (a?-2xy-5(a*-2x) + 6 = 0.
3. a; + 3o^ + 2 = 0.
4. V5o 2 + 4a; + 3(5a; 2 + 4a;) = 24.
5. (3a; + 5)+4V3a; + 5 + 7 = 0.
226 THE ESSENTIALS OF ALGEBRA.
7. 3(2a; + 4)-4V2a; + 4 + l = 0.
8. ^ + 7^-18 = 0.
9. : b 4 + 4^ + 4 + 5(^ + 2)-6 = 0.
10. £^-5^ = 36.
11. a; 4 -9 a: 2 = 400.
12. 2x — 8+V2a; — 5 = — 1.
13. a;— 5— V2x — 11 = 8.
14. a; 2 -2a; + 2Var ! -2a;-5 = l.
15. aj 2 + a; + V^ + a; + 5 = 25.
16. x 2 + 5a; + 4 = 5Va^ + 5a; + 28.
17. x> + 5x — Vx? + 5x + 14: = 42.
18 3 Va + 7 ^ Vx+5
■Vx + 3 Va;
19. Va?-3^'a? = 40.
20. 2a^ — 3a; + 6V2a^-3a; + 2 = 14.
EXERCISES.
1. One half a number plus the square of the number is 150.
Find the number.
2. The sum of two numbers is 15 and their product is 56.
Find the numbers.
3. Find two numbers whose sum is 30 and whose product
is 216.
4. Separate 41 into two parts such that the product of the
part is 330.
QUADRATIC EQUATIONS IN A SINGLE VARIABLE. 227
5. Two numbers differ by 3, and the sum of their squares
is 225. Find the numbers.
6. The sum of the squares of three consecutive numbers is
434. Find the numbers.
7. A rectangular lot is 32 feet longer than it is wide. It
contains 13200 square feet. What are the dimensions ?
8. A man buys a certain number of chickens for $ 6. If
he had paid 10 cents apiece more for each, he would have
gotten 5 fewer for his money. How many chickens did he
buy?
9. Find a number such that if its square be diminished by
1, | of the remainder is 18 more than 10 times the number.
10. There are two numbers whose difference is 8. If 540
is divided by each of these numbers, the difference of the
quotients is 18. Find the numbers.
11. One side of a rectangle is 7 feet longer than the other
and its diagonal is 13 feet. Find the area.
12. The difference of the reciprocals of two consecutive
numbers is t£w Find the numbers.
13. The difference between the reciprocals, two consecutive
odd numbers, is ¥ |^. Find the numbers.
14. By increasing his speed 1 mile an hour a man finds that
he takes 3 hours less than usual to walk 60 miles. What is
his ordinary rate ?
15. The larger of two pipes will fill a cistern in 6 minutes
less time than the smaller. When both pipes are open, the
cistern is filled in 13^ minutes. Find the time required by
each pipe to fill the cistern.
16. A and B have a distance of 150 miles to travel. B
starts 10 hours before A and arrives 10 hours after A. A
travels 2 miles an hour faster than B. What is the rate of
each per hour ?
228 THE ESSENTIALS OF ALGEBRA.
17. A rectangular field is 4 times as long as it is wide. If
the width is increased 20 rods, its area is doubled. Find the
area of the field.
18. What number increased by 4 and squared is equal \ of
itself increased by 10 and squared ?
19. A man sold a horse for $.144, thereby gaining as many
per cent as the horse cost him dollars. What was the cost of
the horse ?
20. A boat's crew can row 9 miles down a river and back
in 4 hours. The rate of rowing in still water is double the
rate of the current. Find the rate of rowing and the rate of
the current.
21. The hypotenuse of a right-angled triangle is 5 feet
longer than the base and 10 feet longer than the perpendicular.
Find the sides of the triangle.
22. The sum of two numbers is a and their product is 6.
What are the numbers ?
23. The perimeter of a rectangular field is 168 rds. and
its area is 9 A. Find the length of the sides.
24. Two men start at the same time from the vertex of a
right angle and walk along its sides at the rate of 3 and 4 miles
per hour, respectively. In how many hours are they 50 miles
apart ?
CHAPTER XVI.
SIMULTANEOUS EQUATIONS INVOLVING QUADRATICS.
168. When simultaneous equations involve quadratics,
they must be solved by methods which depend upon the
form of the equations. The various methods are shown
under the following cases :
Case I.
169. A Linear and a Quadratic. A simultaneous set of
this kind may always be solved. The equations are of the
forms,
(1) ax 2 + bxy + cy 2 — d,
(2) lx+my=k.
(3) x = ! ^=f%, from (2).
(4) a ^^)\h(^fy)y + cf = d,
by substituting in (1).
(5) (am 2 - Mm + cP)y 2 + (bile - 2 ahm)y + ah 2 - dP = 0,
by rearranging (4).
Equation (5) is a quadratic in y, and' therefore has two
roots. The substitution of these roots in (3) will give
two values of x. Hence, the set of equations has two
roots, and only two.
229
230 THE ESSENTIALS OF AtCEBftA. y
In the above we have used general equations and we
have found that the solution depends upon a quadratic in
one variable. Such a quadratic can always be solved.
Hence the simultaneous set can always be solved.
I. Solve the following equations :
(1) x 2 + y 2 = 25,
(2) 7 y-x = 25.
(3) x=7y-25, from (2).
(4) 49 y 2 - 350 y + 625 + y 2 = 25, by substituting in (1) .
(5) 50# 2 -350«/ + 600 = 0.
(6) y 2 - ly + 12 = 0, by dividing (5) by 50.
(7) (y - 3) (y - 4) = 0, by factoring (6).
(8) y = 3 and 4.
(9) x = 7 x 3 - 25 and 7 x 4 - 25,
by substituting in (3).
(10) x = - 4 and 3.
The roots are (- 4, 3) and (3, 4).
Care should be given to the proper association of the
values of x and y. It should be remembered that a root
is a properly associated value of x and of y.
The graphs of x 2 + y 2 = 25 and 7 y — x = 25.
From x 2 + y 2 = 25 we have
y = ±V25-a; 2 -
When £= % y = +5 and —5;
(0, 5), (0, —5) are roots.
When a;=+land -1, y = 2V6and -2V6;
(1, 2V6), (1, -2V6), (-1,2 V6), (-1, -2V6) are roots.
SIMULTANEOUS EQUATIONS INVOLVING QUADRATICS. 231
When a: =2 and -2, y = V21 and -V21;
(2, V21), (2, - V21), (-2, V21), (-2, - V21) are roots.
When x = 3 and — 3, y = 4 and — 4 ;
(3, 4), (3, -4), (-3, 4), (-3, -4) are roots. ^
When x — 4 and — 4, y = 3 and — 3 ;
(4, 3), (4, -3), (-4, 3), (-4, -3) are roots.
When x = 5 and — 5, y = and ;
(5, 0) and ( — 5, 0) are roots.
Locating these roots and drawing a curve through them,
we find the graph to be a circle.
The graph of 7 y — x = 25 is a
straight line. It is the line PQ.
This line cuts the circle in the two
points P and Q. The coordinates of
the two points P and Q, where the
line cuts the circle, are ( — 4, 3) and
(3, 4). These are the two roots of
the given equations. That they should be the roots
appears from the fact that they are the only two points
whose coordinates are the same for the line and the circle.
II. Solve the following equations :
(1) x 2 + y 2 = 25.
(2) Sx + 4y = 25.
25-
Qg^
(3)
x= ■
■■JL
from (2).
(4) 625-200y + 16y' + f= ^ by substitufcing in (1>
y
(5). 625-200^+16 y 2 + 9 y 2 = 225.
(6) 25 y 2 - 200 y + 400 = 0. . >
232
THE ESSENTIALS OF ALGEBRA.
CO
y 2 - 8 y + 16 = 0.
(8)
(y-4)(y-4) = o.
(9)
y = 4 and 4.
(10)
25 -4x4 , 25 -4x4
x = and x = 3
(11)
x = 3 and 3.
The roots are (3, 4) and (3, 4).
These roots are the same. Equations (1) and (2) are
said to have a double, root.
The graphs of x 2 + y 2 = 25 and 3 x + 4 y = 25.
a^ + y 2 =25is the same circle as in (1) . 3 x + 4 y = 25 is
a straight line. The graphs are
shown in the adjacent figure.
In this case the line PQ, which
is the graph of 3 x + 4 y = 25,
^-P does not cut the circle, but just
touches it at the point Q. The
coordinates of the point Q are
(3, 4). (3, 4) is one of the two
equal roots of the given equa-
tions. In the case of equal roots the graph of the linear
equation just touches the graph of the quadratic equation.
III. Solve the following equations :
(1) z 2 + j ! = 25.
(2) x + y = 10.
(3) s = 10-y.
(4) 100-20 y + y»+y«= s 25.
SIMULTANEOUS EQUATIONS INVOLVING QUADRATICS. 233
(5)
(6)
2?/ 2 -20y + 75 = 0.
20 ± V400 - 600
y
20 ±10-
10±5V^2 10±5V2i
CO
as = 10-
10T.5V2t
The roots are
'10-5V2t 10+5 V2 1
10 + 5V2i 10-5V2t
5V2t\
2— /
, and .
2_ 2 7 V '2
These roots are both imaginary.
The graphs of a; 2 + y 2 = 25
and z + y = 10 are shown in
the adjacent figure.
In this case the line P Q, which
is the graph of a; + «/ = 10, neither
cuts nor touches the circle.
In the case of imaginary roots
the graph of the linear equation
neither cuts nor touches the
graph of the quadratic.
170. Graph of the Quadratic in x and /. From the pre-
ceding discussion it must not be inferred that the graph
234 THE ESSENTIALS OP ALGEBRA.
of the quadratic equation in x and y is always a circle.
It may be any one of the following curves :
Circle Ellipse Parabola Hyperbola
Graph of 4 3? + 9 y 2 = 36.
t? 4.1 ■ - /-3b' — 4 x 2
B rom this y =-y
9
When x= 0, y = 2 and - 2 ; (0, 2), (0, -2) are roots.
When a; = 1 and - 1, y = $ V2 and - |V2 ;
(1, |V2), (1, -|V2), (-1, |V2), (-1, -|V_2) are roots.
" When x = 2 and- 2, */ = |V5 and - f Voj
(2, |V5), (2, -|V5), (-2, |V5), (-2, -J V5) are roots.
When a; =3 and -3, y=Q; (3, 0), (-3, 0) are roots.
Locating these points and drawing a curve through
them, we have an ellipse like the following.
SIMULTANEOUS EQUATIONS INVOLVING QUADRATICS. 235
2.
4.
5.
6.
10.
11. {
\3x-2y = 8,
[ ixy = 32.
x + 2y = 9,
.rf-y 2 = 21.
