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COURSE
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IN MEMORIAM
FLOR1AN CAJOR1
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HIGH SCHOOL ALGEBRA
Elementary Course
BY
H. E. SLAUGHT, Ph.D.
ASSISTANT PROFESSOR OF MATHEMATICS IN THE UNIVERSITY
OF CHICAGO
AND
N. J. LENNES, M.S.
INSTRUCTOR IN MATHEMATICS IN THE WENDELL PHILLIPS
HIGH SCHOOL, CHICAGO
oXKc
Boston
ALLYN and BACON
1907
COPYRIGHT, 1907,
BY H. E. SLAUGHT
AND N. J. LENNES.
5<f3
PREFACE
The High School Algebra has been written in two parts, the
one designed solely for the first year and the other for a com-
plete review and advanced course at a later period.
The authors recognize the impossibility of combining in a
single treatment the qualities necessary for beginners with
the more mature point of view suitable for the third or fourth
year student.
The important features of the Elementary Course are :
1. Algebra is vitally and persistently connected with arithmetic.
Each principle in the book is first studied in its applica-
tion to numbers in the Arabic notation. The principles of
Algebra are thus connected with those already known in
arithmetic. Letters are introduced as abbreviations for " a
number" or "any number." Literal expressions are called
number expressions — the vague term "quantity" is not used.
In the exercises, Arabic figures are constantly involved in the
same manner as letters. Checking by the substitution of
parilcular numbers for letters occurs throughout.
2. TJie principles of Algebra used in the Elementary Course
are enunciated in a small number of short statements — eighteen
in all.
The purpose of these principles is to furnish in simple form
a codification of those operations of Algebra which are suffi-
ciently different from the ones already familiar in arithmetic
to require special emphasis. Such a codification has several
important advantages :
[5306144
iv PREFACE
By constant reference to these few fundamental statements
they become an organic and hence permanent part of the
learner's mental equipment.
By their systematic use he is made to realize that the
processes of Algebra which seem so multifarious and hetero-
geneous are in reality few and simple. Invaluable training is
thus afforded in connecting things which are essentially, though
not apparently, related.
Such a body of principles furnishes a ready means for the
correction of erroneous notions, a constant incitement to effec-
tive review, and a definite basis upon which to proceed at each
stage of progress.
No attempt is made at formal demonstration of these prin-
ciples. It is believed that conviction as to their validity is
most effectively brought to a first year pupil by their proper
empirical exhibition. Formal argumentation is reserved for
the Advanced Course.
3. TJie mam purpose of the Elementary Course is the solution
of problems rather than the construction of a purely theoretical
doctrine as an end in itself
While the subject-matter of Algebra is developed in a logical
sequence around the principles, the attempt is made to connect
each principle in a vital manner with the learner's experience
by using it in the solution of a large number and great variety
of simple problems.
In making the problems the following criteria have been
observed :
The subject-matter should be easily within the comprehen-
sion of the pupil, and, so far as possible, the problems should be
such as one would actually need to solve in passing from known
to unknown data by means of given relations. Such are prob-
lems on physical relations in Chapter IV, and those on geomet-
rical relations in Chapters VI and VII.
PREFACE V
However, with the pupil's meagre experience it is impossible
that all problems should be of this kind. Hence a considerable
body of problems has been introduced which involve artificial
relations imposed upon numbers. Such problems are of two
kinds :
(a) Those involving numbers related to concrete things, as,
for example, the problems on pages 44 to 46. The data used in
this class of problems (with very few exceptions) are such as
have a distinct value or real interest in themselves. Every
effort has been made to have the data entirely trustworthy.
(b) Those involving numbers not related to concrete things,
as, for example, problems 12 to 18, page 43. Such problems
are used as the basis for the development of formulas, pages
110 to 114, Chapter IV.
The utmost care has been taken in grading the problems,
both as to difficulty and with reference to the principles upon
which the solutions depend.
4. Tlie order of topics and the inclusion and exclusion of
subject-matter have been determined by the main purpose of the
Elementary Course as just stated.
The equation, therefore, as the instrument for the solution
of problems, occurjies the leading -place. New topics are intro-
duced only as they are needed in extending the use of the
equation. For example, the whole subject of linear equations,
including those with two and with three unknown quantities,
is treated before long division; and quadratic equatiens are
placed before the formal treatment of literal fractions.
The subject of factoring is introduced when quadratic equa-
tions are to be solved, and is immediately used for that
purpose in special cases.
Long division of polynomials is first found necessary as an
introduction to square root, which together with radicals is
essential to the solution of the general quadratic equation.
vi PREFACE
The topics excluded are : (a) complicated factoring, (&) H. C.F.
by the long division process, (c) complicated fractions, (d) si-
multaneous equations in more than three unknowns, (e) cube
root, (/) fractional and negative exponents, (g) equations con-
taining complicated radicals, Qi) simultaneous quadratics except
the case of a quadratic and a linear equation.
These omissions have permitted the introduction of a much
greater number and larger variety of problems than would
otherwise be possible without diminishing the number of drill
exercises.
For example, instead of an extended and purely theoretical
discussion of radicals, only so much is given as is needed in
treating simple quadratic equations; and this is immediately
applied to the solution of a body of interesting problems in-
volving such concrete geometrical relations as are freely used
in modern grammar school arithmetics.
This larger space allotted to problems makes it possible to
pass by slow gradation from extremely simple to more compli-
cated cases which would otherwise be too difficult, and thus to
secure the facility and power in interpreting and solving prob-
lems which is demanded in physics and other sciences.
Attention is further called to the following features :
The simple and scientific treatment of the solution of the
equation as enunciated in Principle VIII, page 36.
The numerous illustrative solutions which have been given
in full for the purpose of exhibiting the best methods of attack
and the most effective arrangement of the work.
The development of negative numbers from concrete rela-
tions, immediately followed by a large number of problems
showing their practical use and interpretation.
The introduction of graphs at the beginning of the chapter
on simultaneous equations, thus making the graph the basis
of the study of simultaneous equations.
PREFACE Vll
The notions of a formula and of substitution in a formula
are developed in connection with subjects already familiar,
such as interest, areas, volumes, etc. (See page 101.) These
notions are then applied to the study of other subjects. (See
page 115.)
Literal equations are in each case introduced as a generaliza-
tion of a series of concrete problems immediately preceding.
(See page 111.) Such equations are then to be solved for each
letter involved. (See page 127.)
Problems involving physical relations are in each case intro-
duced by means of a series of carefully graded problems lead-
ing up to the general case. (See page 120.)
If in any case the problems are found to be too numerous,
the later sections of Chapter IV may be omitted or postponed
until after graphs and simultaneous equations have been
studied. Many of the exercises are simple enough to be read
in class as mental drill exercises.
Keview questions and exercises are given at appropriate
intervals throughout the book.
Ratio and proportion are treated in the chapter on fractions
preparatory to their use in plane geometry.
The authors desire to express their thanks to many who
have given useful suggestions, and especially to Mr. J. A.
Dixon, Wendell Phillips High School, Chicago, and to Miss
Mabel Sykes, South Chicago High School, Chicago, who have
solved all the problems and exercises and whose criticisms
have been of special value. Thanks are also due to Mr. Edwin
Hand, Jr., for helping to prepare the cuts.
H. E. SLAUGHT.
N. J. LENNES.
Chicago, May, 1907.
CONTENTS
CHAPTER I
INTRODUCTION TO THE EQUATION
PAGE
Peinciple I. Addition of Numbers having a Common Factor . 1
Peinciple II. Subtraction of Numbers having a Common Factor 9
Peinciple III. Multiplication of the Product of Several Factors 12
Pbinciple IV. Multiplication of the Sum or Difference of Two
Numbers 14
Peinciple V. Division of the Product of Several Factors . . 20
Peinciple VI. Division of the Sum or Difference of Two
Numbers 23
Peinciple VII. Number Expressions in Parentheses . 28
Peinciple VIII. Identities and Equations 34
Directions for Written Work 38
Equations involving Fractional Coefficients ... 41
CHAPTER II
POSITIVE AND NEGATIVE NUMBERS
A New Kind of Number ....
Peinciple IX. Addition of Signed Numbers
Addition by Counting
Averages of Signed Numbers
Peinciple X. Subtraction of Signed Numbers .
Peinciple XL Multiplication of Signed Numbers
Peinciple XII. Division of Signed Numbers
Interpretation and Use of Signed Numbers in Problems
47
49
53
55
58
62
65
67
PAGE
72
74
82
88
92
X CONTENTS
CHAPTER III
INVOLVED NUMBER EXPRESSIONS
Double Use of the Signs + and —
Addition and Subtraction of Polynomials
Principle XIII. Multiplication of Polynomials
Squares of Binomials ....
Review Questions and Exercises .
CHAPTER IV
SOLUTION OF PROBLEMS
Problems involving Interest 100
Problems involving Areas 105
Problems involving Volumes 108
Problems involving Simple Number Relations . . . 110
Problems involving Motion 115
Problems involving the Simple Lever 120
Problems involving Densities 122
Problems involving Momentum 125
Problems involving Thermometer Readings . . . 128
Problems involving the Arrangement and Value of Digits . 130
Review Questions and Problems 131
CHAPTER V
INTRODUCTION TO SIMULTANEOUS EQUATIONS
Graphic Representation of Statistics 138
Graphic Representation of Motion 142
Graphic Representation of Equations 146
Solution of Simultaneous Equations by Substitution . . 153
Solution of Simultaneous Equations by Addition or Sub-
traction 156
Problems involving Two Linear Equations .... 160
CONTENTS
XI
Linear Equations in Three Variables
Problems involving Three Unknowns
Keview Questions
PAGE
166
168
171
CHAPTER VI
SPECIAL PRODUCTS AND FACTORS
Principle XIV. Products of Powers of the same Base
Principle XV. Products of Monomials
Factors of Number Expressions
Monomial Factors
Trinomial Squares
The Difference of Two Squares
The Sum of Two Cubes
The Difference of Two Cubes
Trinomials of the Form x2 + (a + 6) x + db
Trinomials of the Form ax2 + bx + c
Factors found by Grouping .
Equations and Problems solved by Factoring
172
175
177
178
179
183
185
186
188
191
193
197
CHAPTER VII
QUOTIENTS AND SQUARE ROOTS
Principle XVI. Quotients of Powers of the same Base . . 210
Principle XVII . Division of Monomials ..... 212
Principle XVIII. Square Roots of Monomials .... 214
Division of Polynomials 216
Square Roots of Polynomials 221
Square Roots of Numbers in the Arabic Notation . . 225
Problems involving Square Roots 234
Solution of Quadratic Equations by Means of Square Root 240
Quadratic and Linear Equations 246
Problems involving Quadratics 247
Review Questions 253
xii CONTENTS
CHAPTER VIII
FRACTIONS WITH LITERAL DENOMINATORS
PAGE
Common Factors 255
Common Multiples 257
Reduction of Fractions to Lowest Terms .... 260
Reduction of Fractions to a Common Denominator . . 262
Addition and Subtraction of Fractions .... 267
Multiplication and Division of Fractions .... 270
Complex Fractions 277
Ratio and Proportion 279
Equations involving Fractions 283
Simultaneous Equations involving Fractions . . . 289
Problems involving Fractions . . . . . . 291
HIGH SCHOOL ALGEBRA
ELEMENTARY COURSE
CHAPTER I
INTRODUCTION TO THE EQUATION
ADDITION OF NUMBERS HAVING A COMMON FACTOR
1. Algebra like arithmetic deals with numbers. The num-
bers of algebra include those used in arithmetic, and also other
numbers which are defined as need arises for them. The sym-
bols employed to represent numbers, and the principles used
in operating upon them, are introduced by means of illustra-
tive problems as we proceed.
2. Illustrative Problem. The shortest railway route from
Chicago to New York is 912 miles. How long does it take a
train averaging 38 miles an hour to make the journey ?
Solution in words. The product of the average number of miles per
hour and the required number of hours equals the whole distance
traveled. That is, 38 multiplied by "the required number of hours "
equals 912. Hence " the required number of hours " is one thirty-
eighth of 912, or 24.
Solution using abbreviations. If, instead of the expression "the
required number of hours," we use the word "time," or simply the
abbreviation t, the solution may be written :
38 x t = 912.
Hence t = 912 + 38 = 24.
1
2 INTRODUCTION TO THE EQUATION
Illustrative Problem. If in the above problem a train makes
the journey in 18 hours, find the average number of miles per
hour.
Solution. In words we have, as before, 18 multiplied by " the aver-
age number of miles per hour " equals 912. Hence " the average
number of miles per hour " equals one-eighteenth of 912, or 50§.
Using for the expression " the average number of miles per hour,"
the word " rate," or simply the abbreviation r, the solution reads :
18 x r = 912.
r * 912 + 18 = 50$.
It should be clearly understood that t and r represent num-
bers whose values are unknown at the outset, but which become
known at the conclusion of the solution.
3. Abbreviations representing numbers and operations such as
are used in these problems occur constantly in algebra. In
fact, the systematic use of such abbreviations is one of the
chief distinctions between arithmetic and algebra.
4. The signs +, — , x , ■+-, and = are used in algebra with the
same meaning that they have in arithmetic. However, instead
of x a point written above the line is often used. Thus 2 • 3
means 2x3.
The product of two numbers represented by letters is gener-
ally indicated by writing the letters consecutively with no sign
between them.
Thus rt means the same as r • t or r x t. For example, in such prob-
lems as those just given we would write rt = d, meaning " the number of
miles per hour multiplied by the number of hours equals the distance
or number of miles traveled."
Similarly 2 r means 2 • r, but (as in arithmetic) 25 means 20 + 5,
not 2 • 5.
A fraction is often used to indicate division
Thus ? = 2-^3; - = r-<.
3 t
ADDITION OF NUMBERS 3
PROBLEMS
In the same manner as above solve the following problems,
first by writing out the solution in words and then by using
such abbreviations as may be found convenient.
1. Five times a certain number equals 80. What is the
number ? Use n for the number.
2. Twelve times a number equals 132. What is the number ?
3. A tank holds 750 gallons. How long will it take a pipe
discharging 15 gallons per minute to fill the tank ?
4. The cost of paving a block on a certain street was $ 7 per
front foot. How long was the block if the total cost was $ 4620?
5. A city lot sold for $ 7500. What was the frontage if
the selling price was $ 225 per front foot ?
6. An encyclopedia contains 18,000 pages. How many
volumes are there if they average 750 pages to the volume ?
7. What is the cost of fencing per rod if it requires $ 256
to fence a quarter section of land ?
8. A square court 60 feet on a side is to be paved with
square tiles. How many square feet in each tile if the pave-
ment requires 1600 tiles ?
9. An excavation for a building is to be 130 feet long,
80 feet wide, and 9 feet deep. If 900 cubic feet are removed
each day, how long will it require to complete it ?
10. A railway embankment which contains 48,900 cubic
yards of earth was completed in 163 days. At what rate was
it filled in ?
11. Taking the length of the earth's orbit as 584 million miles,
find how far the earth travels in one day ; also in one hour.
5. Definition. If a number is the product of two or more num-
bers, then these numbers are called factors of the given number.
E.g. The integral factors of 10 are 1 and 10 or 2 and 5. 2\ and
4 are also factors of 10.
4 INTRODUCTION TO THE EQUATION
6. Illustrative Problem. Divide the number 84 into three
parts such that the second part is five times the first and the
third part is eight times the first.
Solution. Using the abbreviation p for " the first part of the
number," we have 1.porp = the first part>
5p = the second part,
8 p = the third part.
Since the sum of the three parts is 84,
p + 5p + &p = 84.
To complete the solution of this problem it is necessary to find the
sum of the numbers p, 5p, and 8p without knowing what number is
represented by p itself. If we suppose this sum to be lip, then
Up a 84
and p = 84 +■ 14 = 6.
This supposition leads to the correct result since, if p = 6, then
/>+5j9 + 8jp = 6 + 5.6 + 8.6=6 + 30 + 48=84. Hence the three
parts into which 84 is divided are 6, 30, and 48.
7. The process here used for adding p, 5p, and 8p is a new
method of adding which is of very great importance in
algebra. This method is further exhibited by the following
examples :
(1) To add 18, 42, 54, and 30, we first factor these numbers so as
to show the common factor 6 and then add the remaining factors 3, 7,
9, and 5, and multiply this sum by the common factor 6. The result
is 24 • 6, or 144. The work may be arranged as follows:
18 a 3-6 Similarly (2)
42= 7-6
54= 9-6
30= 5-6
16 =
1
16 =
2
8 =
4
1
64 =
4
16 =
8
8 =
16
1
32 =
2
16 =
4
8 =
8
4
48 =
3
16 =
6
8 =
12
4
160 =
10
16 =
20
8 =
40
i
144 = 24 . 6
From example (2) we see that in case the numbers to be
added have two or more common factors it does not matter
ADDITION OF NUMBERS 5
which one is selected. If each number is separated into tivo
factors one of which is a common factor, then it is evident
that the sum in any case is found by multiplying the common
factor by the sum of the other factors.
In the above manner add the following sets of numbers and
show in each case by adding in the ordinary way that the result
is correct :
14
36
32
17
39
32
28
72
64
34
52
48
35
48
128
68
78
40
70
12
256
85
91
72
8. Definition. If a number is the product of two factors,
then either of these factors is called the coefficient of the other
in that number.
E.g. In 2 • 3, 2 is the coefficient of 3, and 3 is the coefficient of 2.
In 9 rt, 9 is the coefficient of rt, r is the coefficient of 9 t, and t is the
coefficient of 9 r. In such expressions as 9 rt the factor represented by
Arabic figures is usually regarded as the coefficient.
The preceding examples illustrate the following principle :
9. Principle I. To add numbers having a common
factor, add the coefficients of the common factor and
multiply tJie sum by the common factor.
If n represents some number, then 4»i, in the same dis-
cussion, means four times that number, and 5 n means five
times that number. Hence, by Principle I, 4 n -f- 5 n = 9 n,
which is true no matter what number is represented by n.
By means of this principle perform the following additions,
understanding that each letter represents some number :
1. Sx + 7 a + 16 a +2 a. 4. 7b+8b +& + 6&.
2. 13 n + 8 n + 7 n + 9 n. 5. 8 £ + 7 t + 5t.
3. 3a + a + 2 a + 6 a. 6. 3r + 5r + llr.
6 INTRODUCTION TO THE EQUATION
Show the correctness of the result in each of the above by
letting x = 2, n = 1, a = 4, 6 = 3, t = 6, and r = 7. Try also
other values for the letters.
Such a test for the correctness of an operation is called a check.
10. Definitions. Combinations of Arabic figures or letters, or
both, by means of the signs of operation, +, — , etc., are called
number expressions.
E.g. 38, 18 r, p + 5 p + 8p, are number expressions.
Two number expressions representing the same number, when
connected by the sign =, form an equality.
The expressions thus connected are called the members of the
equality and are distinguished as the right and left members.
Equalities, such as 8t-\-7 1 = 15t, in which the letters may
be any numbers whatever, are called identities. Equalities of
the type p + 5p + 8p = 84 are called equations. (See § § 29-34.)
EXERCISES
1. Add 5 1, 11 t, 20 t, and 47 t. Check the result by letting
t = 3 ; also t = 50, and t = 150.
2. Add 7 r, 23 r, 28 r, 52 r, and 117 r. Check the result for
r = 11, r = 20, and r = 1.
3. Add 3 rt, 7 rt, 65 rt, and 16 rt. Check for r = 1, t = 2.
4. Add 1^ w, 2| w, 3^ n, and check for n = 6.
5. Add 66, 88, 99, and 121 by Principle I.
6. Add 144, 96, 120, and 50 • 12 by Principle I.
7. Add 5 • 40, 8 . 60, and 6 • 20 by Principle I.
8. Add 5 ax, 3 ax, and 7 ax. Check for a = 2, x = 4.
9. Add 7 ax, 3 bx, and 12 ex.
In this case the common factor is x. Hence using Principle I,
7 ax + 3 bx + 12 ex = (7 a + 3 6 + 12 c)x. The parenthesis here indicates
that the numbers represented by x and the expression 7 a + 3 b + 12 c
are to be multiplied.
ADDITION OF NUMBERS
SOLUTION OF PROBLEMS
11. One great object in the study of algebra is to simplify
the solution of problems. This is done by using letters to repre-
sent the unknown numbers, by stating the problem in the form
of an equation, and by arranging the successive steps of the
solution in an orderly manner.
Skill in translating problems into equations depends upon
attention to the following points :
(1) Read and understand clearly the statement of the prob-
lem, as it is given in words.
(2) Select the unknown number, and represent it by a suit-
able letter, say the initial letter of a word which will keep its
meaning in mind. If there are more unknown numbers than
one, try to express the others in terms of the one first selected.
(3) Find two number expressions which, according to the
problem, represent the same number, and set them equal to
each other, thus forming an equation.
These steps are exhibited in the following solution :
A tree 108 feet high was broken off by the wind so that the
part left above the first branch was three times as long as the
part broken off, and the part below the first branch was twice as
long as the part broken off. How long was the part broken off ?
Solution. Let 6 represent the number of feet broken off.
Then 36 is the number of feet left above the first branch,
and 2 6 is the number of feet below the first branch.
Hence, b + 36 + 26 and 108 are number expressions, each repre-
senting the total height of the tree.
Therefore 6 + 36 + 26 = 108. (1)
By Principle I, 66 = 108. (2)
Then 6 = one sixth of 108, or 18. (3)
Hence, the part broken off was 18 feet long.
Equation (3) is derived from (2) by dividing both members by 6.
8 INTRODUCTION TO THE EQUATION
PROBLEMS
1. The greater of two numbers is 5 times the less, and
their sum is 180. What are the numbers ?
2. A number increased by twice itself, 4 times itself,
and 6 times itself, becomes 429. What is the number ?
3. A father is 3 times as old as his son, and the sum of
their ages is 48 years. How old is each ?
4. In a company there are 39 persons. The number of
children is twice the number of grown people. How many
are there of each ?
5. A and B receive $45 for doing a certain piece of work.
If A gets 4 times as much as B, how much does each receive ?
6. The population of Tokio is twice that of Canton, and the
sum of their populations is 2,700,000. How many inhabitants
in each city ?
7. Find two consecutive integers whose sum is 133.
8. The area of Louisiana is (nearly) 4 times that of
Maryland, and the sum of their areas is 60,930 square miles.
Find the (approximate) area of each state.
9. The horse-power of a certain steam yacht is 12 times
that of a motor boat. The sum of their horse-powers is 195.
Find the horse-power of each.
10. There are three circles on the blackboard. The circum-
ference of the second is 5 times that of the first, and the
circumference of the third is 10 "times that of the first. The
sum of their circumferences is 16 feet. Find the circumfer-
ence of each.
11. At a football game there were 2000 persons. The num-
ber of women was 3 times the number of children, and the
number of men was 6 times the number of children. How
many men, women, and children were there ?
SUBTRACTION OF NUMBERS 9
12. The population of Portland, Oregon (estimate of the
Census Bureau, 1904), was twice that of Dallas, Texas, and the
population of Toledo was 3 times that of Dallas. The
three cities together had 300 thousand inhabitants. How
many were there in each city?
Let >i = number of thousands of inhabitants in Dallas.
Then 2ra = number of thousands of inhabitants in Portland,
and 3 n = number of thousands of inhabitants in Toledo.
Hence n + 2n + 3n = 300.
13. It is twice as far from New York to Syracuse as from
New York to Albany, and it is 4 times as far from New
York to Cleveland as from New York to Albany. The sum of
the three distances is 1015 miles. Find each distance.
14. In Maryland there were (census of 1900) 4 times as
many whites as negroes. The total population was 1185 thou-
sand. How many of each were there ?
If n equals the number of thousands of negroes, then the equation
is n + in = 1185.
SUBTRACTION OF NUMBERS HAVING A COMMON FACTOR
12. Numbers having a common factor may be subtracted in
a manner similar to the process exhibited under Principle I.
Thus, from 64 = 8 • 8 From 84 = 12 - 7 From 17 n
subtract 48 = 6-8 subtract 49 = 7-7 subtract 6n
Eemainder 16 = 2 . 8 35= 5-7 liw
In like manner perform the following subtractions :
1. 9.7-3-7. 5. 6.99-5.99. 9. Qn-2n.
2. 10-4-6-4. 6. 20-19-13-19. 10. 6-50-2-50.
3. 8.8-2-8. 7. 8a -3a. 11. 106-4 6.
4. 5-11-3-11. 8. 8-5-3-5. 12. 7a -4a.
10 INTRODUCTION TO THE EQUATION
These examples illustrate the following principle :
13. Principle II. To find the difference of two numbers
having a common factor, subtract the coefficients of the
common factor and multiply the result by tlie common
factor.
Illustrative Problem. If thirteen times a certain number
diminished by eight times the number equals 75, what is the
number ?
Solution. Let n represent the required number.
Then 13 n — 8 n and 75 are expressions representing the same
number.
Hence, 13 n — 8 n =75.
By Principle II, 5n = 75.
Dividing each member by 5, n =15, the required number.
Check. 13 • 15 - 8 • 15 = 195 - 120 = 75.
EXERCISES AND PROBLEMS
1. By means of Principle II subtract 72 from 160 ;
50 from 300; 39 from 78; 34 from 85; 58 from 174; and
69 from 161.
2. Subtract 109 . 87 from 209 • 87 by Principle II.
Check by first finding the products and then subtracting as in
arithmetic.
Perform the following indicated operations and check
those in which letters are involved by substituting convenient
numbers :
3. 6St-llt. 7. 3.4n+5.4w+11.4w-7-4n.
4. 15w+25w-18n. 8. 13rt+16rt+3rt-20rt.
5. 70aj-15»+7aj-23a?. 9. 144-96+50.12-20-12.
6. 18- 7-3- 7- 2- 7+6- 7. 10. lla«-3aa;+4aa;.
SUBTRACTION OF NUMBERS 11
11. 11 ax — dbx + kcx. 19. ar + br — cr.
12. 11- 9 -6 -9 + 3 -9. 20. 3ry-2sy-ty.
13. 20»-6n + 2w. 21. 11.17 + 47-17-8.17.
14. an — bn + en. 22. accy + bxy — 3 ay.
15. 5* + 2(H-3£. 23. 3a&c + 7a6c-2a&c.
16. 8s-3s + 20s. 24. 7 -5 a -3 -5^ + 8. 5 x.
17. 6a-4a + 3a-2a. 25. a«2r + 6-2r — c-2r.
18. llrs — 2rs + 4rs. 26. 2ar + 26r — 2cr.
27. Four times a certain number plus 3 times the number
minus 5 times the number equals 48. What is the number ?
28. One number is 4 times another, and their difference is 9.
What are the numbers ?
29. Find a number such that when 4 times the number is
subtracted from 12 times the number the remainder is 496.
30. The population of Ohio (1901) was twice that of Wis-
consin. The difference of their populations was 2100 thou-
sand. Find the population of each state.
31. The population of Illinois in 1903 was 5 times as great
as that of West Virginia. The difference between their popu-
lations was 4080 thousand. What was the population of each ?
32. There are three numbers such that the second is 11 times
the first and the third is 27 times the first. The difference
between the second and the third is 64. Find the numbers.
33. A cubic foot of asphaltum is twice as heavy as a cubic
foot of light anthracite coal. Seven cubic feet of coal weigh 390
pounds more than 1 cubic foot of asphaltum. Find the weight
per cubic foot of each.
34. Thirty-nine times a certain number, plus 19 times the
number, minus 56 times the number, plus 22 times the num-
ber, equals 12. Find the number.
12 INTRODUCTION TO THE EQUATION
MULTIPLICATION OF A PRODUCT
14. Illustrative Problem. Three men, A, B, and C, invest
together $ 33,000. B puts in twice as much as A, and C 4
times as much as B. How much does each invest ?
Solution. Let d represent the number of dollars invested by A.
Then 2 d represents B's investment, and 4 (2 d) represents C's invest-
ment.
Hence, d + 2d + 4 (2d)= 33000.
To complete the solution of this problem it is necessary to multiply
2 c? by 4 without knowing what number is represented by d. If we
suppose the product to be 8 d,
then d + 2d + 8d = 33000.
By Principle I, 1 1 d = 33000.
Dividing each member by 11, d = 3000, A's investment ;
2c?= 6000, B's investment ;
4-26? = 24000, C's investment.
The supposition that 4 (2 d) = 8 d is justified by the fact that the
numbers thus found satisfy the conditions of the problem.
That is, 6?+2tf+4(2o?)= 3000 + 6000 + 24000 = 33000.
15. The process here used for multiplying 2 d by 4 is of great
importance in algebra.
The following examples further exhibit this new method of
multiplying :
1. 4(3- 5) =4. 15 = 60 2. 2(3-4.5) = 2-60 =120
Also 4(3-5) = 12- 5 = 60 Also 2(3 • 4 . 5)=6 -4 -5 =120
And 4(3 • 5) = 3 • 20 = 60 And 2(3 • 4 • 5) = 3 • 8 - 5 =120
And 2(3 -4- 5)= 3- 4- 10 = 120
In like manner find the following products in two or more
ways :
3. 6(2-3). 5. 2(5-149). 7. 20(5 a- 4). 9. 8(4*. 3).
4. 4(25-99). 6. 4(19-5). 8. 16 (4 s. 7). 10. 9(xy-5).
MULTIPLICATION OF A PRODUCT 13
These examples illustrate the following principle :
16. Principle III. To multiply the product of several
factors by a given number, multiply any one of the factors
by that number, leaving the others unchanged.
EXERCISES AND PROBLEMS
Multiply as many as possible of the following in two or
more ways. Check where letters are involved.
7. 4(19-25). 13. 2(rs-16).
8. '5 (17- 20 -3). 14. 5(xy-5z).
9. 15(7a&). 15. 7(3-4a&).
10. 3 (4 ran). 16. 9(a-5-xy).
11. 5(abc). 17. 4(125-17).
12. 7(2xy). 18. 40(25-29).
19. There are three numbers whose sum is 80. The second
is 3 times the first and the third is twice the second. What are
the numbers ?
20. There are three numbers such that the second is 8
times the first and the third is 3 times the second. If the
second is subtracted from the third the remainder is 48. Find
the numbers.
21. The population of Bridgeport, Connecticut, is twice that
of Butte, Montana. Three times the population of Bridgeport
plus twice that of Butte equals 320 thousand. Find the popu-
lation of each city.
22. It is 4 times as far from New York City to Cincinnati
as from New York to Baltimore. Twice the distance from New
York to Cincinnati minus 5 times that from New York to Bal-
timore equals 567 miles. How far is it from New York to each
of the other cities ?
1.
7(3-4-5).
2.
8(7-2-3).
3.
9(2-3-4).
4.
5(2 ab).
5.
3(5 xy).
6.
12(8-4-20)
14 INTRODUCTION TO THE EQUATION
23. The population of Hartford, Connecticut, is 3 times that
of Oshkosh, Wisconsin. Four times the population of Hart-
ford plus 5 times that of Oshkosh equals 510 thousand. Find
the population of each city.
24. One cubic inch of emery weighs twice as much as 1
cubic inch of ivory. The combined weight of 10 cubic inches
of each substance is 2.1 pounds. Find the weight per cubic
inch of each.
25. It is twice as far from Boston to Quebec as from Boston
to Albany and 3 times as far from Boston to Jacksonville,
Florida, as from Boston to Quebec. How far is it from Boston
to each of the other three cities, the sum of the distances being
1818 miles ?
26. A cubic inch of porcelain china is twice as heavy as
a cubic inch of ebony, and a cubic inch of rolled zinc is. 3
times as heavy as a cubic inch of porcelain. The combined
weight of 1 cubic inch of each substance is .387 pounds. Find
the weight per cubic inch of each.
MULTIPLICATION OF THE SUM OR DIFFERENCE OF TWO NUMBERS
17. Illustrative Problem. The length and width of a rectan-
gle together equal 58 inches. If the width were 5 inches
greater, the length of the rectangle would then be twice its
width. Find its dimensions.
Solution. Let 10 represent the number of inches in the width. If
it were 5 inches wider, it would then be w + 5 inches. Hence the
< length is 2 (w + 5), the parenthesis indicating that the sum of w and 5
is to be found and the result multiplied by 2.
Hence the sum of the length and width is
id + 2 (to + 5) = 58.
The solution of this problem involves the multiplication of the
number (w + 5) by 2 without knowing what number is represented
by w.
MULTIPLYING A SUM OR DIFFERENCE 15
If we suppose that 2 (w + 5) = 2 w + 10,
then we have w + 2 w + 10 = 58. (1)
By Principle I, 3 w + 10 = 58. (2)
Hence 3w = 48, since 58 is 10 more than 3 w, (3)
and w = 16, the width. (4)
Then 58 - 16 =42, the length.
The correctness of the above process is shown by the fact that the
numbers 42 and 16 fulfill the conditions of the problem ;
that is, 2 (16 + 5) = 2 . 21 = 42,
and 16 + 42 = 58.
Equation (3) is derived from (2) by subtracting 10 from each
member.
18. The multiplication of the number (w + 5) by 2 without
knowing in advance what number is represented by w is a new
method of multiplying which is constantly used in algebra. The
following examples further exhibit this method :
(1) 4(2 + 7) = 4-9 = 36,
or 4(2 + 7)=4- 2 + 4- 7 = 8 + 28 = 36.
(2) 3(3 + 8 + 9) = 3- 20 = 60,
or 3(3 + 8 + 9) =3- 3 + 3 -8 +3- 9 = 9 + 24 + 27 = 60.
It is thus seen that in each case the same result is obtained
whether we first add the numbers in the parenthesis and then
multiply the sum or first multiply the numbers in the parenthe-
sis one by one and then add the products.
Multiply each of the following in two ways where possible :
1. 3(2 + 7). 5. 3(a + 6). 9. a(3 + 7 + 10).
2. 5(3 + 4 + 5). 6. ll(h + k). 10. 15(x + y + z).
3. 8(5 + 9+7). 7. 4(5a + 76 + c). 11. 20(ra+w+_p).
4. 7(6+11+9+9). 8. a(5 + 4 + 7). 12. 9(2r+3s+*).
16 INTRODUCTION TO THE EQUATION
19. In a manner similar to the above the difference of two
numbers represented by Arabic figures may be multiplied by a
given number in either of two ways. ■
E.g. 6(8-3) = 6 -5 = 30,
or 6(8-3) = 6-8-6- 3 = 48-18 = 30.
The same result is obtained whether we first perform the sub-
traction indicated in the parenthesis and then multiply the dif-
ference, or first multiply the numbers separately and then
subtract the products. In the case of numbers represented by
letters evidently the second process only is available.
E.g. 6(r-t) = 6r-6t.
Perforin as many as possible of the following multiplications
in two ways :
1. 7(9-2). 4. 17(18-11). 7. 5(3-1). 10. m{r-s).
2. 12(17-7). 5. 9 (a -2). 8. 3(y-2). 11. x{y-z).
3. 5(12-8). 6. 8 (ft -4). 9. a(c-d). 12. t(u~v).
The second of the above methods is needed in problems like
the following :
Illustrative Problem. The rate of an express train plus that
of a freight is 70 miles per hour. If the rate of the freight
were 7 miles less, the express would be going twice as fast as
the freight. Find the rate of each.
Solution. Suppose the rate of the freight is now r miles per hour.
Then r— 7 is the supposed rate of the freight,
and 2 (r — 7) is the present rate of the express.
Hence r + 2 (r - 7) = 70, the sum of the rates, (1)
and r+ 2r- 14 = 70. (2)
By Principle I, 3 r - 14 = 70. (3)
MULTIPLYING A SUM OR DIFFERENCE 17
Then 3 r = 84, since 70 is 14 less than 3 r. (4)
Hence r = 28, the rate of the freight,
and 2 (28 - 7 ) = 2 • 21 = 42, the rate of the express.
Check. 42 + 28 = 70.
Equation (4) is derived from (3) by adding 14 to both members
and observing that 14 — 14 = 0.