(x 2 + 3y 2 = 27,
EXERCISES.
Solve the following sets of equations :
' x + y = 5,
x 2 + y 2 = 25.
ix' + Qy^^Se,
2 x + 3 y = 6.
V + T/^lOO,
I a; - y = 2.
x + y = l5,
xy = 56.
'x~y = 6,
xy == 27.
la? + f = 106.
3x + y = 9,
xy=6. [x + 2y = 7,
f 2 a; + 5 =5,
8. i
I 5 a^ — xy = 2.
Construct graphs for Exercises 1, 2, 3, 5, 13.
12.
13.
14.
15.
I x + y = 10.
a^ + i/ 2 =26,
a: + 5 2/ = 26.
' 3 a; — y = 12,
,x 2 — y 2 = 16.
' 3 a; + 5 = 4,
■ »W + f = 9-
x y
, Case II.
171. Both Equations Quadratic of the Form ax 2 + by 2 = c.
When the equations are of this form, one of the variables
may be eliminated as in simultaneous equations of the first
degree, and the resulting equation is a pure quadratic in
the other variable.
236
THE ESSENTIALS OE ALGEBRA.
I. Solve
f(l) z? + f = 16,
1(2) 4z 2 + 25y 2 = 100.
Multiplying (1) by 4 and subtracting from (2),
(3)
(4)
(5)
(6)
CD
(8)
21y 2 =36.
21
=v-
*/ 2 = ±V^-=±2V|.
a^ + ( ± V^-) 2 = 16, substituting in (1) .
a? = 16 - \ 2 - = 10i.
t 10
"K-W(^ 2 4("f- 2 4
( = , — 2-v- ). There are, as in all solutions under this
case, four roots.
Graphs of a?
+ f = 16 and 4 z 2 + 25 z/ 2 = 100. The
graph of the first equa-
tion is a circle and of the
second an ellipse. The}'
are shown in the figure.
The graphs of the two
equations intersect in
the four points P, Q, B,
and S. The coordinates
of these four points are
the four roots of the set
of equations.
SIMULTANEOUS EQUATIONS INVOLVING QUADRATICS. 237
II. Solve
(1) a 2 + f = 25,
(2) 4z 2 + 25«/ 2 = :100.
Eliminating a; 2 as in Example I,
(3) 21</ 2 = 0.
(4) f = 0.
(5) y = and 0.
(6) ^ + 2 = 25.
(7) z=±5.
The roots are (5, 0), (5, 0), (-5, 0), (-5, 0).
set of equations has two pairs of double roots.
The
Graphs of a? + y*=25 and 4a? + 25# 2 = 100. The
graphs are a circle and ellipse. They are shown in the
adjacent figure.
The graphs of the two equations do
not intersect, but they touch each
other at the points Q and P. The
coordinates of these points are the
roots of the equations. As in Case I,
when the graphs touch, the coordi-
nates of the points where they touch
are double roots.
III. Solve
(1) a? + f=l,
1(2) 4z 2 + 25# 2 = 100.
Eliminating x? as in Example I,
(3) 21r/ 2 =96.
(4) f=n-
(5) «/=±V|f=±4Vf.
238
(6)
THE ESSENTIALS OF ALGEBRA.
^+(±V|f) 2 =l.
x — x 21 — 21'
(7) x=±5V}i.
The roots are (5V|i, 4V|), (5Vji, -4Vf),
(-5Vi*,4Vf), (-5V£t\ -lV|).
These roots are all imaginary.
Graphs of a: 2 + y 2 = 1 and 4 2 2 + 25 ?/ 2 = 100 are a circle
and an ellipse, respec-
tively. They are shown
in the figure.
The graphs of the two
equations neither inter-
sect nor touch. The
circle is entirely within the ellipse. As in Case I, the
roots are imaginary.
EXERCISES.
Solve the following sets of equations : —
-0
x 2 + y 2 = 13,
I2x> + y* = 17.
2x 2 -y 2 = 2,
ar 2 + 2 2 / 2 = 41.
(x 2 + y 2 = 25,
1 3 af + ltf = 180.
(6x 2 -y 2 = 5 >
\x 2 + 2y 2 = 107.
5.
6.
7.
| x 2 + 3y 2 = 52,
[2x 2 + 5y 2 = 19.
x 2 + y 2 = M,
x?-y 2 = 16.
x 2 -9y 2 = 0,
I 3a; 2 + 4^ = 12.
4^9
9"
t
4
= 1,
:1.
Construct graphs for Exercises 1, 6, 8.
SIMULTANEOUS EQUATIONS INVOLVING QUADRATICS. 239
Case III.
172. Both Equations Homogeneous in the Part involving the
Variables. The equations are of the form
ax 2 + bxy + cy 2 = k.
The first step of the solution is the elimination of the
constant terms of the two equations.
I. Solve '
(1) 2x 2 -2>xy + y 2 = 2,
(2) 2^-3^=3.
Multiply (1) by 3, (2) by 2, and subtract.
(3)
2z 2 -9a;y + 9«/ 2 = 0.
(4)
(2a;-3«/)(*-3«/)=0.
(5)
2x=3y.
(6)
x=%y.
. (7)
x = 3 y.
Substitutir
tg x = f y in (2),
(8)
2 (¥) 2 - 3 ^ =3 -
CO
il!_3* 2 = 3.
2 "
(10)
^=3.
2
(11)
(12)
(13)
y 2 = 2.
y = ±V2.
z = f(±2)=±|V2.
Substitutii
lg a; = 3 y in (2),
18 # 2 - 3 # 2 = 3.
15 y 2 = 3.
*/ 2 = f
y = ±iV5.
z = 3(±|V5)=±fV5.
II. Solve
240 THE ESSENTIALS OF ALGEBRA.
The roots are
(fV2, V2), (-|V2, - V2), (|V6,*V5), (-|V6, -|VB").
In this case there will always be four roots. There
may be one or two pairs of double roots. Two or /owr of
the roots may be imaginary.
In this case the equations may be of the form
ax 2 + bxy + cy 2 = dx or ey.
They are solved precisely as the above.
(1) 2s*-3ay + a =2y,
(2) 2z 2 -3y 2 = 3y.
Eliminate the right-hand members by multiplying (1)
by 3, (2) by 2, and subtracting,
(3) 2 x 2 - 9 xy + 9 y 2 = 0.
(4) (2*-3yX*-3y) = 0.
(5) a; = | y and 3 y.
Substituting x= § y in (2),
(6) 2(fy) 2 -3y 2 =3y.
(7) ^=3y, y«-2y = 0, y(y-2)=0.
(8) y=0and2.
(9) x = and 3.
Substituting x = 3 y in (2),
(10) 9y 2 -3y 2 =3y.
(11) 6y2=3y.
(12) y(2y-l) = 0.
(13) y=0andl
(14) «=0and§.
SIMULTANEOUS EQUATIONS INVOLVING QUADRATICS. 241
The roots are (0, 0), (0, 0)), (3, 2), (§, l). (0, 0) is
a double root.
In each of these examples, after finding the value of x in
terms of y, the substitution might have been made in the
first equation instead of the second. The second was
selected because it was of simpler form than the first.
EXERCISES.
Solve the following sets of equations :
2j?-3xy + 2y 2 = 4,
{a? + y 2 = 5.
(x 2 -
[2x>
x 2 + 2xy-y 2 = 7,
3 xy -
= -1.
x 2 +xy = 21 )
2xy-y 2 = 8.
fa; 2 -
tar 2
-2/ 2 = 3,
T 2xy + 2y 2 = 2.
x? — xy + y 2 = 21,
2 xy — y 2 = 15.
■x 2 + 2 xy + 2y 2 = 17,
. 3 x 2 - 9 xy-y 2 = 119.
9.
10.
11.
12.
f 2 x 2 — xy + y 2 = 16,
{x 2 + xy + 2y 2 = 4A.
x 2 + xy = 40,
y 2 + xy = 60.
J 3/ -5 ^ = 70,
[ f - 3 xy = 10.
te 2 + my 2 = m,
aa; 2 + by 2 = c.
£ x 2 + 4 y 2 = 13 — A xy,
8x 2 -12xy=ll-8y 2 .
(a; + 2/) 2 = 73-ar , .
Case IV.
173. When Both Equations are Symmetrical in x and /.
Equations are symmetrical in x and y when the interchange
of these Utters does not change the equations.
242 THE ESSENTIALS OF ALGEBRA.
Examples : (1) a?- 3 xy + y 3 = 27. Interchange x and y,
and we have y s -Zyx + x 3 = 27, which is the same as (1).
(2) x 2 -2xy + y 2 = lQ,
(3) as* + y*=12,
(4) sy = 15,
and (5) a; + «/ = 6,
are all symmetrical equations.
x z _|_ y2._ 25,
xy = 12.
2 a;?/ = 24 to (1), and we have
v 2 + 2xy + y 2 = 49,
x + y = ± 7, from (3).
2 £«/ = 24 from (1), and we have
2xy + y 2 =l,
x — y = ± 1.
From (4) and (6) we have, by adding and subtracting,
x = 4, 3, - 4, - 3,
# = 3,4, -3, -4.
The roots are (4, 3), (3, 4), (-4, -3), (-3, -4).
When both equations are general quadratics, the solu-
tion depends upon a cubic or quartic. The investigation
of such equations is beyond the compass of this book.
I. Solve
1(2)
Add
(3)
X 2 -
(4)
Subtract
(5)
X 2 -
(6)
II. Solve
(1) x 2 + 2xy-y 2 =l,
I (2) x 2 - 2 y 2 + y =2.
Solving (2) for x,
(3) x = ± V2+2y 2 -y.
SIMULTANEOUS EQUATIONS INVOLVING QUADRATICS. 243
Substituting in (1),
2.
(4) 2 + 2^-y±2yV2 + 2y"-y-y
(5) , ±2y^/2 + 2y*-y = 5-y 2 + y.
Squaring,
(6) 8y 2 +8z/ 4 -4y 3 =25+y 1 -9«/ 2 -2«/ 3 + 10y.
(7) 7y 4 -2*/ 3 + 17y 2 -10y = 25.
This equation is a quartic in y, and unless it breaks up
into factors of degree not higjier than two it can not be
solved by our present methods.
Graphs of x 2 + y 2 = 25 and xy = 12.
The graph of x 2 + y 2 = 25 is a circle.
xy = \2.
12
y = —
x
When x == 0, y = oo ; (0, oo) is a root.
When x = + 1 and — 1, y = 12 and — 12 ;
(1, 12), (-1, -12) are roots.
When x = + 2 and — 2, y = 6 and — 6 ;
(2, 6), (-2, - 6) are roots.
When x = + 3 and — 3, y = 4 and — 4 ;
(3, 4), (-3, - 4) are roots. ^
When x = 4 and — 4, y = 3 and — 3 ;
(4, 3), (-4, - 3) are foots.