The foregoing examples illustrate the following principle :
20. Principle IV. To multiply the sum or difference of
two numbers by a given number, multiply each of the
numbers separately by the given number, and add or sub-
tract the products.
21. Principles III and IV should be carefully contrasted, as
in the following example :
2 (2 • 3 • 5).= 4- 3- 5 = 2- 6- 5 = 2- 3- 10,
but 2(2 + 3 + 5) =4 + 6 + 10.
In multiplying the product of several numbers we operate
upon any one of them, but in multiplying the sum or difference
of numbers we operate upon each of them.
EXERCISES AND PROBLEMS
1. Multiply 5 + 7 + 11 by 3 without first adding, and then
check by performing the addition before multiplying.
2. Multiply m + n by 4 and check for m = 5, n = 7.
4 (m + n) = 4 m + 4 n
Check. 4 (5 + 7) = 4 • 12 = 48, also
4 • 5 + 4 • 7 = 20 + 28 = 48.
3. Multiply r + s + x by a and check for r ==s = x = a = 2.
4. Multiply a + 6 + c by m and check for a = 1, b = 3,
c = 5, m = 4.
5. Multiply x + y by r and check for x = 2, y = 4, r = 6.
18 INTRODUCTION TO THE EQUATION
Where letters are involved, check the results in the follow-
ing by substituting convenient values :
6. 8(13-5). 16. 8(2a-3&+4c).
7. 71(12 + 41-36). 17. 37(3x-2y-z).
8. 9(o-6 + 8). 18. 13(5r-3x + t).
9. 3(a + x + y -11). 19. 20(2a + 3x - 5y).
10. ra (a + & — c). 20. 7 (11 s — 2 s + 3 t).
11. a(18-7). 21. 35(x-2y + 3z).
12. 2 (13 + 8 + 9 - 21). 22. 78 (10 m + lln + 2 r).
13. 7(3 + 8 + 9 -a). 23. 4(25a; + 32a;-2/).
14. 5(17+ a -6). 24. 3(13 a? + 14 y - t).
15. 32 (a?— y-M). 25. 7(4a-36 + c).
26. Find two consecutive integers such that 3 times the first
plus 7 times the second equals 217.
27. Find two consecutive integers such that 7 times the first
plus 4 times the second equals 664.
28. There are three numbers such that the second is 17 less
than the first, and the third is 8 times the second. The sum of
the first and third is 89. What are the numbers ?
29. The number of representatives and senators together in
the United States Congress is 476. The number of represen-
tatives is 26 more than 4 times the number of senators. Find
the number of each.
30. The area of Illinois is 6750 square miles more than 10
times that of Connecticut. The sum of their areas is 61,640
square miles. Find the area of each state.
31. The sum of the horse-powers of the steamships Campania
and Mauritania is 102 thousand. The Mauritania has 12
thousand horse-power more than twice that of the Campania.
What is the horse-power of each ship?
MULTIPLYING A SUM OB DIFFERENCE 19
32. The population of Wyoming (census of 1900) was 50
thousand more than that of Nevada, and the population of
Utah was 93 thousand more than twice that of Wyoming.
The population of Utah was 277 thousand. Find the popula-
tion of Nevada and Wyoming.
33. It is 73 miles farther from Newark to Philadelphia than
from New York to Newark, and it is one mile more than 10
times as far from Philadelphia to Chicago as from Newark to
Philadelphia. The sum of the distances from New York to
Newark and from Philadelphia to Chicago is 830 miles. Pind
each of the three distances, and the total distance from New
York to Chicago.
Let . d = number of miles from New York to Newark.
Then d + 73 = number of miles from Newark to Philadelphia,
and 10 (d + 73) + 1 = number of miles from Philadelphia to Chicago.
Hence d + 10 (d + 73) + 1 = 830.
34. In the championship season of 1906 the Chicago National
League baseball team lost 20 games less than New York, and
Pittsburg lost 12 less than twice as many games as Chicago.
Pittsburg and New York together lost 116 games. How many
did each of the three teams lose?
35. Pikes Peak is 3282 feet higher than Mt. ./Etna, and
Mt. Everest is 708 feet more than twice as high as Pikes Peak.
The sum of the altitudes of Mt. iEtna and Mt. Everest is
39,867 feet. Find the altitude of each of the three mountains.
36. The altitude of Chimborazo is 22,820 feet less than
3 times that of Mt. Shasta. What is the altitude of each
mountain if Chimborazo is 6060 feet higher than Mt. Shasta ?
Let x = the number of feet in the altitude of Mt. Shasta. Then
3x -22820 = altitude of Chimborazo, and 3a; - 22820 - x = 6060.
37. The distance of Mars from the sun is 39 million
miles less than 5 times as great as that of Mercury from the
sun. Mars is 105 million miles farther from the sun than
Mercury. What is the distance of each planet from the sun ?
20 INTRODUCTION TO THE EQUATION
38. The population of New York City (estimate of Census
Bureau, 1904) was 5 thousand more than 11 times that of
Pittsburg. If 21 times the population of Pittsburg is sub-
tracted from twice that of New York, the remainder is 363
thousand. Find the population of each city.
39. The standing army of France (1906) was 383 thousand
less than twice that of Germany. If 7 times the number of
men in the German army is subtracted from 6 times the num-
ber of men in the French army, the remainder is 177 thousand.
Find the number of men in each army.
DIVISION OF A PRODUCT
22. The division of the product of several factors by a given
number may be performed in various ways :
E.g. (4- 6 -10) --2 = 240-=- 2 = 120.
Also (4.6.10)h-2 = 2-6.10 = 120,
(4.6.10)-2 = 4.3-10 = 120,
and (4- 6- 10)- 2 = 4- 6- 5 = 120.
In each case only one factor is divided, and since any factor
may be selected, we naturally choose one which exactly contains
the divisor if possible.
Perform each of the following divisions in more than one
way where possible :
1. (5 -8- 3) -=-2. 4. (11 • 20 • 16) -- 4. 7. (10 . 35 . 3) -5.
2. 20 abc-h 4. 5. 14 xyz '-*- 7. 8. 14 xyz -=- x.
3. 12a5e-i-3. 6. 12 abc + c. 9. (12 • 40 • 13) -s- 8.
These examples illustrate the following principle :
23. Principle V. To divide the product of several factors
by a given number divide any one of the factors by that
number, leaving the other factors unchanged.
DIVISION OF A PRODUCT 21
Principle V is already known in arithmetic in the process
called cancellation.
2 • 6 • 9
Thus, in the fraction — , 3 may be canceled out of either 6
o
or 9, giving 2'6'9 = 2 . 2 . 9 or 2 • 6 . 3.
O
Principle V is necessary in the solution of problems like the
following :
Illustrative Problem. There are three numbers whose sum
is 26. The second is 8 times the first, and the third is one-half
the second. Find each of the numbers.
Solution. Let x represent the first number,
then 8 x represents the second number,
and — represents the third number.
8a;
Hence, x + 8 x -\ = 26, the sum of the numbers.
By Principle V, x + 8 x + 4 x = 26, since 8 x + 2 = 4 x.
By Principle I, 13 x = 26.
Hence, x = 2, the first number,
8 x = 16, the second number,
o_ 8.9
and — = — — = 8, the third number.
2 2
EXERCISES AND PROBLEMS
1. Multiply 2 x + 3 y by 5. Check for x = 7, y = 11.
2. Multiply 3 a + y by 8. Check f or a = 2, y = 3.
3. Divide 3 • 7 • 18 by 6 by means of Principle V.
4. Multiply 7 • 56 • 5 by 6 by means of Principle III.
5. Divide 21 • 36 • 42 by 7, leaving the result in two differ-
ent forms.
6. Multiply 5-13-27 by 3, leaving the result in three dif-
ferent forms.
22 INTRODUCTION TO THE EQUATION
7. Divide 7 a • 14 b • 21 c by 7 in three different ways.
8. Add 5 a, 1-^, %2, and 1|^, using Principles V and I.
9. From —p^- subtract x^, using Principles V and II.
, _ ^ 14 a . 10 a i , ,6a
10. Irom — — H — — - subtract——
— ■ o o
i , -pi- j j-v, ,. 16 a? 20 a; 16 a; „. •, Q
11. lind the sum of , , , 7 a, and 3 a:.
8 5 4
,. • ja n 100 rs 90 rs , 25 rs
12. Find the sum of , , and — - — •
10 ' 9 5
13. From 25 xy subtract — — 2—
z
i-i a j a 18 abc OA , -, 30 be
14. Add , 20 be, and
a ' 10
15. Add , , and •
s a u
16. From i^M + 17a subtract ^^.
i M -c -(^n .39 mn , , . 25 am
17. From 179 m -\ ■ subtract
n a
18. From M(a + b) + U(a + b) subtract 7 (a + &).
19. Cleveland had (estimate of Census Bureau, 1904) 8 times
as many inhabitants as Portland, Maine. If twice the popu-
lation of Portland is added to ^ that of Cleveland, the sum is
212 thousand. Find the population of each city.
20. One cubic foot of a certain kind of brick weighs as
much as 3 cubic feet of cedar. The combined weight of ^ of a
cubic foot of brick and 5 cubic feet of cedar is 187 pounds.
Find the weight per cubic foot of each.
DIVISION OF A SUM OR DIFFERENCE 23
21. It is 8 times as far from Philadelphia to Louisville as
from Philadelphia to Baltimore. If \ the distance from Phila-
delphia to Louisville is added to 3 times that from Philadel-
phia to Baltimore, the sum is 485 miles. Find each of the two
distances.
22. Coinage silver weighs 4 times as much per cubic inch as
feldspar. The combined weight of \ a cubic inch of silver and
7 cubic inches of feldspar is .846 pound. What is the weight
of a cubic inch of each ?
23. It is 3 times as far from New York to Washington, D.C.,
as from New York to New Haven, and it is 14 times as far
from New York to Seattle as from New York to Washington.
If \ the distance from New York to Seattle is added to 5 times
that from New York to New Haven, the sum is 836 miles.
Find the distance from New York to each of the other cities.
24. The population of Washington, D.C. (estimate of Cen-
sus Bureau, 1904), was 9 times that of Galveston, Texas, and
the population of Savannah, Georgia, was 33 thousand less
than \ that of Washington. The combined population of Gal-
veston and Savannah was 99 thousand. Find the popidation
of each city.
25. A cubic foot of steel weighs 17 times as much as a cubic
foot of yellow pine. The combined weight of 11 cubic feet of
pine and 3 cubic feet of steel is 1773.2 pounds. Find the
weight of 1 cubic foot of each.
DIVISION OF THE SUM OR DIFFERENCE OF TWO NUMBERS
24. In dividing the sum or difference of two numbers by a
given number, when these are represented by Arabic figures,
the process may be carried out in two ways. Thus,
(1) (12 + 8) --2=20-5- 2 = 10,
or (12 + 8) - 2 = 12 + 2 + 8 + 2 = 6 + 4 = 10.
(2) (20 - 12) + 4 = 8 - 4 = 2,
or (20 - 12) -s- 4 = 20 + 4 - 12 + 4 = 5 - 3 = 2.
24 INTRODUCTION TO THE EQUATION
The same result is obtained in each case whether the num-
bers in the parenthesis are first added (or subtracted) and the
result then divided by the given number; or the numbers in
parenthesis are first divided separately and then the quotients
added (or subtracted).
If the numbers in the dividend are represented by letters,
the division can usually be carried out only in the second
manner shown above.
E.g. (r + t)+5 = r + 5+t + 5,or,r-±l = ^ + ±.
o 5 o
This is read : the sum of r and t divided by 5 equals r divided by 5
plus t divided by 5.
In this manner perform each of the following divisions in
two ways when possible :
1. (16 + 12) -h 4. 8. <» + $ + at) ■*- a
2. (20a- 10 6) - 5. 7. (15« - 3t + 18t)+3.
3. (36 x- 24^/) -s-6. 8. (18 a? + 36 y - 21 1) -s- 3.
4. (108y-72«)-s-12. 9. (m - n - r) -s- a.
5. (50 x + 75 x) -5- 25. 10. (m + n + r) -e- b.
These examples illustrate the following principle :
25. Principle VI. To divide the sum, or difference of
two numbers by a given number divide each number sep-
arately and find the sum or difference of the quotients.
Principle VI is necessary in problems such as the following :
Illustrative Problem. The population of Iceland is 40,500 less
than 10 times that of Greenland. The population of Green-
land plus \ that of Iceland is 27,600. Find the population of
each.
DIVISION OF A SUM OR DIFFERENCE 25
Solution. Let x be the number of inhabitants in Greenland.
Then 10 x — 40500 is the number in Iceland.
u , 10 x - 40500 ___-.
Hence x + =27600.
5
By Principle VI, x + 2 x - 8100 = 27600.
By Principle I, 3 x - 8100 = 27600.
Adding 8100 to each member and observing that 8100 — 8100 = 0,
3 x =35700.
Therefore x = 11900, the population of Greenland,
and 10 x — 40500 = 78500, the population of Iceland.
Check. 1 1 900 + ^^ = 27600.
5
26. Principles V and VI should be carefully contrasted :
Thus : 12 * 1S ' 24 = 4- 18 • 24 = 12 • 6 • 24 = 12 • 18 • 8,
o
whi,e 12 + ^ + 24=4 + 6 + 8.
That is, in dividing the product of several numbers we operate
upon any one of them as found convenient, but in dividing the
sum of several numbers we must operate upon each of them.
EXERCISES AND PROBLEMS
1. Divide 72 + 56 by 8 without first adding.
2. Divide 144 — 36 by 12 without first subtracting.
3. Divide r + t by 5 and check the quotient when r = 15,
t = 25; also when r = 60, t = 75.
4. Multiply 7 + 9 by 3 without first adding 7 and 9.
5. Multiply 25 — 8 by 5 without first subtracting.
6. Find the product of 12 and a + b, checking the result
when a = 5, b — 1.
26 INTRODUCTION TO THE EQUATION
Perform the following indicated operations :
7. 3 (a + b + c + d). Check for a = 1, b = 2, c = 3, d = 4.
8. 7(r— s-\-t — x). Check for r = t = 5, s = x = 4.
9. (m + n + r) -h 4. Check for ra = 64, n = 32, »• = 8.
10. (x + y + z) -e- 5. Check for a? = 100, y = 50, and 2 = 25.
11. Add 6(o + b + c) and 6a + 126 + 24c.
o
12. From 25 (x + y + «) subtract 10° x + 50 ? + 25 z.
13 Add 6 ft6 + 7 ac +ad and 12 h + 15 c + 9 d
a 3
14. Perform the divisions: S3x~^V and ™t-39s,
11 13
15. Divide 7 ma$ + 3 wa;?/ — 2 ram/ by y, and check for ra = 2,
n = 3, x = 4, y = 5.
16 Add 21 CTa? + 49 5y + 56 a& , 24 aa; + 56 by + 64 aft
7 n( 8 '
and check for a=b = c = 2, x = y = 3.
17. If the sum of 4 times a number and 32 be divided by 2
the result is 30. Find the number.
18. The melting temperature of glass is 548 degrees (Centi-
grade) lower than 4 times that of zinc. One-half the number
of degrees at which glass melts plus 7 times the number at
which zinc melts equals 3434. Find the melting point of each.
19. The melting temperature of silver is 496 degrees (Centi-
grade) lower than that of nickel. Five times the number of
degrees at which nickel melts plus 7 times the number at
which silver melts equals 13,928. Find the melting point of
each metal.
DIVISION OF A SUM OR DIFFERENCE 27
20. The population of Paris (1904) was 1360 thousand less
than twice that of Berlin. The sum of their populations was
4730 thousand. Find the population of each city.
21. A cubic foot of nickel weighs 1288 pounds less than
4 cubic feet of tin. One-half a cubic foot of nickel plus 1 cubic
foot of tin weighs 724 pounds. Find the weight per cubic foot
of each metal.
22. A cubic foot of gold weighs 2730 pounds less than 6
cubic feet of silver. One-third of a cubic foot of gold together
with 5 cubic feet of silver weighs 3675 pounds. Find the weight
per cubic foot of each metal.
23. The population of Japan (1904) was 106 million less
than 12 times that of Manchuria. If the population of Man-
churia be subtracted from \ that of Japan, the remainder is 12
million. Find the population of each country.
24. The population of Panama (1900) was 1160 thousand
less than 3 times that of Nicaragua (1905). Three times the
population of Panama plus twice that of Nicaragua is 2020
thousand. Find the population of each country.
25. The population of the Philippines (1903) was 400 thou-
sand less than 50 times that of Hawaii. One twenty-fifth of
the population of the Philippines plus the population of Hawaii
is 464 thousand. What was the population of each ?
26. One gallon of benzine weighs 45.4 pounds less than 8
gallons of alcohol. The weight of \ a gallon of benzine and
2 gallons of alcohol is 16.9 pounds. Find the weight of one
gallon of each liquid.
27. During the season 1906 the American League baseball
team of Chicago won 201 games less than 6 times as many as
Boston. One-third the number of games won by Chicago plus
7 times the number of games won by Boston equals 374. How
many games did each team win ?
28 INTRODUCTION TO THE EQUATION
28. The altitude of Aconcagua is 70,800 feet less than 6
times that of Mt. Blanc. One-sixth the altitude of Aconcagua
added to that of Mt. Blanc equals 19,760 feet. Find the alti-
tude of each mountain.
29. The area of Great Britain is 30,387 square miles less
than 12 times that of the Netherlands, and the area of Japan
is 42,065 square miles less than 15 times that of the Nether-
lands. One-third the area of Great Britain plus £ the area
of Japan is 69,994 square miles. Find the area of each
country.
30. The diameter of the earth is 1918 miles more than
twice that of Mercury, and the diameter of Venus is 1700
miles more than twice that of Mercury. The diameter of the
earth plus \ that of Venus equals 11,768 miles. Find the
diameter of each planet.
31. The diameter of Jupiter is 500 miles more than 20
times that of Mars, and the diameter of Saturn is 4200 miles
more than 16 times that of Mars. One-tenth the diameter of
Jupiter plus \ that of Saturn is 45,150 miles. Find the
diameter of each planet.
32. The diameter of Neptune is 29,000 miles less than
twice that of Uranus. One-half the diameter of Neptune plus
4 times that of Uranus is 145,000 miles. Find the diameter
of each planet.
From the last three problems make a table of the diameters
of all the planets.
NUMBER EXPRESSIONS IN PARENTHESES
27. When a number expression is inclosed in a parenthesis,
it is sometimes possible to perform the operations indicated
within the parenthesis and thus to remove it.
E.g. 6 + (7 + 10 - 9) = 6 + 8 = 14.
REMOVAL OF PARENTHESES 29
When, however, the operations within the parenthesis can-
not be carried out, the parenthesis may sometimes be removed
by Principle IV or VI.
E.g. 5(2 n + 7 r) = 10 n + 35 r, and (20 x - 16 y) + 4 = 5 x - 4 y.
In these examples the operations upon the parentheses are
multiplication or division. The cases in which the operations
on the parentheses are addition or subtraction are considered
in the following examples :
(1) 5 + (3 + 4) = 5 + 7 = 12,
also 5 + (3 + 4) = 5 + 3 + 4 = 8 + 4 = 12.
(2) 10 + (7-3) = 10 + 4 = 14,
also 10 + (7-3)=10 + 7-3 = 17-3 = 14.
(3) 20- (7 +2) =20-9 = 11,
also 20 - (7 + 2) = 20 - 7 - 2 = 13 - 2 = 11.
(4) 20-(7-2)=20-5 = 15.
also 20-(7-2)=20-7+2 = 13 + 2 = 15.
From (1) it appears that the expression, 3 + 4, may be added
to 5 by first adding 3 and then adding 4.
From (2) it is seen that the expression, 7 — 3, may be added
to 10 by first adding 7 and then subtracting 3.
From (3) it is seen that the expression, 7 + 2, may be sub-
tracted from 20 by first subtracting 7 and then subtracting 2.
From (4) it is evident that the expression, 7 — 2, may
be subtracted from 20 by first subtracting 7 and then adding 2,
since 7 is 2 more than the number which was to be subtracted.
In like manner perform each of the following operations in
two ways :
1. 18 + (5 + 2 + 3). 4. 40 -(6 + 7+3). 7. 25n+(22w-3rc).
2. 28 + (7 -3 + 4). 5. 55 -(16 -7). 8. 18*-(6a;-2aj).
3. 32 -(5 + 3). 6. 101 + (73 -22). 9. 16r-(7r + 2r).
30 INTRODUCTION TO THE EQUATION
These examples illustrate the following principle :
28. Principle VII. A number expression consisting of
two or more numbers connected by the signs + or — may
be added to anotiier given number by adding each number
with the plus sign and subtracting each number with the
minus sign. Such a number expression may be sub-
tracted from a given number by subtracting each number
with the plus sign and adding each number with the
minus sign.
Principles IV, VI, and VII are useful in removing parenthe-
ses from number expressions when the values of the numbers
involved are not known.
Instead of a parenthesis, a bracket [ ], or a brace { }, may be used.
Thus, 2 (6 + 4) = 2 [6 + 4] = 2 {6 + 4}.
In expressions involving parentheses the operations within
the parentheses should be performed first if possible. Then
perform the indicated multiplications and divisions, and finally
the remaining additions and subtractions.
E.g. Given 7 + 15 - (9 - 4) - 4 [7 + 11] + (29 - 20).
Performing operations within the parentheses, this reduces to :
7 + 15-5-4-18^9.
Performing multiplications and divisions, we have 7 + 3 — 8.
Performing the remaining additions and subtractions, the given
expression reduces to 2.
EXERCISES
In performing the following indicated operations, state in
each case what principles are used.
1.8(7+4 + 8 + 2). 8. 6(7-3)+2.
3. 3(2a-4)-4(a-6). 4. 14 + 4 (8 + 4) -=- (32-20).
The minus sign preceding 4 (a — 6) indicates that this whole
product is to be subtracted. Hence, using Principle IV, we have
6a— 12 — (4a —24). Then, using Principle VII, this becomes
6a-12-4a + 24 = 2a + 12.
REMOVAL OF PARENTHESES 31
5. 8-3(4-2)+6 + 3[6 + l].
6. 16 -5- [3 + 1]- [12 -3j -4-3.
7. 5 (17 - 5) + 18 -=- (8 - 2) - (21 + 14) -=- 7.
8. 7(a + 6)+6(a + 6).
9. 5(a+6)+ 3 (a + b + c).
10. 16(r + *) + H {r4 0- 11. 15s-3(r + s).
12. 15 (r + s) -3(r-s). 13. 20-(cc + w),
14. 50x-25(x-y). 15. 12 m - 6(* - 2m).
16. ll< + 5(2* -1) -3(2 + 2). Check fori = 2.
17. (12* -6m) -r-2 + 3« — 2m. Check for t = 1, u = 1.
18. 3(5a-7y) + (21a-28M) -e- 7. Check for a; = 2, m = 1.
19. 8(r — s) + 5(2r + s) — 3 (r + s). Check for r = 1, 8 = 1.
20. 10(a; + y) -r- 5 + 6 (a? - y) -*- 3. Check f or cc = 2, m = 1.
21. 5(h+k) + 2(h+k)+3(h + 7c).
Use Principle I, then IV ; also IV, then I.
22. 5(7x-4:y)+9(9x-5y).
23. 8(r-s)-2(r-s). 24. 9(3p- q)-S(3p-q).
Use Principle II, then IV ; also reverse this order.
25. (16 x -12 x) -f- 4. 26. (18 a6 — 12 a6a?) -s- a.
Use first Principle VI, then II ; also II, and then V.
27. 9(6 + 4) -3 (6 -4). 28. (5 axy - 3 ex) -=- x.
29. 9 (6 abc + 4: xyz) — 3(6 abc — 4 an/z).
30. 8(5a;-M + 2z)-ll(3a; + 2M-7z).
31. 3[2(a + 6 + c)-3(a-6 + c)].
First remove the parentheses, then the bracket.
32. 7\Sx-(2y-3x)+(2x-Ay)'i.
32 INTRODUCTION TO THE EQUATION
PROBLEMS
1. There are three numbers whose sum is 80. The second
is 3 times the first, and the third twice the second. What are
the numbers ?
2. A man has three buildings whose total value is $46,800.
The second building cost $ 800 less than the first, and the third
cost twice as much as the second. What is the cost of each
building ?
3. The population of Connecticut (Census of 1900) was 50
thousand more than twice that of Rhode Island, and the popu-
lation of Massachusetts- was 81 thousand more than 3 times
that of Connecticut. The population of Massachusetts minus
that of Connecticut was 1897 thousand. Find the population
of each state.
4. The population of Indiana (Census of 1900) exceeded that
of Iowa by 284 thousand, and the population of Illinois was
210 thousand less than twice that of Indiana. The population
of Illinois minus that of Indiana was 2306 thousand. Find
the population of each state.
5. The melting point of copper is 250 degrees (Centigrade)
lower than 4 times that of lead. Ten times the number of
degrees at which lead melts minus twice the number at which
copper melts equals 1152. What is the melting point of each
metal ?
6. The melting point of iron is 450 degrees higher than 5
times that of tin. Three times the number of degrees at
which iron melts plus 7 times the number at which tin melts
equals 6410. Find the melting point of each metal.
7. In 1904 the gold product of Africa was 11 million dol-
lars more than 3 times that of Russia, and the gold product of
Australia was 84 million less than twice that of Africa.
Russia and Australia together produced 113 million. How
much did each country produce ?
REMOVAL OF PARENTHESES 33
8. In 1904 the value of the silver produced in the United
States was 5 million dollars more than 14 times as much as
that of Canada, and the product of Mexico was 1 million less
than 16 times that of Canada. Twice the product of the
United States minus that of Mexico equals 71 million. How
much did each country produce ?
9. The Nile is 100 miles more than twice as long as the
Danube. Ten times the length of the Danube plus 4 times the
length of the Nile equals 3400 miles. How long is each river ?
10. The number of first-class battleships in the United States
navy (1906) was 22 less than 5 times the number of protected
cruisers. Twelve times the number of cruisers less twice the
number of battleships equals 60. Find the number of each.
11. The number of torpedo boats in the United States navy
(1906) was 77 less than 7 times .as great as the number of
torpedo boat destroyers. Three times the number of torpedo
boats minus 4 times the number of destroyers equals 41. Find
the number of each.
12. The weight of a cubic foot of spruce is 16 pounds more
than that of a cubic foot of cork, and the weight of a cubic
foot of dry live oak is 5 pounds more than twice that of spruce.
One cubic foot of oak weighs 36 pounds more than one cubic
foot of spruce. Find the weight of a cubic foot of each.
13. In 1904 Canada produced 3 million dollars more of
gold than Mexico, and the United States produced 17 million
more than 4 times as much as Canada. The combined product
of Mexico and the United States was 94 million. How much
did each country produce ?
14. The value of the copper produced in the United States
in 1904 was 5 million dollars more than the value of the crude
petroleum, and the value of the bituminous coal was 94 million
more than twice the value of the copper. Nine times the value
of the copper minus twice the value of the coal equals 342
million. Find the value of each.
34 INTRODUCTION TO THE EQUATION
15. The pressure developed in the chamber of a modern
twelve-inch gun is 21 thousand pounds per square inch more
than that developed in an ordinary hunting rifle. The
maximum pressure which gunpowder can develop is 18
thousand pounds per square inch less than 3 times as great
as that developed in the twelve-inch gun. The sum of
the three pressures is 151 thousand pounds. Find each
pressure.
16. The standing army of Australia (1906) was 14 thousand
more than that of Canada, and the standing army of Great
Britain was 47 thousand more than 4 times that of Australia.
Great Britain's army contained 287 thousand men. How many
men were there in the armies of Canada and Australia ?
IDENTITIES AND EQUATIONS
29. In the preceding pages equations have been freely
used, and certain simple methods have been employed in
solving problems by means of them. We now proceed to a
more detailed study of these methods and of the equation
itself.
30. Equalities in which letters are used as number symbols
are of two kinds as shown by the following examples:
(1) 3n + 4n = 7»isa true statement no matter what number is
represented by n. 3 (5 x + 6y) = 15 x + 18 ?/ holds for all values
which may be assigned to x and y.
(2) to + 2 (w + 5) = 58 is a true statement if, and only if, w = 16. If
w is replaced by any number less than 16, the number expression on
the left is less than 58, and if w is replaced by any number greater
than 16, the result is greater than 58.
31. The equality w -f 2 (w -f 5) = 58 is said to be satisfied by
w=16, because this value of w reduces both members to the
same number, 58.
IDENTITIES AND EQUATIONS 35
32. Definition. An equality which is satisfied, no matter
what numbers are substituted for one or more of its letters,
is called an identity with respect to those letters.
E.g. 3n + 4n = 7nisan identity with respect to n.
a (b + c) = ab + ac is an identity with respect to a, b, and c.
When it is especially desired to distinguish an equality as an
identity, the equality sign is written =.
Each of the Principles I to VII may be thus stated in symbols as
identities. Thus,
I, 4n + 6n = 10n; II, 12 n - 5n=7n;
III, 5 (4 ab)=20 ab ; IV, 5 (a ± b)=5 a±5b;
V, 30a&^-6 = 5aZ>; VI, (16 a ± 206) + 4=4 a ± 56;
VII, x + (a - b) - (c - d)=x + a-b-c + d.
33. Definition. An equality which is satisfied only when
certain particular values are given to one or more of its letters
is called a conditional equality with respect to those letters.
E.g. 3 x + 5 = 35 is an equality only on the condition that x = 10.
x + y = 10 is an equality for certain pairs of letters like 1 and 9, 2 and 8,
3 and 7, 5 and 5, but certainly not for all values of x and y ; for instance,
not for 3 and 8.
34. A conditional equality is called an equation, and a letter
whose particular value is sought to satisfy an equation is called
an unknown number in that equation, or simply an unknown.
The equations at present considered contain only one un-
known.
35. To solve an equation in one unknown is to find the value
or values of the unknown which satisfy it.
The following example exhibits the process of solving an
equation which contains one unknown and which is satisfied
by one value only of the unknown :
%
36 INTRODUCTION TO THE EQUATION
Given w + 2 (w + 5) = 58. (1)
By Principle III, to + 2 to + 10 = 58. (2)
By Principle I, 3 to + 10 = 58. (3)
Subtracting 10 from both members, 3 to = 48. (4)
Dividing both sides by 3, w = 16. (5)
Check. Putting w = 16 in (1), 16 + 2 (16 + 5) = 16 + 2 . 21 = 58.
Explanation. Any value of to which satisfies (1) also satisfies (2)
and (3), since the expressions in the left members of (1), (2), and (3)
represent the same number expressed in different forms. Any value
of to which satisfies (3) also satisfies (4), for if 10 more than 3 to is
58, then 3 to must be 48. Any value of to which satisfies (4) also
satisfies (5), for if 3 to is 48, then to is | of 48.
By similar considerations the value of to which satisfies (5) may be
shown to satisfy (4), (3), (2), and (1). Hence to = 16 is the solution
of the given equation.
The above solution illustrates the following principle :
36. Principle VIII. An equation may be changed into
another equation such that any value of the unknown
which satisfies one also satisfies the other, by means of any
of the following operations :
(1) Adding the same number to both members.
(2) Subtracting the same number from both members.
(3) Multiplying both members by the same number.
(4) Dividing both members by the same number.
(5) Changing the form of either member in any way
which leaves its value unaltered.
The operations under Principle VIII are hereafter referred
to in detail by means of the initial letters, A for addition, S for
subtraction, M for multiplication, D for division, and F for
form changes.
Note. — It is not permissible to multiply or divide the members of
an equation by any expression which is equal to zero. See Advanced
Course.
SOLUTION OF EQUATION H 37
37. The operations involved in Principles I to VII are
all form changes which leave the value of the expression
unaltered.
E.g. 4 (vi + n) by Principle IV has the same value as 4 m + 4 n.
There are other form changes which are already familiar in
arithmetic.
E.g. 2 + 4 = 4 + 2, or in general a + b = b + a. Likewise 2-7 = 7-2,
or in general ab = ba. That is, the order in which numbers are added
or multiplied is immaterial.
Again, 3 + 4 + 6 = (3 + 4) + 6 = 3 + (4 + 6), or in general a + b + c
= (a + 6) + c = a + (6 + c), and likewise a -b • c = (a • b) • c = a • (6 • c) ;
i.e. numbers to be added or multiplied may be grouped in any manner
desired. Still other form changes will be learned later.
38. Principle VIII is further illustrated as follows :
On the scale pans of a common balance are placed objects of
uniform weight, say tenpenny nails. The scales balance only
when the weights are the same in both pans ; that is, when the
number of nails is the same.
If now the scales are in balance, they will remain so under
two kinds of changes in the weights :
(a) When the number of nails in the two pans is in-
creased or diminished by the same number ; correspond-
ing to the operations A, S, M, D, on the members of an
equation.
(6) When the number in each pan is left unaltered but
the nails are rearranged in groups or piles in any manner ;
corresponding to the operations F on the members of an
equation.
The equation, then, is like a balance, and its members are
to be operated upon only in such ways as to preserve the
balance.
38 INTRODUCTION TO THE EQUATION
DIRECTIONS FOR WRITTEN WORK
39. The solution of an equation consists in deducing, by
means of Principles I to VIII, another equation whose first
member contains the unknown only and whose second member
does not contain the unknown. The successive steps should
be written down as in the following example :
Solve 6 (4 n - 3) + 25 (n + 1) = 50 + 31 ft + 2 (3 - ft)- 9. (1)
By F, using III and IV, we obtain from (1)
24 n -18 + 25 n + 25 = 50 + 31n + 6-2 n- 9. (2)
By F, I and II, we obtain from (2)
49 » + 7 = 29 n + 47. (3)
Subtracting 7 and 29 n from each member of (3) and using Prin-
ciple II, we have 20n = 40. (4)
Dividing each member of (4) by 20,
n=2. (5)
Check. Substitute n = 2 in Equation (1).
For convenience this work can be abbreviated as follows:
6(4n - 3) + 25 (n + 1) = 50 + 31 n + 2(3- n) - 9. (1)
By F, III, IV, 24 n - 18 + 25 n + 25 = 50 + 31 n + 6 - 2 n - 9. (2)
By F, I, II, 49n + 7 = 29n + 47. (3)
By S 1 7, 29 n, 20n = 40. (4)
ByD|20, n = 2. (5)
S | 7, 29 n means that 7 and 29 n are each to be subtracted from
both members of the preceding equation. D 1 20 means that the
members of the preceding equation are to be divided by 20.
Similarly, in case we wish to indicate that 6 is to be added to each
member of an equation, we would write A I 6, and if each member is
to be multiplied by 8, we would write M\ 8. It is important that the
nature of each step be recorded in some such manner.
+ 4 n - 6 = 20 + 5n - 5 + 3 n.
(2)
21 n + 2 = 15 + 8 n.
(3)
21n-8n = 15-2.
(4)
13 n = 13.
(5)
n = 1.
(6)
SOLUTION OF EQUATIONS 39
(2) Solve 17 n + 4(2 + ft) - 6 = 5(4 + n) - 5 + 3 w. (1)
By F, IV, 17 n
By F, I,
By S | 2, 8 n,
By F, II,
By Z> 1 13,
Check. Substitute n = 1 in equation (1);
then 17 + 12 - 6 = 25 - 5 + 3,
or 23 = 23.
An equation may be translated into a problem. For example,
the equation 21a; + 2 = 15 + 8ic may be interpreted as follows :
Find a number such that 21 times the number plus 2 is 15
greater than 8 times the number.
EXERCISES AND PROBLEMS
Solve the following equations, putting the work in a form
similar to the above and checking each result. Translate the
first twenty into problems.
1. 13a-40-tt = 8.
2. 3z + 9 + 2a; + 6 = 18 + 4:C.
3. 5 x + 3 — x = x-^-18.
4. 13y + 12 + 5y=32+8y.
5. 4m + 6ra + 4 = 9m + 6.
6. 7m-18 + 3m = 12 + 2m + 2.
7. 3y-4; + 2y-6 = y + 7 + y + 3 + 10.
8. 5« + 3 + 2a; + 3 = 2a; + 5 + 3a; + 3 + a;.
9. 2a; + 4a;4-9 + a; + 6 = 20 + 3a; + 5 + 2£c.
10. 18 + 6m + 30 + 6m = 4m + 8+12 + 3m + 3 + m + 29.
40 INTRODUCTION TO THE EQUATION
11. y + 72 + 45y = 106 + 12y.
12. 42 a -56 = 20 a: + 10.
13. 6x + 8-3x + 4: + 5x = 7x + 32 + x-20.
14. 32x-4 + 7x = B8 + 3x + 5x.
15. 12m-3-3ra = 32 + 2m.
16. 15ra + 3-2m + 7 = 3m + 60.
17. a + 7+3a = 2a + 45.
18. 5 6-30 + 66 = 36 + 90.
19. 3c + 18+14c = 6c + 51.
20. 17x- + 4 + 3x = 7a; + 30.
21. 7(m + 6) + 10 m = 42 - 8(2 m + 2) + 181.
22. 20 - 3(x - 4) + 2 x = 2 x + 17.
23. (8x + 4)-=-2 + 7a; = 4z + 9.
24. 6 a + 4(4 x + 2) = 85 - 3(2 x + 7).
25. 8 + 7(6 + 6 n) + 2 w = 2(4 n + 5) + 18 n + 49.
26. 5(9a: + 3) + 6a: = 24a:-4(3a; + 2) + 36.
27 7(12 «; +8) + 13 + 5 a; - 6 = 47. Apply V and VI.
28. 12(5 + 4o;)_5(6 + 4a;)+5-()==a; + 18j
29 15 , 21(3 + *) 2(6 + 18^)^3(9^+12) 2S
7 3 3 "*"
__ ll(5z + 25), 3(6^-2) 7(4a; + 8),12a;+36jW
30. ^ + — — -J = r-^ + _ +35.
SOLUTION OF EQUATIONS 41
EQUATIONS INVOLVING FRACTIONS
40. Solve w + ^ + ^ = 88. (1)
First Solution. The coefficients of n are 1, f, and J.
Applying Principle I, If n = 88. (2)
By D I If , n = 88 + If = 48. (3)
Second Solution. Multiply both members of (1) by 6.
That is, by 1/ 1 6, 6n+^ + ^= 528. (2)
By F, V, 6n+3n+2n= 528. (3)
By F, I, 11 n = 528. (4)
By D | 11, n = 48, as before. (5)
The object is to multiply both members of the equation by such a
number as will cancel_each denominator. Hence the multiplier must
contain each denominator as a factor.
Evidently 12 or 18 might have been chosen for this purpose, but not
8 or 10. 6 is the smallest number which will cancel both 2 and 3, and
hence this was chosen as the multiplier.
41. The process explained in the second solution above is
called clearing of fractions.
As another illustration solve the equation
Here the smallest multiplier available is 36.
Hence by M|36, §^L* + ?<L£ + Hl&£ - 89 • 65i (2)
By V, each denominator is cancelled by the factor 36,
9 • 3 x + 18 x + 4 • 5 x = 36 • 65. (3)
By III, 27 x + 18 x + 20 x = 36 • 65. (4)
By I, 65 x = 36 . 65. (5)
ByD,V, x = §6^65 = 36< (6)
o5
Check. Substitute x = 36 in (1).
42 INTRODUCTION TO THE EQUATION
EXERCISES AND PROBLEMS
Solve the following equations, indicating the principles
used at each step. Check each solution by substituting in
the original equation the value obtained. Translate each
into a problem.
For instance, from equation 2: Find a number such that when
increased by its half, its third, and its fourth, the sum is 25.
1. V- = 5.
2^3
3. ^ + 4n-^=^ + 26.
4 3 2
2 3 4 10
lb 5 10
6. 4n + ^ = 3ft + 2
7 2
7. 12 | 4(9a; + 6) 2(3 + lla;)_5(4a; + 4) | 11
o o o
8. l3& + 7-^±^ = ?A±9 + 38.
7 5
5 a -f 7 . 2a + 4 = 3a + 9 ^ .
2 3 4
10 17a~ 5 10a + 2 = 5a + 7 g
3 4 " 2
11. ^±3 + 5(2a;-fl0) = 2a; + 24
— 5
SOLUTION OF EQUATIONS 43
12. The sum of two numbers is 12, and the first number is
\ as great as the second. What are the numbers ?
13. The smaller of two numbers is f of the larger. If their
sum is 66, what are the numbers ?
14. Find two consecutive integers such that 4 times the
first minus 3 times the second equals 9.
15. Find three consecutive integers such that \ of the first
plus the second minus ^ the third equals 5.
16. Find three consecutive integers such that 3 times
the first plus 9 times the second minus 4 times the third
equals 73.
17. There are three numbers such that the second is 4 more
than 9 times the first, and the third is 2 more than 6 times
the first. If -|- of the third is subtracted from \ of the second,
the remainder is 3. Find the numbers.
18. There are three numbers such that the second is 2 more
than 9 times the first and the third is 7 more than 11 times the
first. The remainder when 4 times the third is subtracted from
13 times the second is 144. Find the numbers.
19. The population of Philadelphia (estimate of Census
Bureau, 1904) was 508 thousand less than 20 times that of
Dayton, Ohio. The population of St. Louis was 245 thousand
more than 4 times that of Dayton. One-half the population of
Philadelphia plus ^ that of St. Louis was 851 thousand. Find
the population of each city.
20. The shortest railway route from Boston to Chicago
is 166 miles more than 4 times that from Boston to New
York ; and the shortest route from Boston to Atlanta is
196 miles less than 6 times that from Boston to New York.
The distance from Boston to Chicago is 481 miles more than
\ the distance from Boston to Atlanta. Find each of the three
distances.
44 INTRODUCTION TO THE EQUATION
Solution. Let d — distance in miles from Boston to New York ;
then 4 d + 166 = distance in miles from Boston to Chicago,
and 6 d — 196 = distance in miles from Boston to Atlanta,
and 4rf + 166 = 6J~196 + 481.
By VI, 4 d + 166 = 3 d - 98 + 481.
By S, II, d = 481 - 98 - 166 = 217 = distance from Boston
to New York.
4 d + 166 = 1034 = distance from Boston to Chicago,
and 6 d — 196 = 1106 = distance from Boston to Atlanta.
21. The railway mileage in the United States in 1900 was
8 thousand greater than twice that of 1880, and in 1904 it was
66 thousand less than 3 times that of 1880. One-half the mile-
age of 1900 plus \ that of 1904 equals 168 thousand. Find
the railway mileage in 1880, 1900, and 1904.
22. The number of newspapers in the United States in 1880
was 381 less than 4 times that in 1850. The number in 1905
was 700 more than twice that of 1880 and 7990 more than
6 times that of 1850. Find the number of newspapers in 1850,
1880, and 1905.
23. The number of telephones in the United States in 1900
was 40 thousand less than 30 times as great as the number in
1880. Half the number in 1905 was 660 thousand more than
that in 1900 and 80 thousand more than 40 times that in 1880.
Find the number of telephones in use in 1880, 1900, and 1905.
24. The displacement of the battleship Kearsarge is 1232
tons greater than that of the Oregon, and the displacement
of the Connecticut is 4480 tons greater than that of the
Kearsarge. One-third the displacement of the Kearsarge
minus \ that of the Connecticut equals 640. Find the dis-
placement of each ship.
25. The distance of the fixed star Vega from the sun is 11.7
light-years greater than the distance from the sun to Sirius.
The distance of Regulus is 8 light-years less than twice that of
SOLUTION OF EQUATIONS 45
Vega. What is the distance of each star from the sun if \ the
distance of Vega minus \ that of Regulus is 2 light-years ?
Light travels at the rate of 186,000 miles in one second. A light-
year is the distance traveled by light in one year.
26. The distance from the sun to the fixed star 61 Cygni is
4 light-years more than that from the sun to Alpha Centauri
(the star nearest the sun), and the distance from the sun to
Polaris (the pole star) is 6 times that of 61 Cygni. One-
eighth the distance of Polaris minus \ the sum of the dis-
tances of the other two equals 2 light-years. How many
light-years is each star from the sun?
27. In 1905 the wheat crop of Russia was 47 million bushels
less than 4 times that of Argentina, and the crop of the United
States was 77 million bushels less than 5 times that of Argentina.
The crop of the United States exceeded that of Russia by 124
million bushels. What was the wheat crop of each country ?
28. The total imports of the United States in 1900 were 60
million dollars less than 10 times the imports in 1800. The
imports in 1906 were 473 million less than twice the imports of
1900, and 135 million more than 12 times the imports of 1800.
Find the imports of the United States in 1800, 1900, and 1906.
29. The total exports of the United States in 1900 were 26
million dollars less than 20 times the exports in 1800. The
exports in 1906 exceeded those of 1900 by 350 million and were
31 million less than 25 times those of 1800. Pind the exports
of 1800, 1900, and 1906.
30. The gross tonnage of the German merchant marine was
(1906) 548 thousand more than that of the United States, and
the tonnage of the British marine was 2662 thousand greater
than 4 times that of the German marine. The British marine
was greater by 930 thousand tons than twice that of Germany
and 3 times that of the United States combined. Find the
tonnage of each marine.
46 INTRODUCTION TO THE EQUATION
31. The population of San Francisco (estimate of Census
Bureau, 1904) was 289 thousand more than that of Springfield,
Massachusetts, and the population of Chicago was 132 thousand
more than 5 times that of San Francisco. The population of
Springfield minus -£-$ that of Chicago was 2 thousand. Find
the population of each city.
32. The per capita tax of New York City (1905) was $ .49
less than twice that of Chicago and the per capita tax of Boston
was $ 3.56 less than 3 times that of Chicago. The combined
per capita tax of Boston and New York was $ 52.15. What
was the per capita tax of each city ?
33. The number of pupils attending public high schools in
Chicago (1905) was 1237 more than twice as great as in Phila-
delphia. One-third the number of pupils in Chicago minus \
the number in Philadelphia was 2524. How many pupils
attended public high schools in each city?
34. The number of pupils in public high schools in Boston
(1905) was 1574 less than 4 times as many as in Baltimore.
Three times the number in Boston minus 6 times the number
in Baltimore equals 5268. How many pupils in each city ?
35. The distance from San Francisco to Yokohama is 2031
statute miles less than 3 times that from San Francisco to
Honolulu, and the distance from San Francisco to Hongkong is
2219 more than twice that from San Francisco to Honolulu.
One-fifth the distance to Hongkong plus \ the distance to
Yokohama equals 3152. Find the distance from San Francisco
to each of the other three cities.
36. The distance from New York to Singapore is 542 statute
miles more than 3 times that from New York to London, and
the distance from New York to Manila is 5342 miles less than
5 times that from New York to London. Three times the dis-
tance from New York to Manila plus 6974 miles equals 4 times
the distance from New York to Singapore. Find the distance
from New York to each of the other three cities.
CHAPTER II
POSITIVE AND NEGATIVE NUMBERS
42. Thus far the numbers used have been precisely the same
as in arithmetic, though their representation by means of
letters and some of the methods used in operating upon them
are peculiar to algebra.
43. A new kind of number will now be studied in connection
with problems of the following type.
Illustrative Problems. 1. If a man gains' $1500, and then
loses $ 800, what is the net result ? Answer, $ 700 gain.
2. The assets of a commercial house are $ 250,000, and the
liabilities are $ 275,000. What is the net financial status of
the house ? Answer, $ 25,000 net liabilities.
3. The thermometer rises 18 degrees and then falls 28
degrees. What direct change in temperature would produce
the same result ? Answer, 10 degrees fall.
4. A man travels 700 miles east and then 400 miles west.
What direct journey would bring him to the same final desti-
nation ? Answer, 300 miles east.
In the statement of each of these problems the numbers are
applied to things which are opposite in quality ; namely, gain
and loss, assets and liabilities, rise and fall, distances measured
in two opposite directions, as east and west.
The answer in each case not only gives the proper arithmeti-
cal number, but also connects with this number one of the two
opposite qualities involved in the problem.
E.g. The answer in (1) is $700 gain and not simply f 700; in (2)
it is not $25,000, but $25,000 liabilities; in (3) it is not 10°, but 10°
fall ; and in (4) it is not 300 miles, but 300 miles east.
47
48 POSITIVE AND NEGATIVE NUMBERS
These problems are illustrations of the sense in which the
word " opposite " is here used.
E.g. Gain and loss are opposite in that they annul each other
when taken together ; rise and fall of temperature are opposite in that
one counteracts the other.
44. In all problems involving one or both of two qualities
which are opposite in this sense there is constant need to dis-
tinguish which one is meant.
E.g. On a day in winter it is not sufficient to say the temperature
is 5° ; we must specify whether it is above or below zero.
In the case of the thermometer we call the temperature positive if
it is above zero and negative if it is below zero.
These words positive and negative are used to describe all pairs of
qualities which are opposite in the sense here understood.
The signs + and ~ stand respectively for the words "posi-
tive " and " negative."
E.g. $ + 700 is read positive $ 700, and means either $700 gain or $700
assets, according to the problem in which it occurs. Likewise $~700
is read negative $700, and means either $700 loss or $700 liability.
Referring to the thermometer, +18° is read positive 18°, and means
either a temperature of 18° above zero or a rise of the mercury 18° from
any point. The latter is the meaning in Problem 3. Similarly ~28°
means either a temperature 28° below zero or a, fall of the mercury 28°
from any point.
It is commonly agreed to call gain positive and loss negative,
assets positive and liabilities negative, above zero positive and
below zero negative, motion upward positive and downward
negative, motion to the right positive and to the left negative.
45. Numbers marked with the sign + are called positive and
those marked with the sign ~ are called negative.
The positive sign maybe omitted; that is, a number with
neither sign written is understood to be positive.
ADDITION OF SIGNED NUMBERS 49
ADDITION OF POSITIVE AND NEGATIVE NUMBERS
46. While in each of the problems on page 47 the result
was obtained by subtracting one number from the other,
yet they are not properly subtraction problems, but addition
problems.
E.g. In Problem 1, we are not asking for the difference between
$ 1500 gain and $800 loss, but for the net result when the gain and the
loss are taken together ; that is, the sum of the profit and loss. Hence
we say $1500 gain + $800 loss = $700 gain, or, using positive and
negative signs, + 1500 + ~ 800 = +700. .
Similarly in Problem 2,
$250,000 assets + $275,000 liabilities = $25,000 net liabilities,
Or + 250,000 + - 275,000 = - 25,000.
In Problem 3, 18° rise + 28° fall = 10° fall,
Or +18 +-28 = -10.
In Problem 4,
700 miles east + 400 miles west = 300 miles east,
Or +700 + -400 = +300.
Note. — In the last case east is called + and west -, since, as a map
is usually held, east is to the right and west to the left. Likewise
north or up is called + and down or south is called -.
PROBLEMS
1. A balloon which exerts an upward pull of 460 pounds is
attached to a car weighing 175 pounds. What is the net
upward or downward pull? Express this as a problem in
addition, using positive and negative numbers.
Solution. 460 lb. upward pull + 175 lb. downward pull = 285 lb. net
upward pull. Using positive numbers to represent upward pull and
negative numbers to represent downward pull, this equation becomes
+ 460 + -175 = +285.
50 POSITIVE AND NEGATIVE NUMBERS
2. How would the morning paper print the following
thermometer readings from the weather report ? Jacksonville
38° above zero, Seattle 5° below zero, Nashville 28° above zero,
Chicago 13° below zero.
3. The temperature rises 15° and then falls 24°. What
direct change in temperature would produce the same final
result ? Express this as an example in addition.
4. A vessel on the equator sailed north 3° and was then
forced south 5° by a hurricane. It then resumed its course
northward 8°. Express this as a sum and show the final posi-
tion with reference to the equator. (See note, § 46.)
In each of the following translate the solution into the language of
algehra by means of positive and negative numbers, as in Problem 1.
5. A 450-pound weight is attached to a balloon which
exerts an upward pull of 600 pounds. What is the net up-
ward or downward pull ?
6. A man's property amounts to $45,000 and his debts to
$ 52,000. What is his net debt or property ?
7. The assets of a bankrupt firm amount to $245,000 and
the liabilities to $ 325,000. What are the net assets or liabili-
ties ?
8. A weight exerting a downward pull of 280 pounds
when submerged in water is attached to a float which will just
support a weight of 240 pounds. What is the net pressure
upward or downward when both are submerged ?
9. A man on the deck of a steamer is walking at the rate
of 4 miles an hour toward the stern. If the boat is sailing
eastward in a river at the rate of fifteen miles per hour, what
is the actual motion of the man with respect to the bank of
the river ?
10. A man can row a boat at the rate of 6 miles per hour.
How fast can he proceed against a stream flowing at the rate
of 2^ miles per hour ; 7 miles per hour ?
ADDITION OF SIGNED NUMBERS 51
11. A steamer which can make 12 miles per hour in still
water is running against a current flowing 15 miles per
hour. How fast and in what direction does the steamer
move ?
12. A dove capable of flying 40 miles per hour in calm
weather is flying against a hurricane blowing at the rate of 60
miles per hour. How fast and in what direction is the dove
moving ?
13. If of two partners, one loses $ 1400 and the other gains
$ 3700, what is the net result to the firm ?
14. A man's income is $ 2400 and his expenses $ 1500 per
year. What is his saving ?
15. A man loses $ 800 and then loses $ 600 more. What is
the combined loss ? Indicate the result as the sum of two
negative numbers.
16. A man gains $ 500 and then gains $ 700 more. What
is the combined gain ? Express the result as the sum of two
positive numbers.
17. A tug which can steam 9 miles per hour in still water is
going down a stream whose current is 6 miles per hour. How
fast is the tug moving ?
18. The thermometer falls 8° and then 17° more. Express
the combined result of the two changes as a negative number.
47. Definitions. Positive and negative numbers are some-
times called signed numbers, because each such number consists
of an arithmetic part, together with a sign of quality.
The arithmetic part of a signed number is called its absolute
value.
Thus, the absolute value of + 3 and also of ~ 3 is 3. Two
signed numbers are of like quality when they have the same
signs, and of opposite quality when one is positive and the
other negative.
52 POSITIVE AND NEGATIVE NUMBERS
In the one case we say they have like signs, in the other op-
posite or unlike signs. The preceding exercises illustrate the
following principle :
48. Principle IX. To add two signed numbers of like
quality, find the sum of their absolute values, and
prefix to this the common sign of quality.
To add two signed numbers of opposite quality, find
the difference of their absolute values, and prefix to
this the sign of that one whose absolute value is the
greater.
In case their absolute values are equal, their sum is
zero.
The sum of two signed numbers thus obtained is called their
algebraic sum.
49. This principle may be further illustrated by means of
the following graphic representation of signed numbers.
On an unlimited straight line call some starting point zero,
and lay off from this point equal divisions of the line indefi-
nitely both to the right and to the left, as shown in the figure.
■i o +i
-HH-
In order to describe the position of any one of these division
points, we require not only an integer of arithmetic, to specify
how far to count from the starting point (the point marked
zero) in order to reach the given point, but also a quality sign,
to indicate the direction of the counting.
E.g. + 7 marks the division point 7 units to the right of zero, and
- 5 marks the point 5 units to the left of zero. Such a diagram is
called the scale of signed numbers.
The part of the scale to the right, taken alone, would need no
signs, and would picture the integral numbers of arithmetic, while the
ADDITION BY COUNTING 53
two parts together require the distinguishing signs of quality, and
picture the integral numbers of algebra.
Fractions would of course be pictured at points between the inte-
gral division points, on the right or the left of the scale, according
as the fractions are positive or negative.
ADDITION BY COUNTING
50. In arithmetic two numbers are added by starting with
either and counting forward as many units as there are in the
other.
E.g. To add 3 and 5 we start with 5 on the number scale and
count 6, 7, 8 ; or start with 3 and count 4, 5, 6, 7, 8.
51. In algebra two signed numbers are added in the same
manner except that the direction, forward or backward, in which
we count, is determined by the sign, + or ~, of the number
which we are adding.
Thus, to add + 7 and +5, begin at 7 to the right of the zero point in
the scale of signed numbers and count 5 more toward the right, or
begin at 5 to the right and count 7 more to the right, in either case
arriving at + 12.
To add _7 and ~5, we may begin at 7 to the left and count 5 more
toward the left, or begin at 5 to the left and count 7 more in that di-
rection, in either case arriving at _ 12.
To add + 7 and _ 5, begin at 7 to the right and count 5 toward the
left, or begin at 5 to the left and count 7 toward the right, in either
case arriving at +2.
To add _ 7 and + 5, begin at 7 to the left and count 5 toward the
right, or begin at 5 to the right and count 7 toward the left, in either
case reaching - 2.
I.e. +7 + +5 = +12, -7 + -5 = ~12, +7 + "5 = +2, ~7 + +5 = -2.
52. In adding several signed numbers, we reach the same
result whether we add them in the order in which they happen
to be given, or in any other order. Thus we may add first all
54 POSITIVE AND NEGATIVE NUMBERS
the positive numbers and then all the negative numbers, and
finally combine these two results.
E.g. + 5 + ~8 + +7 + -6, taken in order from left to right, gives -3
as the sum of the first two, +4 as the sum of the first three, and ~2 as
the final result. But +5 + +7 s + 12, -8 + "6 = "14, and +12 + -14 =
-2, the same result as before.
53. Addition by counting makes it clear that two numbers
of opposite signs tend to cancel each other when added.
E.g. In adding ~o to +8 we start with +8, which we reach by count-
ing from zero eight units to the right, and then add ~5 by counting
five units to the left, thus retracing five of the units just counted.
That is, -5 annuls five of the +8 units, leaving the sum +3.
In case two numbers have opposite signs and equal absolute
values one completely cancels the other.
E.g. +8 +-8 = 0.
EXERCISES AND PROBLEMS
Perform the following indicated additions :
1. -31 + +42. 3. +68 + ~46. 5. +104 + ~245.
2. "17 + -13. 4. +34 + "46. 6. -llf + +8£.
7. +16 + ~3 + +7 + +4 + -19. 8. +42 + +74 + -92 + -7 + -3.
9. -13n + +7n = +7n + -13n = -6n.
Solution. By Principle IX, to add ~13 n to +7 n or to add +7 n to
-13 n is to "find the difference of their absolute values and prefix to
this the sign of that one whose absolute value is the greater." That is,
-13n + +7n = -(13 n— 7n) = _6n, since by Principle II, 13/i — 7n = 6n.
10. +14* + -8* 4- "3* + ~45*. 11. +68x++3±x+-16x+-3x.
12. 296r* + -367r* + -119rt. 13. ~3 (a + 6) + +4 (a + 6).
14. +7(x — y) + -5(x — y). 15. "81 (r + s) + ~91 (r + s).
16. A man travels 10 miles east, then 23 miles west, and
finally 5 miles east. Express his final distance and direction
from the starting point as the sum of three signed numbers.
AVERAGES OF SIGNED NUMBERS 55
17. The temperature falls 35°, rises 24°, falls 3°, rises 17°,
and finally falls 5°. What is the net change in temperature
between the last reading and the original reading ?
18. A real estate firm gains $5000 on one transaction, loses
$2500 on a second, loses $1400 on a third, and gains $200 on a
fourth. What is the aggregate result of the four transactions ?
AVERAGES OF SIGNED NUMBERS
54. Half the sum of two numbers is called their average.
Thus 6 is the average of 4 and 8. Similarly, the average of
three numbers is one-third of their sum, and in general the
average of n numbers is the sum of the numbers divided by n.
Find the average of each of the following sets :
1. 10, 12, 14, 16, 18. 4. 7, 9, 11, 13, 15.
2. 5, 9,20,30, 3. 5. 7,10,21,29,30.
3. 11, 10, 4, 6, 5. 6. 13, 8, 9, 10, 4.
The average gain or loss per year for a given number of
years is the algebraic sum of the yearly gains and losses di-
vided by the number of years.
Illustrative Problem. A man lost $400 the first year, gained
$300 the second, and gained $1000 the third. What was the
average loss or gain ?
Solution. -400 +. + 300 + +1000 = +900 = +30Q
3 3
That is, the average gain is $ 300.
Illustrative Problem. During four years a certain business
shows an average annual gain of $ 5000. What was the loss
or gain the first year, if for the remaining years there were
gains of $8000, $9000, and $7500 respectively?
Solution. Let x represent the number of dollars gained or lost the
first year. Then the average for the four years is
x + +8000 + +9000 + +7500
56 POSITIVE AND NEGATIVE NUMBERS
But the average for the four years is given as $ 5000.
Hence « + + 8000 + +9000 + + 75Q0 = +5000.
4
By If, x + + 8000 + +9000 + +7500 = +20000.
By IX, x + + 24500 = + 20000.
By A, x + + 24500 + - 24500 = + 20000 + -24500.
By IX, a; =-4500.
Hence there was a loss of S$ 4500 the first year.
PROBLEMS
1. Find the average of $1800 loss, $3100 loss, $6800 gain,
$10,800 loss, and $31,700 gain.
2. Find the average of $180 gain, $360 loss, $480 loss,
$100 gain, $700 gain, $400 gain, $1300 loss, $300 gain, $4840
gain, and $ 12,000 gain.
3. A merchant gained an average of $ 2800 per year for 5
years. The first year he gained $3000, the second $1500, the
third $4000, and the fourth $2400. Did he gain or lose and
how much during the fifth year ?
4. A certain business shows an average gain of $4000
per year for 6 years. During the first 5 years the results
were, $8000 loss, $10,000 gain, $7000 gain, $3000 gain,
and $12,000 gain. Find the loss or gain during the sixth
year.
5. Find the average of the following temperatures : 7 a.m.,
"4°; 8 a.m., -2°; 9 a.m., -1°; 10 a.m., +1°; 11 a.m., +5°;
12 m., +7°.
6. During the 12 hours ending at 6 a.m., January 19, 1892,
the U. S. Weather Bureau at Helena, Montana, recorded the
following temperatures: "9°, "8°, "8°, "9°, "9°, ~9°, "8°, +36°,
+ 37°, +40°, +20°, +16°. Find the average temperature for the
12 hours.
AVERAGES OF SIGNED NUMBERS 57
Find the average yearly temperatures at the following
places, the monthly averages having been recorded as given
below :
7. For New York City: +29°, +33°, +39°, +46°, +53°, +63°,
+ 67°, +67°, +61°, +52°, +47°, +41°.
8. For Singapore, Straits Settlement : + 81°, + 84°, * 85°, + 85°,
+85°, +86°, +86°, +87°, +85°, +84°, +83°, +81°.
9. For St. Vincent, Minnesota: "5°, 0°, +15°, +35°, +55°,
+ 60°, +66°, +63°, +55°, +40°, +22°, +5°.
10. For Nerchinsk, Siberia: "23°, -13°, "10°, +35°, +55°,
+ 70°, +70°, +64°, +50, +30°, +5°, ~15°.
If the latitude of a place is midway between the latitudes of
two given places, then its latitude equals one-half the algebraic
sum of the two given latitudes.
Thus, if the latitude of one place is +16° and that of the
other " 56°, then the latitude of the place midway between them
is"56° + +16O = ^==-20°.
2 2
The data used in some of the following problems vary
slightly from the published records, but in no case more
than 5'.
11. The latitude of New Orleans, Louisiana, is +30°, and of
Toronto, Canada, +43° 40'. Find the latitude of Norfolk, Vir-
ginia, which is midway between these latitudes. (Positive
numbers represent north latitude and negative numbers repre-
sent south latitude.)
12. The latitude of Alexandria, Egypt, is +31° 10', and of
Christiania, Norway, +59° 50'. That of Venice, Italy, is mid-
way between these latitudes. Find the latitude of Venice.
13. The longitude of Edinburgh, Scotland, is "3° 10', and of
Warsaw, Poland, +21°. Find the longitude of Bremen, Ger-
many, which is midway between these. (Positive numbers
represent east longitude and negative numbers west longitude.)
58 POSITIVE AND NEGATIVE NUMBERS
14. The longitude of Vienna, Austria, is +16° 20', and of
Splugen Pass, Switzerland, +9° 20'. The longitude of Splilgen
Pass is midway between the longitudes of Vienna and Paris,
France. Pind the longitude of Paris.
15. The longitude of Brussels, Belgium, is +4° 20', and of
Jena, Germany, +11° 40'. The longitude of Brussels is mid-
way between those of Jena, and Liverpool, England. Find
the longitude of Liverpool.
16. The longitude of Berlin, Germany, is +13° 20', and of
St. Petersburg, Russia, +30° 20'. The longitude of Berlin is
midway between those of St. Petersburg and Madrid, Spain.
Find the longitude of Madrid.
17. The latitude of Eome, Italy, is +41° 55', and of Cux-
haven, Germany, +53° 55'. The latitude of Rome is midway
between those of Cuxhaven and Cairo, Egypt. Find the
latitude of Cairo.
18. The longitude of Calais, France, is +1° 55' and of Luck-
now, India, +80° 55'. The longitude of Calais is midway be-
tween those of Lucknow and Washington, D.C. What is
the longitude of Washington?
SUBTRACTION OP SIGNED NUMBERS
55. In arithmetic the accuracy of subtraction is tested by
showing that the remainder added to the subtrahend equals
the minuend.
E.g. We say 8 - 5 = 3 because 5 + 3 = 8.
Indeed, we may, and often do perform subtraction by start-
ing with the subtrahend and counting until we reach the
minuend.
E.g. A clerk in changing a dollar bill after a purchase of 63 cents
might count out two pennies, a dime, and a quarter, saying "65, 75,
one dollar." That is, he counts from 63 cents (the subtrahend) up
to 100 cents (the minuend).
SUBTRACTION OF SIGNED NUMBERS 59
56. In like manner, signed numbers also may be subtracted,
but we must find not only the number of units, but also the
direction from the subtrahend to the minuend. This is most
easily done by reference to the number scale, page 52.
Ex. 1. +8 — +5 = +3, since in passing from +5 to +8 on the
number scale we count 3 units in the positive direction.
Ex. 2. -8 — -5 = ~3, since from ~5 to "8 we count 3 units in
the negative direction.
Ex. 3. "8 — +5 = "13, since from +5 to "8 we count 13 units
in the negative direction.
Ex. 4. +8 — ~5 = +13, since from ~5 to +8 we count 13 units
in the j>ositive direction.
57. In each of these examples we see that the result fulfills
the test of arithmetic ; namely,
Subtrahend + Difference = Minuend
Hence _8 — ~5 = -3 is checked by finding that _5 + -3 = ~8.
+8 - -5 = +13 is checked by finding that "5 + +13 = +8.
EXERCISES AND PROBLEMS
Perform the following indicated subtractions by finding the
distance and direction on the number scale from subtrahend
to minuend, and apply the check to each result.
1. -10- "5 6. "17 --20 11. +93 -+22
+6 --14 12. +17 --13
+7 - -9 13. -78 - -37
11- +6 14. +57 -+84
-21- "6 15. "48 --31
16. How many degrees, and in what direction, must the
temperature change in order to vary from 12° below zero to
38° above zero ? This is an example in subtraction, since we
are required to find a number which added to "12° gives +38°.
Hence we write it +38° - ~12° = what ?
2.
-15- +5
7.
3.
+20 - -15
8.
4.
+11- +3
9.
5.
-11- +5
10.
60 POSITIVE AND NEGATIVE NUMBERS
17. How many degrees, and in what direction, does the ther-
mometer change in passing from 27° above zero to 3° below
zero ? That is, ~3° - +27° = what ?
18. What must be added to $35 loss to make the sum $30
gain ? That is, +30 - -35 = what ?
19. What must be added to $ 15 gain to make the sum $ 8
loss ? That is, "8 - +15 = what ?
58. Subtraction always possible. In arithmetic subtraction is
possible only when the subtrahend is less than or equal to the
minuend.
E.g. 5 from 8 leaves 3, 5 from 5 leaves 0 ; but we cannot take 5
from 2, since when we have subtracted 2 units from 2 there are no
more to take away.
However, by means of negative numbers we can as easily
perform the subtraction, 2 minus 5, as 5 minus 2.
Thus, 2 - 5 = -3, since -3 + 5 = 2; and 5 - 2 = +3, since, +3 + 2
= 5.
It thus appears that, in terms of signed numbers, a — b has
a meaning no matter what numbers are represented by a and
b ; that is, a — b means the number which added to b gives a.
59. A short rule for subtraction. Since +8 — ~5 = +13, and
since +8 +• +5 = +13, it follows that subtracting ~5 from +8
gives the same result as adding +5 to +8. Similarly ~8 — _5 =
-3 and "8+-+5 = -3.
Hence subtracting a negative number is equivalent to adding
a jiositive number of the same absolute value.
Since +8 -+5 = +3, and since +8 + -5 = +3, it follows that
subtracting +5 from +8 gives the same result as adding ~5 to +8.
Similarly ~8 - +5 = ~13 and "8 + ~5 = "13.
Hence subtracting a positive number is equivalent to adding
a negative number of the same absolute value.
SUBTRACTION OF SIGNED NUMBERS 61
These statements are illustrated by such facts as : Removing
a debt is equivalent to adding property and removing property is
equivalent to adding debt.
Perform the following subtractions as explained in this
paragraph, by changing the sign of the subtrahend and adding:
1. "5 - -2. 5. +57 - -32. 9. +37 -+50.
2. "4 - +1. 6. "32 - +34. 10. -23 - +57.
3. -5- +2. 7. -52 -"32. 11. -16 a- +4 a
4. +3 _ -5. 8. "16 - "12. 12. +13 * - ~20 1
The preceding exercises illustrate the following principle :
60. Principle X. To subtract one signed number from
another signed number, add the subtrahend with its sign
changed to the minuend.
The change in the sign of the subtrahend may be made men-
tally without re-writing the problem. The results are to be
checked by showing that the difference added to the subtrahend
equals the minuend.
EXERCISES
1. Prom +6 + ~2 subtract "14.
2. From "6 a -f +2a subtract "14 a.
3. From 17 ab+ 8 ab subtract "35 ab.
4. From 5 ax + 4 ax subtract 7 ax + 2 ax.
5. From 54 abc +~47 abc -f 36 abc subtract 80 abc.
6. From 54 . 13 +~47 .13 + 36-13 subtract 80 . 13.
7. From 29 • 3 • 11 + 37 • 3 • 11 subtract "34 . 3 • 11.
8. From 29 xy + 37 xy subtract ~34 xy.
9. Solve x + 8 = 4.
Solution. Subtract +8 from each member (which is equivalent to
adding -8).
62 POSITIVE AND NEGATIVE NUMBERS
Then x = +4 - +8 = ~4, which is correct, since ~4 + +8 = +4. This
is a problem in subtraction, since one of two numbers, 8, and their
sum, 4, are given, and we are to find the second number, which is
represented by x.
Solve the following equations :
10. x + -'3 = 7.
16.
x+ 9= 3.
11. a; + -9 = 1.
17.
-4 + x = 7.
12. 3+ x = 0.
18.
-5 + a = 4.
13. x + -1 = 2.
19.
-20+ * = -12.
14. a: +13 =7.
20.
8+ n=-l&
15. x + 4=2.
21.
ft + -40 = -65.
MULTIPLICATION OF SIGNED NUMBERS
61. The multiplication of signed numbers is illustrated by
the following problems.
Illustrative Problem. A balloonist, just before starting, makes
the following preparations : (a) He adds 9000 cubic feet of gas
with a lifting power of 75 pounds per thousand cubic feet ; (b)
He takes on 8 bags of sand, each weighing 15 pounds. How
does each of these operations affect the buoyancy of the
balloon ?
Solution, (a) A lifting power of 75 lbs. is indicated by + 75, and
adding such a power 9 times is indicated by + 9. Hence + 9 • +75 =
+ 675, the total lifting power added.