When a; = 5 and -5, y=2f and -2f;
(5, 2f), (-5, - 2|) are roots.
When a; = 6 and - 6, y = 2 and - 2 ;
(6, 2), (-6, - 2) are roots.
244
THE ESSENTIALS OF ALGEBRA.
When x = 7 and -
(7, L&), (-7,
When x = 8 and ■
(8, 1J), (- 8, ■
When a; = 9 and ■
(Mi), (-9,-
When a; = 12 and
(12,1), (-12,
When x = 24 and
(24,1), (-24,
— 7, y = If and - If ;
— 1 j^) are roots.
-8,^ = 1$ and -11;
— 1J-) are roots.
— 9, 2/ = 1| and -1J;
— 11) are roots.
— 12, y = 1 and — 1 ;
— 1) are roots.
— 24, «/ = \ and — ^ ;
— J) are roots.
The graph of a;^ = 12 is the above hyperbola intersect-
ing the graph of x 2 + «/ 2 = 5 in the points P, #, i2, and
S, whose coordinates are the four roots of the equations.
In solving this example we get the equations x+y= ±7
and x — y = ± 1. The graph of x + y = 7 is the line P§,
and that of x+y=—l is the line RS. The graph of
x — y = \ is the line #<S, and that of x — y = — 1 is the
line Pi?. These four lines intersect in the four points
SIMULTANEOUS EQUATIONS INVOLVING QUADRATICS. 2-15
P, Q, It, and S. The four lines have precisely the same
intersections as the circle and hyperbola. This is why the
set jz + y=±7,
\x-y = ±\,
is equivalent to the set
"z 2 + y 2 =25,
I ay =12,
as was used in the solution.
EXERCIJSES.
Solve the following sets of equations :
ra* + jf = 13»
M
3.
4. •(
a;?/ = 6.
as» + »* = 34,
xy = 15.
a; + y — 11,
*?/ = 24.
f as + 1/ = 6,
!CT/:
5 _^-2/ = 5,
xy = 14.
* + 2/ = 8,
a; 2 + 2/ 2 = 34.
x-y = l,
x 2 + y 2 = 13.
6.
a
4^5 '
* + *=*
16 25
9.
M = 7,
x y
ar y'
10.
a; j/
1-1 = 7.
9 y
. k 2/
11.
- + - = H'
a; y o
11 13
x 2 y 2 36
t3x + 5y = 2xy,
' \ xy = 15.
13.
x 2 + y 1 = 500,
,x +y =30.
14.
[xy = \(a*-b*).
Construct graphs for Exer-
cises 4
and 7.
246 THE ESSENTIALS OF ALGEBRA.
174. Special Methods; Higher Degrees. Simultaneous
equations of higher degree than the second can frequently
be solved by special methods. This is particularly true
when they are symmetrical.
A few of the special methods will be illustrated. In
such problems the student is expected to devise his own
methods.
I. Solve fCD *•-*•= 211,
1(2) x-y=l.
(1) -h (2) = (3) x* + x s y + x 2 y 2 + xy 3 + f= 211.
(2)*= (4) z* - 4 x*y + 6 x 2 y 2 -ixyZ + y*^ 1.
(3) - (4), (5) 5 afy - 5 x 2 y 2 + 5xy s = 210.
(6) a?y — x 2 y 2 + xy z = 42.
(T) xy(x 2 + y 2 )-x 2 y 2 =42.
From (2), x 2 -2xy + y 2 =l.
x 2 + y 2 = 2 xy + 1.
Substituting in (7),
xy(2 xy + i~)- x 2 y 2 = 42.
xhj 2 + xy - 42 = 0.
(asy + 7)(sy-6) = 0.
xy = and — 7.
x = - and
y y
Substituting in (2),
6 1 j -7
--y=l, and -y-V=U
S-y 2 = y, -7-y 2 = y,
y 2 + y-6 = 0, ^ + + 7 = 0,
(y-2)(y + 3) = 0, -1±V^2 7
# = 2 and - 3 ; 2
SIMULTANEOUS EQUATIONS INVOLVING QUADRATICS. 247
The corresponding values of x are 3, — 2, — ~~ > '
Li
The roots are (3, 2), (- 2, - 3),
n+v.
27
II. Solve
1+V- 27\ (\ -V-27 -1-V327>
d)Vrf.
ar y* 4
(2)
1
x
1
2/
9
4'
(2)2==(3) I A+i:
a;^ a;z/ ^
(l)-(3) = (4) - 2 - = §6.
#2/ 4
1 9 1 81
(6)1 + 1-^-
Combining (2) and (6) by addition,
= 3 am rl
2
,2.
- = 3 and
a;
= J and
= i and
III. Solve
Put
y
(1) ^ + ^ = 641,
(2) x-y = l.
X=!U + V,
y = u — v.
x — y = 2v = 7.
v — 2*
^ + / = 2m 4 + 12mV + 2w 4 =641.
248 THE ESSENTIALS OF ALGEBRA.
Putting in the value of v,
2 u* + 12 y? (i ? 9-) + 2 (i ? 9 -) 2 = 641 -
2 m 4 + 147 m 2 - -2^=0.
m 2 = f and - &%i.
3 , ^J-303
u=±- and ±\— -j
, „ 7^ 1-303
u + v = b, 2, -±^/— —
, 7 1-303
^ = M - v = -2, -5, --±yj— —
IV. Solve
(1) z 2 + *, 2 =10,
(2) xy-x-y = -\.
Multiply (2) by 2 and add to (1),
(3) xZ+2xy + f-2(x + y-)=8.
(4) ( a; + y) 2_2( a; + ^)_8 = 0.
(5) + 2/-4)<>+z/ + 2)=0.
(6) a; + y = 4 or — 2.
From (2) by substituting the value of x + y,
(7) xy = 3, or - 3.
Multiply (7) by 2 and subtract from (1),
(8) x 2 -2xy + y* = i, or 16.
(9) a;— y=±2, or ±4.
Combine (6) and (9),
x=S, 1, 1, -3;
y = 1, 3, - 3, 1.
The roots are (3, 1), (1, 3), (1, - 3), (- 3, 1).
SIMULTANEOUS EQUATIONS INVOLVING QUADRATICS. 249
EXERCISES.
Solve the following sets of equations :
M
i?
2.
5.
7.
10.
_1
x y
10 ait/ = 1,
y
1,
-| = 3,
a;?/ — j/ 2 = 4.
.b + 7 y = 15,
v? + f = %
x + y = 3. •
x s -y s = 19,
a;-y = l.
^ + 2/ 3 = 91,
a; + y = 7.
^ + ^ = 28,
ai 2 — xy + y 2 = 7.
x*y + xf = 30,
x + y = a.
(x + yf+4: (x + y) =45,
a; - ?/ = 1.
r(a-y)* + 30c-y)=18,
a; + ?/ =7.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
a^-f-i/ 2 + ai + ?/ = 32,
I xy = 12.
■ «* + »* = 97,
. x-y = l.
p 4 + 2/ 4 = 82,
1 ai + z/ = 4.
fa; 2 + 22/ 2 = 54,
1 xy + y 2 = 35.
f 3a^ + 5?/ 2 =17,
ia^ + 4ai?/ + 3?/ 2 = 15.
' ai 3 + z/ 3 = 28 ail/,
x + y = 12.
x* + tf = %5,
. a* + y = ll.
a; 2 + ^ = 37,
a; + y + xy = 13.
a; 3 — t/ 3 = 7 xy,
x — y = 2.
0^ + ^ = 35,
(a ; + 2/)(^ + 2 / 2 ) = 65.
EXERCISES.
1. Find two numbers whose difference is 5 and the differ-
ence of whose squares is 145.
2. The difference of two numbers multiplied by the greater
= 100, but multiplied by the less = 84, Find the numbers.
250 THE ESSENTIALS OF ALGEBRA.
3. The sum of two numbers is 7, and the sum of their cubes
is 91. Find the numbers.
4. The product of the sum and difference of two numbers
is 96, and the sum of their squares is 146. Find the numbers.
5. The sum of two numbers multiplied by their product is
120 ; and their difference multiplied by their product is 30.
Find the numbers.
6. The difference of two numbers is 3, and the difference of
their cubes is 117. Find the numbers.
7. The sum of the areas of two square fields is 2500 square
rods ; the sides of the fields are to each other as 3 to 4. Find
the area of each field.
8. If the length and width of a rectangular field are each
increased 10 rods, the area is. increased 5 acres. But if the
dimensions are each decreased 10 rods, the area will be 2\
acres. Find the dimensions of the field.
9. The diagonal of a rectangle is 130 feet; the length
of the rectangle is 2| times the width. Find the dimen-
sions of the rectangle.
10. Find two numbers such that their product is 16 times
their difference, and one of the numbers is double the other.
11. A rectangular lot containing 13200 square feet is sur-
rounded by a walk- 6 feet wide. The walk contains 3336
square feet. Find the dimensions of the lot.
12. In a certain number of two digits the sum of the squares
of the digits is one more than twice their product, and the dif-
ference of the squares of the digits is 7. Find the number.
13. The fore wheel of a carriage makes 12 revolutions more
than the hind wheel in going 240 yards ; but if the circumfer-
ence of each wheel, is increased 1 yard, then the fore wheel will
make only 8 revolutions more than the hind wheel in the
same distance. Find the circumference of each wheel.
SIMULTANEOUS EQUATIONS INVOLVING QUADRATICS. 251
14. If a man had worked 5 days less and had received $1
a day less, he would have earned f> 30. If he had worked 10
days less, and had received $2 a day more, he would have
earned $ 50. How many days did he work, and what were
his wages a day ?
15. If the numerator of a fraction be increased by 3 and the
denominator be decreased by 3, the resulting fraction is the
reciprocal of the first. If \% be added to the fraction, the sum
is \ the reciprocal of the fraction. Find the fraction.
EXERCISES- MISCELLANEOUS.
1. Extract the square root of x 2 y 2 — axy 2 — (4 x — \ y)a 2 y
+ 2 a s y + 4 a\
• 2. Find the roots by factoring :
(a) a 2 -7 a; = 30.
(6) a 2 + 7x = 60.
(c) 2/ 2 -9c«/ + 20a 2 = 0.
(d) (3y + 4)(2y-3)-39 = 0.
( e ) 2/ 2 _(c-a)(c-6) = (a-6)a;.
3. Determine whether 1, — 1, f , or any one of them, is a
root of 9 x 2 - 3 x = 2.