(b) A weight of 15 lbs. is indicated by ~15, and adding 8 such
weights is indicated by + 8. Since the' total weight added is 120 lbs.,
we have + 8 • "15 = - 120.
Illustrative Problem. During the course of his journey this
balloonist opens the valve and allows 2000 cubic feet of gas to
escape, and later throws overboard 4 bags of sand. What
effect does each of these operations produce on the balloon ?
MULTIPLICATION OF SIGNED NUMBERS 63
Solution, (a) The gas, being a lifting power, is positive, but the
removal of 2000 cubic feet of it is indicated by ~2, and the result is a
depression of the balloon by 150 lbs. ; that is, _2 • + 75 = _150.
(b) The removal of 4 weights is indicated by ~4, but the weights
themselves have the negative quality of downward pull. Hence to
remove 4 weights of 15 lbs. each is equivalent to increasing the
buoyancy of the balloon by 60 lbs; that is, ~4 • _15 a +60.
62. These illustrations of multiplying signed numbers are
natural extensions of the process of multiplication in arith-
metic.
E.g. Just as 3 • 4 = 4 + 4 + 4 = 12, so 3 . -4 = -4 + -4 +~4 = -12,
and since 3 • 4 is the same as + 3 • + 4, we write + 3.- + 4 = +12.
Again, just as we take the multiplicand additively when the
multiplier is a positive integer, so we take it subtractively when
the multiplier is negative integer.
E.g. -3-+4 means to subtract +4 three times; that is, to sub-
tract + 12. But to subtract +12 is the same as to add ~12. Hence
-3- +4 =~12. Again, -3 • _4 means to subtract - 4 three times;
that is, to subtract -12. But to subtract -12 is the same as to add
+12. Hence -3- -4 = +12.
EXERCISES AND PROBLEMS
Explain the following indicated multiplications and find the
product in each case :
1.
-3 • -10.
5.
43 • -192.
9.
71 • -x.
2.
-3- +10.
6.
-27 • -235.
10.
-112 • ~t.
3.
-5 • +50.
7.
~5-+r.
11.
-U-y.
4.
-75 • -89.
8.
+ 16- -r.
12.
-20 • -v.
13. A man gained $212 each month for 5 months, then lost
$175 per month for 3 months. Express his net gain or loss
as the sum of two products.
64 POSITIVE AND NEGATIVE NUMBERS
14. A man gained $2100 during a certain year. For the
first 4 months he lost $125 per month. During the next 5
months he gained $500 per month. Find his gain or loss dur-
ing the remaining 3 months of the year. Express the net gain
as the sum of two products.
15. A raft is made of cork and iron. What effects are pro-
duced upon its floating qualities by the following changes ?
(a) Adding 4 braces, each weighing (under water) 5 lbs. (b) Ee-
moving 3 pieces of cork, each capable of sustaining 3 lbs.
(c) Adding 10 pieces of cork, each capable of sustaining 7 lbs.
16. What are the effects on a shipwrecked man's ability to
float ? (a) If he holds fast to 3 bags of gold, each weighing 1 0 lbs.
(b) If he ties on two life preservers, each capable of supporting
15 lbs. (c) If he throws away his two boots, each weighing 2 lbs.
The preceding exercises illustrate the following principle :
63. Principle XI. If two signed numbers are of the same
quality, their product is positive; if they are of opposite
quality, their product is negative. The absolute value of
the product is the product of the absolute values of the
factors.
In applying this principle observe that the sign of the prod-
uct is obtained quite independently of the absolute value of
the two factors.
E.g. f • -5 = -('£) = -3f ; "12 • ~3.5 =+42.
64. Principle XI is also stated in symbols as follows :
+a • +6 = +ab, ~a • ~b —+ab, +a • ~b =~ab, ~a • +b =~ab.
The product of several signed numbers is found as illustrated
in the following :
-2 • +5 • -3 • -4 • +6 = -10 . -3 • -4 • +6 = +30 . ~4 • +6 =
_120 • +6 = ~720. That is, any two factors are multiplied
together, then this product by another factor, etc., until all the
factors are multiplied.
DIVISION OF SIGNED NUMBERS 65
65. Evidently the factors in such a product may be taken in
any desired order. Let the student try other orders in the
above example.
Since the product of all positive factors is positive, the final
sign depends upon the number of negative factors. If this
number is even, the product is positive ; if it is odd, the product
is negative.
E.g. If there are 5 negative factors, the product is negative; if
there are 6, it is positive.
EXERCISES
In the following exercises determine the sign of the product
before finding its absolute value. State each principle used in
the reduction to the final form.
1. -4- +3- ~6 .~7. 9. «(3a+"4a-+5a).
2. -50 • -20 • "30 • -40. 10. 8 (16 x + "20 x) -s- 4.
3. ~a- ~b -+c-+d-+e. 11. 5x — 3("2a; + +3:»— -4a>).
4. a>-b-+c--d--x. 12. 6r + 4(3r — "5r+-7r).
5. -5 ("3 +"7). 13. -5 ("4 • +3 . -2).
6. a(-b — ~c). 14. 6x— "14 x— (~5 x +~1 x).
7. -c{x-~y). 15. 8 y -16 y +(4 y + "11 y).
8. 1 a{x+-y—~z). 16. lit — 20 (* +' -3 • -5 *).
DIVISION OF SIGNED NUMBERS
66. In arithmetic we test the correctness of division, by
showing that the quotient multiplied by the divisor equals the
dividend.
E.g. 27 -r- 9 = 3, because 9 -3 = 27.
Hence division may be defined as the process of finding one
of two factors when their product and the other factor are given.
66
POSITIVE AND NEGATIVE NUMBERS
This definition also applies to the division of signed numbers.
In dividing signed numbers, however, we must determine the
sign of the quotient as well as its absolute value.
E.g. -42- +6 = -7, because "7- + 6 = -42;
also -42 -=- -6 = +7, because +7- -6 = -42.
So in every case the test is :
Quotient x Divisor
Dividend.
In like manner perform the following :
1. -26 + 5. 4. -9rs-=-+3.
2. -ab + a. 5. + 75y+~15.
3. + 5xy+~x. 6. -121 x + +11.
The preceding exercises illustrate the following principle :
67. Principle XII. The quotient of two signed numbers
is positive if the dividend and divisor have the same sign,
negative if they have opposite signs. The absolute value of
the quotient is the quotient of the absolute values of divi-
dend and divisor.
EXERCISES
Perform the following indicated divisions. Check by multi-
plying quotient by divisor.
-99 x -=--25.
• 87 y ■+■ -400.
8*-f-4y)-f--2.
16 a + -20 b) -=- -4.
6r + 9s--120n--3.
7 aa;+-14 ay-~21 &z)-=-+7.
12 xy — ~3 ax) •+■ ~x.
27 ab - 36 ac) -=- "9.
3-4?/ + 6-8a)-f--3.
x — 3 y — z) -f- a.
1.
-28 -s- + 7.
11. 1
2.
-42 -s- -6.
12. 1
3.
51 -=--17.
13. (<
4.
21xy + 3.
14. ('
5.
"16a&-s--4.
15. (
6.
" 15 ax -=- cc.
16. (
7.
-32 (a- 6) +
_(a-
-&).
17. C
8.
21(x + y) +
9.
18. (!
9.
4.-9z-=--3.
19. (.
10.
3-8*-s--4
20. 2
INTERPRETATION OF SIGNED NUMBERS 67
68. While Principles I-VIII were studied in connection
with unsigned, or arithmetic numbers only, it is now very
important to note that they all apply to signed numbers as
well. The form changes described in § 37 also apply to signed
numbers just the same as to arithmetic numbers.
In the statement of these principles the word number will
from now on be understood to refer either to the ordinary num-
bers of arithmetic or to the signed numbers, as occasion may
require. It should also be noticed that the numbers of arith-
metic are used as freely in algebra as in arithmetic. It is only
when we wish to distinguish them from negative numbers that
they are called positive numbers.
The number system of algebra, as far as we have studied it,
consists of the numbers of arithmetic together with the negative
numbers.
INTERPRETATION AND USE OF NEGATIVE NUMBERS
69. Illustrative Problem. Divide 34 units into two parts such
that one part is equal to the remainder when 3 times the other
part is subtracted from 46.
Solution. Let the two numbers be represented by x and 34 — x.
Then 34-x = 46-3x. (1)
S|34, -x = 12-3z. (2)
,4|3ar, 8*-x=12. (3)
Principle II, 2 x = 12. (4)
D\2, x = Q. (5)
Substituting x = 6 in (1), 34 - 6 = 46 - 18, or 28 = 28.
In the above solution, equation (2) was obtained from (1) by sub-
tracting 34 from each member. This would clearly be impossible
without the use of negative numbers.
In this case the problem itself does not involve negative numbers,
but in the course of its solution they naturally occur. If the nega-
tive number could not be used, we should be compelled to keep the
68 POSITIVE AND NEGATIVE NUMBERS
members of each equation positive or zero. This would be impossible,
since we do not know what numbers are represented by the letters
involved, and hence cannot tell by inspection whether a given term
is positive or not.
70. We have seen how naturally the use of signed numbers
has arisen in problems where things of opposite qualities have
to be distinguished.
In solving a problem, a negative result may have a natural
interpretation or it may indicate that the conditions of the
problem are impossible.
A similar statement holds in reference to fractional answers in
arithmetic. For example, if we say there are twice as many girls as
boys in a schoolroom and 35 pupils in all, the number of boys would
be 35 -^- 3 = llf, which indicates that the conditions of the problem
are impossible.
71. Illustrative Problem. The crews on three steamers to-
gether number 94 men. The second has 40 more than the first,
and the third 20 more than the second. How many men in
each crew ?
Solution. Let n = number of men in first crew.
Then n + 40 = number of men in second crew,
and n + 40 + 20 = number of men in third crew.
Hence n + n + 40 + n + 40 + 20 = 94,
and 3 n + 100 = 94.
n s= -2.
Here the negative result indicates that the conditions of the
problem are impossible.
72. Illustrative Problem. A real estate agent gained $ 8400 on
four transactions. On the first he gained $6400, on the second
he lost $ 2100, on the third he gained $ 5000. Did he lose or
gain on the fourth transaction ?
INTERPRETATION OF SIGNED NUMBERS 69
Solution. Since we do not know whether he gained or lost on that
transaction, we represent the unknown number by n, which may be
positive or negative, as will be determined by the solution of the
problem.
Thus we have 6400 + -2100 + 5000 + n = 8400. (1)
Hence by IX, F, + 9300 + n = 8400. (2)
By S, n = 8400 - 9300. (3)
ByX, n = -900. (4)
In this case the negative result indicates that there was a loss on
the fourth transaction.
PROBLEMS
In the following problems give the solutions in full and state all
principles used, together with the interpretation of the results :
1. A man gains $2100 during one year. During the first
three months he loses $125 per month, then gains $500 per
month during the next five months. What is the gain or loss
per month during the remaining four months ?
2. A man gained $ 1250 during four months. During the
second month he gained $600 more than the first month, the
third month he gained $300 less than the second, and the
fourth he gained $200 more than the third. Find the gain
or loss for the first month.
3. A box containing a Christmas toy weighed 25 oz. When
the toy and the packing were removed, the box weighed 20 oz.
The packing weighed 7 oz. What kind of a toy was it ?
4. A man rowing against a swift current rows 8 miles in 5
hours. The second hour he rows one mile less than the first,
the third two, miles more than the second, and the fourth and
fifth one mile more each than he rowed the third hour. How
many miles did he row each hour ?
5. There are three trees the sum of whose heights is 108
feet. The second is 40 feet taller than the first, and the
third is 30 feet taller than the second. How tall is each tree ?
70 POSITIVE AND NEGATIVE NUMBERS
Find the average yearly temperature at each of the follow-
ing places, the average monthly temperatures being as here
given :
6. Port Conger, off the northwest coast of Greenland ; 37°,
-43°, "32°, -15°, +14°, +18°, +35°, +34°, +25°, +4°, -17°, - 30°.
7. Franz Joseph's Land ; "20°, "20°, "10°, 0°, 15°, 30°, 35°,
30°, 20°, 10°, 0°, "10°.
8. Western Baffin Land ; "30°, "30°, ~20°, 0°, 20°, 35°, 40°,
35°, 25°, 10°, -10°, -20°.
Find the average yearly loss or gain in each of the following :
9. $1600 gain, $8000 loss, $24,000 gain, $40,000 loss.
10. $32,000 gain, $45,000 loss, $24,000 gain, $42,000 loss.
11. The average yearly temperature of north central Siberia
is ~5°. The average monthly temperatures beginning in Febru-
ary are: "60°, "30°, 0°, 15°, 40°, 40°, 35°, 30°, 0°, "30°, -50°.
Find the temperature for January.
12. The business transactions of a certain firm averaged
$ 1500 loss for 4 years. For the first year there was a gain of
$ 800, the second year a loss of $ 1800, the third year a loss of
$300. What was the loss or gain for the fourth year ?
13. A commercial house averaged $15,000 gain for 5
years. What was the loss or gain the first year if the remain-
ing years show : $8000 gain, $24,000 gain, $2000 loss, $20,000
gain, and $ 50,000 gain, respectively ?
14. The longitude of Boston, Massachusetts, is "71° 10', and
that of Chicago, Illinois, is ~87° 35'. Find the longitude of
Lake Chautauqua, which is midway between these.
15. The longitude of New Haven, Connecticut, is ~72° 58',
and that of Bombay, India, is +72° 48'. The longitude of St.
Paul's Cathedral, London, is midway between these. Find the
longitude of the cathedral.
INTERPRETATION OF SIGNED NUMBERS 71
16. The longitude of Cincinnati, Ohio, is ~84° 30', and that
of Indianapolis, Indiana, is "86° 5'. The longitude of Cincin-
nati is midway between those of Indianapolis, and Columbus,
Ohio. Find the longitude of Columbus.
17. The longitude of Bristol, England, is ~2° 30', and that of
Minneapolis, Minnesota, is ~93° 20'. The longitude of Bristol
is midway between those of Minneapolis and Calcutta, India.
Find the longitude of Calcutta.
18. The latitude of Columbus, Ohio, is +40° and that of
Winnipeg, Canada, is +50°. The latitude of Columbus is mid-
way between those of Winnipeg and Houston, Texas. Find
the latitude of Houston.
19. The longitude of Montreal, Canada, is "73° 40' and that
of Baltimore, Maryland, "76° 40'. Find the latitude of Phila-
delphia, which is midway between these.
20. The latitude of Lima, Peru, is ~12° and that of Buenos
Ay res, Argentina, "34° 35'. The latitude of Lima is midway
between those of Buenos Ayres and Caracas, Venezuela. Find
the latitude of Caracas.
21. The longitude of Providence, Rhode Island, is _71° 75'
and that of Fargo, North Dakota, -96° 50'. The longitude of
Fargo is midway between those of Providence and Seattle,
Washington. Find the longitude of Seattle.
I
CHAPTER III
INVOLVED NUMBER EXPRESSIONS
73. Double Use of the Signs + and—. In the preceding
chapter it has been found that the negative quality may be
regarded as implying subtraction and the positive quality as
implying addition. It was for this reason that + and ~ were
selected as symbols for the words " positive " and " negative."
74. It is now possible to dispense with these special signs
of quality. For, by Principle X, a — +b and a + ~ b are the same
in effect, and likewise a—~b and a-\-+b. Hence, omitting
the positive signs (§ 45), we may write
a -f- ~& = a — b and a — ~b = a + b.
One set of signs is, therefore, sufficient as symbols both of
operation and of quality.
E.g. 5 — 7 means either 5 +~7 or 5 — +7, and in either case equals
~2, which we now write — 2. Thus 5 — 7 equals — 2, which is read,
5 minus 7 equals negative 2.
Ex. 1. "5.-4 = + 20 is now written - 5 • - 4, or (— 5)(-4),
= + 20, V_ '
and -5-+4 = -20is written -5 • +4 or, (-5) ( + 4),= -20.
Ex.2. 5(9 a +"2 a) = 5(9 a— 2 a).
By IV, =45 a- 10a.
By II, = 35 a.
Ex.3. 5a+-4(-3a-+76) = 5a-4(-3a-7&).
By IV, XI, =5a + 12a + 28&.
By I, =17 a +28 6.
72
POLYNOMIALS 73
EXERCISES
Rewrite the following, using one set of signs, and then per-
form the indicated operations.
1. 7- +3 + "8. 5. 27 abc + ~35abc- 2 abc.
2. -9 --3 + -12. 6. 5 ("2 + -3) + 4(5 -~7).
3. -4a + -5a + "6a. 7. '8 (7 - 2) - 8(6 + "9).
4. -3-5a; + 4.7a;-8.8;K. 8. 3(4a-"5 b) -11(~2 a+~36)
Rewrite the following expressions, using special signs of
quality so that all signs of operation shall indicate addition :
9. 5-8-14 = 5 +-8 +-14. 13. 56 ay — 72 ay + 7ay.
10. -7 + 8-18. 14. 3(2-6) + 6(3-7).
11. -4a + 5a-17a. 15. -3(8 - 6) -4(6-9).
12. -7-4x+7-4x-- 8 -4a:. 16. 8(4* -5?t)-5(— t + 4n).
POLYNOMIALS
75. We have found that the solution of problems leads
us to build involved number expressions out of single number
symbols.
E.g. If a; is a number representing my age in years, then 2 (x — 10)
is double the number representing my age 10 years ago, and
2 l(x — 10) + (x+ 15)] is the number representing twice the sum of my
ages 10 years ago and 15 years hence.
Number expressions are now to be studied more in detail.
76. Definition. A number expression composed of parts con-
nected by the signs + and — is called a polynomial. Each of
the parts thus connected together with the sign preceding it is
called a term.
E.g. 5 a - 3 xy — f rt + 99 is a polynomial whose terms are 5 a,
-Zxy, - | rt, and + 99. The sign + is understood before 5 a.
74 INVOLVED NUMBER EXPRESSIONS
77. Definitions. A polynomial of two terms is called a bino-
mial, one of three terms is called a trinomial. A term taken
by itself is called a monomial.
E.g. 5 a — 3 xy is a binomial; 5 a — 3 xy — § rt is a trinomial ; 5 a,
— 3 xy, —%rt are monomials.
According to the above definition x + (6 + c) may be called
a binomial notwithstanding it is equivalent to the trinomial
x + b + c.
In this case x is called a simple term and (b + c) a compound
term. Likewise we may call 3t + £x — 5(a-\- b)y a trinomial
having the simple terms 3 t, -f 4 x, and the compound term
-5(a + b)y.
It should be clearly understood that a negative or positive
sign before a compound term (as well as before a simple term)
applies to the number represented by the whole term.
78. Definition. Two terms which have a factor in common
are called similar with respect to that factor.
E.g. 5 a and — 3 a are similar with respect to a ; — 3 xy and — 7 x
are similar with respect to x ; 5 a and — 5 b are similar with respect
to 5 ; 7 abc and — f abc are similar with respect to abc.
Similar terms may be combined by Principles I, II, and IX.
E.g. 5a-3a = (5-3)a = 2a; -3xy-7x= -x (3 y+7) ; 5a-5b =
5(a-b).
ADDITION AND SUBTRACTION OF POLYNOMIALS
79. In adding or subtracting polynomials the work may be
conveniently arranged by placing the terms in columns, each
column consisting of terms which are similar.
Ex. Add 5x-6y + 4:z + 5at, — 3 x + 11 y -16z- 96*,
and —7y-\-8z.
ADDITION OF POLYNOMIALS 75
Arranging as suggested and applying Principles I, II, and IX,
we have
5x— 6y + 4 2 + 5 at
-3x + lly-16z-9bt
- iy+ 8z
2x — 2y- 4:Z + t(5a-9b)
5 x and — 3 x are similar with respect to their common factor
x. Hence by Principle I we add the other factors 5 and — 3,
obtaining (5 — S)x = 2x.
Likewise we add + 5 at and — 9 bt with respect to the com-
mon factor t, obtaining (5 a — 9 b)t. In the second column the
sum is (— 6 + 11 — l)y= — 2y, and in the third column the
sum is ( + 4 — 16 + 8) z = — 4 z.
Check by giving convenient values to x, y, z, t, a, and b.
EXERCISES
1. Add76-3c + 2o'; _26 + 8c-13d\
2. Add 6x— 3 y + ±t — 7 z; x — 5 y-3t; 4 x-±y + 8 t.
3. Add 5 ac + 3 6c -4c + 86; 26 + 3c-2 6c -3 ac; 46 +
4 c + 6c — ac; 2 6c + 4ac + c; 36— 4c.
4. Add 3-4.7 — 5x + 5abx; 3aby + 3x — 5 -4 ■ 7; 7 x +
2-4-7; baby — 3x — 5-4- 7; 45# + a6cc — 4 -7.
5. Add 3(x — 5) + a(c + 6) + 6(a-?/); 6(c + 6) — a(x — y)
+ 8(a>-5); 7(c + 6)-4(a- </); 3(a>-y) + (a-5).
6. Add 16(a+6-c)-3(x-2/)+2(a-6); 2{x-y)-
3(a-6) + (a+ 6— c); 7 (a — 6) + a(x — y) — 6(a + 6 — c).
7. Add a(a — 6) — c(# + ?/) + d(x — z)— 4a6c; c(# — 2)—
d(x+y) + (a— 6)+2a6c; e(a — 6)+ma6c+3(a — z)+8(a+y).
8. Add 7a — 4a+12 2; 6a — 3 x + C2 ; 2 6a + 4 a — 3 cz.
76 INVOLVED NUMBER EXPRESSIONS
9. Add (a — &) — 3(c — d)+m(a + d) ; c(a—b)+a(c — d).
10. Add 34 ax + 4 by — 3 z; 2by + 5z; 3bx — 7y + 5dz.
11. Add 3b + 4:cd—2ae; ab—3cd + 3ae; 3cd — 2ab.
12. Add 7ax-13by + 5; 9ax + 8by-4; 3 b -12 ax.
13. Add5(a + 6)-3(c-d); 3(c-d)-8(a + 6); — 2(a + 6).
14. Add 3 + 4(c -d)-5(a-6-c) ; 4 (a— 6 - c) + 5(c-d);
3(a_6_c)-9(c-d) + 12.
15. Add U(c — 9) + 3(x + y) + 21ivu; —71wu — 5(x + y)
-13(c-9).
16. Add 5a& - 3- 7 -9 + 5(a-l); 5- 9- 7 + 3a&- 2 (z-1);
3(»-l)-4.7.9 + 2a&.
17. Add 31 • 50 - 43 • 74 + 2 • 18 ; 21 • 74 + 7 • 18 - 56 • 50 ;
- 12 • 18 + 42 • 50 - 6 • 74.
18. Add 7(a5-y)-4(aj + y) + 4-7; 9 (a; 4- 2/) + 3(x-y) -
9-7; 6(^-^)4-2.7-3(^ + 2/).
19. Add 16 xy - 13 • 64 ; 15 a& - 2 a# ; 34 • 64 - 3 xy 4- 2 a& ;
14- 64 -3a?/- 2 a&.
80. The subtraction of polynomials is illustrated by the
following example :
From 15 ab — 17 xy + 11 rt subtract — 5 ab 4- 4 xy — 5 nt.
Arranging as on page 75 and applying Principles II and X,
15 ab —17xy + 11 rt
— 5ab + 4 xy — 5 n<
20 ab -21 xy+ t(Ur + 5 n)
As suggested in § 60, it is sufficient to change the signs of the sub-
trahend mentally, rather than to rewrite them before adding to the
minuend.
SUBTRACTION OF POLYNOMIALS 77
EXERCISES
1. From 9x+3y— 11 z subtract — 5x+8y— 3z.
2. From 12 ab — 3 cd + 12 xy subtract 3 ab + 2 cd — 11 xy.
3. From 9 xc + 4 ad — 3 cz + 2 ?/ subtract 3 ?/ — 3 ad + 5 cz.
4. From 13 t + 5 mx — 5 cu subtract 2 £ — 4 ma + 3 cv.
5. From 3 v — 2 w + 5 ?nn — 4 asz subtract — -y + 5 w — 3 raw.
6. From 31 b + 4 ary -f 10 a.c — 4 subtract 8 6 — 5 xy — 3 ax.
7. From 4 — 3 a — 5xz— 3vy-x subtract 7 a + 2 az 4- 4 vy.
8. From 8 .ry — 3 x + 4 y subtract — 2xy + 13w + 4x—2y.
9. From 2 a& — 5 + 7 v + 13 abc subtract 3 ab + v + 8 a&c.
10. From 8 ca;a — 4yb—3yc subtract 4 fom + 2 yb + 4 yc — 49.
11. From 31 • ir> — 7xy subtract 12 • 45 + 9 a*/.
12. From 3 abc - 4 -28 + 2 (x + y) -3 xy subtract 6 • 28
+ 4 xy — 3 (x + y) + 8 a&c.
13. From 7 • 3 • 5 + 9 (»y — z) -f 4 • 3 (a + 6) subtract 8(a# — z)
-8-3(a + b)+8-3-5.
14. From 5 ax — 3 by + 4 ax+ 5 by subtract 5 by— 3ax+ 7 by.
15. From 19 (r - 5 s) + 13 (5 x — 4) + 7 (a? — y) subtract
17(5 x _ 4) - 5 (x- y) - 11 (r - 5 s).
16. From 16 - 15 • 30 + 14(* - 5 yz) — 13 (5 y — z) subtract
32-16- 30 + 8(5y-*).
17. From -41.3 + 13-4-16 subtract 7- 4- 16-8-3.
18. From a (b + c) + 4 (ra + ») —16 c subtract 9 (ra + n)
+ 31c-d(6 + c).
19. From 5(7a;-4) + 3(5t/-3a;) + 5 - 7 subtract 8-7
-9(7 x - 4) + 8(5 y— 3*).
20. From 15 • 48 + 8 a& + 49 a; subtract 7 • 48 - 9 a& - 14 x.
78 INVOLVED NUMBER EXPRESSIONS
EXERCISES IN ADDITION AND SUBTRACTION
1. Add 5x — 3y — 7 r + 8t, —7 x + 18y — ±r — 7t, —20 a;
— 24 y + 18 r - 15 1, and 13 x + 15 y + 11 r + 6 t.
Check the sum by substituting a; = 1, y = 1, r = 1, f as 1.
2. Add 17a — 9 6, 3c + 14a, 6 — 3 a, a — 17c, and a — 36
-f 4 c. Check for a = 1, 6 = 2, c = 3.
3. Add 2x + 3y — t, —6y + 8t, —x + y — t, — 4* + 7z,and
3 y. Check f or x = 2, y = 3, £ = 1.
4. Add 17r + 4s-£, 2t + 3u, 2r-3s-f4£, 5w-6«, 7r
— 3s + 8u, and 8r — 2f + 6w. Check by putting each letter
equal to 1 ; also equal to 2.
5. Add 3 ft + 2 £ + 4 u and ft + 3 1 + 3 w. Check by putting
ft = 100, t = 10, rt = 1 ; i.e. 324 + 133 = 457.
6. Add 4 ft + 3 £ + u and 3 ft + 2 £ -f 7 u. Check as in 5.
7. Write 247, 323, 647, 239, and 41, as number expressions
like those in 5 and 6 and then add them.
8. Add At — u, bt — u, 6t — u, 7 t — u, and 8 1 — u. Check
for t = 10, u = 1 ; also t = 1, u = 1.
9. Add 647, 391, 276, and 444 as in example 7.
10. Simplify: 3 xyz — 2 xyz + 5 xyz — 4 xyz + xyz — xyz.
11. Subtract 5 a — 3 6 + 6 c from — 8a-f 76 — lie and check.
12. From 7 xy + 8 xz + 9 yz take 17 a;y — 19 xz — 20 yz.
13. From 6 a; — 3 y take 8 y — 3 z.
14. From 3p — 4 g + 8 r take 7j9 — 11 r + 11 5-
15. From a + b + c take a; — y + z. Suggestion : By Principle
VII, a + 6 + c-(a; — y + z)=za + b + c-x + y — z.
16. From 2a; — 3y take 5. r + 7 y + 2 a — 3 6.
17. From the sum of 18 a6c — 27 xyz + 13 rst and — 11 a6c
-f 16 xyz — 52 rst take 67 rst — 39 a6c.
ADDITION AND SUBTRACTION 79
18. To the difference between the subtrahend 15 a; — 18 y
+ 27 z and the minuend 117 x + 07y — 81 z add 4 a: — 6 y + 3z.
19. Add ll(a?-y) + 15 (a -6) and -20(a5-y)-37(o-6)
and from the sum subtract 135 (a; — y)— 213 (a — b).
20. Add 6ax +7 bx — 8cx, —11 ax — 18 &# + 25 ex, and 19 ax
-16cx + 24:bx.
21. From 13 mn — 25 mp + 36 mq subtract 18 mn + 23 mp.
22. Add by Principle I, 6 • 3 • 9 - 11 • 5 • 7 + 16 ■ 9 • 11 and
- 8 • 3 . 9 + 24 • 5 • 7 - 23 • 9 . 11.
23. From 83-9 + 78 -13 subtract 57-9-93.13+85.17.
24. From 3 h + 4 1 + 2 u subtract h + 5t+3u. Check.
25. Subtract 7(a— x)— 10 (b — y) from 13(a — x)+5(b — y).
81. In solving problems it is often unnecessary to arrange the
work of addition and subtraction in columns as above. In most
cases the operations can be readily indicated by means of paren-
theses as illustrated in the following example :
From the sum of 3 a + 4 b and 5 a — Sb subtract 7 a — 6 b.
Indicate these operations thus : 3 a + 4 b + (5 a — 8 6) — (7 a — 6 b).
Applying Principle VII, we have 3a + 46 + 5a — 86 — 7 a + Qb.
Collecting similar terms, 3a + 5a — 7a + 46 — 86 + 6 6.
Finally, applying Principles I and II, we obtain a + 2 6.
After a little practice the last two steps can be taken at once.
EXERCISES AND PROBLEMS
Perform the following indicated operations by collecting
similar terms at once without arranging in columns :
1. 15 + (7 -9 a;)- (-7* -9) +9.
2. 7 + 5y-(3y + 2) + (8-4:y).
3. 2a + 3 + (4a-5)-(lla-14).
4. 326-(176-12)-(46-13).
80 INVOLVED NUMBER EXPRESSIONS
5. 16c-(41-7c) + (15-8c).
6. - (5 a - 3 c) - (2 c - 8 a) + 3 a.
7. _(_ 12 a _7 ?/ -15 a:) -(-9 # + 8 a + 3?/).
8. (19 x + 4 y - 32 x - 17 x) - 12 a - (49 y + 18 a> - 70 a).
9. 17a-3-(7a-2) + (6a-5).
10. 5a- (8-4x + 7y) + (5x + 3)-(5y + 3x-99).
11. -(3a + 56-7c) + (8a-4c)-(9c-46 + 4a)-91a.
12. 7-(4-4c + 2d-2a)+31c-(4-2a-5d)-(-8c).
13. (41a&-21c+4)-(36c + 15-78a6) + (13c-90ao-8).
14. 9 by - (4 c - 8 by - 13) - 2c - 16 - (34 by -12 c + 8 by).
15. 6 mn+(— 9m — 7 w +14) -8 n + (13 mn—11 m)+34 raw.
16. 34 ax— (— 17 a# + 42) + 8 a -(14 a + 24 ax — 7).
17. 19 - (+ 2 - 7a - 4 b + 11 oft) - (- 2 6 + 8 ab + 4 a).
18. 41 % - (4 b - 13 y + 17 by) - (- 5 b - 17 6y + 13 y).
19. 39 rs - 20 s - 19 r - (7 »•* + 8 s - 19 r) - (15 r - 5 s - 56).
20. a(3 x — 2 y — z) — (5 ax — ay + 3 z) + az.
21. 5(4 h + 3bk-7br)- 6(15 & - 35 r) - 20 h.
22. The altitude of Popocatepetl is 1716 feet less than that
of Mt. Logan, and the altitude of Mt. St. Elias is 316 feet
greater than that of Popocatepetl. Find the altitude of each
mountain, the sum of their altitudes being 55,384 feet.
23. The Ganges River is 1800 miles shorter than the Ama-
zon, and the Orinoco is 300 miles shorter than the Ganges.
The sum of their lengths is 6900 miles. How long is each ?
24. A cubic foot of red oak weighs 35 pounds less than 2
cubic feet of cherry wood, and 21 pounds more than a cubic
foot of chestnut ; while a cubic foot of chestnut weighs 100
pounds less than 3 cubic feet of cherry. Find the weight of
each kind of wood per cubic foot.
ADDITION AND SUBTRACTION 81
25. Lead weighs 259 pounds more per cubic foot than cast
iron, and 166 pounds more than bronze ; while a cubic foot of
bronze weighs 807 pounds less than 3 cubic feet of iron. Find
the weight per cubic foot of each metal.
26. Green glass weighs 60 pounds per cubic foot less than
dense flint glass, and 8 pounds more than crown glass ; while a
cubic foot of crown glass weighs 293 pounds less than 2 cubic
feet of flint glass. Find the weight per cubic foot of each.
27. Europe has 12 million inhabitants less than 10 times as
many as South America, and North America has 29 million
more than twice as many as South America. If 3 times the
population of North America be subtracted from that of Eu-
rope, the remainder is 65 million. How many inhabitants has
each continent ?
28. The length of the Rio Grande River is f that of the
Volga, and the Mississippi is 600 miles less than twice as long
as the Volga. If \ the length of the Mississippi be subtracted
from that of the Rio Grande, the remainder is 400 miles.
Find the length of each river.
29. In 1900 the total wealth of the United States was 1532
million dollars more than 13 times as great as in 1850, and
9016 million more than twice as great as in 1880. The total
wealth in 1880 was 174 million less than 6 times as great as
in 1850. What was the wealth in each of the three years ?
30. The money circulation of the United States in 1880
was 13 million dollars more than 60 times that in 1800 and
in 1905 it was 188 million more than 150 times that in 1800.
One-seventh of the amount in 1880 plus \ the amount in 1900
was 786 million. Find the circulation for each year.
31. The total bank deposits in the United States in 1905
were 3127 million dollars less than twice as great as in 1900
and 681 million more than 5 times as great as in 1880. The
deposits in 1880 were 5105 million less than in 1900. Find
the deposits for each of the three years.
82 INVOLVED NUMBER EXPRESSIONS
32. The amount of deposits in savings banks in the United
States in 1905 was 703 million dollars greater than in 1900,
and 183 million less than 4 times that in 1880. The amount
in 1900 was 67 million less than 3 times as great as in 1880.
Find the deposits for each of the three years.
33. The total value of the farms in the United States in
1880 was 280 million dollars more than 3 times their value in
1850, and 8333 million less than their value in 1900. The
value of the farms in 1880 was 1924 million less than ^ their
value in 1900. Find the value in each of the three years.
34. The value of the manufactures in the United States in
1900 was 811 million dollars more than 12 times that in 1850,
and 3071 million less than 3 times that of 1880; while the
value in 1880 was 744 million less than 6 times that in 1850.
Find the value of the manufactures for each year.
MULTIPLICATION OF POLYNOMIALS
82. Illustrative Problem. A rectangular field is 12 rods
longer than it is wide ; a second rectangular field which is 4
rods shorter and 2 rods wider than the first has an area of 80
square rods less. What are dimensions of the first field ?
Solution. Let w = number of rods in the width of first field.
Then w + 12 = number of rods in the length of first field,
and w (to + 12) = width x length, or area of first field ;
also w + 2 = width of second field,
and w + 8 = length of second field.
Hence (w + 2) (w + 8) = width x length, or area of second field.
But the area of the second field is 80 square rods less than that of
the first.
Hence (w + 2) (w + 8) = w (w + 12) - 80. (1)
To solve this equation it is necessary to obtain the product
of the two binominals w -+- 2 and w + 8 without first combining
the terms of each binominal. In order to determine how these
3-5
3-8
5-5
5
5-8
8
MULTIPLICATION OF POLYNOMIALS 83
are to be multiplied, let us consider two binomials in each of
which the terms can be combined if desired.
E.g. Consider the product of the binomials 5+3 and 5 + 8. This
may be represented by the area of a
rectangle 8 feet wide and 13 feet
long. That is, (5 + 3) (5 + 8) = 8 . 13
= 104 square feet. But such a
rectangle may be divided into four
rectangles, as in the figure.
Hence the area may be expressed
as the sum of four areas. Thus,
(5 + 3)(5 + 8) = 5 • 5 + 5 • 8 + 3 • 5 + 3 • 8 = 104, as before.
83. The product 5 • 5 is abbreviated to 52, the small figure
indicating that 5 is to be used as a factor twice. It is read
the square of 5, or 5 squared, since it represents the area of a
square whose sides are each 5.
The second method here used for multiplying (5 +-3)(5 +8)
is applicable to any similar case, and does not depend upon the
possibility of first combining the terms of the binomials.
Applying this method to the binomials in equation (1) of the
problem in § 82, we have
w2 + 8 to + 2 w + 16 = to2 + 12 w - 80. (2)
Subtracting to2 from both sides and applying Principle I,
10 w + 16 = 12 w - 80. (3)
Subtracting 10 to from both sides and adding 80 to both sides,
96 = 2 w. (4)
Hence by D, 48 = to, the width of the field ;
and 60 = w + 12, the length of the field.
Check by substituting to = 48 in equation (1), and also by showing
that the numbers 48 and 60 satisfy the conditions stated in the
problem.
84
INVOLVED NUMBER EXPRESSIONS
84. In a manner similar to that just illustrated we may mul-
tiply two trinomials.
E.g. The product of a + b + c and m 4- n + r, in which the letters
represent any positive numbers, may represent the area of a rectangle,
divided into small rectangles as follows :
a
am
an
ar
b
bm
bn
br
c
cm
en
cr
Hence, the product is :
(a + b + c) (m 4- n + ?•) = am + bm + cm + an + bn + en + ar + br + cr,
in which each term of one trinomial is multiplied by every term of
the other, and the products are added.
Evidently the same process is applicable to the product of
two such polynomials each containing any number of terms.
EXERCISES AND PROBLEMS
Find each of the following products in two ways
1. (3 + 7 + 10)(2 + 6).
2. (5 + ll)(13 + 10 + 5).
3. (6-f-ll)(6 + 7).
4. (15 + 8)(15 + 4).
5. (7 + 13)(a + 6).
6. (ra + ft)(ll + 5+4).
7. 42 -36 = (40 + 2)(30 + 6).
8. 28- 73 = (20 + 8) (70 + 3).
Find as many as possible of the following indicated products
in two ways :
9. (a + b)(c + d).
10. (a; + 4)(x + 3).
11. (x + y + z)(a + b + c).
12. (ll + 13)(r + f).
13. (5 + z)(5 + 7).
14. (3 + 8)(2 + 4 + 6).
15. (aj + 7)(3a? + 4).
16. (a + 4)(3a + l).
17. (3 + x)(2-\-5x).
18. (a + 6)(3a + 7&).
19. (x + y)(2x + 3y).
20. (7x + 4x)(x + 8).
MULTIPLICATION OF POLYNOMIALS 85
21. A rectangle is 7 feet longer than it is wide. If its
length is increased by 3 feet and its width increased by
2 feet, its area is increased by 60 square feet. What are its
dimensions ?
22. A field is 10 rods longer than it is wide. If its length
is increased by 10 rods and its width increased by 5 rods, the
area is increased by 630 square rods. What are the dimen-
sions of the field ?
23. A farmer has a plan for a granary which is to be
12 feet longer than wide. He finds that if the length is
increased 8 feet and the width increased 2 feet, the floor
space will be increased by 160 square feet. What are the
dimensions ?
24. If the length of a rectangular flower bed is increased
3 feet and its width increased 1 foot, its area will be increased
by 19 square feet. What are its present dimensions, if its
length is 4 feet greater than its width ?
85. Polynomials with Negative Terms. The polynomials mul-
tiplied in the foregoing exercises contain positive terms only.
The same process is applicable to polynomials containing nega-
tive terms, as is seen in the following examples :
Ex. 1. Find the product of (7 - 4) and (3 + 5). This
product, written out term by term, would give
(7 + -4)(3 + 5) = 7- 3 + 7- 5 + -4- 3 + "4. 5
= 21 + 35-12-20 = 24.
Also (7-4)(3 + 5) = 3-8 = 24.
Ex. 2. Multiply 7 - 4 and 8 - 3.
(7 _ 4)(8 _ 3) = (7 + "4)(8 + ~3) = 7 • 8 + 7 • "3 + "4 • 8 + "4 . "3
= 56 - 21 - 32 + 12 = 15.
Also (7-4)(8-3)=3-5 = 15.
86 INVOLVED NUMBER EXPRESSIONS
EXERCISES
Find each of the following products in two ways, as in the
above examples :
1. (ll-7)(6 + 5). 5. (2-3)(46-76).
2. (22 - 13) (3 + 7). 6. (3 + aj)(7*-3aj).
3. (8-5)(7-3). 7. (8-3)(8-2).
4. (17-9)(29-4). 8. (9-13)(9-17).
Perform the following indicated operations :
9. (a-6)(c + d). 13. (a — 6)(7a + 3 6).
10. (a — 6)(c — d). 14. (5 — y)(5 x + 3 y).
11. (a - 4) (a — 5). 15. (2 a — 3 b + c)(w + n).
12. (a + 6 — c)(m — n). 16. (y — t)(7v — 5t).
86. The preceding exercises illustrate the following principle :
Principle XIII. The product of two polynomials is found
by multiplying each term of one by every term of the other,
and adding these products.
In case some of the partial products are negative, these are
combined by Principle IX.
If there are similar terms in either polynomial, these should
usually be added first, thus putting each polynomial in as sim-
ple form as possible.
E.g. (3x + 2 - 2x)(4 x + 3 - 3x) = (x + 2)(x + 3)
= a;2 + 2z + 3x + 6=a;2 + 5ar + 6.
EXERCISES AND PROBLEMS
Perform the following indicated operations :
1. (a-7)(3a;-4 + 8).
2. (l-2a? + x-3){2x + ±a + lx).
3. (4a — x — 3a)(2x + 4:a + 7x).
MULTIPLICATION OF POLYNOMIALS 87
4. (5x + 3y-4:X-2y)(6y + 3x-2y + y).
5. (13a-6-12o)(2&-3a).
6. (xy-5xy + 4:y)(Sy-3-7y).
7. (llao + 3a)(2&-3& + 5).
8. (6-4aj + 3*)(7aj + y — 3« + l).
9. (13a — 12x-y + 3)(5x-3y + xy + 5).
10. (37 — 13w + a)(a — w + 8).
11. (x-2+y)(4:y-3x).
12. (ll&-a-10&)(6a-3&-2a).
13. (7 + y-x)(2y + x-l).
14. (5as + 3y-l)(a;-2).
15. (—8a- 1 + 7 a) (5a -8 — 3a).
Solve the following equations, in each case verifying the
solution by substituting in the given equation the result found
for the unknown number.
16. (a? + 2) (a; + 3) = (x - 3) (x + 10) + 10.
17. (5a;-4)(6-aj)-97 = (a;-l)(6-5a;).
18. (3 n -1) (18 - ft) = (ft + 6) (16 - 3 n).
19. (7-a)(9a-8) = 31 + (36-9a)(a + 2).
20. (4 a + 4) (a - 3) = (4 a + 1) (a + 7) - 13 a + 221.
21. (n + 6)(3 n - 4) - 14 = (« + 8) (3 n - 3).
22. (8 n + 6) (10 - w) + 150 = (1 - w) (8 n + 3).
23. (a-l)(13-6a) = (6a-3)(8-a)-21.
24. (7 a; - 13)(6 - a;) - (a; + 4)(3- 7 a;) = 70.
25. A club makes an equal assessment on its members each
year to raise a certain fixed sum. One year each member
pays a number of dollars equal to the number of members of
the club less 175. The following year, when the club has 50
more members, each member pays $ 5 less than the preceding
year. What was the membership of the club the first year
and how much did each pay ?
88 INVOLVED NUMBER EXPRESSIONS
26. There are two numbers whose difference is 6 and whose
product is 180 greater than the square of the smaller. What
are the numbers ?
27. There are four consecutive even integers such that the
product of the first and second is 40 less than the product of
the third and fourth. What are the numbers ?
28. There are four consecutive integers such that the prod-
uct of the first and third is 223 less than the product of the
second and fourth. What are the numbers ?
29. There are four numbers such that the second is 5
greater than the first, the third 5 greater than the second, and
the fourth 5 greater than the third. The product of the first
and second is 250 less than the product of the third and
fourth. What are the numbers ?
30. Prove that for any four consecutive integers the prod-
uct of the first and fourth is 2 less than the product of the
second and third.
SQUARES OF BINOMIALS
87. Just as x2 is written instead of x • x, so (a + 6)2 is written
instead of (a + b) (a + 6). The square of a binomial is found
by multiplying the binomial by itself as in § 86.
E.g. (a + b)2 = (a + b)(a + b) = a2 + ab + ab + b\
Hence (a + 6)2 = a2 + 2 ab + b2.
This product is illustrated in the accompanying figure, and
is evidently a special case of the type exhibited in the figure,
page 83.
Translated into words this identity is :
The square of the sum of any two numbers is
equal to the square of the first plus twice the
product of the two numbers plus the square of
the second.
ba
b3
an-
ab
SQUARES OF BINOMIALS 89
88. Similarly we obtain the square of the difference of two
numbers. («_*)»s «»-*«* + *•
Translate this identity into words.
While these squares are ordinary products of binomials and can
always be obtained according to Principle XIII, they are of special
importance and should be studied until they can be reproduced from
memory at any time.
EXERCISES AND PROBLEMS
Perform the following indicated operations :
1. (a + 6)2. 4. (a-2)2. 7. (4 + 9)2.
2. (17 -3)2. 5. (21 -bf. 8. (c-f)2.
3. (6 + a)2. 6. (z-7)2. 9. (x-\)\
10. (r — s)2 — (r-f s)2+(r — s)(r + s).
Check each of the above by substituting special values for
the letters and combining the terms of the binomial before
squaring.
11. Find the square of 42 by writing it as a binomial, 40 + 2.
12. Square the following numbers by writing each as a
binomial sum : 51, 53, 93, 91, 102, 202, 301.
13. Find the square of 29 by writing it as a binomial, 30 — 1.
14. Square the following numbers by first writing each as a
binomial difference : 28, 38, 89, 77, 99, 198, 499, 998, 999.
Solve the following equations, verifying each solution :
15. (a + 4)2 + (a - 1) (2 a + 5) = (a + 4) (3 a + 2) .
16. (a - 1) (3 a - 1) - (a + 1)2 = 2 a2- 18.
17. (6 - a)2 -f (a - 3)(2 a - 5) = (3 a + 1) (a - 3) + 84.
18. (7 a -18)(a -f 4) - (a - 1)2 = 6 (a + 1)2 - 79.
90 INVOLVED NUMBER EXPRESSIONS
19. (2&-30)(&-l)-5&2 = 6 6-3(7> + 5)2 + 65.
20. (5-&)(6& + 5)+4(6-3)2 = 20-2(7> + l)2 + 3 + 16&.
21. (5 - c)2 + (7 - c)2 + (9 - c)2=(c - 1)(3 c-58) - 93.
22. (5 c - 3) (2 + c) - 4 (c - l)2 = (c + 1)2+ 54.
23. (8 - 4 c) (5 - c) = (c + 1)2 + (c + 3) (3 c - 8) + 218.
24. (4y-9)(y-5)-5(2/-4)2=(8-2/)(4 + 2/)-82.
25. (y + 6)"-3(y-l)»+(4-y)(5--2y) = 25y + 9.
26. (y-l)2 + 4(i/ + l)2 + (l-2/)(5y + 6) = 15y-29.
27. a(a; + 3) + (a; + l)(a; + 2) = 2a;(a:-f-5)+2.
28. aj2=(*-3)(o: + 6)-12.
29. (5 + 5x)(3-x)+2(x + l)2+3(x + l)(x-7) = 17(x+l).
30. (8 + 3 *)(4 - x) + (a? - l)(aj - 2) + 2 (as + 5)2 = 105.
31. There is a square field such that if its dimensions are
increased by 5 rods its area is increased 625 square rods. How
large is the field ?
Suggestion : If a side of the original field is w, then its area is to2,
and the area of the enlarged field is (w + 5)2.
32. A rectangle is 9 feet longer than it is wide. A square
whose side is 3 feet longer than the width of the rectangle is
equal to the rectangle in area. What are the dimensions of
the rectangle?
33. A boy has a certain number of pennies which he
attempts to arrange in a solid square. With a certain num-
ber on each side of the square he has 10 left over. Making
each side of the square 1 larger, he lacks 7 of completing it.
How many pennies has he ?
34. A room is 7 feet longer than it is wide. A square room
whose side is 3 feet greater than the width of the first room
is equal to it in area. What are the dimensions of the first
room ?
SQUARES OF BINOMIALS 91
35. Find two consecutive integers whose squares differ
by 51
36. Find two consecutive integers whose squares differ
by 97.
37. Find two consecutive integers whose squares differ by a.
Show from the form of the equation obtained that a must be
an odd integer.
38. There are four consecutive integers such that the sum
of the squares of the last two exceeds the sum of the squares
of the first two by 20. What are the numbers ?
39. Two square pieces of land require together 360 rods of
fence ? If the difference in the area of the pieces is 900 square
rods, how large is each piece ? (Hint : x2 — (90— a;)2 = 900.)
40. There is a square such that if one side is increased by
12 feet and the other side decreased by 8 feet the resulting
rectangle will have the same area as the square. Find the
side of the square.
41. A regiment was drawn up in a solid square. After 50
men had been removed the officer attempted to draw up the
square by putting one man less on each side, when he found
he had 9 men left over. How many men in the regiment ?
42. There is a rectangle whose length exceeds its width
by 11 rods. A square whose side is 5 rods greater than the
width of the rectangle is equal to it in area. What are the
dimensions of the rectangle ?
89. Thus far the processes of algebra have all been based
upon thirteen fundamental principles, together with certain
obvious form changes indicated in § 37. These latter are of
general use in elementary arithmetic and need no special em-
phasis here. The principles, however, for the most part refer
to methods not common in arithmetic. These should now be
carefully reviewed and a list of them made for convenient
reference.
92 INVOLVED NUMBER EXPRESSIONS
REVIEW QUESTIONS
1. How would 3 • 5 and 7 • 5 be added in arithmetic ? Why
cannot 3n and In be added in the same manner? State in
full the principle by which 3n and In are added. In this
example what number is represented by n ? Test the identity
3n + 7w=10nby substituting any convenient value for n.
2. How is 5 • 9 subtracted from 11 • 9 in arithmetic ? In
what different manner may this operation be performed?
Why is it sometimes necessary to perform subtraction in the
second way ? State in full the principle by which 12 x is sub-
tracted from 31 x. In the identity 31 x — 12 x = 19 x, what
number is represented by #? Test the equality by substituting
any convenient number for x.
3. How is the product 2-3-5 multiplied by 4 in arithmetic ?
In what different way may this multiplication be performed ?
Why should it ever be performed in the second way ? State
in full the principle by which 2 ax is multiplied by 3.
4. How is 11 + 3 multiplied by 4 in arithmetic ? In what
different way may this operation be performed? Why is it
sometimes necessary to multiply in the second way ? State in
full the principle by which a + 8 is multiplied by 7.
What principles are used in performing the following indi-
cated operations ?
ax + bx + cx, aby+cby, c(4a + 2&),
3y-by + cy, 41 a -32 b- 17 a + SOb.
5(16a-36), -3(6x-7y), a (11 -6).
5. Divide 2-4-6-20 by 2 without first performing the
multiplication indicated in 2-4-6-20. Do this in several
ways and show that all the quotients obtained are equal.
State in full the principle used.
What principles are used in the following operations ?
3 • 5 ab = 15 ab, c • 5 b = 5 (be), 20 ab + 4 = 5 ab,
16xy + x = l&y, 3* + 15* = 18<, 78^-41 A = 37 ft.
REVIEW QUESTIONS 93
6. How is 12 + 18 divided by 6 in arithmetic ? In what
different way may this division be performed ? Why is it
sometimes necessary to perform division in the second way ?
State in full the principle used in performing the operation
(6 x + 9 y) -5- 3. What principle is used in performing each of
the following indicated operations ?
5(a + x), 3y — 4:y, (24 x + 9 y) -r- 3, 5x+ax.
7. Define equality ; equation ; identity. State in detail
how the equation and the identity differ. Give an example
of each.
8. In what ways may an equation be changed into another
equation such that any number which satisfies either also
satisfies the other ?
Describe some of the operations which change the form of
the members of an equation, but not their value.
State Principle VIII in full.
9. Name several pairs of opposite qualities all of which
are conveniently described by the words " positive " and " nega-
tive." What symbols are used to replace these words when
applied to numbers ?
10. When loss is added to profit, is the profit increased or
decreased ? What algebraic symbols are used to distinguish
the numbers representing profit and loss ?
11. Give an illustration by means of the number scale to
show that a number may represent either a change of position
in one direction or the other, or & fixed position with respect to
the zero point.
12. Why do we call positive and negative numbers signed
numbers ? What is meant by the absolute value of a number ?
13. State Principle IX in full.
14. By means of the number scale describe the "counting"
method of adding signed numbers
94 INVOLVED NUMBER EXPRESSIONS
15. Make a list of pairs of opposite qualities to which posi-
tive and negative numbers apply. State in each case what is
represented by positive and what by negative numbers.
16. How is the correctness of subtraction tested in arith-
metic ? Is the same test applicable to subtraction in
algebra ?
17. Explain subtraction by counting on the number scale.
18. How do negative numbers make subtraction possible in
cases where it is impossible in arithmetic ?
19. What is a convenient rule for subtracting signed num-
bers ? State Principle X.
20. Give examples of equations which could not be solved
without negative numbers and show that such equations can
be solved by means of negative numbers.
21. Give an example in which positive and negative num-
bers are multiplied. State Principle XI.
22. Define division. How do we obtain the law of signs in
division ? State Principle XII. What is the test of the cor-
rectness of division ?
23. What may be the significance of a negative number
when obtained as a result of solving a problem ?
24. Explain how one set of signs + and — can be used to
indicate both quality and operation.
Show that a -f- ~b = a — b and a — +b = a — b.
25. What is a polynomial ? A term ? How are poly-
nomials classified? What are similar terms? By what prin-
ciple are similar terms added ? By what principle are they
subtracted ? In adding or subtracting polynomials, how may
the terms be arranged for convenience ? What is the prin-
ciple for removing a parenthesis when preceded by the sign + ?
By the sign — ? State Principle VII in full.
REVIEW EXERCISES 95
26. Make a diagram to show how to multiply (7 + 4) by
(11 + 8) without first uniting the terms of the binomials.
Multiply (a + 6) by (c + cT) in the same manner. Multiply
(12 — 3) by (9 — 7) in two ways and compare results. State
the principle by which two polynomials are multiplied.
27. Explain why x2 is called the square of x or x squared.
■ State in words what is the square of the binomial (x + a) ; of
the binomial (x — a).
28. Show by an example how negative numbers may be used
in solving a problem, even though the answer to the problem
is positive and the statement of the problem does not involve
negative numbers.
REVIEW EXERCISES
In the following exercises perform the indicated operations,
remove all parentheses, solve all equations, verify the results,
and in each case state the principles used.
1. Add 3x + 4y — 3z, 5x — 2y — z, and 3y — 5 a; + 7 z.
2. From 15 a + 4 6 — 13 be subtract 3 a — 8 b + 2 be.
3. Subtract 7 x — 5y — 7 a from 6x + 5y + 3a.
4. (3x-4:y-z)(x + y + z). 5. (6 -5)(2 a2b -3 ab- b).
6. Add 11 axy + 13 x — 14 y, 2 y — 4 x, and 3 y + x — 8 axy.
7. (5 x - 3 b) + (2 x + b) - (4 x - 2 b - x + 5 6).
8. Add 19 b + 3 c, 2 b - 7 c, 2 c - 14 b, and c + 8 6.
9. -(a — 3 6— c) — (2 c— a- 5 6) + (a- c + 6).
10. Subtract 2 a; -f- 4 y + z from 13 a — 3y — 5z + 8.
"• 3 + 6 + 12 + T = "
12. 5(^-7) -3(14-z) +60= l-10ar.
13. 13 (!-»)- 6 (2 a; -5) =80 + 12 a:.
96 INVOLVED NUMBER EXPRESSIONS
14 x-3 s-4 a?-8 a?-6 = 18
7 5 2^6
15. Add 7aj-3y-4, 5a; + 2y-f-5, and 3y-Sx-6.
16. Add 13 a + 4 b - 9 c, 2 c - 8 b - 16 a, and 8 a - 5 6 - 8 c.
17. (13-4-2)(5*-3 + 4aj).
18. 5(x-y)2 -5(y-xf + (x-y)(x + y).
19. Square 73, 79, 92, 98, 1003, and 995 as binomials.
20. From 17 6 -4 a -2 c - 19 subtract 8 c-5 a — 8 6 + 4.
2i. 3 -(3 -2 + 6 + 8 -3) + 8- (9 -3 + 8).
22. 3(4-a)-2(5-6a;) = 8<c + 4.
Aft 0* - />• /y» iM
23. _4--4-___4-_ = — 2.
4 8 16 2 32
24 y-3 y + 9 = y-ll 2
* 4 10 4
25. ^i + ^±J + ^+8 = 2y-20.
3 3 3 *
26. 5a;-(3a;-2 + 2i/4-a;) + 13y-(6-3a; + 4).
27. From 3 — 4a — 5c + 8ar* subtract 2a?2— 2a — 4c + 8.
28. 12 + (2o-3c-46)-(36-c-a-8).
29. y+y±^+y±5=25.
30. r-fciJ!+fc=»+«hI8.iBL
3 5 5 2
31. Add y - 20, 4 y + 6, 2 y + 4 x - 13, and 2 a - 8 y - 40.
32. (x - 1)(2 * - 2) + (a? - 5)2 = (3 - a) (24 - 3 a) - 7.
33. Subtract 16 — a; + 2z — 4y from 3x — 5z — 8y.
34. 19 + (2 a; - 7) - (31 - 4 a? - 8 - 2 a?) = 5 x + 7.
35. (4 aft — 6ac — 5 ad)(b — c + a").
36. (17 x + 8)(* - 1) + 8 = (2 - x)(6 - 17 aj) + 19.
REVIEW EXERCISES 97
37. 5 - (a + b - c - d + 8) + (3 + a + c - d) - 5.
38. h- — ! = a — ^o.
10 10 10
39. Add 6 a + 9, 8 a - 13, 46 a - 8, and 6 - 54 a.
40. (a-2)(6a-4)+2(a-l)2=(6-a)(30-8a) + 4.
41. From 6 (a+ 2) + 3(c + 4) - 2 (6 - d) subtract 2 (a + 2)
-2(c + 4)+3(6-d).
a , a + 7 a — 3 _ a + 227 1
42' 3 + _4 3~~^ 1'
43. (a-2-3c-8 + 2 6)(6-a-c-& + 8).
44. ^ll + ^±l + lip3=2
2 2 12
45. a2& - (3 6 - 8 a2 - 7) + 3 a&2 - (4 ab2 + 8 - 2 a2).
46. ^+l + ^Z^ + ^lI = 2a-26.
4 4 4
47. Add 12 a262c + 8 ax, 6 ax -8 a2b2c, and 2 ax + 3 a262c.
48. Add 5 a#2 + 3 a?y -\- <ixy, 2 x2y — 6 xy2 — 3 ay, and 4 xy.
49. Add 6ab — 3c— 2 a, 2c — 4 a& — 5 a, 5c— a + a&, and
3 + 5a-2c-3a6.
CA fl + l.« + 3.»-l w + 13 ,n-2
5U. — _ 1 . 1 . — — _ [_
3 4 4 3 3
51. (n-4)(6-3n)-(6-w)2-10 = -4n(n-4).
52. From 35 ab - 8 x — 9 z + 13 subtract 16a6 - 4z + 5 x + 8.
53. (n + 2)2 + (n-l)2+(n + l)2 = 3n(?i + 2) + 60w + 130.
54. Subtract 5a — 8x— 6y from 13 x + 14 y — 15 z — 4 a.
55. From 9y — 4a: — 6 z — 3 & subtract 8— 9 y — 3# — 2 z.
56. 2 <e + 4 - 6 (5 x- 8 - 7 a) +2 - 4 a = 6 (2 -3 a;) -42.
57. _ (7 + 4a;-8-2a;)+4-2a; = 6a; + 25.
98 INVOLVED NUMBER EXPRESSIONS
58. 16 + 5x-(8x + 9-4x + 17)=8x-3.
59. 6x-3-(4x+8-9x)-(5x-2)=x + ll.
60. (a — 1 + 6 — c — cf)(4a + 5& + 3c — 2d).
61. (4ax — 3a?/-f 5az — 8)(x + ?/ — z + 2).
62. (3ac-2ad + 4ed)(2 + 3 + 4 + 5).
63. 7-(3a-2&-4a)+6 + 2a-(3&-2a-a).
64. 8x + (5y-8) + (2y-l)-(13y + 8x-17).
65. 16 ax + 4 - (8 - 8 ax - a) - (12 ax - 13 - ax).
66. Add 15 ax2 + 3 6c2, 2 6c2 - 7 ax2, and 5 + 2 ox2- 5 6c2.
67. Addl6-7a6-2a2 + 5a&, 4a2-2a6, and5a&-8.
68. Add 51 x2y - 35 + 12 a2, 41 - 17 a2 - 57 x*y, and 3x*y.
69. Add 35 b2 - 13 c2, 8 c2 - 3 b2c2, and 6 &2 - 8 c2 - 9 62c2.
70. Add 19-2x + 3x26 + 26, 4 x2& + x + 5 b + 8, and 4x.
71. Subtract 2 a - 6 tfb-3x+2l from 19 - 2 * + 3 x2b - 7 a.
72. From 6 a - 45 -f 8 b - 3 c + 82 c& subtract 7 & -f 18 + 6 c.
73. (17 + 2a-3&-4&c)(2-a + &-c).
74. (13c-4d + 8e-3)(c-d).
75. (4xy-2y-3x-2)(y-x + y + 5).
76. (9 ax — 3 op - 5 a — 2 x + 4) (5 — x).
77. From 41 a2 + 7 a& - 5 c2- 9 ab subtract 8 ab + 7 c2+ 50 a2.
78. From 15 + 2x — 9xy — 3y subtract 5 xy — 4 a; — 2 ?/.
79. Subtract 12 + 8 x — 4 a —6 c — 18 abc from 5 x— 2 a + 3 c
80. 2-(71z + 42y-15x-64)-(5-91x-2y-13xy).
81. (31-2* + 3y-5)(aj + y).
82. 8x + (13 + 18x-6)-(5-6x) = 16x+10.
REVIEW EXERCISES 99
83. (9z-3)(4-aO+(z-3)2= -8 (z + 2)2 +94.
84. (x + l)2 + (a; + 2)2 + (a; + 3)2=(3a;-l)(a;+12)-43.
85. (2 x + 5) (a; -7)-(x- 1)2 = (* + l)(x + 2) - 28.
86. 3 (o-xf- (2x - l)(a? - !)=(*- 7) (a; + 10) + 17 * + 50.
87. 5 - (7 - 4 x + 2 ^ - 4 b) - (8 a? - 6 y - 9 + 2 6) + 8 a.
88. 8a>-(-3-2-4-7) + 5a; + (2 + 6 + 4)-(-3a; + 2).
89. 5y + 2a?-(6-4aj-5a;)-3y-(4a;-2y)-(-7y+8).
90. 35 y - (41 x - 16 - 12 y) + 5 a; + ( - 6 + 46 y - 18 a?) .
91. (.T-l)2-(a?-8)(2a;-l)=-a;2 + 98.
92. (32 + a)(4a-l) + (5-a;)2 + (a;-l)2 = 6(a: + l)2 + 194.
93. (2a?-7)(5-aj)-(2-5a;)(l - a;) = - a; (7 a? - 34) - 17.
94. (7 + x)(x-4) + (l-x)2=-23 + 2x2.
95. (12-4a)(2-a;)-4(l + :z)2 = 5a; + 119.
96. (x - 17)(59 - 2 x) - (1 - a;)2 = (6 - 3a;) (* - 2) + 384.
97. (3 x- 2) + (x - 1)2 + (x - 2)2 = 2 (a; - l)(x - 2) + 5.
98. (6-3 a>)(2 + x) + 16 (x - 1)2 = 13 (x + 4)2 + 364.
.. x+8 x- 9 . x- 17 4a-7 , 2a; + 6 . 5-31a;
12 6 2 3 12
100. 3j^_3jE±3 a^=aL±_5 20.
6 3 2 6 3
CHAPTER IV
THE SOLUTION OF PROBLEMS
90. Some of the advantages of algebra over arithmetic in
solving problems have been pointed out in the preceding chap-
ters. For instance, brevity and simplicity of statement secured
through the use of letters to represent numbers ; the transla-
tion of the words and sentences of problems into number ex-
pressions and equations ; and the clear and logical solution of
the equation, step by step.
91. A still greater advantage is set forth in the present
chapter ; namely, the opportunity offered by the symbols and
processes of algebra to summarize a whole class of problems
under one solution, called the formula, which is thereafter used
to solve all problems of the class.
PROBLEMS INVOLVING INTEREST
92. A class of problems already within the pupil's experi-
ence will illustrate this point. The different cases of percentage
or interest have been studied in arithmetic, and a large number
of isolated problems have been solved according to the rules.
In this instance, therefore, we proceed at once to summarize
all of this work in a few short statements.
T,etp = any principal, i.e. a number of dollars at interest.
i = the interest, i.e. the number of dollars accrued.
r = the rate, to be expressed in hundredths.
t = the time, to be expressed in years and fractions of a
year.
100
PROBLEMS ON INTEREST 101
Then the rule of arithmetic for finding the interest when the
principal, rate, and time are given is
interest = principal x rate x time,
i.e. i=prt. (1)
If in this equation i = 150, r = .05, t = 6, find p. If i= 190.5, /> = 635,
r = .03, find t. If i = 665, p = 1000, t = 17, find r. Substitute other
values for any three of these letters and find the value of the remain-
ing letter.
Solve equation (1) for t in terms of i, p, and r ; also for p in
terms of i, r, and t, and for r in terms of p, t, and i.
It follows that if any three of the four numbers, p, r, i, and
t, are given, the remaining one may be found.
Let the student state four rules of interest by translating
these formulas into words.
Note the simplicity of these equations compared with the corre-
sponding rules in arithmetic.
Solve each of the following problems by substituting in the
proper formula :
1. What is the simple interest at 5 % on $400 for 5 years
and 9 months ?
Solution. i=p-r-t = 400 • T^ • 5| = 4 • 5 • \s = 115.
2. In what time will the simple interest on $750 at 3 %
amount to $225? Substitute in the formula t = —
pr
3. What is the semi-yearly income from an endowment of
$2,700,000, the rate being 4| of0 per annum ? (Here t = \.)
4. A father invested $1500 at 5^%, the simple interest on
which was to go to his eldest son on his 21st birthday. The
young man received $1240. How old was the son when the
investment was made ?
102 THE SOLUTION OF PROBLEMS
5. What is the amount of money invested if it yields
$ 787.50 interest per annum, the rate being 5\%? (Here t = 1.)
6. A certain investment yields $8160 in 8 years. What is
the principal, if the rate of interest is 4 °fo ?
7. A $45,800 investment yielded $13,396.50 interest (sim-
ple) in 6\- years. What was the rate of interest ?
8. The endowment of a small college is $ 750,000, the yearly
income from which is $45,000. What is the average rate at
which the endowment is invested ?
9. A capitalist has investments amounting to $ 360,000, the
total income from which amounts to $ 1800 per month. What
is the average rate at which the money is invested ? (t = ^.)
10. If $ 700 is invested at 5±- % simple interest, what is the
amount at the end of 5 years 6 months ? This problem calls
for the amount, which is the sum of principal and interest.
If a = amount, then a= p +i = p + prt.
11. Solve the equation a—p+prt for p in terms of a, r,
and t, and translate the result into words.
12. Solve a = p +prt for r in terms of a, p, and t, and trans-
late the result into words.
13. Solve a =p +prt for t in terms of a, p, and r, and trans-
late the result into words.
14. Seven years ago I invested a certain sum of money at 6 %
simple interest. The amount at present is $5680. How much
did I invest ? (Use the formula obtained in Problem 11.)
15. Six years ago A invested $470 at simple interest. The
amount at present is $ 611. What is the rate per cent ?
16. Some years ago I invested $500 at 7 % simple interest,
which now amounts to $815. How many years ago was the
investment made ?
17. A merchant bought goods for $ 250, and some months
later sold the goods for $ 300, making a profit of 1 % per
month. How many months between the purchase and the sale ?
PROBLEMS ON INTEREST 103
18. A real estate dealer sold a house and lot for $ 7500,
for which he received a commission of 4%. What was the
profit ?
In this case the element of time does not enter. The word
commission is then used for interest and the principal is called
the base. Letting c = commission, b = base, and r = rate, we
have
c = br.
Applying this formula, c=br = 7500 • y^ = 75 • 4 = 300.
19. Solve the equation c = br for b in terms of c and r, and
translate the result into words.
20. Solve c = br for r in terms of c and b, and translate the
result into words.
21. A merchant's commission for selling a carload of peaches
was $ 18.75. What was the rate of commission if the peaches
brought $ 375 ? (Use the formula found in 19.)
22. A broker received $25.50 commission for selling $850
worth of bonds. What was his rate of commission ?
23. An agent received 3 % commission for buying and 3| %
for selling some property. He paid $ 5750 for it and sold it
for $ 7200. What was his total commission ?
24. How much must I remit to my broker in order that he
may buy $ 600 worth of bonds for me and reserve 5 °J0 for his
commission ?
I must send him both base and commission. Calling this the
amount and representing it by a, we have
a = b + c = b + br = b (1 + r).
Hence a = 600 + 600 . fa = 600 + 6 • 5 = 630.
25. Solve the equation a=b + br for b and translate the
result into words.
26. Solve a = b + br for r and translate the result into
words.
104 THE SOLUTION OF PROBLEMS
27. A merchant received $ 918 with which to buy corn after
deducting his commission of 2 °J0 on the price of corn. How
much was his commission and how much was used to buy
corn ?
Here a = 980. Find b by the formula under 25, which gives the
sum paid for corn.
28. A broker received $790, of which he invested $750
in stocks, reserving the balance as his commission. Find
the rate of his commission by means of the formula obtained
in 26.
29. An agent received $ 945 with which to buy lumber after
deducting his commission of 5 % on the cost of the lumber.
How much was his commission ? (First find the base by sub-
stituting in the formula of 25.)
30. A dealer sold berries for $ 18.95, and after deducting a
commission of 2 % sent the balance to the truck gardener.
How much did he remit ?
The sum he sent was the difference between the base and the com-
mission ; calling this d, we have
d=b-c = b-br = b(l-r).
Hence in this case d = 18.95 (1 — Tfo) = 18.57.
31. Solve the equation d = 6(1 — r) for b in terms of d and r
and translate the result into words.
32. Solve the equation d = b — br for r in terms of b and d
and translate the result into words.
33. After deducting a commission of 3 % for selling bonds,
a broker forwarded $ 824.50. What was the selling price
of the bonds ? (Solve by means of the formula under
Problem 31.)
34. A broker sold stocks for $ 1728 and remitted $ 1693.44
to his principal. What was the rate of his commission ?