4. Make an equation whose roots are 3 + a/7 and 3 — a/7.
5. Simplify 8 a/3 + 13 V243 - 5 Vl21 + 4 a/27.
6. Multiply (x + Vx^) by af-Var*.
1-a/5
7. Rationalize the denominator of
8. Eationalize the denominator of
2a/5-3a/6
3 + a/^4
6-V-16
9. Divide a — b by a*
i
252
THE ESSENTIALS OF ALGEBRA.
10. Solve
x y
M = 4.
x y
11. Solve dx 2 — x + c = 0.
12. By means of the discriminant, tell what kinds of root*
each of the following equations has :
(a) 3x>-5x + 2 = 0. (c) 5 x 2 - 5 x + 10 = 0.
(6) 2 ik 2 + 11 a; -10 = 0. (d) - x 2 + ix + 2 = 0.
13. Find two consecutive numbers whose product is 1260.
14. A number consisting of two digits which differ by 3, is
6 less than 7 times the sum of the digits. Find the number.
15. What value must a have to make the roots of 5 x 2 — 11 x
+ a = equal ?
(3 2
16.
y
= o,
^+i=i
x 2 y 2 2
Find x and y.
17.
Find x and y.
' a?+x — y = 10,
3i/-a; 2 + 3 = 0.
18. Two trains stait at the same time to go 320 miles. One
goes 8 miles an hour faster than the other and reaches its
destination 2 hours sooner than the other. Find the rate of
each train.
19 . Solve 5+1 + 5±2 + 5±| = 3 .
x + 2 a; + 3
_4
x + 5
20. Express x~ s y * + 2 *V* without negative or fractional
exponents.
21. 12^-17^ + 6 = 0. Find re.
22. j/ 3 - 3^ = 88. Find?/.
SIMULTANEOUS EQUATIONS INVOLVING QUADRATICS. 253
23. A number consists of two digits. If its digits be
inverted, the sum of the new and original number is 77 and
their product is 1300. Find the number.
24. 36 a,- 2 + 29 era + 5 a 2 = 0. Find a.
25. Make an equation whose roots are a + ^ 2 an a ft — V 2 .
3 3
26. Vx + 2-\/x = 8. Find*.
x + a b — a x + c 6 — c -„. ,
2v. = - Find x.
b-\-a x — a b + c x—c
28. Find the roots of (y - 2) (tf - 12 y + 20)0/ - 1) = 0.
29. 5V^x3V z ^x2V-9a 2 x3V" ::: T& 5 = what?
30. Multiply V3 — -y/x — Vy by V3 + Vx — Vy.
31. Perform the indicated operations and simplify the
result :
^{bd
a + l
j + bj {a — b
32. Square a* + &* — cK
33. Solve 11 x -U =<£+!.
2
34. Solve 22 - 35 a; + 2 a; 2 = 0.
q+5 2 a +3
... a — 5 2 a — 3
3?- Slm P llf y a + 5 q-5 -
2a+3 2a-3
36. A path around the outside of a rectangular garden is 6
feet wide and 4224 square feet in area. The area of the gar-
den is 28000 square feet. Find the dimensions of the garden.
I 1
--2
37. Simplify 1
y+JZ — l
254
THE ESSENTIALS OF ALGEBRA.
38. Solve ^t} 6 +
25
39.
25 ' a^ + 16
3y 1 -5xy + 2a? = U,
2y 2 -5xy + Sx 2 = 6.
Find x and y.
40. Solve ^ + A = ^ + -
.-1 2 y-6
41. Solve (a 2 - 6 2 )<V - 1) = 2 ic(a 2 + ft 2 ).
a: — ^ a; '
42. Solve |
i .
~5
X —
X-
= 53.
J)
43. What must be the value of x in order that
Q + 3) 2
3a; 2 + 9a;-5
5 a? -12 ar* -13 x -7, when
may equal — 1 ?
44. Find the value of x*
45. Two rectangular fields each contain 10 acres. The
perimeter of one is \ longer than that of the other. One of the
fields is a square. What are the dimensions of each field ?
46. If ab + be + ca = 0, prove that
(a) (a + & + c) 2 = a 2 + 6 2 + c 2 .
(6) (a + b + c) 3 = a s + b 3 + c s -Sabc.
(c) (a + b + c) 4 = a 4 + b 4 + c 4 - 4 abc (a + b + c).
CHAPTER XVII.
RATIO, VARIATION, AND PROPORTION.
I. RATIO.
175. The ratio of a quantity A to a quantity B is the
quotient of A by B.
This quotient may be written in any one of the forms,
A
A -5- B, — , A/B, or A : B, each of which is read, the
B
ratio A to B.
176. Ratio can exist only between two abstract num-
bers, or between two concrete numbers of the same kind.
5
The ratio 5 .to 7, or - , has a meaning, so does the ratio
6 bushels to 15 bushels, but not so with 6 bushels to 15
inches. Ratio merely expresses the part one magnitude is
of another.
177. The terms of a ratio are the numbers compared,
the numerator being called the antecedent, the denominator
the consequent.
178. The ratio of antecedent to consequent is called a
direct ratio ; the ratio of consequent to antecedent is called
an inverse ratio.
Thus, 14 : 28 is direct, while 28 : 14 is its inverse ratio.
b ■ ,, t a
- is the inverse ot -■
a b
255
256 THE ESSENTIALS OF ALGEBRA.
179. A compound ratio is the product of two or more
single ratios.
Thus, - x - x - is the compound ratio of the single
b d f
ape
ratios, r , -, -■
oaf
180. Laws of Ratios. Since a ratio is a fraction, the
operations which may be performed upon fractions may
likewise be performed upon ratios. Below are enumerated
the more important laws relating to ratios :
(1) A ratio is unchanged by multiplying or dividing
both antecedent and consequent by the same number.
5_ 5x4 _ 5h-2 . A _ mA _ A-^m
8 _ 8x4 - 8-5-2' B~ mB~ B + m
This law shows that the ratio of two concrete magni-
tudes of the same denomination is independent of the unit
of measurement. The ratio of 2 miles to 5 miles is - ; the
J 5
ratio of 2 miles, expressed in feet, to 5 miles, expressed in
feet, is — — - = -■ The introduction of a multiplier in
5 x 5280 5
numerator and denominator of a ratio merely changes the
denomination of the terms, if they be considered as con-
crete quantity, the ratio of the two magnitudes remaining
unchanged.
(2) A ratio is changed by extracting the same root of
each term of the ratio, or by raising each to the same
power.
This law is true except in the case when antecedent and
A -VA
consequent are equal. If — = m, then will — — = Vm, and
RATIO, VARIATION, AND PROPORTION. 257
A k i—
-— = m k ; but Vffl + m, and m k =£ m. Hence, the ratio has
B
been changed in all cases except that in which m = 1.
(3) A ratio is changed by performing unlike operations
upon antecedent and consequent.
5 5x2.
-=£= — „ 5 in which the antecedent has been multiplied
by 2 and the consequent by 3. — ^= ^— , for ^— = ( — ) — .
J H J B nB' nB \nj B
(4) A ratio is changed by adding the same quantity to
antecedent and consequent, except when the ratio is unity.
2 2 + 4 . A A + x
5^5 + 4 ; B*B + x
The truth of this law may be easily proven. Take the
A *\ I x
two fractions, — and , where x is any number what-
B B+x
ever. By division,
A + x = A x(B-A~)
B + x B B(B + xy
A A A- T
which shows that — =£ ,, , except when A = B.
B B + x
(5) A ratio is made more nearly equal to unity by
adding any positive number to each of its terms.
Let — be any ratio, and x any positive number. Then
B
A + x_i_ k A^-B
B + x B + x'
, A , A-B _
and 5 - 1-— g-
The first of these two differences is seen to be smaller
than the second. Why? Hence, the truth of the law
js established,
258 THE ESSENTIALS OF ALGEBRA.
(1) Compare - and ^-±^-
'2 = 20 2 + 3 = 5 _ 35
7 _ 70' 7 + 3 10 70"
35 . , , ,, 20
— is more nearly 1 than — ~
70 J 70
(2) Compare - and ^-±^.
5 = 35 5 + 4 = 9 _ 27
3 21' 3 + 4 7 21*
27 35
— - is more nearly 1 than — •
21 J 21
From these illustrations we may see that if a ratio be
less than unity, the addition of the same positive number
to the antecedent and consequent increases its value toward
unity; and if the ratio be greater than unity, the addi-
tion of the same positive quantity to both antecedent and
consequent diminishes the ratio toward unity.
181. The terms ratio of less inequality, ratio of equality,
and ratio of greater inequality are sometimes employed to
describe ratios less than unity, ratios equal to unity, and
ratios greater than unity, respectively.
182. Limit. The result shown above,
A + x 1= A-B
B + x B + x'
indicates that the difference between — ^t-5 and unity
A-B B+x
is a fraction — whose value may be made as small
B + x J
as we please by making x sufficiently large. Hence, the
RATIO, VARIATION, AND PROPORTION. 259
value of the ratio — - 1 — , as x becomes infinitely large,
B + x J 6
approaches unity, which is called the limit of the ratio.
The valme that any algebraic expression continually
approaches but never reaches is called its limit.
EXERCISES.
Write in their simplest forms the ratios of :
1. 625 to 125. 4. a?-{y + zf to x + y + z.
2. 480 x to 120 x 1 . 5. a?-tf to x 2 + xy + y 2 .
3. x^+Sx^tox + S. 6 ( (a A +bf -l) to a-b.
\ iab J
a . X s + 3 x 2 xHx + 3") 2
Suggestion. — — = — ^ — ! — - = x 2 .
x+S x+S
7. a 2 -12a+20 to a-10.
8. 6z 2 + 23a:e + 20a 2 to 3z + 4a.
9. x* + x'y 2 + y* to a? — xy + 2/ 2 -
10. (x 2 + y 2 ) 2 - 4 x 2 y 2 to (a 2 - y 2 ) 2 . '
Write the compound ratios of the ratios :
11. 3 to 5 and 10 to 15.
12. x + y to x — y and x 2 — y 2 to (* + yf.
13. 25 to a? and 3 a; 2 to 50.
14. a 3_27& 3 to (a-3 6) 2 and a -3b to a 2 + 3a& + 9& 2 .
15. (x + l) 2 :(x 2 + 2x + l),(x i + l):(x + l) !
and (»— 1) : (x'—x+l).
Find the value of x for which the ratio of :
16. 128 to x 2 is 2. 18. a; + 5 to a; — 1 is 7.
17. 625 to x 9 is 5. 19. ce 2 + 12a; + 5 to * 2 + 5 is 3.
20. (as + 4):(3as + l)=f
260 THE ESSENTIALS OF ALGEBEA.
Arrange the following ratios in descending order of magni-
tude : ,, 021 15 16 14 8 4 1
*"■■ TiT) 3 2' T5> TT> 2 5' TW> T5> Tf-
a + 5 ct + 1 a + 3 a+7 a + i a
' 6 + 5' 6 + 1' 6 + 3' 6 + 6' 6 + 4' &'
II. VARIATION.
183. The term variation has little use in ordinary alge-
bra, but its use is so frequent in physics that a brief treat-
ment of the subject will be introduced here.
In physics we say " the weight of a uniform mass varies
as the volume." This means that if W is the weight, and
V the volume, then is W= k x V, where k is a constant, the
weight of a unit volume of any given substance.
In mensuration the circumference varies as the diameter.
This means, that if be the circumference and D the
diameter of any circle, then will
0=kxD,
k being a fixed constant for all circles. This constant is
usually denoted by the Greek letter w ; its numerical value
is an incommensurable number, 3.14159
184. In general a variable y is said to vary as another
variable x, when
V
- = a constant.
x
The phrase, y varies as x, is sometimes written
yccx,
but is to be interpreted to mean
- = k, or y = kx.
From this definition we see that a variation as here con-
sidered is equivalent to an equation,
RATIO, VARIATION, AND PROPORTION. 261
185. Variations may be classified as follows :
(1) Direct, y varies directly as x, when
y = kx, k = a constant.
The circumference of a circle varies directly as the
radius.
C= kit, where k = 2 ir.
(2) Inverse, y varies inversely as x, when
h
y=x-
The volume of a gas varies inversely as the pressure,
k
V= — ,. where ~V= volume and n = pressure.
(3) Joint, y varies jointly with x and z, when
y = kxz.
The weight of a rectangular parallelopiped of metal of
unit height varies as the product of the length by the
width,
W=k(lxb).
k = weight of unit volume of the substance, I = length, and
i = width of the rectangular solid.
(4) Quadratic, y varies as the square of x when
y = kx 2 .
An example of such variation is found in the law of
falling bodies; i.e., the space fallen through by any body
starting from Vest equals a constant times the square of
the time expressed in seconds.
S= kt 2 , k = \g, g= 32 feet, 2 inches.
S= space described, i = time in seconds.
262 THE ESSENTIALS OF ALGEBRA.
(5) Direct and Inverse, y varies directly as x and in-
versely as z when
x
y = k~-
3 z
An example of this form of variation is found in
Newton's Law of gravitation. If M, m, be the masses of
two attracting bodies, D their distance apart, and Q- the
force of gravitation, then
Mxm
a = k-
D 2
EXERCISES.
1. If yccx, and y = b when x = a, find the value of y when
x = c.
Solution.
If y cc x, then is y = kx. But y = b when x= a;
hence, h = ka, or k = -.
a
.: y = -x for any value of x.
Hence, y = - . c when x = c.
2. If yoca, and if y = 10 when a; = 2, find the value of y
when x = 12.
3. If y ccx, and if a; = 16 when y = 64, find the value of x
when y = 15-
4. The circumference of a circle varies as the radius
(Oocfi). If 0=3.1416 when B = \, find the circumference
of a circle whose radius is 12.
5. If xccy and wacz, prove -oc^.
w z
6. If xxy and vcct, prove xvccyt.
RATIO, VARIATION, AND PROPORTION. 263
7. If xecy, prove that x"ccy n .
8. If xcc y and zazy, prove that (x^ — z^xy^.
9. If y varies inversely as x 2 , and if y = 16 when x = 4, find
a; when y = 10.
10. The volume of a sphere varies as the cube of its radius.
If the volume of a sphere whose radius is 3 be 113.1, find
the volume of a sphere whose radius is 20.
III. PROPORTION.
186. Proportion. The equality of two ratios is called
a proportion.
A
Thus, — = — is a proportion.
B B
Various forms have been employed in writing a propor-
tion, the following being the ones more frequently used :
A:B=O.B, A : B :: C : B,
A + B=C+ D , | = |
Each form is read A is to B as is to D.
187. Proportionals. The terms of the ' two ratios are
called proportionals.
In the proportion A : B=C: B, the terms A and B are
called extremes, the terms B and are called means, of the
proportion. y
In the proportion A : B = B : B, B is called the mean
proportional to A and B; B is called a third proportional
to A and B.
188. Theory of Proportion ; Theorems. A Theorem is a
statement of a truth to he proved.
A Corollary is a truth derived from the proof of a theorem.
264 THE ESSENTIALS OE ALGEBRA.
The following theorems apply to proportions in which
the terms of each ratio are considered abstract numbers.
Theorem I. In any proportion the product of the ex-
tremes equals the product of the means.
Given A : B = O : D, or, more simply,
A = C
B B'
Then is AB = BG. Why?
Corollary. The mean proportional to two numbers
equals the square root of their product.
This corollary results from the above theorem by letting
C=B, 4=1, or AB = W, whence B = VAB.
B 1)
Theorem II. If the product of two numbers equals the
product of two other numbers, then either pair may be taken
as extremes, and the other pair as means, of a proportion.
(Inverse of Theorem I.)
Given AB=BO.
Divide by BB, (1) ^=-^.
Divide by OB, (2) ^ =^.
J < v J C B
Divide by A C, (3) - = ^.
J V J A
In each of the proportions (1), (2), (3), we have taken
one pair of factors, A, B, or B, 0, as extremes, the other
a* means.
RATIO, VARIATION, AND PROPORTION. 265
Theorem III. If four numbers be in proportion, they
are in proportion by inversion.
Expressing this theorem algebraically,
A . , • , B D
- = -, from winch I = -.
Proof is left to the student. Result is easily shown
true from Theorem I.
Theorem IV. If four numbers be in proportion, they
will be in proportion by alternation; that is, the first is to
the third as the second is to the fourth.
Algebraically, if A : B = : D, then is A : 0= B : D.
See (2) under Theorem II.
Theorem V. If four numbers are in proportion, they
are in proportion when taken by composition ; that is, the
sum of the first and second is to the second as the sum
of the third and fourth is to the fourth.
This theorem stated algebraically is,
., A C ., . A + B O+D
lf _ = _,thBina_— p-
Proof is easily derived by adding 1 to each member
of the given proportion and reducing 5 each member to a
fractional form.
Theorem VI. In a series of equal ratios the sum of
all the antecedents is to the sum of all the consequents as
any antecedent is to its consequent.
Proof. Let the equal ratios be
A=C = l=<l-=...=r
B D F H
where r is the common value of the ratios.
266 THE ESSENTIALS GF ALGEBRA.
Then — = r, or A = r x B,
C
— = r, or C=r X i>,
— = r, or .#=r x F,
— = r, or Gr = r x IT,
Ji
Adding the equalities,
A + C+H+&+ — = rx(B + D + F+ff+-).
Hence A + C + E + a + - A _
if A = 1 = § = ± = A
15 3 9 12 18'
then is
1 _ 5+1+ 3+4+6 _ 3 + 4 5+1+6
3 15 + 3 + 9 + 12 + 18 9+12 15 + 3 + 18'
and so on.
EXERCISES.
Find the value of the variable for which each of the following
proportions is true :
1. 5:20 = x:45.
2. x : 37 = a? : 26.
3. a; : 45 = 5 : x.
4. (Saj + 4):(!B + 5) = (5a! + l):(|x-4).
5. (4a:-3):(2a: + l) = (7a;-4):(3a! + 2).
6. Find a /cwrtfc proportional to 12, 16, and 40 ; also to a,
b, and c.
RATIO, VARIATION, AND PROPORTION. 267
7. Find a mean proportional to 16 and 49 ; also to I and m.
8. Find a third proportional to 25 and 35 ; also to x 2 and xz.
Are the following proportions true for all values of the
letters :
9. (9-x 2 ):(B + x) = (3x-x s ):x?
10. (^^-l\-.2xy(x-y) = (x-y):±x*tf?
\ if /
11. (Jx + yf -z*): (x + y + z) = {x + y -z):x?
lx + mz x
12. If x : y = 2 : w, show
ly + mio y
Suggestion. Let - = r, — = r, then x = ry, z = rw, also to = Iry,
k = mi'io. Add, Za; + mz = r(my + mz), etc.
13. If- = -, show - 1 JT_ i =_ = -.
y w y + mt 2/io 2/
14. The rates of walking of two travelers are to each other
as « to b. If one walk c miles in a given time, how far does the
other walk in the same time ?
15. The rear wheel of a wagon is a feet in circumference,
the fore wheel is b feet in circumference. How often does the
fore wheel rotate while the rear wheel makes m revolutions ?
CHAPTER XVIII.
PERMUTATIONS AND COMBINATIONS.
I. PERMUTATIONS.
189. This subject can best be understood by introduc-
tion through a few concrete examples.
(1) How many different numbers of two digits each
can be formed by using in every way any two of the five
digits 5, 6, 7, 8, 9?
By writing any one digit first and each of the remaining
four digits after it, we have the following five rows, each
composed of four numbers :
56,
57,
58,
59,
65,
67,
68,
69,
75,
76,
78,
79,
85,
86,
87,
89,
95,
96,
97,
98.
In all there are 5 x 4 = 20 different numbers.
(2) How many different numbers of two digits each
can be formed by using in every way any two of the four
digits 5, 6, 7, 8?
Here we select any one of the four digits as the first,
and place after it successively every one of the remaining
three digits.
268
PERMUTATIONS AND COMBINATIONS. 269
This gives the following numbers:
56, 57, 58,
65, 67, 68,
75, 76, 78,
85, 86, 87.
In all there are four selections of the first digit, and four
less one selections of the second, giving 4 x 3 = 12 different
numbers.
(3) How many different numbers of three digits each
can be formed by using in every way any three of the five
digits 5, 6, 7, 8, 9?
The first digit can be any one of the five ; hence, the
first place of each number can be filled in five different
ways. Four digits remain to fill the other two places of
each number. But we have just seen that two digits can
be selected from four in 4 x 3=12 ways. Hence, with
each of the five selections of the first digit, can be placed
twelve selections of the digits filling the two remaining
places. Hence, there are 5x4x3 = 60 different numbers.
190. Definitions. (1) The number of ways of selecting
three things from a collection of five things is called the
permutations of five things taken three at a time.
(2) The number of ways of selecting r objects from a
collection of n distinct objects, regard being had for the order
of selection, is called the permutations of n things taken r
at a time.
In this general case, n may be any number of objects,
and r may be any integral number from 1 to n.
270 ME ESSENTIALS OE ALGEBRA.
191. Symbol. Instead of writing the permutations of
n things taken rata time, the symbol n P r is generally-
used.
Illustrations. (1) 5 P 2 = 5 x 4, the permutations of five
things taken two at a time.
(2) 4 P 2 = 4x3, the permutations of four things taken
two at a time.
(3) 1& P 3 = 10 x 9 x 8, the permutations of ten things
taken three at a time.
192. Examples: Let the pupil construct tables, if
necessary, to verify the following results :
(1) 3 P 2 = 3x2.