(Solve by means of the formula under Problem 32.)
PROBLEMS ON AREAS 105
35. In how many years will $200 double itself at 5 % ?
In this case i =p. Hence, using formula (1), page 101, we have
200 = 200 x r§5 x t.
Hence t = 20.
36. In how many years will any sum, p, double itself at any
rate, r?
Here p = prt. Hence, solving, t = 1 + r.
37. In what time will a sum of money double itself at 6% ?
7%? 4i%? 3|%?
38. Collect all the formulas worked out in this set of
problems. Translate each into words. Which were used
as the original ones from which the others were deduced ?
How many and which ones are needed in order to derive
all the others ? Why are such formulas better than rules
expressed in words ?
PROBLEMS INVOLVING AREAS
93. Another class of problems already well known to the
pupil in arithmetic concerns the areas of rectangles and tri-
angles.
If in a rectangle we let the number of units of length be de-
noted by I, the number of units of width by w, and the number
of square units in the area by a, then area = length x width;
i.e. a = Iw. (1)
If in equation (1) a = 144, I = 16, find w. If a = 1116, w = 31, find I.
Substitute other values for any two of these letters and find the value
of the remaining one.
Solve this equation for to in terms of I and a, and also for I in
terms of w and a.
106 THE SOLUTION OF PROBLEMS
Also if b is the base of a triangle, h its altitude (height), and
a its area, then area = \ (base x altitude) ;
i.e. . .=**. (2)
Substitute particular values for any two of these letters and find
the value of the remaining one.
Solve (2) for h in terms of a and b, and also for b in terms
of a and h.
Let the student translate each of these equations into words.
Use these formulas in the solution of the following problems :
1. How many tiles each 3 by 4 inches are needed to cover
a rectangular floor 18 by 22 feet ? Use formula (1).
2. How long a space 25 feet in width can be covered by
340 square feet of roofing ?
3. The base and altitude of a triangle are 8 and 6 respect-
ively. Find its area.
4. The base of a triangle is 12 and its area 72. Find its
altitude.
5. The altitude of a triangle is 16 and its area 144. Find
its base.
6. A rectangle is 5 feet longer than it is wide. If it were
3 feet wider and 2 feet shorter, it would contain 15 square feet
more. Find the dimensions of the rectangle.
Let w equal the width; then construct number expressions for the
length, width, and area under the supposed conditions.
7. A rectangle is 6 feet longer and 4 feet narrower than a
square of equal area. Find the side of the square and the
sides of the rectangle.
8. The base of a triangle is 8 inches greater than its alti-
tude. If the base is increased by 4 inches and the altitude
decreased by 2 inches, the area remains unchanged. Find the
base and altitude of the triangle.
PROBLEMS ON AREAS 107
9. A rectangle is 14 inches longer than it is wide. If the
width is increased by 5 inches and the length decreased by 4
inches, the area is increased by 70 square inches. Find the
dimensions of the rectangle.
10. A rectangle is 15 rods longer and 10 rods narrower than an
equivalent square. What are the dimensions of the rectangle?
11. The altitude of a triangle is 16 inches less than the base.
If the altitude is increased by 3 inches and the base by 2 inches,
the area is increased by 52 square inches. Find the base and
altitude of the triangle.
12. The width of a rectangular field is 20 rods less than its
length. If each side is decreased by 10 rods, the area is de-
creased by 900 square rods. What are the dimensions of the
field ?
13. A picture is 4 inches longer than it is wide. Another
picture, which is 12 inches longer and 6 inches narrower, con-
tains the same number of square inches. Find the dimensions
of the pictures.
14. A picture, not including the frame, is 8 inches longer
than it is wide. The area of the frame,
which is 2 inches wide, is 176 square inches.
Find the dimensions of the picture.
15. A picture, including the frame, is
10 inches longer than it is wide. The area
of the frame, which is 3 inches wide, is
192 square inches. What are the dimensions of the picture ?
16. The base of a triangle is 11 inches greater than its alti-
tude. If the altitude and base are both decreased 7 inches, the
area is decreased 119 square inches. Find the base and alti-
tude of the triangle.
17. The base of a triangle is 3 inches less than its altitude.
If the altitude and base are both increased by 5 inches, the
area is increased by 155 square inches. Find the base and alti-
tude of the triangle.
— * — ' — ■ — — - — —.
■!.■
x + 8
I Vol/,',/!,,/, ■/,,„,/,/,/,/, w,i„/>r/,W,
'■■..,■:,-) _^__/ Uiu.
108 THE SOLUTION OF PROBLEMS
18. A commander in attempting to draw up his men in the
form of a solid square finds that there are 80 men more than
enough to complete the square. If he places 2 more men on
each side of the square he needs 84 more men to complete
it. How many men in his command ?
PROBLEMS INVOLVING VOLUMES
94. Still another class of problems for which the funda-
mental formulas are already well known concerns the volumes
of rectangular solids and of pyramids.
If the number of units of length, width, and height of a
rectangular solid be denoted by I, w, and h respectively, and the
number of units of volume by v, then
volume = length x width x height ;
i.e. v = lwh. (1)
If w is the width, I the length of the rectangular base of a
pyramid, h its altitude, and v its volume, then
volume = ^ (area of base x altitude) ;
Iwh /ox
i.e. " = ~3~' ()
In equations (1) and (2) substitute particular values for any three
of the letters and find the value of the remaining one.
Use these formulas in solving the following problems :
1. Solve equations (1) and (2) for I, w, and h respectively
and translate each equation into words.
2. How many cubic feet of earth are removed in digging
a cellar 18 feet long, 12 feet wide, and 9 feet deep ? Solve by
substituting in formula (1).
3. If a cut in an embankment is 500 yards long and 4 yards
deep, how wide is it if 18,760 cubic yards are removed ?
PBOBLEMS ON VOLUMES 109
4. How deep is a rectangular cistern which holds 500 cubic
feet of water if it is 6 feet wide and 8 feet long ?
5. The base of a pyramid is 16 inches long and 12 inches
wide. Its altitude is 30 inches. Find its volume.
6. A pyramid whose volume is 72 cubic feet has a base
whose area is 24 square feet. Find the altitude.
7. A pyramid whose volume is 91 cubic inches has an
altitude of 21 inches. Find the area of its base.
8. A rectangular room which is 10 feet high is 4 feet
longer than it is wide. If it were 5 feet longer and 2 feet
wider, it would contain 950 cubic feet more than it does.
Find its length and width.
9. A city building 50 feet high extends back 25 feet
more than its frontage. If the building were 8 feet wider
and extended 10 feet farther back, its capacity would be in-
creased 59,000 cubic feet. What are the ground dimensions
of the building ?
10. A pyramid whose altitude is 12 inches has a rectangu-
lar base 5 inches longer than it is wide. If the length of the
base is decreased by 1 inch and
the width increased by 2 inches, /\K
the volume of the pyramid is in- //'(,\
creased by 72 cubic inches. Find / 1 I \ \
the dimensions of the base of the / / I » \
pyramid. / / I \ \
11. The base of a rectangular / I j \ \
pyramid whose altitude is 15 \* / *^~s \
inches is 12 inches longer than \/ x+5 ^-\
it is wide. If the length and
width of the base are both decreased by 3 inches, the volume
is decreased by 675 cubic inches. Find the dimensions of the
base of the pyramid.
110 THE SOLUTION OF PROBLEMS
12. The base of a rectangular pyramid is 15 inches wide.
The altitude of the pyramid is 4 inches less than the length
of the base. If the altitude is increased by 6 inches and the
length of the base by 3 inches the volume is increased by 1155
cubic inches. Find the altitude of the pyramid and the
length of its base.
13. The altitude of a pyramid is 7 inches less than the
length of the base. The width of the base is 13 inches. If
the altitude and the length of the base are both decreased by
3 inches, the volume is decreased by 364 cubic inches. Find
the altitude of the pyramid and the length of its base.
14. The width of the base of a pyramid is 2 inches greater
than its altitude. The length of the base is 34 inches. If
the altitude and the width of the base are both increased by
6 inches, the volume is increased by 2992 cubic inches. Find
the altitude and the width of the base.
PROBLEMS INVOLVING SIMPLE NUMBER RELATIONS
95. In the problems thus far considered in this chapter
the formulas are already known from previous study or ex-
perience. Algebra affords a means of deriving such formulas
in many cases where they are not already known.
1. The sum of two numbers is 35 and their difference is 5.
AY hat are the numbers ?
Let g = the greater number, then 35 — g = the lesser.
2. The sum of two numbers is 48 and their difference is
24. What are the numbers ?
3. The sum of two numbers is 41^ and their difference is
23^. What are the numbers ?
4. The sum of two numbers is 8590 and their difference
is 3480. What are the numbers ?
PROBLEMS ON NUMBER RELATIONS 111
The four problems just preceding are similar in character.
A simple rule can be found for solving all problems of this
kind. Consider problem 1.
We have g — (35 — g) = 5. (1)
By VII, 2-35 + g = 5. (2)
By I, A, 2 g = 35 + 5. (3)
By D, VI, *=? + f' (4)
By F, g = 20, the greater number, (5)
and I = 35 — g = 15, the lesser number. (6)
The results in the final form tell nothing more than the
answers to this particular problem ; but the value of g in the
35 5
form g = — 4- -, if examined closely, tells much more. Stated
in full it means :
the sum of the numbers . their difference
the greater = h ^
Examine now the solution of the other three problems to see
whether this same expression will give the greater number
in each case.
The great advantage in not adding 35 and 5 before dividing
35 5
by 2, is that the expression 1-- preserves the original num-
bers as given in the problem, so that we see how each enters
into the result.
If the sum of the two numbers is called s and their differ-
ence d, then we are compelled to keep these letters separate
to the end of the solution.
Thus, the lesser is 1 = s — g,
and g-(s-g)=d.
By VII, g-s + g = d.
By I, A, 2g = s + d.
112 THE SOLUTION OF PROBLEMS
s d
Hence by D, VI, the greater is g = - + - , and the lesser is
l = S-g = S-(j + f)
s d _ 2 s _ s_d_ s _d
2~2~T 2 2~2 2*
Hence J =| + g, am/ / = | - J-
These results put into words are as follows :
77*e greater of any two numbers is half their sum plus half
their difference, and the lesser is half their sum minus half their
difference.
Any problem of this kind is solved by substituting in this
formula the particular values given to s and d and simplifying
the results.
E.g. In problem 4, page 110, s = 8590 and d = 3480. Hence the
greater number is §590 + §M0= 4295 + 1740 = 6035, and the lesser
number is — - — = 4295 - 1740 = 2555.
2 2
5. Find by this formula two numbers whose sum is 17,540
and whose difference is 11,240.
6. Find two numbers whose sum is 40 and whose difference
is 52. Solve also without the formula.
Evidently one of these must be a negative number in order to
make the difference more than the sum, but the formula applies even
in such cases.
7. Find two numbers whose sum is 38 and whose difference
is 50. Solve also without the formula.
8. The sum of two numbers is 48, and one is 3 times the
other. Find the numbers.
9. The sum of two numbers is 108, and one is 6 times the
other. Find the numbers.
PROBLEMS ON NUMBER RELATIONS 113
10. The sum of two numbers is s, and one is k times the
other. Find the numbers.
Let
n = one of the numbers,
Then
s — n — the other number,
and
kn = s — n.
By ,4,
kn 4- n = s.
By I,
(1 + 1) n = s.
By A
n — , one of the nu
L l 1
k + 1
and kn =k , the other number.
* + l
Problems 8 and 9 may be solved by substitution in this
formula. State this formula in words.
11. The sum of two numbers is 195, and one is 14 times the
other. Find the numbers. Solve also without the formula.
12. The sum of two numbers is 75, and one is f of the other.
Find the numbers. Solve also without the formula.
13. The sum of two numbers is — 52, and one is 12 times the
other. Find the numbers. Solve also without the formula.
14. If 9 be added to a number and the sum multiplied by 4,
the product equals 7 times the number. What is the number ?
15. If 21 be added to a number and the sum multiplied by 5,
the product equals 12 times the number. What is the number ?
16. If a be added to a number and the sum be multiplied by
b, the product is c times the number. What is the number ?
If n = the number. Show that
ab
c-b'
17. If 24 be added to a number and the sum multiplied by 3,
the product is 9 times the number. Find the number. Solve
by use of the formula of 16, and also without it.
114 THE SOLUTION OF PROBLEMS
18. If 3 be added to a number and the sum multiplied by
16, the product is 10 times the number. Find the number by-
use of the formula, and also without it.
19. The sum of three numbers is 108. The second is 16
greater than the first, and the third 25 greater than the sec-
ond. What are the numbers ?
20. The sum of three numbers is 98. The second is 7
greater than the first, and the third is 9 greater than the sec-
ond. What are the numbers ?
21. The sum of three numbers is s. The second is a greater
than the first, and the third is b greater than the second.
What are the numbers?
. n l
If n = the first number, show that n= - — — - — -•
3
Translate the result of problem 21 into words. It should be real-
ized that if in problems 19 and 20 the numbers had been kept iincora-
bined, as they were necessarily in 21, those results would translate
into words, exactly as in 21.
Use the formula derived in 21 to solve the following, and
also solve each one without the formula.
22. The sum of three numbers is 198. The second is 28
larger than the first, and the third is 25 larger than the second.
What are the numbers ?
23. The sum of three numbers is 31. The second is 3 more
than the first, and the third 2 less than the second. What are
the numbers ?
Since the third number is 2 less than the second, this means that
b of the formula is negative ; i.e. b = — 2.
24. The sum of three numbers is 91. The second is 11 less
than the first, and the third is 12 more than the second. What
are the numbers ?
25. The sum of three numbers is 69. The second is 9 less
than the first, and the third is 6 less than the second. What
are the numbers ? (How does the formula apply in this case ?)
PROBLEMS INVOLVING MOTION 115
26. Divide the number 248 into two parts, such that 7 times
the first is 42 more than twice the second.
27. Divide the number 645 into two parts, such that 13 times
the first part is 20 more than 6 times the other.
28. Divide the number a into two parts, such that b times
the first part is c more than d times the second part.
If x is the first part, show that x = fl •
b+d
29. Translate the formula in 28 into words.
30. Divide the number 1240 into two parts such that 5 times
the first is 200 more than the second. Solve by use of the
formula, and also without it.
PROBLEMS INVOLVING MOTION
96. Problems like those in the preceding class are useful
chiefly in cultivating skill in deducing formulas, and so mak-
ing rules. Those, however, in this and most of the following
classes are extremely important in themselves, because of the
simple laws of nature which they exemplify, and of which they
afford a wide range of application.
97. In scientific language the distance passed over by a
moving body is called the space, and the number of units of
space traversed is represented by s. The rate of motion, that
is the number of units of space traversed in the unit of time, is
called the velocity, and is represented by v. The number of
units of time occupied is represented by t.
E.g. At a certain temperature sound travels 1080 feet per second.
Hence, in 5 seconds it will travel 5 • 1080 = 5400 feet. In this case,
s = 5400 feet, v = 1080 feet per second, t = 5 seconds.
We then have the formula
* = Kf (1)
116 THE SOLUTION OF PROBLEMS
Solve the equation s = vt for t in terms of s and v, and for v
in terms of s and t.
Translate each of these formulas into words.
In equation (1) give particular values to any two of the letters and
find the value of the remaining one.
It is to be understood in all problems here considered that the veloc-
ity remains the same throughout the period of motion ; e.g. sound
travels just as far in any one second as in any other second of its
passage.
1. If sound travels 1080 feet per second, how far does it
travel in 6 seconds ?
2. If a glacier moves 450 feet per year, how far does it
move in 7 years?
3. If a transcontinental train averages 35 miles per hour,
how far does it travel in 2£ days ? (Given v = 35, t = 2\ • 24,
to find s.)
4. A hound runs 23 yards per second and a hare 21 yards
per second. If the hound starts 79 yards behind the hare,
how long will it require to overtake the hare ?
If t is the number of seconds required, then by formula (1) during
this time the hound runs 23 1 yards and the hare runs 21 1 yards. Since
the hound must run 79 yards farther than the hare, we have :
23 t = 21 1 + 79.
5. An ocean liner making 21 knots an hour leaves port when
a freight boat making 8 knots an hour is already 1240 knots
out. In how long a time will the liner overtake the freight ?
6. A motor boat starts 7| miles behind a sailboat and runs
11 miles per hour while the sailboat makes 6^ miles per hour.
How long will it require the motor boat to overtake the sail-
boat?
7. A freight train running 25 miles an hour is 200 miles
ahead of an express train running 45 miles an hour. How long
before the express will overtake the freight ?
PROBLEMS INVOLVING MOTION 117
8. A bicyclist averaging 12 miles an hour is 52 miles ahead
of an automobile running 20 miles an hour. How soon will the
automobile overtake him ?
9. A and B run a mile race. A runs 18 feet per second
and B Yl\ feet per second. B has a start of 30 yards. In
how many seconds will A overtake B ? Which will win the
race ?
10. Two objects, A and B, move in the same direction, A at
vx* feet per second and B at v2 feet per second. If A has n feet
the start, in how many seconds will B overtake him ?
If t is the number of seconds, then during this time A moves vxt
feet and B moves v<£ feet. Since B must move n feet farther than A,
wehave v2t=v1t + n. (2)
The solution of (2) for t gives the time sought.
It is of the utmost importance that formulas (1) and (2) be
clearly understood since they are fundamental in every motion
problem in this chapter.
11. A fleet, making 11 knots per hour, is 1240 knots from
port when a cruiser, making 19 knots per hour, starts out to
overtake it. How long will it require ?
12. In how many minutes does the minute hand of a clock
gain 15 minute spaces on the hour hand ?
Using one minute space for the unit of distance and 1 minute as
the unit of time, the rates are 1 and TJ2 respectively, since the hour hand
goes fa of a minute space in 1 minute. Letting t be the number of
minutes required, we have, just as in problem 10,
1 • t = fa t + 15.
13. In how many minutes after 4 o'clock
will the hour and minute hands be together ?
(Here the minute hand must gain 20 minute
spaces.)
* Vi and v2, read v one and v two, are used instead of two different letters to
represent the velocities of the first and second respectively.
118 THE SOLUTION OF PROBLEMS
14. At what time between 5 and 6 o'clock is the minute
hand 15 minute spaces behind the hour hand ? At what time
is it 15 minute spaces ahead ?
Since, at 4 o'clock, it is 25 minute spaces behind the hour hand, in
the first case it must gain 25 — 15 = 10 minute spaces, and in the
second case it must gain 25 + 15 = 40 minute spaces. Make a dia-
gram as in the preceding problem to show both cases.
15. At what time between 9 and 10 o'clock is the minute
hand of a clock 30 minute spaces behind the hour hand ? At
what time are they together ?
In each case, starting at 9 o'clock, how much has the minute hand
to gain ?
16. A fast freight leaves Chicago for New York at 8.30 a.m.
averaging 32 miles per hour. At 2.30 p.m. a limited express
leaves Chicago over the same road, averaging 55 miles per
hour. In how many hours will the express overtake the
freight?
If the express requires t hours to overtake the freight, the latter
had been on the way t + 6 hours. Then the distance covered by the
express is 55 /, and the distance covered by the freight is 32 (t + 6).
As these must be equal, we have 55 1 = 32 (t + 6).
17. In a century bicycle race one rider averages 19^ miles
per hour, while another, starting 40 minutes later, averages
22^ miles per hour. In how long a time will the latter over-
take the former ?
18. A sparrow flies 135 feet per second and a hawk
149 feet per second. The hawk in pursuing the sparrow
passes a certain point 7 seconds after the sparrow. In how
many seconds from this time does the hawk overtake the
sparrow ?
19. A courier starts from a certain point traveling vx miles
per hour, and a hours later a second courier starts, going at
PROBLEMS INVOLVING MOTION 119
the rate of v2 miles per hour. In how long a time will the
second overtake the first, supposing v2 greater than Vi ?
If the second courier requires t hours to overtake the first the latter
had been on the way t + a hours. Thus the distance covered by the
second courier is v%t and by the first Vi(t + a). As these numbers are
equal we have rrf«r,(f-M) (3)
20. In an automobile race A drives his machine at an
average rate of 53 miles per hour, while B, who starts \ hour
later, averages 57 miles per hour. How long does it require B
to overtake A ? Use formula (3). Solve also by finding how
far A has gone when B starts and then use formula (2).
21. A freight steamer leaves New York for Liverpool aver-
aging 10^ knots per hour, and is followed 4 days later by an
ocean greyhound averaging 20^ knots per hour. In how long
a time will the latter overtake the former ?
22. One athlete makes a lap on an oval track in 26 seconds,
another in 28 seconds. If they start together in the same
direction, in how many seconds will the first gain one lap on
the other ? Two laps ?
Let one lap be the unit of distance. Since the first covers one lap
in 26 seconds his rate per second is fa. Likewise the rate of the other
is fa. If t is the required number of seconds the distance covered by
the first is ^g t and by the second fa t. If the first goes one lap farther
than the second the equation is -fat = fat + 1; if two laps farther it
is fa t = fa t + 2.
23. Two automobiles are racing on a circular track. One
makes the circuit in 31 minutes and the other in 38£ minutes.
In what time will the faster machine
gain 1 lap on the slower ? y^ ,>v.
24. The planet Mercury makes a cir- / jr >v \
cuit around the sun in 3 months and / / sun^.^A.lvenua
Venus in 1\ months. Starting in con- \ V / /
junction, as in the figure, how long be- \ v^, ^S /
fore they will again be in this position ? ^s.^ ^y^
120 THE SOLUTION OF PROBLEMS
25. Saturn goes around the sun in 29 years and Jupiter in
12 years. Starting in conjunction, how soon will they be in
conjunction again ?
26. Uranus makes the circuit of its orbit in 84 years and
Neptune in 164 years. If they start in conjunction, how long
before they will be in conjunction again?
27. The hour hand of a watch makes one revolution in 12
hours, and the minute hand in one hour. How long is it from the
time when the hands are together until the}' are again together ?
28. One object makes a complete circuit in a units of time
and another in b units (of the same kind). In how many units
of time will one overtake the other, supposing b to be greater
than a ?
29. At what times between 12 o'clock and 6 o'clock are the
hands of a watch together ? (Find the time required to gain
one circuit, two circuits, etc.)
PROBLEMS INVOLVING THE LEVER
98. Two boys, A and B, play at teeter. They find that the
teeter board will balance when equal products are obtained by
multiplying the weight of each by his distance from the point
of support.
Thus, if B weigh?
U-!0-^ _ _ A____ __ ywih) 80 pounds and is t
• « £5 " ~IJ5 ' feet from the poinfc
of support, then A, who weighs 100 lbs., must be 4 feet from this
point, since 80 x 5 = 100 x 4.
The teeter board is a certain kind of lever; the point of
support is called the fulcrum.
In each of the following problems make a diagram similar
to the above figure :
1. A and B weigh 90 and 105 lbs. respectively. If A is
seated 7 feet from the fulcrum, how far is B from this point ?
PROBLEMS INVOLVING THE LEVER 121
2. Using the same weights as in the preceding problem, if
B is 6^ feet from the fulcrum, how far is A from that point ?
3. A and B are 5 and 7 feet respectively from the fulcrum.
If B weighs 75 pounds, how much does A weigh ?
4. A and B weigh 100 and 110 pounds respectively. A
places a stone on the board with him so that they balance
when B is 6 feet from the fulcrum and A 5^ feet from this
point. How heavy is the stone ?
5. If the distances from the boys to the fulcrum are respec-
tively dv and d2, and their weights w^ and w2, then
dlwl = d2w2. (1)
If any three of these four numbers are given, the fourth may be
found by means of this equation. Solve dxivx = d2w2 for each of the
four numbers involved in terms of the other three. Problems 1 to 4
can be solved by substitution in the formulas thus obtained.
6. A and B are seated at the opposite ends of a 13-foot teeter
board. Using the. weights of problem 1, where must the ful-
crum be located so that they shall balance ?
If the fulcrum is the distance d from A then it is (13 — d) from B.
Hence 90tf = 105 (13 -d).
7. A and B together weigh 212^ pounds. They balance
when A is 6 feet, and B 6f feet, from the fulcrum. Find the
weight of each.
8. A, who weighs 75 pounds, sits 7 feet from the fulcrum, and
B, who weighs 105 pounds, sits on the other side. At what dis-
tance from the fulcrum should B sit in order to make a balance ?
9. If in the preceding problem C, weighing 48 pounds, sits on
the side with A and 4 feet from the fulcrum, where must D,
who weighs 64 pounds, sit so as to maintain a balance?
From problem 8, when the teeter balanced it was found that A 'a
weight acted like a lever with a downward force of 7 • 75 pounds and
B's on the other side with a force of 5 • 105 pounds.
7 • 75 = 5 • 105.
122 THE SOLUTION OF PROBLEMS
Then in problem 9, C and D must sit so as to keep the board in
balance, that is, so as to add the same downward force to both sides.
Hence, as 4-48 is added on the side of A, 3 • 64 must be added on
the side of B, since 4 • 48 = 3 • 04. Therefore, adding these two
equations, we still have the balance, namely,
7 • 75 + 4 • 48 = 5 • 105 + 3 • 64.
B (105 lbs.) D(Ulbs.) (7(18 lbs) (75 lbs.) A.
L | 3-feer A J-Tevr ' ]
6 feet. ^m 7 feet
The weight of the boy multiplied by his distance from the
fulcrum is called his leverage. The sum of the leverages on
the two sides must be the same. Hence, if two boys, weighing
respectively wx and w2 pounds, are sitting at distances dx and
d2 on one side, and two boys, weighing w3 and wt pounds, sitting
at distances d3, dt on the other side, then
wxdx + wJ, = Ms + »A- (2)
10. If two boys weighing 75 and 90 pounds sit at distances
of 3 and 5 feet respectively on one side and one weighing 82
pounds sits at 3 feet on the other side, where should a boy
weighing 100 pounds sit in order to make the board balance ?
11. A beam carries a weight of 240 pounds 1\ feet from the
fulcrum and a weight of 265 pounds at the opposite end which
is 10 feet from the fulcrum. On which side and how far from
the fulcrum should a weight of 170 pounds be placed so as to
make the beam balance ?
PROBLEMS INVOLVING DENSITIES
99. If a cubic foot of a certain kind of rock weighs 2.5 times
as much as a cubic foot of fresh water, the density of this rock
is said to be 2.5. If a cubic foot of oak weighs .85 times as
much as a cubic foot of fresh water, the density of the oak is
.85. The density of water is taken as the standard with which
the densities of other substances are compared.
PROBLEMS INVOLVING DENSITIES 123
Thus, when we say that the density of a certain kind of iron is
7.25, we mean that any given volume of the iron weighs 7.25 times as
much as a like volume of fresh water ; and when we say that the
density of cork is .24, we mean that a given volume of cork weighs
.24 times as much as a like volume of fresh water.
A cubic centimeter of distilled water at the freezing point
which weighs one gram is used as a standard of comparison.
We therefore say that the density of any substance is equal to
the number of grams which a cubic centimeter of it weighs.
E.g. if a cubic centimeter (ccm.) of a certain kind of marble
weighs 2.5 grams, then it weighs 2.5 times as much as the same vol-
ume of water, and hence its density is 2.5.
The weight of an object in grams is, therefore, the product
of its volume in cubic centimeters multiplied by its density.
E.g. if the density of cork is .24, this means that a cubic centi-
meter of cork weighs .24 grams. Any volume of cork, say 10 ccm.,
weighs 10 x .24 = 2.4 grams. Hence, if we represent the weight of an
object in grams by w, its volume in cubic centimeters by v, and its
density by d, we have the relation,
w = vd. (1)
1. What is the weight of an object whose density is 4.3 and
whose volume is 250 ccm. ? (Here v = 250, d =4.3. Find w.)
2. What is the density of an object whose weight is 23.5
and whose volume is 17 ccm. ?
3. What is the volume of an object whose weight is 24
grams and whose density is .65 ?
4. If 500 ccm. of alcohol, density .79, is mixed with 300
ccm. of distilled water, what is the density of the mixture ?
The volume of the mixture is the sum of the volumes, and the
weight of the mixture is the sum of the weights, of the water and
alcohol. Hence, from formula (1) :
500 x .79 + 300 x 1 = d x 800. To find d.
124 THE SOLUTION OF PROBLEMS
5. If 1200 ccm. of cork, density .24, are combined with 64
ccm. of steel, density 7.8, what is the average density of the
combined mass ? Will it float or sink ? (A substance sinks
if its density is greater than that of water.)
In solving problems of this kind find the expressions representing
weight, density, and volume, and substitute in the equation w = dv.
6. How many cubic centimeters of cork, density .24, must
be combined with 75 ccm. of steel, density 7.8, in order that
the average density shall be equal to that of water, i.e. so that
the combined mass will just float ?
Let v = volume of cork to be used. Then the total volume is
75 + v, the total weight is 75 x 7.8 + .24 v, and the density is 1.
Hence, 75 x 7.8 + .24 v = 1 • (75 + r). Solve this equation for v.
7. Brass is an alloy of copper and zinc. How many cubic
centimeters of zinc, density 6.86, must be combined with 100
ccm. of copper, density 8.83, to form brass whose density is
8.31?
8. Coinage silver is an alloy of copper and silver. How
many ccm. of copper, density 8.83, must be added to 10 ccm. of
silver, density 10.57, to form coinage silver, whose density is
10.38 ?
9. The density of pure gold is 19.36 and of nickel 8.57.
How many ccm. of nickel must be mixed with 10 ccm. of pure
gold to form 14 karat gold whose density is 14.88.
10. How much mercury, density 13.6, must be added to 20
ccm. of gold, density 19.36, so that the density of the compound
shall be 16.9 ?
11. What is the average density of 40 ccm. of water, den-
sity 1, and 180 ccm. of alcohol, density .79 ?
12. How many cubic centimeters of water must be mixed
with 350 ccm. of alcohol, so that the density of the mixture
shall be .97?
PROBLEMS ON MOMENTUM 125
13. The density of copper is 8.83. 500 ccm. of copper is
mixed with 700 ccm. of lead, whose density is 11.35. What is
the density of the combined mass ?
14. "When 960 ccm. of iron, density 7.3, is fastened to 8400
ccm. of white pine, the combination just floats, i.e. has a density
of 1. What is the density of white pine ?
PROBLEMS ON MOMENTUM
100. The force with which a moving body strikes another
depends both upon its weight and upon its rate of motion.
The product of the weight and velocity of a moving body is
called its momentum. The weight of a body is also commonly
called its mass.
What is the momentum of a body whose weight is 10 pounds and
which moves 15 feet per second ?
What is the velocity of a body whose weight is 50 and whose
momentum is 350 pounds ?
What is the weight of a body whose momentum is 500 and whose
velocity is 25 feet per second ?
A bullet weighing \ of a pound, moving 2250 feet per second,
has a momentum equal to that of a stone weighing 50 pounds, which
is hurled at the rate of 9 feet per second, since \ • 2250 = 50 • 9.
By careful experiment it has been found that when a mov-
ing body strikes a body at rest but free to move, the two will
move on with a combined momentum equal to the momen-
tum of the first body before the impact.
Thus, if a freight car, weighing 25 tons and moving at the rate of
12 miles per hour, strikes a standing car weighing 15 tons, the two will
move on with the original momentum of 12 • 25. But as the combined
weight is now 25 + 15, the rate of motion has been decreased to 7\
miles per hour, since 12 • 25 = 1\ (25 + 15). In this case the automatic
coupler connects the cars and they move on together. Even if the
126 THE SOLUTION OF PROBLEMS
two bodies after impact do not cling together, as two croquet balls,
still the momentum of the one plus that of the other equals the
original momentum.
E.g. if a croquet ball weighing 8 ounces and moving 20 feet per
second, strikes another weighing 7 ounces and starts it off at the
rate of 18 feet per second, then if the diminished velocity of the
first ball is called v, we have
8-20 = 7- 18 + 8 v,
and solving, v — 4.25.
This indicates that the first ball is nearly stopped, which coincides
with common observation.
1. In a switch yard a car weighing 40 tons and moving 8
miles per hour strikes a standing car weighing 24 tons. What
is the velocity of the two after impact ?
2. A billiard ball weighing 6 ounces and moving 16 feet
per second strikes another ball which it sends off at the rate of
10 feet per second. The rate of the first ball is reduced to 9
feet per second by the impact. What is the weight of the
second ball ?
Since the momentum before impact equals the sum of the mo-
mentums after impact, we have 6 • 16 = 9 • 6 + 10 w, w being the
unknown weight of the second ball.
3. A bowler uses a 16-ounce ball to take down the last pin.
The ball sends the pin off at a velocity of 6 feet per second,
the weight of the pin being 48 ounces, while the velocity of
the ball is reduced to 4 feet per second. With what velocity
did the ball strike the pin ?
Since the momentum before impact equals the sum of the mo-
mentums after impact, we have 16 1> = 6 -48 + 4- 16, v being the
unknown velocity of the ball before impact.
4. In each of these problems we have considered the weight
of two bodies, which we may call wx and w2. If we call vx the
velocity with which the first strikes the second, v2 the velocity
PROBLEMS ON MOMENTUM 127
imparted to the second, and v/ the resulting decreased velocity
of the first,* we have
■W = Wjft' + w2 v2 (1)
Translate this equation into words. It contains 5 different num-
bers, toi, w2, vi, vi', and v2. If any four of these are given, the fifth
may be found by solving this equation for that one in terms of the
other four.
5. Solve the equation (1) for vx in terms of w1} w^ v/, and
v2 . Translate into words.
6. Solve the equation (1) above for Wy in terms of vi} vx', w2>
and v2. Translate into words.
7. Solve the equation (1) above for vx' in terms of wv iv2,
vv and Vjj. Translate into words.
8. Solve equation (1) for w2 in terms of wv v1} V> v2? an(i
translate the result into words.
9. Solve equation (1) for v2 in terms of w1} w2, vlt v^, and
translate the result into words.
10. In the result of the last exercise substitute w1 = 50,
w2 = 40, vx = 10, Vy = 2, and find the value of v2. Make a
problem to fit this case.
11. In the result of problem 8 substitute wx = 1000,
V!=75, Vi = 25, v2 = 50, and find the value of w2. Make a
problem to fit this case.
12. In the result of problem 6 substitute vx = 60, V = 10,
w2 = 80, v2 = 25, and find the value of wx. Make a problem
to fit this case.
13. In the result of problem 7 substitute v\ = 250, w2 = 125,
Vy = 50, and v2 = 50, and find the value of vx\ Make a prob-
lem to fit this case.
* v-l is read v one prime and is used rather than a new letter for the
purpose of recalling that this is another rate of the first body.
128
THE SOLUTION OF PROBLEMS
PROBLEMS ON THERMOMETER READINGS
101. There are two kinds of thermometers in use in this
country, called the Fahrenheit and Centigrade, the former for
common purposes, and the latter for scientific
records and investigations. Hence it frequently
becomes necessary to translate readings from
one kind to the other.
The freezing and boiling points are two
fixed temperatures by means of which the com-
putations are made. On the Centigrade these
are marked 0° and 100° respectively and on the
Fahrenheit they are marked 32° and 212° respec-
tively. See the cut. Hence between the two
fixed points there are 100 degrees Centigrade
and 180 degrees Fahrenheit.
That is, 100 degree spaces on the Centigrade cor-
respond to 180 degree spaces on the Fahrenheit.
Hence 1° Centigrade corresponds to |° Fahrenheit,
or 1° Fahrenheit corresponds to |° Centigrade.
All problems comparing the two thermometers are solved by
reference to these fundamental relations.
1. If the temperature falls 15 degrees Centigrade, how
many degrees Fahrenheit does it fall ?
2. If the temperature rises 18 degrees Fahrenheit, how
many degrees Centigrade does it rise ?
3. Translate + 25° Centigrade into Fahrenheit reading.
25° Centigrade equals | • 25° = 45° Fahrenheit.