(2) 3 P 1= 3.
(3) 4 P 3 = 4x3x2.
(4) 4 P 4 = 4x 3x2x1.
(5) 6 P 1 = 5.
(6)
5 A =
= 5x4.
co
S ^3 =
= 5x4x3.
(8)
5^4 =
= 5 x 4 x 3 x
2.
(9)
5^5 =
= 5 x 4 x 3 x
2x
1.
(10)
P -
= 10 x 9.
The above examples indicate that there is a law govern-
ing the formation of permutations. It will be noted that
the number of factors giving the permutations in each
case equals the number of objects in each selection ; the
highest factor is the number to be permuted, and each
succeeding factor is one less than the preceding. This
PERMUTATIONS AND COMBINATIONS. 271
observation should lead one to some conclusion regarding
the value of the general symbol
p .
We should expect the number of permutations of n things
taken r at a time to be expressed by a product of r of the
natural numbers beginning with n. Hence, we should find
(1) n Pi = n.
(2) n P 2 = rc(>-l).
(3.) ,P,= n(»-l)(n-2).
(4) „P 4 = »(n-l)(n-2)(»-8).
(5) re P 5 = w(>- 1)0-2)0-3)0 -4).
O) n P r = »(» - 1)(» - 2)0 - 8) - (» - r + 1).
193. Value of „P,. To determine the number of permu-
tations of n things taken r at a time, we may proceed as
follows :
(1) Let the n distinct things be represented by n letters
of the alphabet.
(2) Select any one letter to stand first in a set of words
of two letters each. Then there would remain n — 1 letters
to fill the second place ; but the first letter may be selected
in n ways, and with each of these selections any one of
the n — 1 remaining letters may be placed.
(3) Hence, for the number of permutations of n things
taken 2 at a time we have
B P 2 = «0-1).
(4) Let the first two letters of a three-lettered word be
selected from the n letters ; this selection can be made in
272 THE ESSENTIALS OF ALGEBRA.
n(n — 1) ways, as shown in (3) above. Now we may-
select any one of the remaining n — 2 letters to fill the
third place.
(5) Hence, the formation of a three-lettered word from
n letters can be accomplished in
n P s = n(n - 1) (n - 2) ways.
(6) In a similar manner we may show that
„P 4 = „P 3 x (n - 3)= n(n - l)(n - 2)(n - 3),
A = ,^i x (»- 4)= n(»-l)(»- 2)(»- 8)(»-4),
and in general
n P r = „P r _! x(n-r + V)
= m(« — l)(w — 2)(»— 3) ••■ (w — r + 1).
This general result may not be understood at the first
reading of this subject, but its truth may be assumed until
the pupil has had more experience.
Corollary I. Whenr = w,the general formula becomes
„P, = n(n- l)(n-2)(»- 8) -4x3x2x1,
a result easily remembered.
194. The Factorial Symbol. In the value of „P n above,
we have the product of the natural numbers from 1 to n.
This product is often spoken of as " factorial »," and is
written for brevity
[n, or n\.
Either \n or n ! is to be read factorial w, aDd means
the product
n(n - 1)0 - 2)(n - 8) •■■ 5 x 4 x 8. x 2 x 1.
PERMUTATIONS AND COMBINATIONS. 273
125
Find values for :
EXERCISES,
l. [6.
2 H
[16
5 ' M
[5
110
3
6. ^.
115 14
" |5 x |4
120
4 ' -. .
120
v ' — .
10.
[28 [2
[40
J86]T
[8^0
[16 x [2 ' [10 112 ' [16
195. Corollary II. When n objects are permuted all
together, but are not all different, the number of distinct
permutations is given by n P n + \s, where s is the number
of objects which are alike.
Illustration. Required the number of different numbers
obtainable from the five digits 5, 6, 6, 7, 8, taking five
at a time.
If all digits be different, the number of selections would
clearly be 5 P 5 = [5. But the two sixes, when permuted,
give no new numbers ; hence, all the permutations of the
two sixes, i.e. |2, must be excluded (divided out) from
the total.
. J >_.J > K _[5_ 5x4x3x[g _
" [2 [2 [2
EXERCISES.
1. How many different numbers of three digits can be made
from 1, 2, 3, 4, 5, 6 ?
2. How many different permutations can be made by taking
4 of the letters of the word working ? By- taking all of them ?
274 THE ESSENTIALS OF ALGEBRA.
3. Find the value of ]6 P 3 ; ^Pt ; 20^5-
4. How many permutations can be made from the 26 letters
of the alphabet, taking 4 at a time ?
5. How many six-place numbers can be formed from the
Arabic numerals ? (Include 0.)
6. In how many ways can a class of 6 be seated in a row of
6 chairs ?
7. In how many ways can the front row of 6 chairs be
filled from a class of 20 ?
8. How many different permutations can be made from the
letters of the word Indiana f Mississippi 9
9. How many even numbers of 6 places can be formed from
the digits 1, 3, 4, 5, 7, and 9 ?
10. How many numbers between 50,000 and 60,000 can be
formed from the digits 3, 4, 5, 6, 7 ?
11. In how many ways can 10 books be arranged on a shelf
provided 2 particular books are always to be at the ends of
the shelf ?
12. In how many ways can 12 balls be arranged, if 5 are
red, 4 white, and 3 blue ?
II. COMBINATIONS.
196. (1) How many products of two factors each can
be made from the five digits 5, 6, 7, 8, 9?
We have seen that the number of ways of selecting two
things out of five is the permutations of five things taken
two at a time. But in the case of products, 5x6 = 6x5;
hence, each arrangement of two digits is the result of a
permutation of two things taken two at a time. These
must all be excluded. Hence the number of products, is
5 x4-r[2 = 5x 4-5-2 = 10,
PERMUTATIONS AND COMBINATIONS. 275
(2) How many products of three factors each can be
made from the five digits 5, 6, 7, 8, 9 ?
Since any three factors may be arranged in 3x2x1
different ways, each arrangement giving the same product,
we shall have to divide out [3 of the permutations of
five things taken three at a time. Hence, the total number
of different products is
5 x 4 x 3 -=- 1 3 = 20.
197. Definition. (1) The number of ways of selecting
three things from a group of five, no regard being had for
the order of selection, is called the combinations of five things
taken three at a time.
(2) In general, the number of ways of selecting r things
from a group of n things, no regard being had for the order
of selection, is called the combinations of n things taken r
at a time.
198. Symbol. Instead of the phrase, combination of n
things taken rata time, the symbol „C r is usually em-
ployed.
Thus, 5 C 2 is read, the combinations of five things taken
two at a time ; 10 C 4 is read, the combinations of ten things
taken four at a time, etc.
199. Relation between „C, and „P r . It is easy to see that
if we select from a given number of things any specified
number, and do this in every possible way, having no
regard to the order of selection, and then permute all the
objects in each group in every way, we shall have the total
permutations of the n things taken r at a time. The selec-i
276 THE ESSENTIALS OF ALGEBRA.
tions of the groups are the combinations, and the objects
of each group are permuted r at a time ; hence,
P — C v P
n L r nyr ^ r L ?••
. /7 _ „P r _ n(w-l)(n-2~)-(n-r + l)
■ • »W — -; — — ;
\r \r
A second form for n C r may be had by multiplying the
numerator and denominator of the fraction on the right
by \n— r . This multiplier makes the numerator \n, and
the symbol JJ r becomes
\r \n —
r
C- ^
|12
12 7 [7J12-
From the first form
7 [7|5
n 12-11
12°7 —
12-11
•10-9-8-7-6
II
- 10 • 9 • 8 - 7 - 6 -|5
12-11
1111
• 10-9-8-7- 6-5-4-3. 2-1
LI [fi
12
Note. When r = n, the second factor of the denominator \n-r
becomes [0, a symbol whose value is to be taken as unity. To show -
10 = 1,
we take the equality [m = mx|m-l,
and put m = 1 ;
PERMUTATIONS AND COMBINATIONS. 277
then [1 = 1 x 1 0,
and a s II = 1, [0 must be 1.
,.[0 = 1.
The form „c„ = =L_
is more easily remembered than the form
nCr = n(n - l)(n - 2)(n - S)...(n -* r + 1)
lr
but the latter is especially useful in many applications.
EXERCISES.
1. How many combinations can be made from 9 things 3
at a time ? 5 at a time ?
2. Find the values of i„C 4 , 12 C 9 , 12 C 3 .
3. In a meeting of 20 people, in how many ways can a
committee of 5 be selected ?
4. A school is composed of 19 boys and 25 girls. In how
many ways can a committee consisting of 1 boy and 1 girl
be selected ?
5. From the above school, how many committees consisting
of 2 boys and 1 girl can be selected ?
6. From 15 persons, how many committees of 5 can be
formed, provided one particular person is to be a member of
every committee ?
7. If out of 9 candidates there are to be 5 officers elected,
how many different tickets can be formed ?
8. From 4 vowels and 8 consonants, in how many ways can
5 letters be chosen, provided exactly 2 of them are vowels ?
Provided at least 2 of them are vowels ?
9. How many even numbers of 4 places can be formed from
the digits 1, 2, 3, 4, 5, 6, 7, 8?
10. „C 6 <=,Ao; findn.
CHAPTER XIX.
SERIES.
200. General Definitions. (1) Any set of numbers is an
array, or succession.
(2) A series is a succession of numbers arranged ac-
cording to some law.
Thus, 1, 2, 3, 4, 5, 6, •••, is a series, the law of formation
being that any number is to be had from the preceding
by adding 1.
We may also define a series as a succession of numbers,
the knowledge of two or more successive ones being suffi-
cient to determine all.
Thus, 5, 7, 9, 11, •••, form a series, since by inspection
of any two we see their difference to be 2 ; hence, any
number of the series may be had from the preceding by
adding 2.
(3) The numbers forming a series are called the terms
of the series.
EXERCISES.
What law of formation exists in each of the following ?
i. 2, 4, 6, 8, 10, .... 3. i, 1, f, 2, f, 3, £, 4, ....
2. 5, 10, 20, 40, 80, .... 4. 3, §, f, |, JL ....
278
SERIES. 279
5. -1,1,3,5,7,9, ....
6. a, a + d, a + 2d, a + 3 d, • ■•, a + (n-V)d.
7. a, ar, ar 2 , ar 3 , •••, ar"~ l .
8. 5, -15, 45, -135, 405, -1215, ....
9. X -± + t-± + *!_...
[3 |6. [[ |9
(4) When the number of terms of a series is finite, the
series is called a finite series.
Thus, 2, 5, 8, 11, 14, is a finite series.
(5) When the number of terms is infinitely great, the series
is called an infinite series.
Thus, if a series be formed by making any term the
half of the preceding term and this process be continued
indefinitely, as,
1 l i l _i l
x ' 2' 4' IT' 16' 32' '
we have an infinite series.
(6) If the sum of n terms of an infinite series can be
shown to approach some finite number as n is made to
approach infinity, the series is called convergent. If this
sum can not be shown to approach some finite quantity, the
series is called divergent.
In the discussion of the subject of series, we shall
examine only three special forms, the arithmetical series,
the geometrical series, and the binomial series.
280 THE ESSENTIALS OF ALGEBRA.
I. ARITHMETICAL SERIES (Arithmetical Phogkession).
201. Definition. An arithmetical series is a series in
which the difference of any two successive terms is a constant.
Illustrations. (1) 5, 9, 13, 17, 21, •••, is an arithmetical
series, since the difference of any two successive terms is 4.
(2) 3, 3.5, 4, 4.5, 5, 5.5, •■•, is arithmetical, since the
difference is .5.
(3) a, a + d, a + 2d; a + S d, a + 4 d, •■•, a + (n — V)d, is
arithmetical, since the difference of any two succes-
sive terms is d. In this illustration a is taken as any
algebraic number, commensurable or not, and d likewise
as any algebraic number.
202. Notation. We shall denote by a the first term of
any arithmetical series, by d the constant difference (com-
mon difference} between any two successive terms, by I
the last term or nth. term, and by S the sum of n terms of
the series.
203. Fundamental Formulas.
(1) l=a+(n-l)d.
(2) S=±±±xn
2a+(n-l)d
= — — — xn.
In these two relations five letters are involved, any two
of which may be unknown.
The first of the above formulas is easily seen to be true
from the manner of formation of the general arithmetical
series shown in illustration (3) above.
SERIES. 281
To derive formula (2), we write the series
S = a + (a + d) + (a + 2 d) + ••• + (I - d) + I.
Then reverse the series,
S = I + (I - d) + (I - 2d) + - + (a + 2<T) + (a + d)+ a,
and add the two equalities, giving
28 = (a + I) + (a + I) + (a + I) + (a + I) + -
+ (a + I) + (a + I)
= n(jx + l~), since there are n terms in the series.
a a + I
The second form of formula (2) is derived by replacing
I by its value from formula (1).
204. Arithmetical Mean. If a, b, c be three successive
terms forming an arithmetical series, b is called the arith-
metical mean of a and c.
b = |- (a + c), for by the definition of the arithmetical
series, . ,
o — a = c — b ;
(transposing), 2 b = a + e,
or 5 = l(a + a).
205. Arithmetical Means. In an arithmetical series a, b,
e, d, e,f, ••-,?, the terms b, c, d, e,f, •••, are called the arith-
metical means of a and I.
206. To insert k arithmetical means between any two
numbers.
"Let a and b be any two numbers. After k means have
been inserted, the whole series will consist of k + 2 terms.
282 THE ESSENTIALS OF ALGEBRA.
Hence, b is the last or (k + 2) th term of an arithmetical
series of which a is the first, and d an unknown common
difference.
Hence, b = a + (1c + 2 — 1) d
= a + (k + 1) d,
j b — a
The common difference d being known, the series may
be easily written thus :
, b — a . 2 (b — a) ,
207. An arithmetical series is determined when two of its
terms are known.
Let a be the kth and b be the with term of an arith-
metical series. Let x be the first term and y the common
difference.
Then a = x + (k — V)y,
b = x + (m — 1) y.
By subtraction
b — a=(m — k)y,
or y = (the common difference),
7YI — fC
and x = a — (k — 1)
b-a 1 = (m-Y)a-(k-V)b
m — k) m — k
The first term x and the common difference y being
known in terms of a, b, k, m, the series may be written
down.
Determine the arithmetical series in which the 5th term
' is 17, and the 12th term is 38.
SERIES. 283
Solution.
Let x = first term, and let y = common difference.
■Then i = a + (n — 1) d becomes respectively,
rl7 = z + (o-l)y,
l38 = z+(12-l)y.
By subtraction, 21 = 7 y, or j? = 3 ;
when y = 3, a; = 5.
Then the series is 5, 8, 11, 14, 17, 20, ••-, 35, 38.
EXERCISES. '
1. Find the 18th -term of 2, 5, 7, 10, etc.
2. Sum 4, 7, 10, etc., to 9 terms.
3. Insert 5 arithmetical means between 10 and 34.
4. Find the 15th term of an arithmetical series whose 2d
and 7th terms are 9 and 21, respectively.
5. Which term of the series 1, 6, 11, 16 is 96 ?
6. Find the sum of the natural numbers from 91 to 187.
7. Show that if any four numbers are in arithmetical pro-
gression, the sum of the 1st and 4th is the same as the sum of
the 2d and 3d.
8. Find the 18th term of 27, 21, 15, 9, etc.
9. Find the sum of 12 terms of 3, 4|, 6, 7J-, etc.
10. How many terms of 1, 2, 3, 4, etc., will make 465 ?
11. How many terms of 7, 11, 15, 19, etc., will make 297?
12. How many strokes does a clock strike in 12 hours ?
13. Find the sum of all the even numbers from 100 to 200
inclusive.
14. Find the sum of all the numbers from 48 to 135 inclu-
sive which are divisible by 3.
15. What debt could he paid in a year by the payment of
10 f the 1st week, 40 ^ the 2d week, 70 ^ the 3d week, etc. ?
284 THE ESSENTIALS OE ALGEBRA.
16. Determine the series whose 10th term is 51 and whose
20th term is 101.
17. Determine a series whose 15th term is and whose 31st
term is 64.
18. Find the sum of all numbers from 105 to 361 inclusive,
which, when divided by 4, leave a remainder of 1.
II. GEOMETRICAL SERIES (Geometrical Progression).
208. Definitions. A geometrical series is a series in which
the ratio of any two successive terms is a constant.
Illustrations. (1) 2, 4, 8, 16, 32, is a geometrical series
in which the ratio 16-=-8 = 8-s-4=i32-j-16 = 2 is a constant.
(2) 1, £, i, jj, ^L, •■•, is a geometrical series with ratio
equal to ^.
(3) 1, x,a?, X s , x\ ■••, is a geometrical series with ratio
equal to x.
(4) a, ar, ar 2 , ar 3 , ar*, •••, ar n ~\ is a geometrical series
in which the ratio is r.
209. Notation. The illustration (4) above- suggests a
notation for the geometrical series.
(1) a = first term.
(2) r = constant ratio.
(3) I = last term, or nth. term.
(■4) S = sum of n terms.
210. Formulas.
(1) I =ar n -\
(3) S= - , when r <1, and n=ao.
1 — r
SERIES. 285
The first of these formulas results from the law of forma-
tion of the series as indicated in illustration (4) above.
The second formula we may derive as follows :
(1) S = a + ar + ar 2 + ar 3 + ••• + ar n ~ 2 + ar n ~ l .
Multiply by r,
(2) Sr= ar + ar 2 + ar 3 + ■■• + ar n ~ 2 + ar"- 1 + ar n .
Subtracting (2) from (1),
(3) S — Sr = a — ar":
(4) S(l - r) = a (1 -?•").
(5) *-.xg=£)
Another method of derivation is worthy of attention.
By actual division we know that
l-J-3
1-r
1-r 4
1-r
1-r 5
1-r
and so on ; for the general case
= 1 + r+r 2 ,
= l + r + r 2 + r 3 ,
= l+r + »- 2 + r 3 + r 4 ,
. = ! + »• + r 2 +r 9 + r i + r 5 -{ \-r n
1-r
Now by writing the value of S again,
$ = a + ar + ar 2 + ar 3 + ar* + ••• + ar n ~ 2 + ar n ~ r ,
and factoring out a from each term on the right, we have
S<= a (1 + r + r 2 + r 3 + ••■ + r n ~ 2 + j-*" 1 ).
The value in this bracket is the same as the value of
above ; hence, /^ _ r n
S =a x
286 THE ESSENTIALS OF ALGEBRA.
211. Sum to Infinity when r < 1. If the ratio r be less
than unity, r" < 1, and when n = go, r" = 0. Hence, s= --— ,
when r < 1, and n = oo.
The sign = is read approaches.
Illustration. Find the sum ofl + J + ^ + |- + etc.
Here, a = 1, r = -, and S= -=2.
^ 1 — ^
212. The Geometrical Mean. 7/" a, b, c, be three successive
terms of a geometrical series, then b is equal to the square
root of the product of a by c, and is the geometrical mean of
a and c.
By the definition of a geometrical series,
- = - ; whence, b 2 = ac, or b = ^/ac.
a b
213. Geometrical Means. In a geometrical series the
terms lying between any two terms are called the geometrical
means of those two terms.
Thus, 5, 10, 20, 40, 80, 160, are six terms of a geometrical
series with ratio 2. The terms 10, 20, 40, 80, are the four
geometrical means between 5 and ISO.
214. Insertion of Geometrical Means. Any number of geo-
metrical means may be inserted between any two numbers.
Proof. Let a and b be any two numbers, and let k be
the number of means to be inserted. Then b is the
(k + 2y h term of a geometrical series, whose first term
is a. If r be the unknown ratio,
b = ar(*+ 2 -« = ar ( * +1) .
SERIES. 287
Insert five geometrical means between 128 and 2.
These means may be written down if we know the value
of the ratio. This is given by
r = ^-i where k = 5, b = 128, a = 2.
* a
e/128 _ f/S
Hence, the series is 2, 4, 8, 16, 32, 64, 128, and the
means are 4, 8, 16, 32, 64.
215. A Geometrical Series Known. A geometrical series
is known when any two terms are known.
Proof. Let a be the &th, and I the rath terms of a geo-
metrical series ; and let r be the unknown ratio, with x
as first term.
Then a = xr k ~ l ,
and I = xr m ~ l .
These two equations are sufficient to determine the first
term x and the ratio r.
By division
or
(1)
Also, (2) x = -^ =
xr k
-1
a
r m
-*_
I
a
•'•
r =
m-ifl
a
m-
-k\ a m ~ x
fm
-*/r
\k-l
\^T
Equations (2) and (1) give the first term and ratio,
respectively, of the required series.
288 THE ESSENTIALS OF ALGEBRA.
EXERCISES.
1. Find the 10th term of 1, 2, 4, 8, etc.
2. Find the sum of the 10 terms in (1) above.
3. The 3d and 6th terms of a geometrical series are 27 and
729. Find the 8th term and the sum of the 8 terms.
4. Find the sum of 10 terms of 1, \, \, \, etc.
5. Sum 3, - 3 2 , 3 3 , - 3" to 8 terms.
6. A house with 8 windows was sold for $1 for the 1st
window, $ 2 for the 2d, $ 4 for the 3d, etc. What was received
for the house ?