45° above the freezing point = 45° + 32° above 0° Fahrenheit.
Hence, calling the Fahrenheit reading F, we have F = 32 + f • 25.
4. Translate + 14° Centigrade into Fahrenheit reading.
Eeasoning as before, F = 32 + § • 14.
PROBLEMS ON THERMOMETER READINGS 129
From the two preceding problems we have the formula
F = 32 + |C. (1)
Translate this into words, understanding that F and C stand
for readings on the respective thermometers.
5. Solve the above equation for C in terms of F and find
C = f(F-32). (2)
Translate this into words. Verify the solution of each of
the following by referring to the cut on page 128.
6. Translate -f 41° Fahrenheit into Centigrade reading.
Substitute in the proper formula.
7. Translate + 98° Fahrenheit, blood heat, into Centigrade
reading.
8. Translate — 20° Fahrenheit into Centigrade reading.
9. Translate — 40° Centigrade, the freezing point of mer-
cury, into Fahrenheit reading.
10. Translate 0° Fahrenheit into Centigrade by use of the
formula. Also 0° Centigrade into Fahrenheit.
11. Translate -f- 212° Fahrenheit into Centigrade, and also
+ 100° Centigrade into Fahrenheit.
12. What is the temperature Centigrade when the sum of
the Centigrade and Fahrenheit readings is 102° ?
13. What is the temperature Fahrenheit when the sum of
the Centigrade and Fahrenheit readings is zero ?
14. What is the temperature Centigrade when the sum of
the Centigrade and Fahrenheit readings is 140°?
15. What is the temperature in each reading when the
Fahrenheit is 50° higher than the Centigrade ?
16. The Fahrenheit reading at the temperature of liquid
air is 128 degrees lower than the Centigrade reading. Find
both the Centigrade and the Fahrenheit reading at this
temperature.
130 THE SOLUTION OF PROBLEMS
PROBLEMS ON THE ARRANGEMENT AND VALUE OF DIGITS
102. If we speak of the number whose 3 digits, in order
from left to right, are 5, 3, and 8, we mean 538 = 500 -f 30 + 8.
Likewise, the number whose three digits are h, t, and u is writ-
ten 100 h + 10 t + u.
Hence, when letters stand for the digits of numbers written
in the decimal notation, care must be taken to multiply
each letter by 10, 100, 1000, etc., according to the position it
occupies.
Illustrative Problem. A number is composed of two digits
whose sum is 6. If the order of the digits is reversed, we
obtain a number which is 18 greater than the first number.
What is the number ?
Solution. Let x = the digit in tens' place.
Then 6 — x = the digit in units' place.
Hence, the number is 10 x + 6 — x. Reversing the order of the
digits, we have as the new number 10(6 — x) + x.
Hence 10(6 - x) + x = 18 + 10 x + 6 - x.
1. A number is composed of two digits, the digit in units'
place being 2 greater than the digit in tens' place. If 4 is
added to the number, it is then equal to 5 times the sum of
the digits. What is the number?
2. A number is composed of two digits, the digit in tens'
place being 3 greater than the digit in units' place. The num-
ber is one more than 8 times the sum of the digits. What
is the number ?
3. A number is composed of two digits whose sum is 9.
If the order of the digits is reversed, we obtain a number
which is equal to 12 times the remainder when the units' digit
is taken from the tens' digit. What is the number ?
REVIEW QUESTIONS 131
4. A number is composed of two digits whose difference
is 4. If the order of the digits is reversed, we obtain a num-
ber which is 3 less than 4 times the sum of the digits. What
is the number ?
5. The digit in tens' place is 1 more than twice the digit
in units' place. If 36 is subtracted from the number, the order
of the digits will be reversed. What is the number?
6. The digit in units' place is 2 less than twice the digit in
tens' place. If the order of the digits is reversed, the number
is unchanged. What is the number ?
7. The digit in tens' place is 12 less than 5 times the digit
in units' place. If the order of the digits is reversed, the num-
ber is equal to 4 times the sum of the digits. What is the
number ?
8. A number is composed of three digits. The digit in
units' place is 3 greater than the digit in tens' place, which in
turn is 2 greater than the digit in hundreds' place. The num-
ber is equal to 96 plus 4 times the sum of the digits. What is
the number ?
REVIEW QUESTIONS
1. In order that a problem may be solved by means of a
formula, how many of the letters in the formula must be given
by the problem ? Illustrate this by the formula i =prt.
2. State the formulas involving areas and volumes which
have been used in this chapter. Solve each formula for each
of its letters in terms of all the others.
3. The area of a circle is found by squaring its radius and
multiplying by 3.1416. State this rule as a formula, using the
Greek letter tt for 3.1416.
4. The volume of a circular column is found by multi-
plying the area of its base by its height. State this rule as
a formula, using r for the radius of the base and h for the
height.
132 THE SOLUTION OF PROBLEMS
5. State formulas for finding two numbers when their sum
and difference are given.
Find a number such that if 1G be subtracted from it, 5 times
this difference equals the number.
Construct other problems like this and then make a formula
for the solution of all such problems. (See page 110.)
6. State three important formulas used in the solution of
motion problems. Translate each into words. Solve each for-
mula for each of its letters.
Of the motion problems, 4 to 29, which ones involve formula
(1) ? Which formula (2) and which formula (3) ?
7. State the formulas used in this chapter in working
problems on the lever. Solve each formula for each of
its letters.
Using a teeter board 12 feet long with the fulcrum in the
middle, how may three boys weighing 50, 75, and 100 lbs. re-
spectively be seated on it so as to make the board balance ? Is
there more than one solution ? Make a diagram for each of
your results.
8. What is meant by the density of a substance ? What is
used as a standard (unit) of density with which the densities of
other substances are compared ?
What is the relation between the weight of a substance, its
volume and its density ?
9. Define momentum. When a moving car strikes a stand-
ing car, free to move, what can you say of the momentum of
the two after impact?
10. Describe the Fahrenheit and Centigrade thermometers.
State the formulas for the reading of each thermometer in terms
of the other.
11. Write 347 as a trinomial. Write the number whose
three digits in order are a, b, c ; also the number whose three
digits in order are c, b, a.
REVIEW PROBLEMS 133
REVIEW PROBLEMS
103. Most of the following problems can be solved by sub-
stitution in some of the formulas developed in this chapter.
Solve as many as possible in this way.
1. One boy runs around a circular track in 26 seconds, and
another in 30 seconds. In how many seconds will they again
be together, if they start at the same time and place and run
in the same direction ?
2. Divide the number 144 into two parts, so that ^ of the
greater is 9 more than \ of the smaller.
3. The sum of two numbers is 2890. Seven times one
is 266 less than 5 times the other. What are the numbers ?
4. What is the simple interest on $ 400 at 6| % for 7 years
and 9 months ?
5. The number of telegraph messages sent in the United
States in 1905 was 5 million less than three times as great as
in 1880, and 69 million less than twice that in 1900 ; while in
1900 it was 16 million more than twice as great as in 1880.
Find the number of messages in each of these years.
6. The same number is added to each of the numbers 8, 9,
10, 12. What is this number if the product of the first and last
sums is equal to the product of the second and third sums ?
7. Find the time between 4 and 5 o'clock when the hands
of the clock are 30 minute spaces apart.
8. A man buys a house for $6500. His yearly tax on the
property is $57. The coal costs $60 per year, repairs $50,
and janitor service $108. To what monthly rental are his
expenses equivalent if money is worth 5 % ?
9. A boatman rowing down a river makes 23 miles in 3
hours and returns at the rate of 3^ miles per hour. How swift
does the river flow ?
134 THE SOLUTION OF PROBLEMS
10. A bird flying with the wind goes 65 miles per hour, and
flying against a wind twice as strong it goes 20 miles per hour.
What is the rate of the wind in each case ?
11. A steamer going with the tide makes 19 miles per hour,
and going against a current \ as strong it makes 13 miles per
hour. What is the speed of the steamer in still water ?
12. A boatman trying to row up a river drifts back at the
rate of 1^ miles per hour, while he can row down the river at
the rate of 12 miles per hour. What is the rate of the current?
13. A boatman rowing with the tide makes n miles per
hour ; rowing against a tide k times as strong, he makes m miles
per hour. At what rate does he row, and what is the velocity
of the stream ? State the result as a formula.
14. A beam is 16 feet long. • At what point must it be sup-
ported if it is to carry, when balanced, 460 pounds at one end
and 690 at the other, its weight being disregarded ?
15. Two numbers differ by 2 and the difference of their
squares is 100. What are the numbers ?
16. A beam is 12 feet long. It carries a 40-pound weight
at one end, a 60-pound weight 3 feet from this end, and a
70-pound weight at the other end. Where is the fulcrum if
the beam is balanced ?
17. There is a number composed of 2 digits whose tens' digit
is 2 less than its units' digit. The number is 1 less than 5 times
the sum of its digits. What is the number ?
18. A farm containing 240 acres can be rented at $3 per
acre. The renter finds that if he borrows money at 5 °J0 to
buy the farm he will save $125 per year. Find its value.
19. Two trains start from the same point at the same time
and in the same direction, one making 25 miles per hour and the
other 42 miles per hour. When will they be 238 miles apart ?
REVIEW PROBLEMS 135
20. The number of national banks in the United States on
March 1, 1906, was 1322 less than twice as many as on
March 1, 1900, and the number in 1903 was 1009 greater than
in 1900. If the number in 1906 be subtracted from 3 times
the number in 1903 the remainder is 7736. Find the number
in each of the three years mentioned.
21. The earth and Mars were in conjunction July 12, 1907.
When are they next in conjunction if the earth's period is
365 days and that of Mars 687 days ? (See figure, page 119.)
22. $ 7500 is invested at 4 % simple interest. Seven years
and three months later the amount is used to build a house.
What is the cost of the house if $ 735 has to be added to com-
plete it ?
23. There is a number composed of two digits whose sum
is 12. If the order of the digits is reversed, the number is
increased by 18. Find the number.
24. A slow steamer sails from New York to Liverpool, mak-
ing 9 knots per hour. A swift liner follows 62 hours later,
making 20| knots per hour. In how many hours will the latter
overtake the former?
25. A man invested a certain sum of money at 5 % simple
interest. The amount 3| years later was $ 950. What was the
investment ?
26. The capital stock of the Bank of France is 35.6 million
dollars less than that of the Bank of England and 6.3 million
greater than that of the Imperial Bank of Germany. The
combined capital stock of the three banks is 134.9 million
dollars. Find the capital stock of each.
27. The capital stock of the Bank of Italy is 27.7 million
dollars less than twice that of the Imperial Bank of Kussia,
and the capital of the Bank of Austria-Hungary is 13.6 million
greater than that of the Bank of Russia. Their combined
capital is 99.1 million dollars. Find the capital of each bank.
136 THE SOLUTION OF PROBLEMS
28. In a bicycle race A starts 32 minutes ahead of B.
B rides at the rate of 20^ miles per hour, while A rides 18|
miles per hour. How many miles from the starting point
does B overtake A?
29. A squadron of warships sails 13 knots per hour. A
torpedo boat making 27f knots leaves port 19 hours later to
overtake the squadron. In how many hours after leaving port
will the torpedo boat overtake it ?
30. A man takes out a life insurance policy for which he
pays in a single payment. Thirteen years later he dies and the
company pays $ 12,600 to his estate. It was found that his
investment yielded 2 °J0 simple interest. How much did he
pay for the policy ?
31. A merchant bought goods for $600 and some months
later sold them for $648, making a profit of 2% per month.
How many months elapsed between the purchase and the
sale?
32. In a building there are at work 18 carpenters, 7 plumbers,
13 plasterers, and 6 hod carriers. Each plasterer gets $ 1.90
per day more than the hod carriers, the carpenters get 35
cents per day more than the plasterers, and the plumbers
50 cents per day more than the carpenters. If one day's
wages of all the men amount to $183.45, how much does each
get per day ?
33. A train running 46 miles per hour leaves Chicago for
New York at 7 a.m. Another train running 56 miles per hour
leaves at 9.30 a.m. Find when the trains will be 15 miles
apart. (Two answers.)
34. Divide the number 280 into two parts so that \ of one
part is 48 less than \ of the other.
35. A merchant bought goods and sold them 5 months later
for $ 2687.50, making a gain on his investment of 1\ % per
month. How much did he pay for the goods ?
REVIEW PROBLEMS 137
36. A bicyclist starts out riding 12 miles per hour, and is
followed 40 minutes later by another riding 16 miles per hour.
Find when they will be 5 miles apart. (Two answers.)
37. A father invested $ 1000 at 6^ % interest. When the
principal and simple interest amounted to $ 2235, it was given
to the son for the expenses of his college education. How
long had the money been invested ?
38. Two trains start west at the same time, one from New
York and the other from Philadelphia. If the New York train
runs 55 miles per hour and the Philadelphia train 47 miles per
hour, how long before they are 15 miles apart, the distance
from New York to Philadelphia being 90 miles?
39. A man bought a tract of coal land and sold it a month
later for $ 93,840. If his gain was at the rate of 24 % per
annum, what did he pay for the land ?
40. There is a rectangle whose length is 60 feet more,
and whose width is 20 feet less, than the side of a square
of equal area. Find the dimensions of the square and the
rectangle.
41. A number is composed of two digits whose sum is 14.
If the digits are interchanged, the number is decreased by 18.
What is the number ?
42. What time between 7 and 8 o'clock are the hour and the
minute hands of a clock together ?
43. Change 104 degrees Fahrenheit (fever heat) to Centi-
grade reading.
44. A steamer leaves Liverpool for New York Saturday,
9 a.m., averaging 18 knots per hour. Seven hours later an-
other steamer leaves New York for Liverpool, making 20£
knots per hour. What time (Liverpool time) will the steamers
meet if the trans-Atlantic distance by their course is 2940
knots ?
CHAPTER V
INTRODUCTION TO SIMULTANEOUS EQUATIONS
104. Graphic Representation of Statistics.
A graphic representation of the temperatures recorded by the
U. S. Weather Bureau at Chicago on December 6 and 7, 1906,
is shown on the next page. The readings were as follows :
3 P.M.
29°
9 P.M.
21°
3 A.M.
12°
9 A.M. 12°
4 P.M.
29°
10 P.M.
20°
4 A.M.
11°
10 A.M. 13°
5 P.M.
28°
11 P.M.
17°
5 A.M.
10°
11 A.M. 16°
6 P.M.
26°
12 m't.
16°
6 A.M.
10°
12 Noon 17°
7 P.M.
24°
1 A.M.
14°
7 A.M.
10°
1 P.M. 18°
8 P.M.
22°
2 A.M.
12°
8 A.M.
10°
2 p.m. 20°
In the graph each heavy dot represents the temperature at a cer-
tain hour. The distance of a dot to the right of the heavy vertical
line indicates the hour of the day counted from noon December 6th,
and its distance above the heavy horizontal line indicates the ther-
mometer reading at that hour. The lines joining these dots complete
the picture representing the gradual changes of temperature from
hour to hour.
Graphs of this kind are used in commercial houses to represent
variations of sales, fluctuations of prices, etc. They are used by
architects and engineers to show the comparative strength of mate-
rials, stresses to which they are subjected, and to exhibit multitudes
of other data. They are used by the historian to represent changes
in population, fluctuations in mineral productions, etc. In algebra
they are used in solving many practical problems and in helping to
understand many difficult processes. In the succeeding exercises the
cross-ruled paper is essential.
138
GRAPHING OF STATISTICS
139
+
+
+
a,
3
p
±
Ifi
2
a
\
+
s
f
IS
M.
I :
..M
12
3 A
,H
1
;
P.M.
uTimi
Line
EXERCISES
Make a graphic representation of each of the following
tables of data:*
1. The population of the United States as given by the
census reports from 1790 to 1900:
1790 . . 3.9 (million) 1830
1800 . . 4.3 1840
1810 . . 7.2 1850
1820 . . 9.6 1860
12.9
1870 .
. 38.6
17.1
1880 .
. 50.2
23.2
1890 .
. 62.6
31.4
1900 .
. 76.3
*In each case the number to be represented by one space on the cross-ruled
paper should be chosen so as to make the graph go conveniently on a sheet.
Thus in (1) let one small horizontal space represent two years and one vertical
140 INTRODUCTION TO SIMULTANEOUS EQUATIONS
2. The population of New York city since 1800 :
1790 . . 33 (thousand) 1830 . . 202 1870 . . 942
1800 . . 60 1840 . . 312 1880 . . 1206
1810 . . 96 1850 . . 515 1890 . . 1530
1820 . . 123 1860 . . 813 1900 . . 1850
3. The population of Chicago since 1850 :
1850 .
. 30 (thousand)
1870 .
. 306
1890
. . 1100
1860 .
. 109
1880 .
503
1900
. . 1698
4. The world's
yearly production of gold since 1872 :
1872 .
99.6
(million)
1883 .
102.4
1894
. . 181.5
1873 .
96.2
1884 .
101.7
1895
. . 194.0
1874 .
90.7
1885 .
108.4
1896
. . 202.3
1875 .
97.5
1886 .
106.0
1897
. . 236.1
1876 .
103.7
1887 .
105.8
1898
. . 286.9
1877 .
114.0
1888 .
110.2
1899
. . 306.7
1878 .
119.0
1889 .
123.5
1900
. . 254.6
1879 .
109.0
1890 .
118.9
1901
. . 262.5
1880 .
106.5
1891 .
130.7
1902
. 296.0
1881 .
103.0
1892 .
146.3
1903
. 325.5
1882 .
102.0
1893 .
157.2
1904
. 346.8
5. Thei
world's
yearly productic
n of silver
since
1872:
1872 .
65.0
(million)
1883 . .
115.3
1894
. 214.5
1873 .
81.8
1884 .
105.5
1895
. 208.0
1874 .
71.5
1885 .
118.5
1896
. 203.0
1875 .
80.5
1886 .
120.6
1897
. 207.0
1876 .
87.6
1887 .
124.3
1898
. 218.6
1877 .
81.0
1888 .
140.7
1899
. 217.6
1878 .
95.0
1889 .
155.4
1900
. 224.0
1879 .
96.0
1890 .
163.0
1901
. 223.7
1880 .
96.7
1891 .
177.0
1902
. 208.6
1881 .
102.0
1892 .
197.7
1903
. 220.4
1882 .
111.8
1893 .
209.1
1904
. 217.7
space a million of population ; and in (3) let one horizontal space represent
one year and one large vertical space one hundred thousand of population.
GRAPHING OF STATISTICS
141
6. Temperature, at Port Conger, New York, and Singapore.
The average temperature for each half month is given.
Poet
New
Singa-
Port
New
Singa-
Month
Conger
YOKK
pore
Month
Conger
York
pore
Jan. 1-15
-35°
30°
80°
July
1-15
+ 32°
66°
86°
16-31
-40°
28°
82°
16-31
+ 37°
68°
86°
Feb. 1-15
-45°
32°
84°
Aug.
1-15
+ 37°
68°
87°
16-28
-40°
35°
84°
16-31
+ 34°
66°
86°
Mar. 1-15
-35°
38°
85°
Sept.
1-15
+ 27°
62°
85°
16-31
-30°
40°
85°
16-30
+ 20°
60°
85°
April 1-15
-20°
45°
85°
Oct.
1-15
+ 8°
55°
85°
16-30
-10°
48°
85°
16-31
- 2°
50°
84°
May 1-15
0°
50°
85°
Nov.
1-15
-15°
48°
83°
16-31
+ 8°
56°
86°
16-30
-20°
45°
82°
June 1-15
+ 16°
60°
86°
Dec.
1-15
-28°
42°
82°
16-30
+ 20°
65°
86°
16-31
-32°
40°
81°
When the temperature is below zero, the distance is measured
downward from the heavy horizontal line, as in the figure, page 149.
7. Average monthly rainfall at San Francisco, Valparaiso,
Chili, and Quebec :
(a) San Francisco
(b) Valparaiso
(c) Qttebec
January .
February .
March . .
. 5.5 (inches)
. 4.5
. 3.5
0.2
0.3
1
(inches)
3.2 (inches)
6.4
4.4
April . .
May . .
June . .
. 2.5
. 1.5
. 0.5
2
3
4.2
6.6
5.1
2.5
July . .
Aiagust . .
September
October
. 0.2
. 0.2
. 0.2
. 1.5
2.9
1.8
1.8
0.5
1.4
1.8
1.8
0.5
November .
. 3.5
0.4
0.4
December .
. 5.5
0.2
0.2
Use 10 small vertical spaces for 1 inch of rainfall and five horizon-
tal spaces for 1 month.
142 INTRODUCTION TO SIMULTANEOUS EQUATIONS
8. After making the graphs in exercise 6, each on a separate
sheet, put all three on the same sheet, using a different colored
ink or pencil for each. In this way the relative average tem-
peratures are simultaneously pictured.
9. After graphing (a), (6), and (c) of example 7, each on a
separate sheet, combine all three, using different colors, as
in 8.
10. Observe the weather reports in a daily paper and make
a graph representing the hourly change of temperature for
twenty-four hours.
GRAPHIC REPRESENTATION OF MOTION
105. A useful picture of the distance traversed by a moving
body can be made by a graph similar to the preceding.
E.g. suppose a man is walking 3 miles per hour. We mark units
of time from the starting point to the right along the horizontal ref-
erence line, and indicate miles traveled by the number of units meas-
ured vertically upward from this line. (See the figure on the oppo-
site page.)
In the figure each horizontal space represents 1 hour, and each ver-
tical space 3 miles. Then in 1 hour he goes 3 miles; in 5 hours,
15 miles ; in 10 hours, 30 miles ; etc. The dots representing the
distances are found to lie on a straight line.
The graph shows at a glance the answers to such questions as :
How many miles does he travel in 4 hours? in 13 hours? How long
does it take him to go 18 miles? 23 miles?
Again, suppose 24 hours later a second man starts out on a bicycle
to overtake the first man, and travels 9 miles an hour. The line
drawn from the 24-hour point shows the distance the wheelman trav-
els in any number of hours counting from his time of starting. The
points marked in this line show how far he has gone in 1, 2, 3, 4, 5, 6
hours, etc., namely, 9, 18, 27, 36, 45, 54, etc.
The point where these two lines intersect shows in how many
hours after starting the pedestrian is overtaken and also how far he
has gone.
GRAPHIC SOLUTION OF PROBLEMS
143
i-<o ^ r <? ■
i£ jZ J
-,**
.? j
In like manner solve the following by means of graphs, and
in each case suggest other questions which may be answered
from the graph :
1. In a century bicycle race A averages 17 miles per hour;
B, who starts 20 minutes later, averages 19 miles per hour. In
how many hours will B overtake A ? Who will win the race
and where will the loser be when the winner finishes ?
2. A starts for a town 12 miles distant, walking 3 miles per
hour, li hours later B starts for the same place, driving 1\
miles per hour. When does B overtake A ? Where is A when
B reaches town ?
144 INTRODUCTION TO SIMULTANEOUS EQUATIONS
3. In a mile race A runs 6 yards per second and B 5 yards
per second. B has a start of 250 yards. Who will win
the race ? How far in the lead is the winner at the finish ?
Let 1 vertical small space represent 20 yards, and 1 horizontal space
represent 5 seconds.
106. Illustrative Problem. A man rides a bicycle into the
country at the rate of 8 miles per hour. After riding a certain
distance the wheel breaks down and he walks back at the rate
of 3 miles per hour. How far does he go if he reaches home
11 hours after starting ?
\J
.•■>a &._ _ >__ __ 3^K _ _ _ __ _
3- G -f -Sw
:**:£ — zztzz *% ~ ~
\ j f **».
In this graph each large horizontal space represents 1 hour, and
each small vertical space represents 1 mile. The problem is solved
as follows :
(1) Construct the line representing the outward journey at the
rate of 8 miles per hour, extending this line indefinitely.
(2) Beginning at the point corresponding to 11 hours, find the
points representing his position at each pi-eceeding hour. The line
connecting these points represents the homeward journey at the rate
of 3 miles an hour. Extend this line until it meets the first line.
The point where the lines meet represents 3 hours and 24 miles,
which is the answer required in the problem.
GRAPHIC SOLUTION OF PROBLEMS 145
PROBLEMS
Solve the following problems by means of graphs. In each
case prepare a list of questions which may be answered from
the graph.
1. A man rows 18 miles per hour down a river and 2 miles
per hour returning. How far down the river can he go if he
wishes to return in 10 hours ?
2. A man goes from Chicago to Milwaukee on a train run-
ning 42^ miles per hour, and returns immediately on a steamer
going 17 miles per hour. Find the distance, if the round trip
requires 7 hours.
3. A pleasure trip from New York to Atlanta by steamer
and return by rail occupied 77 hours. Find the distance, if
the rate going was 16 miles per hour and returning 40 miles
per hour.
Let one small horizontal space represent one hour and one small
vertical space 16 miles.
4. A invests $1000 at 5% and B invests $5000 at 4%.
In how many years will the amount (principal and interest) of
A's investment equal the interest on B's investment?
Let one large horizontal space represent one year and one small
vertical space $50. Then the line representing ^4's amount starts at
the point marked $1000, and rises one small vertical space each year.
The line representing B'a interest starts at the zero point and rises
four small vertical spaces each year.
5. In how many years will the interest on $ 6000 equal the
amount on $ 2000 if both are invested at 5 % ?
6. A invests $500 at 6% and B invests $ 1000 at 5%. In
how many years will A's interest differ by $300 from .B's ?
Let one small vertical space represent $20. In this case both lines
start from the zero point. Find the point on one line which is three
large spaces vertically above the corresponding point on the other
line.
146 INTRODUCTION TO SIMULTANEOUS EQUATIONS
7. Construct a graph representing the relation between the
Fahrenheit and Centigrade thermometer readings.
Let the hue at the bottom of the sheet and the vertical line four
large spaces from the left margin be the reference lines. Let one
small horizontal space represent a degree C, and one smal] vertical
space a degree F.
From F = 32 + | C (page 129), we find that if C = 0° F = 32°, if
C = 10° F=50°, if C = 20° F = 68°, if C = 30° F = 86°, etc. Mark
the point representing each pair of readings, and draw the straight
line connecting these points. This is the required graph.
8. From this graph read the answers to the following
questions :
Find C when F= 41°, F = 59°, F= 79°, F = 14°.
Find F when C= 35°, C = 40°, C= - 5°, C= - 15°.
From the graph it is possible to find any Fahrenheit reading when
the corresponding Centigrade is given, and also to find any Centigrade
reading when the corresponding Fahrenheit is given. Thus the graph
shows to the eye all the information contained in the equation F = 32
+ 1 C. In what follows we shall consider in detail how to represent
equations in this manner.
GRAPHIC REPRESENTATION OF EQUATIONS
107. In all the graphs thus far constructed two lines at right
angles to each other have been used as reference lines. These
lines are called axes. The location of a point in the plane of
such a pair of axes is completely described by giving its distance
and direction from each of the axes. The direction to the
right of the vertical axis is denoted by a positive sign, and to
the left, by a negative sign j while direction upward from the
horizontal axis is positive, and downward, negative.
The horizontal line is usually called the jr-axis and the ver-
tical line the /-axis. The perpendicular distance of any point
P from the ?/-axis is called the abscissa of the point, and its dis-
tance from the x-axis is called its ordinate. The abscissa and
ordinate of a point are together called its coordinates.
GRAPHS REPRESENTING EQUATIONS
147
+
.
1
>
=r>
<W
7J
P
(
8^
M
+
<
•f
_H
i
_
3
_
)
1
*
l
,
+ :;
t
i
I
C
Zx
)f
X
j
1
—
,
-•
I
-
J
I
*;
E.g. the abscissa of point P in the above figure is 3 and its
ordinate 2, or we may say the coordinates of P are 3 and 2, and indi-
cate it thus : P : (3, 2), writing the abscissa first. In like manner for
the other points we write Q: (- 1, 3), R: (- 2, 0), S : (- 3, - 4),
and T : (2, — 3). (For convenience upper signs are used in the
figure.)
We see that in this manner every point in the plane corre-
sponds to a pair of numbers and that every pair of numbers cor-
responds to a point. This scheme of locating points by two
reference lines is already familiar to the pupil in geography,
where cities are located by latitude and longitude, that is, by
degrees north or south of the equator and east or west of the
meridian of Greenwich.
148 INTRODUCTION TO SIMULTANEOUS EQUATIONS
EXERCISES
1. With any convenient scale, locate the following points :
(2,6), (-3,5), (0,1), (1,0), (0,0), (0,-1), (0,-5),
(-5,0), (2i, 5ft, (-4,-8), (3,-10), (-10,3).
2. Calling north + , south — , east + , west — , so that
(— 5°, 8°) means the point on a map whose longitude is 5°
west and whose latitude is 8° north, verify the following on a
map: New York (- 73° 57', 40° 48'), Chicago (-87° 35', 41°
51'), Peking (116° 30', 39° 50'), Sydney, New South Wales,
(151° 15', -33° 51'), Santiago, Chili, (- 70° 39', - 33° 25').
3. On a map of South America give approximately the lo-
cation of the following cities, using the notation of the pre-
ceding exercise : Caracas, Rio de Janeiro, Bogota, Valparaiso,
Lima, and Panama.
4. On a map of Africa locate in similar manner the cities or
islands situated at the following points: (— 5° 40', — 16°),
(- 14° 30', - 8°), (30° 18', 30° 1'), (- 5° 40', 36°), (40° 40',
- 15°), (18° 20', - 33° 55').
5. Locate the following series of points and then see if a
straight line can be drawn through them : (0, 0), (1, 1), (2, 2),
(3, 3), (4, 4), (- 1, - 1), (- 2, - 2), (- 3, - 3). Name still
other points lying on the same line.
6. Locate the following and connect them by a line: (1,0),
(1, 2), (1, 3), (1, 4), (1, 5), (1, - 2), (1, - 3), (1, - 4), (1, - 5).
Name other points in this line.
7. Draw the line every one of whose points has its hori-
zontal distance — 2, also the line every one of whose points has
its vertical distance + 3.
8. Locate the following points and see if a straight line can
be passed through them : (1, 0), (0, 1), (2, - 1), (3, - 2),
(4-3), (-1+2), (-2, 3), (-3, 4), (-4, 5), & *), & f), (f, *).
Can you name other points on this line ?
GRAPHS REPRESENTING EQUATIONS 149
Temperature chart for Port Conger. (See page 141.)
150 INTRODUCTION TO SIMULTANEOUS EQUATIONS
108. In the preceding exercises, in certain cases, a series
of points has been found to lie on a straight line as in
examples 6, 7, and 8. Evidently this could not happen
unless the points were located according to some definite
scheme or law.
Illustrative Problem. Locate a series of points whose co-
ordinates are values of x and y which satisfy : 3 x + 4 y = 12.
We see that x = 0, y = 3, also x = 4, y = 0 are pairs of such values.
Evidently as many pairs of values as we please may be found by
giving any value to x and then solving the equation to find the corre-
sponding value of y. A table may thus be constructed as follows :
Let x = 0, 4, 8, 12, -4,-8, etc.
Then y - 3, 0,-3,-6, 6, 9, etc.
-i
n
-t-
-
'
4
4
+
*
+
+
f
-.
T
-(
It
h|
"■
n
—
_
p
^
'
These pairs of values for x and y correspond to the points as
plotted in the figure, and they are found to lie on a straight
line. This line is called the graph of the equation.
GRAPHS REPRESENTING EQUATIONS 151
Let the student find other pairs of numbers which satisfy this
equation and see if the corresponding points lie on this line. Also
find the numbers which correspond to any chosen point on this line
and see whether they satisfy the equation.
109. Since an equation like 3 x -f 4y = 12 is satisfied by in-
definitely many pairs of values of x and y, it is common to call
the unknowns in such an equation variables. Indeed if a point
be thought of as moving along the graph of this equation the
values of x and y corresponding to the moving point continually
vary, but always so that 3 x + 4y = 12.
110. Definitions. An equation is said to be of the first degree
in x and y if it contains each of these letters in such a way that
neither x nor y is multiplied by itself or by the other.
E.g. 13 x — 5 y = 14 is of the first degree, while 2 xy — x = 5 and 3 x
— 5 y2 = 13 are not of the first degree in x and y.
Every equation of the first degree in two variables has for
its graph a straight line ; hence such an equation is commonly
called a linear equation.
111. To graph an equation of the first degree it is only
necessary to find two points on the graph and draw a straight
line through them.
E.g. In graphing the equation x — y = 5, we choose x = 0 and find
y = — 5, and choose y = 0 and find x = 5 and plot the points (0, — 5)
and (5, 0). The line through these points is the one required.
EXERCISES
Construct the graph for each of the following equations :
1. 3x + 2y = l. 5. 5-2y=12. 9. 3a-4?/=-7.
2. 5x-3y=-3. 6. 3x + 5y= — 15. 10. 3x-4y=-12.
3. 7x + 10y=2. 7. 2x-y = 0. 11. 7y = 9»— 63.
4. x + 2y = 0. 8. 3x-4y = 7. 12. x = 5y + 3.
152 INTRODUCTION TO SIMULTANEOUS EQUATIONS
112. Illustrative Problem. Graph on the same axes the two
equations x + y = 4 and y — x = 2.
/
/
/
/
t
/
f
Lt
i
^
'
h
1
1
-1
t
-
.
-
n
+
t
2
+
:i
-1
i
~J
1
—
Solution. The two graphs are found to intersect in the point (1, 3).
Since the point lies on both lines, its coordinates should satisfy both
equations, as indeed they do. Since these lines have only one point
in common, there is no other pair of numbers which, when substituted
for the variables x and y, can satisfy both equations.
Hence x = 1, y = 3, which is written (1, 3), is called the solu-
tion of this pair of equations.
113. Definition. These two equations are called independ-
ent because their graphs are distinct. They are called simulta-
neous because there is at least one pair of values of x and y
which satisfy both.
ELIMINATION BY SUBSTITUTION 153
114. Since two straight lines intersect in but one point, it
follows that two linear equations which are independent and
simultaneous have one and only one solution.
EXERCISES
Graph the following and thus find the solution of each pair
of equations :
L f2x-3y = 25, 6 ty + 3x = 7,
Lie -f- 3/ = 5. \2y-\-x= —6.
2 |5* + 6y = 7, ? \x-2y = 2,
\2x-y = -A. ' V2x-y=-2.
g f5*+3y»l, 8 f5x-7y = 21,
i2x + y = — 4. (x — £y=— 1.
4 J6a; + 8y = 16, Q \5x + 2y = 8,
\2x-3y = ll. ' (2x-3y=-12.
5 \'3x-4:y = l, 1Q tx + 3y=-6,
\2x-7y = 5. ' \2x-±y=-12.
SOLUTION OF SIMULTANEOUS EQUATIONS BY SUBSTITUTION
115. A pair of linear equations may be solved without con-
structing their graph.
Illustrative Problem. The sum of 'two numbers is 35 and
their difference is 5. What are the numbers ?
Solution. Let x = the greater number and y the smaller.
Then fz + y = 35, (1)
and 1 x — y = 5. (2)
From (1) by 5, y = 35 - x. • (3)
Substituting 35 — x for y in (2), x — (35 — x) = 5. (4)
Solving, x = 20. (5)
Substituting x = 20 in (1), y = 15. (6)
154 INTRODUCTION TO SIMULTANEOUS EQUATIONS
116. In most problems solved heretofore there have been two
or more unknown numbers. In forming the equations the
object has been to express all but one of the unknowns in
terms of that one. In the case of two unknowns this is now
done more systematically as follows :
(a) State two equations involving the two unknowns, as (1)
and (2) above.
(6) Solve one of these equations for one unknown in terms
of the other as in (3) above.
(c) Substitute in the other equation the value of this un-
known as thus expressed, obtaining as in (4) one equation in
one unknown.
This equation (4) is the same as equation (1) on page 111,
where this problem was solved by means of one unknown.
The solution given here amounts to a formal tabulation of the
steps there taken in obtaining the equation, g—(35—g) = 5.