7. If you receive $ 5 Jan. 1, $10 Feb. 1, $ 20 March 1, and
so on for each month of the year, what is the total amount you
will receive during the year ?
8. Find the sum to infinity of ^, -j-j^, T 3 , etc.
9. Find the sum to infinity of 1, \, \, ^ T , etc.
10. The arithmetical mean of two numbers is 13 and their
geometrical mean is 12. Find the numbers.
11. Show that the series of alternate terms of a geometrical
series is also a geometrical series.
12. If every term of a geometrical series is divided by the
same quantity, the quotients form a geometrical series.
13. The reciprocals of the terms of a geometrical series form
a geometrical series.
14. The difference between the 1st and 4th of four numbers
in geometrical progression is 208, and between the 2d and 3d
is 48. Find the numbers.
15. The sum of 3 numbers in geometrical progression is 14
and the sum of their reciprocals is |. Find the numbers.
SERIES. 289
III. BINOMIAL SERIES.
216. Definition. The series defined by
(a + by = a" + a- 1 * + n< ^ n ~ *) a n ~ 2 b 2 + ...
If
+ »(«-l)(»-2)(n-3) - Q-r + 1) ^^ | _
lr
is called the binomial series.
Note. In higher algebra it is shown that this series gives a true J
value of (a + i)™ for all values of a and b provided n is an integer. It
also defines (a + b) n properly when n is a negative number, or a
fraction, provided - be a proper fraction. The general proof of these
a
assertions will not be attempted in this development.
EXERCISES.
1. (1 + a;) 80 . In this n = 30, a = 1, b = x.
2. Write out the first 6 terms of (1+a;) 18 ; (l-y) n ; (a+2/) 25 -
3. Find the coefficient of a; 15 in (1+a;) 30 . (Eef erring to the
binomial series, we see that r — 15.)
4. (1 - xY = 1 + (- 3)(- a) + (-3K-3-1) ( _ ^
If
( _3X-3-l)(-3-2) ( _ a>y
^ [3 ^ ;
== l + 3 £( .+6a! 2 + 10a! 3 + etc.
5. Write out the first 5 terms of
(a) (I-*)- 1 ! (6) (l + *)*i («) (I-*)" 1 5 (<*) d+.aO - *- :
290 THE ESSENTIALS OF ALGEBRA.
EXERCISES- MISCELLANEOUS.
1. Expand (x + 2 yf; {2x-yf; (2 a -3 b) 7 .
2. Find the sum of 8 terms of 1, 2x, 4a; 2 , •••.
3. Find the sum of 30 terms of 7, 11, 15, •■•.
4. Find the 6th term of (1-2 aj)" 3 -
5. How many arithmetical means must be inserted between
10 and 40 so that the sum of the series may be 275 ?
6. Divide 26 into three parts which are in geometrical pro-
gression, and such that when 4 is added to the second part, the
three parts are in arithmetical progression.
7. Insert a geometrical mean between 5 and 45. Explain
the double sign of the result.
8. Sum to infinity f, f, ■£?, •••.
9. Find the coefficient of ' afy 5 in (2 x — 3 y) w .
10. Find the coefficient of w 8 in (1 — .2x)~\
11. In how many ways can 9 persons be selected from a
party of 21 people ?
12. How many committees consisting of 4 men and 3 women
can be formed from 12 men and 10 women ?
13. Each member of a baseball nine, except pitcher and
catcher, can play in any position. In how many ways can the
team be arranged upon the field ?
14. In how many ways may a baseball nine be selected from
16 candidates, if 2 are pitchers, 3 are catchers, and the remain-
der can play in any position ?
15. Expand (4 — 2)i to five terms, and thus get an approxi-
mate value of V2.
SERIES. 291
16. Expand (100—1)* to five terms,, and thus get an approxi-
mate value of V99.
17. Expand
MI-
18. Find the sum of all the even numbers between 205 and
341.
111
19. Find the sum to infinity of 1 + -==• + - + — — 4 .
V2 2 2V2
20. The sum of the first three terms of a geometrical series
is 21, and the sum of their squares is 273. Find the series.
21. A debt is to be paid by 10 payments which form an
arithmetical progression. The third payment is $220, and the
seventh is $ 360. Find the last payment and the total debt.
22. How many consecutive odd numbers beginning with 11
must be taken to make a sum of 759 ?
23. Prove that the squares of the terms of a geometrical
series also form a geometrical series. *
24. (Va~+l-Va-l) 4 =what?
25. Expand (l+x + x 2 ) 5 .
INDEX.
Abscissas, axis of, 148, 149.
Addition, 20-28, 35 ; identities, 26,
27 ; of fractions, 117, 118 ; in
elimination, 161.
Algebraic expression, 4 ; signs used
in, 5, 6 ; addition, 20, 21 ; sub-
traction, 28, 29 ; fraction, 109.
Arithmetic, numerical, 3 ; literal, 3,
4 ; addition in, 20 ; square root
in, 180, 181 ; cube root in, 184.
Arithmetical mean, 281.
Arithmetical means, 281.
Arithmetical series (progression),
280.
Associative law, definition of, 23 ;
of factors, 45.
Axioms, 16.
Binomial, definition of, 25 ; identi-
ties, 70, 71, 74, 75 ; theorem, 76 ;
factors, 100 ; series, 289.
Coefficients, definition of, 7, 8 ; de-
tached, 63 ; law of, 77 ; of a
quadratic, 222.
Combinations, examples of, 274 ;
definition of, 275 ; symbol of,
275.
Commutative law, definition of, 22 ;
of factors, 43.
Constants, 15, 125.
Coordinate axes, 148, 149.
Cube root, 182, 184.
Distributive law, of factors, 45.
Division, 58-67 ; identities, 66, 81 ;
of fractions, 122.
Elimination, definition of, 157 ; in
substitution, 158 ; by compari-
son, 160 ; by addition or subtrac-
tion, 161 ; illustrative examples
in, 162-164.
Equation, definition of, 14 ; sign of
15 ; root of, 16, 126 ; conditional,
125; linear, 125, 147, 148, 170,
229 ; of second or higher degree,
140 ; fractional, 141, 143 ; inte-
gral, 143 ; simultaneous, 157-
173 ; quadratic, 207-228 ; homo-
geneous, 239 ; symmetrical, 241 ;
in higher degrees, 246.
Evolution, 174-186.
Exponents, definition of, 5 ; law of,
77 ; integral, 187.
Factoring, iii, 83-103.
Factors, definition of, 42, 83 ; laws
governing, 43-51 ; degree and
number of, 43; monomial, 84;
by rearrangement and grouping,
97-99 ; binomial, 100 ; rational-
izing, 198.
Fractions, in number system, 2 ;
algebraic, 109; signs of, 110,
111 ; reduction of, 111, 115, 116;
proper and improper, 114 ;. addi-
293
294
INDEX.
tion and subtraction of, 117, 118 ;
multiplication of, 119, 120 ; divi-
sion of, 122 ; complex, 123.
Geometrical series, 284-287.
Graph, the, iii ; of the linear equa-
tion, 148 ; of two linear equa-
tions, 155 ; of surds, 192, -193 ;
of the quadratic, 218-221.
Highest common divisor, 104, 105.
Highest common factor, 104.
Identity, definition of, 14, 125 ; sign
of, 15 ; in addition, 26, 27 ; in
subtraction, 32 ; in multiplica-
tion, 55, 70-73 ; in division, 66,
81.
Imaginaries, 202 ; operations with,
203 ; graph of, 205.
Index laws, 47, 58, 59, 187.
Involution, 56.
Lowest common multiple, 106, 107,
143.
Monomials, definitions of, 22 ; addi-
tion of, 23, 24 ; subtraction of,
30 ; multiplication of, 48, 49 ;
division of, 59, 60 ; H. C. D. of,
104 ; L. C. M. of, 107.
Multiplication, 39-57 ; of monomi-
als, 48 ; of polynomials, 49, 51 ;
identities in, 55, 70-73 ; of frac-
tions, 119, 120.
Notation, 280, 284.
Number system, the, iii, 1-3.
Numbers, incommensurable, 3, 192;
general, 4, 9 ; literal, 4 ; oppo-
site, 10; negative, 11; positive
and negative, 11, 12, 13 ; alge-
braic, 13 ; commensurable, 191 ;
irrational, 193 ; rational, 200 ;
complex, 205 ; ratio of, 255.
Ordinates, axis of, 148, 149.
Parentheses, 7 ; removal of, 33 ; in-
sertion of, 34.
Pascal's triangle, 78, 79.
Permutations, examples of, 268,
270 ; definitions of, 269 ; symbol
of, 270 ; factorial symbol of, 272.
Polynomials, definition of, 25 ; ad-
dition of, 25 ; subtraction of, 30 ;
multiplication of, 49, 51 ; divi-
sion of, 60, 61 ; square of, 73 ;
H. CD. of, 105; L. C. M. of,
107 ; square root of, 176 ; cube
root of, 182, 184.
Proportion, definition of, 263 ; theo-
rems of, 263-265.
Quadratic, the pure, 208 ;see Equa-
tions.
Radicals, definition of, 191.
Ratio, definition of, 255 ; terms of,
255 ; compound, 256 ; laws of,
256 ; limit of, 258.
Root, of equations, 16, 126, 140,
147, 222 ; even and odd, 175 ; of
decimals, 182 ; the double, 213 ;
irrational, 214 ; complex, 215.
Signs, algebraic, 5, 6, 13 ; of aggre-
gation, 7; in addition, 20, 21;
in subtraction, 28, 29; in multi-
plication, 42, 43 ; in division, 58 ;
in factoring, 87 ; in fractions,
INDEX.
295
110, 111; radical sign, 174; in
roots, 175.
Square root, 174, 176, 177, 179-181.
Substitution, i, 8 ; in elimination,
158.
Subtraction, 28-38 ; of fractions,
117, 118 ; in elimination, 161.
Surds, definition of, 192 ; expressed
graphically, 192 ; forms, 193 ;
kinds of, 194 ; quadratic, 194,
196 ; polynomial, 197 ; conjugate
binomial, 198 ; trinomial, 200 ;
and rational numbers, 200 ;
square root of binomial, 201.
Trinomial, definition of, 25.
Type forms, iv ; in multiplication,
51, 57, 70-76; in. division, 59,
81 ; in factoring, 85, 86, 88, 92,
94 ; in linear equation, 137, 147 ;
in square root, 176 ; in cube
root, 182 ; in index laws, 187,
188 ; in quadratic equations, 207,
208, 209, 211, 216; in permuta-
tions, 271 ; in combinations, 275,
276 ; in arithmetical series, 280,
281; in geometrical series, 284,
285.
Variable, definition of, 15 ; equation
in one, 125-146 ; linear equa-
tions in two, 147-156 ; quadratic
equations in a single, 207-228.
Variation, 260-263.
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