Since equation (4) contains but one of the unknowns, the
other is said to be eliminated.
The process here used is called elimination by substitution.
Illustrative Problem. Solve the equations :
\2x + 3y = 13.
\5x-6y=-8.
(1)
(2)
From (1) by
S and D,
13-2*
V= 3 *
(3)
Substituting
in (2)
5x 6Q3-2*)= 8.
3
(4)
ByF,V,
5 x- 2(13 - 2x)=-8.
(5)
By IV, VII,
5x-26+ ix = -k
(6)
By I, A,
9 x = 18.
(7)
By A
x = 2.
(8)
From (3),
13 - 2 • 2 -
y = ., = a-
(9)
Verify this by drawing the graphs and also by substituting these
values of x and y in (1) and (2).
ELIMINATION BY SUBSTITUTION
155
EXERCISES
Solve the following pairs of linear equations by eliminating
one of the variables by substitution, and check by substituting
the results in the original equations.
i.
2.
4.
5.
7.
8.
9.
10.
11.
x — y = 10.
x-y = -3,
a; + 4 y = 12.
2x + 3y = 5,
7x-5y = 33.
3 a; - 4 y = 8,
4 a; + 3 y = - 6.
2 x - 4 y = 8,
3 a? + 2 y = 4.
a; + 2y = 4,
2 a; + y = - 1.
3 x — 4 y = 8,
2aj + 3y = ll.
5 a; + 9 y = 19,
3 x — y = 5.
4 y - 2 x = 3,
2 y + 5 a; = 6.
3 x - 7 y = - 11,
2 a; + y = 4.
5<c — 3y = 4— 2a;+7y,
5 y + a; = 7.
12.
13.
14.
15.
16.
17.
18.
19.
20.
21.
22.
|3y + 5a; = 12 + 2a;,
tl7a;-y = 4y-20.
f6y-a; = 7 + 4y,
\5x + 8y = l.
(6 + x + y = 2x-l,
l3y + z = 6y + 9.
x-y = 37,
+ 3y = 314 + 13y.
2z-3y = y + 6,
a? + 2y = 4y + 3.
y + 5a; = 2a; + 5,
y - 3 x = 19.
\2x
l2y-
J5a; + 3y = 0,
I2aj + y = l.
2x + 3y=6x-l,
|2a; + 3y =
I3a;-2v =
r
y = 3.
5 x - 3 y = 0,
2a; + 2-6y = 2-a;.
|6a;+2y=23,
Il0aj-5v=21.
{
,y:
3a-7y=15,
5aj + 4y = 11.
Solve 21 and 22 by means of graphs and also by elimination
by substitution. Notice that the latter method gives a more
accurate solution than can be obtained from the graph.
156 INTRODUCTION TO SIMULTANEOUS EQUATIONS
SOLUTION OF SIMULTANEOUS EQUATIONS BY ADDITION OR
SUBTRACTION
117. Illustrative Examples. Solve
<x + 2y = 7, (1)
(Sx-2y = 5. (2)
Adding the members of these equations, + 2 y and — 2 y cancel.
Hence, 4x=12, (3)
x = 3. (4)
Substituting in (1), 3 + 2 y = 7, (5)
y=2. (6)
Verify this by drawing the graphs, and also by substituting x = 3,
y = 2, in (1) and (2).
If one of the variables does not already have the same co-
efficient in both equations, the solution may be obtained as
follows :
Solve the equations
[7a?-f3y = 4-y + 4a>, (1)
L3cc — 2/ = 4?/ — 2 — x. (2)
From (1) by A, S, 3x + iy = 4. (3)
From (2) by A, S, 4x-5y = -2. (4)
From (3) by M, 12 x + 16 y = 16. (5)
From (4) by M, 12x-15y = -Q. (6)
Subtracting (6) from (5), 31 y = 22. (7)
From (7) by D, y = f { . (8)
Substituting in (3), x = \\. (9)
Hence, the solution is Jf, f|,
118. The process used in the solution just given is called
elimination by addition or subtraction.
This method is usually simpler than elimination by substitu-
tion, since the latter frequently involves fractions.
ELIMINATION BY ADDITION OR SUBTRACTION 157
EXERCISES
Solve the following pairs of equations by addition or sub-
traction. Substitute the results in the given equations in each
case to test the accuracy of the solution.
L j2a; + 3?/ = 22, 6 i5x+10y=-7,
\x — y = l. \2x+5y = — 2.
2 \5x-2y = 2l, ?> l5x + 3y=-2,
\x — y = 6. \3x + 2y= — 1.
3 iGx+30 = 8y, 8 |3a + 7& = 7,
l3y + 17=2-3a;. l5a + 3& = 29.
4 1 8 x — 4 y = 12 x, 9 fr = 3s-19,
l4a + 2?/ = 3 + 4y. U = 3r-23.
5.
tx + 6y = 2x-16, 1Q J2p = 5g-16,
13 a -2?/ = 24. l7g= — 3i>-|-5.
Solve the following by either process of elimination :
lt j7m = 2n-3, 6 |15 & = 10-20Z,
ll97i = 6m + 89. l25A:-30Z = 80.
2.
19n = 6m + 89
|6c + 15d=-6, ?
l21d-8c=-74. 1
6c + 15d=-6, 7 \28x+Uy = 23,
14 # — 14 y = 1.
5.
I2x-3y = 4:, 8 i5x + 2y = x + 18,
12?/ — 3aj= —21. l2a + 3i/ = 3:B + 27.
(u + v = 27, 9 J7?/ — # = a; — 17,
4v = 19-|m. 12 y + 3 a; = 38.
|7a = l + 10y, 1Q |6a; + 22/=-2,
ll6?/=10a-l. \x-4y = -35.
158 INTRODUCTION TO SIMULTANEOUS EQUATIONS
(3x-y = 2x-l, 16 J'
11.
3x-y = 2x-l, 16 |7o5 — 42/ = 3,
+ y = 14. I5x + 8y = 6.
12.
13.
|2ar + 32/ = 5, 1? (12y - 10a = - 6,
l7a-2y = 74. l5.v-9a; = l
{
14 f6*/ + 2z = ll, 19 {7x + 4:y = 3,
lB f4y+9x = -5, 2Q f34z + 702/ = 4,
3?/ + 12 a = 18. 12 a; + 3?/ = 25.
f 34 x + 70 ?/ =
ls + y = -5. l5a-8y = -36.
119. The equations thus far given have for the most part
been written in a standard form, ax + by = c, in which all the
terms containing x are collected, likewise those containing y,
and those which contain neither variable. When the equations
are not given in this form, they should be so reduced, as in the
second solution on page 156, before applying any method of
elimination, and also before solving by means of graphs.
EXERCISES
After reducing each of the following pairs of equations to
the standard form, solve by graphing, or by means of either
process of elimination, as seems best available.
1.
ix-U = 7y, 3 jr + l=-4s,
Uy + l = x. l2s = 13-5r.
f _l-8n
tl6x-3y = 7x, 4. m 5 '
Uy = 7cc + 5. l3ra + 5n = l.
ELIMINATION BY ADDITION OR SUBTRACTION 159
s rm — n = 16,
l3» = 8-27i.
Ta?- 15
= 2/,
6. 3
1.2 a: — y = 3
fa; — 3
= -2,
7. 5y
la; + 7 y = 6.
8 |3^-5=-y,
\8y + 76 = 5x.
fa + 45 = 14,
l3a-6 = 14.
10. 2 ^ 2 '
[2x-y = 16.
11. <
5 +~3~-8'
2x_-y_3y-x_
2 4 *'
7m + 8 7« — 1
12.
2 m— 4 rt — 1 x
2 +~3~~~*'
(x - 3 | y +_
13. <
a + 7 2y-4__5
14.
12 7
f8a-3 55-2
9 + 3
2a+7 35+10
13,
10
= -34.
15.
Ty-4 2a?-3
5 "*" 2 ~ '
6»-3 2y+l
5^5
16. -
3y + 7 5a:-7
2 3
2a;-4 2y-l
10,
3 4
f5 + 3p 5g-2
17. J 7 4
Up + 8? = 108.
(3x-2y = ±,
18. J2te — 1 7y-4
I 5 3
-2i-
-2,
= -19.
f 5 x + 7 y = 89$,
19. \2x-4 6y-l
[ 3 +~5~ ~16*-
20.
f32a;-9 2/ = 299,
2jc-5 _ 3j/-jL
7 2
= -16.
160 INTRODUCTION TO SIMULTANEOUS EQUATIONS
PROBLEMS
Solve the following problems, using two variables in each
case.
1. A rectangular field is 32 rods longer than it is wide.
The length of the fence around it is 308 rods. Find the di-
mensions of the field.
2. Find two numbers such that 7 times the first plus four
times the second equals 37 ; while 3 times the first plus 9
times the second equals 45.
3. A certain sum of money was invested at 5% interest
and another sum at 6%, the two investments yielding $980
per annum. If the first sum had been invested at 6% and
the second at 5%, the annual income would be $1000. Find
each sum invested.
4. The combined weight of 3 cubic centimeters of platinum
and 50 cubic centimeters of poplar is 84 grams, and the weight
of 1 cubic centimeter of platinum and 150 cubic centimeters of
poplar is 80 grams. Find the weight of 1 cubic centimeter
of each.
5. The combined distance from the sun to Jupiter and
from the sun to Saturn is 1369 million miles. Saturn is 403
million miles farther from the sun than Jupiter. Find the
distance from the sun to each planet.
6. The sum of the distances from the sun to the fixed
stars Altair and Capella is 45.3 light-years. Twice the dis-
tance of Altair plus 3 times that of Capella is 119.6 light-
years. Find the distance from each star to the sun.
7. Find two numbers such that 7 times the first plus 9
times the second equals 116, and 8 times the first minus 4
times the second equals 4.
8. The sum of two numbers is 108. 8 times one of the
numbers is 9 greater than the other number. Find the
numbers. .
PROBLEMS IN TWO VARIABLES 161
9. Two investments of $24,000 and $16,000 respectively
yield a combined income of $840. The rate of interest on
the larger investment is 1 % greater than that on the other.
Find the two rates of interest.
10. A father is twice as old as his son. Twenty years
ago the father was six times as old as his son. How old is
each now?
11. If the length of a rectangle is increased by 3 feet and
its width decreased by 1 foot, its area is increased by 3 square
feet. If the length is increased by 4 feet and the width de-
creased by 2 feet, the area is decreased by 3 square feet.
What are the dimensions of the rectangle ?
Let I = the original length and w the width,
then , (I + 3) (w - 1) = ho + 3, (1)
and (Z + 4)(w-2) = Zw-3. (2)
From (1) by XIII, Iw + 3 w - I - 3 = Iw + 3. (3)
From (2) by XIII, Iw + 4 w - 2 I - 8 = Iw - 3. (4)
From (3) and (4) 3w- 1 = 6, (5)
and 4 w - 2 I = 5. (6)
(5) and (6) may now be solved in the ordinary manner.
12. The greatest distance from the earth to Venus is 160
million miles and the shortest distance is 26 million miles.
How far from the sun are Venus and
the earth, assuming that they move
around the sun in concentric circles
with the sun at the center ?
13. Two weights, 35 and 40 pounds
respectively, balance when resting
on a beam at certain unknown
distances from the fulcrum. If
15 pounds is added to the 35-lb.
weight, the 40-lb. weight must be moved 2 feet farther
from the fulcrum in order to maintain the balance. What
162 INTRODUCTION TO SIMULTANEOUS EQUATIONS
was the original distance from the fulcrum to each of the
weights ?
If the distances from the fulcrum to the weights are di and di re-
spectively, then by the formula, w\dx = w2d2, given on page 121,
we have 35 dx = 40 d2 and 50 di = 40 (d% + 2).
14. A steamer on the Mississippi makes 6 miles per hour
going against the current and 19^ miles per hour going with
the current. What is the rate of the current and at what
rate can the steamer go in still water ?
15. A man starts at 7 a.m. for a walk in the country. At
10 a.m. another man starts on horseback to overtake the pedes-
trian, which he does at 1 p.m. If the rate of the horseman had
been two miles per hour less, he would have overtaken the
pedestrian at 4 p.m. At what rate does each travel ?
16. A camping party sends a messenger with mail to the
nearest post office at 5 a.m. At 8 a.m. another messenger is
sent out to overtake the first, which he does in 2\ hours. If
the second messenger travels 5 miles per hour faster than the
first, what is the rate of each ?
17. There are two numbers such that 3 times the greater is 18
times their difference, and 4 times the smaller is 4 less than
twice the sum of the two. What are the numbers ?
18. Roy is three times as old as Fred was 8 years ago. Five
years from now Roy will be 16 years less than twice as old
as Fred. How old is each now ?
19. A picture is 3 inches longer than it is wide. The frame 4
inches wide has an area of 360 square inches. What are the
dimensions of the picture ?
20. The difference between 2 sides of a rectangular wheat
field is 30 rods. A farmer cuts a strip 5 rods wide around the
field, and finds the area of this strip to be 1\ acres. What
are the dimensions of the field ?
PROBLEMS IN TWO VARIABLES 163
21. The sum of the length and width of a certain field is
260 rods. If 20 rods are added to the length and 10 rods
to the width, the area will be increased by 3800 square rods.
What are the dimensions of the field ?
22. In a number consisting of two digits the sum of the
digits is 12. If the order of the digits is reversed the number
is decreased by 36. What is the number ?
23. A bird attempting to fly against the wind is blown
backward at the rate of 1\ miles per hour. Flying with a
wind \ as strong, the bird makes 48 miles an hour. Find the
rate of the wind and the rate at which the bird can fly in
calm weather.
24. There is a number whose two digits differ by 2. If the
digit in units' place is multiplied by 3 and the digit in tens'
place is multiplied by 2, the number is increased by 44. Find
the number, the tens' digit being the larger.
25. In a number consisting of two digits one digit is equal
to twice their difference. If the order of the digits is reversed,
the number is increased by 18. Find the number, the units'
digit being the larger.
26. If the length of a rectangle is doubled and 8 inches added
to the width, the area of the resulting rectangle is 180 square
inches greater than twice the original area. If the length and
width of the rectangle differ by 10, what are its dimensions ?
27. The Centigrade reading at the boiling point of alcohol
is 96° lower than the Fahrenheit reading. Find both the
Centigrade and the Fahrenheit reading at this temperature.
Use C and F as the unknowns. Then one of the equations is the
formula connecting Fahrenheit and Centigrade readings obtained on
page 129, and the other is C + 96 = F.
28. The Centigrade reading at the boiling point of mercury
is 312° lower than the Fahrenheit reading. Find both the
Fahrenheit and the Centigrade reading at this temperature.
164 INTRODUCTION TO SIMULTANEOUS EQUATIONS
29. There is a number consisting of three digits, those in
tens' and units' places being the same. The digit in hundreds'
place is 4 times that in units' place. If the order of the digits
is reversed, the number is decreased by 594. What is the
number ?
30. A man rowing against a tidal current drifts back 2\
miles per hour. Rowing with this current, he can make 121
miles per hour. How fast does he row in still water and how
swift is the current ?
31. Flying against a wind a bird makes 28 miles per hour,
and flying with a wind whose velocity is 2| times as great, the
bird makes 46 miles per hour. What is the velocity of the
wind and at what rate does the bird fly in calm weather ?
32. A freight train leaves Chicago for St. Paul at 11 a.m.
At 3 and 5 p.m. respectively of the same day two passenger
trains leave Chicago over the same road. The first overtakes
the freight at 7 p.m. the same day, and the other, which runs
10 miles per hour slower, at 3 a.m. the next day. What is
the speed of each?
33. Two boys, A and B, having a 30-lb. weight and a teeter
board, proceed to determine their respective weights as fol-
lows : They find that they balance when B is 6 feet and A 5
feet from the fulcrum. If B places the 30-lb. weight on the
board beside him, they balance when B is 4 and A 5 feet from
the fulcrum. How heavy is each boy ?
34. C is 6£ feet from the point of support and balances D,
who is at an unknown distance from this point. O places a
33-lb. weight beside himself on the board and when 4| feet from
the fulcrum, balances D, who remains at the same point as be-
fore. C's weight is 84 pounds. What is D's weight and how
far is he from the fulcrum ?
35. E weighs 95 pounds and F 110 pounds. They balance
at certain unknown distances from the fulcrum. E then takes
PROBLEMS IN TWO VARIABLES 165
a 30-lb. weight on the board, which compels F to move 3 feet
farther from the fulcrum. How far from the fulcrum was each
of the boys at first ?
36. A fast freight leaves New York for Chicago at 8 a.m.
At 4 p.m. the same day an express train leaves New York for
Chicago and passes the freight 12 hours later. Another ex-
press leaving New York at 6 p.m. of the same day overtakes
the freight 10 hours after starting. Find the rate of each
train if the second express goes 8 miles per hour faster than
the first.
37. The Centigrade reading at the melting point of silver
is 796° lower than the Fahrenheit reading. Find both Centi-
grade and Fahrenheit readings at this temperature.
38. The Fahrenheit reading at the melting point of gold is
992° higher than the Centigrade reading. Find both Centi-
grade and Fahrenheit readings at this temperature.
39. $10,000 and $8000 are invested at different rates of
interest, yielding together an annual income of $ 820. If the
first investment were $ 12,000 and the second $ 6000, the
yearly income would be $ 840. Find the rates of interest.
40. In a switch yard a car weighing 50 tons and going at a
certain rate strikes a standing car, whereupon both cars move
off at the rate of 4 miles per hour. If the second car, moving
at the same rate as the first before impact, were to strike the
first car when standing still, they would move off at the rate
of 2 miles per hour. How fast did the first car move before
impact and what is the weight of the second car ? (Set up the
equations by means of the formula iv^x = w-px + w2v2, ob-
tained on page 127.)
41. 200 ccm. of white oak is fastened to 25 ccm. of steel,
making a combination whose average density is 1.56. If
250 ccm. of oak is fastened to 20 ccm. of steel, the average
density of the combination is 1.3. Find the density of white
oak and also of steel.
166 INTRODUCTION TO SIMULTANEOUS EQUATIONS
SIMULTANEOUS EQUATIONS IN THREE VARIABLES
120. Illustrative Problem. Three men were discussing their
ages and found that the sum of their ages was 90 years. If
the age of the first were doubled and that of the second trebled,
the aggregate of the three ages would then be 170. If the ages
of the second and third were each doubled, the sum of the three
would be 160. Find the age of each ?
Solution. Let x, y, and z represent the number of years in their
ages in the order named.
Then, x + y + z = 90, (1)
2x+S y+z= 170, (2)
and x+2y+2z= 160. (3)
Since by supposition x represents the same number in all three
equations, and likewise y and z, if we subtract (1) from (2), we obtain
a new equation from which z is eliminated.
I.e. x + 2y=80. (4)
Again, multiplying (2) by 2 and subtracting (3),
3 x + 4 y = 180. (5)
(4) and (5) are two equations in the two variables x and y. Solving
these by eliminating y, we find x = 20. (6)
Substituting x = 20 in (4), y = 30. (7)
Substituting x and y in (1), z = 40. (8)
Check by substituting the values of x, y, and z in all three given
equations and also by showing that they satisfy the conditions of the
problem.
The values of x, y, and z as thus found constitute the solution
of the given system of equations.
Evidently x could have been eliminated first, using (1), (2) and
(1), (3), giving a new set of two equations in y and z. Let the
student find the solution in this manner.
Also find the solution by first eliminating y, using (1), (2) and (2),
(3), getting two equations in x and z, from which the values of x and
z can be found.
EQUATIONS IN THREE VARIABLES
167
121. Definition. An equation is said to be of the first degree
in three variables if no one of the variables is multiplied by
itself or by one of the others (§ 110).
The fact that the solutions are found to be the same no
matter in what order the equations are combined, indicates
that a system of three independent and simidtaneous eqtiations of
the first degree in three variables has one and only one solution.
As in the case of two equations, each should be first reduced
to a standard form in which all the terms containing a given
variable are collected and united and all fractions removed by M,
Principle VIII.
EXERCISES
Solve the following systems of equations, and check the
results by substituting the values found for each variable in
the given equations :
2 x — y + z = 18,
x-2y + 3z = 10,
3 x + y - 4 z = 20.
5x — 3y -\-z = 15,
ix + 3y— z — 3,
2x — y + z = 8.
4:x + 2y + z = 13,
x—y + z = 4:,
x + 2y — z — \.
6»+4y — 4z = — 4,
4a-2y + 8z = 0,
x + y+z = 4:.
(x + 2y + 3z = 5,
5. <4:X — 3y — z = 5,
[x + y + z = 2.
9.
10.
\2x-8y-£3z = 2,
|X-4y + 5z = l,
[3x-10y-z = 5.
x + y + z = l,
x + 3y + 2z = 8,
2x + 8y-3z=:15.
\2x — 3y+z = 5,
3a + 2y-z = 5,
{x + y + z = 3.
\x+ y + z == 6,
3 x _ 2 y - z = 13,
[2x-y + 3z = 26.
lx + y + z = 6,
Ax — y — z= — 1,
[2x + y-3z=-G.
1G8 INTRODUCTION TO SIMULTANEOUS EQUATIONS
PROBLEMS INVOLVING THREE VARIABLES
122. Illustrative Problem. A broker invested a total of
$15,000 in the street railway bonds of three cities, the first
investment yielding 3 %, the second 3| %, and the third 4 %,
thus securing an income of $535 per year. If the second
investment was one-half the sum of the other two, what was
the amount of each ?
Solution. Suppose x dollars were invested at 3 %, y dollars at 3 \ %,
and z dollars at 4 %.
Then, x + y + z = 15000,
(1)
.03 x + .035 y + .04 z = 535,
(2)
and x + z = 2 y.
(3)
From (3), ar-2^ + z = 0.
(4)
Subtract (4) from (1), 3 y = 15000,
(5)
and y = 5000.
(6)
From (1), by M, .035 x + .035 y + .035 z = 525.
(7)
Subtract (7) from (2), - .005 x + .005 z = 10.
(8)
Divide (8) by .005, -x + z = 2000.
(9)
Substitute (6) in (4), x + z = 10000.
(10)
Add (9) and (10), 2 z = 12000.
(11)
z = 6000.
(12)
Substitute (5) and (12) in (1), x = 4000.
(13)
Hence, $4000, $5000, and $6000 were the sums invested.
Solve the following problems, using three unknowns :
1. The sum of three angles A, B, and O of a triangle is 180
degrees. | of A+\ of B+\ of C is 48 degrees, while I of A+ \
of B-\-\ of C is 30 degrees. How many degrees in each angle ?
2. The combined weight of 1 cubic foot each of compact
limestone, granite, and marble is 535 pounds. 1 cubic foot of
limestone, 2 of granite, and 3 of marble weigh together 1041
pounds, while 1 cubic foot of limestone and 1 of granite together
weigh 195 pounds more than 1 cubic foot of marble. Find the
weight per cubic foot of each kind of stone.
PROBLEMS IN THREE VARIABLES 169
3. A number is composed of 3 digits whose sum is 7. If
the digits in tens' and hundreds' places are interchanged,
the number is increased by 180 ; and if the order of the digits
is reversed, the number is decreased by 99. What is the
number ?
4. The sum of the angles A, B, and C of a triangle is 180
degrees. If B is subtracted from \ of A the remainder is
\ of C, and when C is subtracted from twice A the remainder
is 4 times B. How many degrees in each angle ?
5. The sum of the three sides a, b, c of a certain triangle is
35, and twice a is 5 less than the sum of b and c, and twice c
is 4 more than the sum of a and b. What is the length of
each side ?
6. The combined number of students at Harvard, Yale,
and Columbia during the year 1905-1906 was 13,390. The
number at Harvard minus the number at Columbia plus twice
the number at Yale was 6893, and the number at Columbia
plus 4 times the number at Yale minus twice the number
at Harvard was 7258. What was the number at each
university ?
7. The total number of students at the universities of
Illinois, Michigan, and Wisconsin during the year 1905-1906
was 12,216. Twice the number at Illinois plus 3 times that at
Michigan plus 4 times that at Wisconsin was 36,145. If the
number at Michigan is subtracted from the sum of the num-
bers at Illinois and Wisconsin, the remainder is 3074. Find
the number at each university.
8. The total number attending schools in the United States
in 1904-1905 was 17,953,844. If the number in secondary
schools and colleges combined be subtracted from the number
in elementary schools, the remainder is 15,924,656; while if
twice the number in colleges be added to the number in ele-
mentary and secondary schools combined, the sum is 18,092,388.
Find the number of students of each kind.
170 INTRODUCTION TO SIMULTANEOUS EQUATIONS
9. The combined foreign trade in 1905 at the three ports,
London, Liverpool, and New York, was 3598 million dollars.
If the amount at London is subtracted from the combined
amounts at New York and Liverpool, the remainder is 988
million ; and if the amount at New York is subtracted from
the combined amounts at London and Liverpool, the remainder
is 1384 million. Find the amount of foreign trade at each
port.
In the following three examples find the number of seconds
in each record and reduce the results to minutes and seconds.
10. If x is the number of seconds in the Eastern inter-
collegiate record for a mile run, y the number in the Western
intercollegiate record, and z the number in the world's record,
then
x + y + z = 781.15,
—x+2y+z= 519.35,
2 x — y + 1 = 514.55.
11. If a; is the number of seconds in the Eastern inter-
collegiate record for a half mile run, y the number in the West-
ern intercollegiate record, and z the number in the world's
record, then
2x + 3y + z = 697.7,
3x + 2y + 2z = 809.8,
2x-y + z = 22S.l.
12. If x — number of seconds in the world's mile trotting
record in 1806, y = number of seconds in the world's record in
1885, and z = number of seconds in the. world's record in 1903,
then
* + y + z = 426.25,
2z + 4?/ + 6z = 1584,
- a> + y + 2 « = 186.75.
REVIEW QUESTIONS 171
REVIEW QUESTIONS
1. How may a point in a plane be located by reference to
two fixed lines ? What are these fixed lines called ? What
names are given to the distances from the point to the fixed
lines ? Why are negative numbers needed in order to locate
all points in this manner ?
2. Draw a pair of axes in a plane and locate the following
points: (5,0), (-2,0), (0,3), (0,-1), (0,0).
3. State a problem involving motion and solve it by means
of a graph.
4. How many pairs of numbers can be found which satisfy
the equation x — 2 y = 6 ? State five such pairs and plot the
corresponding points. How are these points situated with
respect to each other ? What can you say of all points corre-
sponding to pairs of numbers which satisfy this equation?
What is meant by the graph of an equation ?
5. How many pairs of numbers will simultaneously satisfy
the two equations 3x+2y=7 and x +y = 3 ? Show by means
of a graph that your answer is correct.
6. Describe elimination by the process of substitution ; also
by the process addition or subtraction. Under what conditions
is one or the other of these methods preferable ?
7. Why is the solution by elimination in some cases pref-
erable to the solution by means of the graph ?
8. Describe the solution of a system of three linear
equations in three unknowns. Is it immaterial which of the
three variables is eliminated first ?
9. Can you find a definite solution for two equations each
containing three unknowns ? Illustrate this by means of the
equations 4 a; — 3y — z = 5 and x + y + z = 2.
CHAPTER VI
SPECIAL PRODUCTS AND FACTORS
123. Repeated Factors. Number expressions containing re-
peated factors have already been considered in Chapter III.
x-x was written x2 and called the square of x, or x square;
similarly, (a + b) (a -f b) was written (a + 6)2 and read the
square of the binomial a + 6 or the binomial a + b squared.
124. Definitions. Any number written over and to the right
of a number expression is called an index or exponent and, if
a positive integer, shows how many times that expression is to
be taken as a factor.
A product consisting entirely of equal factors is called a
power of the repeated factor. The repeated factor is called the
base of the power.
E.g. x% means x • x -x and is read the third power of x or x cube; x6
means x • x • x • x • x, and is read the fifth power of x or briefly x fifth,
(x — y)s = (x — y)(x — y)(x — y) and is read (x —y) cubed or the
cube of the binomial (x — y).
Notice two important differences between an exponent and
a coefficient.
(1) 5a = a+a-r-a + a + a, while a5 = a • a • a • a • a.
(2) In 5 abc the coefficient 5 applies to the product abc,
while in abc5, the exponent 5 applies only to the factor c. In
order to make it apply to the product it is necessary to use a
parenthesis, thus, (abc)5 means the product abc taken five
times as a factor.
172
PRODUCTS OF POWERS OF SAME BASE
173
EXERCISES
Perform the following indicated multiplications
1.
23
, 24, 25,
26.
9.
102, 103, 104, 105.
17.
(x — y — 5)2.
2.
32
33, 3*,
35.
10.
(a + bf.
18.
(w + z — 4)2.
3.
42
43, 4«,
45.
11.
(c-d)\
19.
(2—z — x + w)2.
4.
52
53, 54,
55.
12.
982 = (100-2)2.
20.
(x — y — 5)2.
5.
62
63, 64.
13.
(a + b + cy.
21.
(a - y)\
6.
72
7s, 74.
14.
(a + 6 - c)2.
22.
(x + y)3.
7.
82
83, 8*.
15.
(3 - a)2.
23.
(a + b)\
8.
92
9s, 94.
16.
(3 _ 6 _ C)2.
24.
(c - d)4.
125
. In the case
of factors expressed
in Arabic figures mul-
tiplications like the following may be carried out in either of
two ways.
E.g. 32 • 3* = 9 • 81 = 729.
or 32 • 34 = ( 3 • 3)(3 • 3 • 3 • 3) = 32+4 = 36 = 729.
But with literal factors the second process only is possible.
E.g. a2. a* = (a • d)(a • a • a • a) = a?+* = a6.
The process in which the exponents are added applies only
when the factors are powers of the same base.
E.g. 28 . 32 = 8 • 9 = 72 cannot be found by adding the exponents.
21 or 2 is called the first power of 2.
Thus 2 • 28 = 21 • 28 = 21+8 = 24.
EXERCISES
In the following exercises carry out each indicated multipli-
cation in two ways in case this is possible :
1. 25.26. 3. 2-24. 5. 3-34. 7. 42-43.
2. 22-23. 4. 32-33. 6. 32-35. 8. 4-44.
174
SPECIAL PRODUCTS AND FACTORS
9. 5-52.
12. 7-73.
15. x7-x\
18. 23-22.24.
10. 52-58.
13. a2 -a3.
16. *3-t4.
19. 3-32-33.
11. 62-62.
14. tf-x2.
17. J2-*3-*4.
20. 22.23-22.2
126. Illustrative Problem. To multiply 2* by 2n, k and n
being any two positive integers.
Solution. 2* means 2 • 2 . 2 • 2, etc., to k factors,
and 2n means 2 • 2 • 2 • 2, etc., to n factors.
Hence 2* . 2» = (2 . 2 . 2 • . . to n factors) (2 • 2 • • • to i factors)
= 2.2.2.2... to fc + n factors in all.
That is, 2*.2»=2*+».
The preceding examples illustrate the following principle :
127. Principle XIV. The product of two powers of the
same base is found by adding the exponents of the powers
and making this sum the exponent of the common base.
EXERCISES
Perform the
following
indicated
multiplications by means
of Principle XIV :
1. 23-27.
8.
32« . 3»
15. wx-wy+Sx.
2. a3 • a7.
9.
52+» , 52-»
16. n2-n3c+ih.
3. 34-35.
10.
J2x+v . J2x
~". 17. cx • c2_x.
4. xA • Xs.
11.
am • an. .
18. ar-z40.
5. 3*. 3".
12.
/2a . fb+a
19. r30-?'40.
6. sc*.zB.
13.
y*a - y3"-
20. s^-s™.
7. 4a • 4J
14.
qAb , rtfa+ib
21. -y2*"1 . <y3*+2#
Perform the following multiplications by means of Princi-
ples IV, III, and XIV :
22. 23(22 + 24). 25. aX^b-ab3).
23. 22(3 • 24 + 5 • 2s). 26. a^a(4 of + 3 a88).
24. 44(3-43-5.42). 27. ^(5^-3^).
PRODUCTS OF MONOMIALS 175
Multiply the following and state all principles used :
28. a2m2 - b2m3 by m4. 37. 32 ■ 42n - 63 • 44n by 43\
29. 4.32-5-7-2by 24. 38. aV-ac^byafc.
30. 2-3-4-32-i-7.3by 3. 39. 62 • 43-74 • 42 + 44 by 43.
31. 4z2-3ar3 + 6a4by x2. 40. 12 x2f - 6 x?y2 + 3 xy by x3.
32. ox-3^ + 2»4by x4. 41. 2 a36 - 3 a2c - 4 a45 by a2*.
33. 3 y2 + 4 y* - y3 by y. 42. 6 ar3"-25 + 8 x2"-3* by x2a+2i.
34. 7x4-5x?-2xby &. 43. y3m+2n - y2n 3m+l by ^m+3n.
35. 3 a2&4 + 2 d2b - 4 a262 by a3. 44. ar5c-3a + ark+2a + a;2abyz4a-2c.
36. 4 a26 — a;?c + a2d by a26. 45. a"*-** — abx-,lx —ax by a5*+n*.
PRODUCTS OF MONOMIALS
128. In multiplying two numbers each of which is in the
factored form, if the factors are all expressed in Arabic figures,
the operation may be carried out in two ways.
E.g. (2 • 3) (2 • 3 • 5) = 6 • 30 = 180.
Also (2 • 3) (2 • 3 • 5) = 22 • 32 • 5 = 4 • 9 • 5 = 180.
In the second process one of the factors, 2 or 3, is multiplied in
and this product is then multiplied by the other factor, 2 being
combined with the factor 2 and 3 with 3 by Principles III and XIV.
In the case of literal factors the second process only is available.
E.g. (a262) (5 a*b»c) = 5 a4+2 62+» c = 5a%5c.
EXERCISES
Perform the following indicated multiplications, each in two
ways where possible :
1. (5 • 72)(53 • 7s). 4. (3.42)(4-32). 7. 6t(S$P).
2. (32-5)(3.23.52). 5. 3a6(5a262). 8. 7m(4mV).
3. (4 • 53)(5 • 43). 6. 4 x(3 xy). 9. 6 k2(62tf).
By definition (§ 77), expressions in the factored form, such
as 3 ab, 5 a2b2, 3 • 23 • 52, etc., are monomials.
176 SPECIAL PRODUCTS AND FACTORS
The preceding examples illustrate the following principle :
129. Principle XV. The product of two monomials is
found by multiplying either one by each factor of the
other in succession.
Each factor of the multiplier is associated with any desired factor
of the multiplicand according to Principle III, and where the bases
are the same the exponents are added according to Principle XIV.
If there are no factors common to multiplier and multiplicand, the
product can only be indicated by writing all the factors of both in
succession.
Multiply : exercises
1. 4- 7 -8-9 by 2 -3 -5. 5. 3 *V by 4 ay*.
2. 3 .8- 2 by 2- 5- 6. 6. 5 a2b3c by ab*c.
3. 2 xyz by 3 z?yz. 7. 2 a^&V by 5 xb*c.
4. 63 • 2* • x5 by 6 • 23 . x. 8. 9 x° yV by 4 x2ay2hzc.
9. 6 x*n+lfn~hn by 3 xl-*ny5~n^~n-
10. 3 n2x+4my-5r21-1 by w^m'-V*.
In each of the following exercises state which of the Prin-
ciples I-XV are used :
11. 32-23(24.32-4.22-2.32).
12. 6.34(42-22.32+43.24-5.34).
13. 2x(4: + 7xi-3x2y), 16. 4*y(3 <m— 4 bg +2 xy).
14. 4yz(3y2x3 — fx2 -{- y*x*) . 17. 2xa(xa — xb — 3c).
15. 5 a2b\as - b3 + a2b2). 18. 2 yaf (y - as* + 3 y).
19. 4a^,+X^n+1-4y2M+5 + 5a2ny4).
20. 4 a?!n+y,+2m(y,2/m — ccy + £"* + yn).
21. (a + by(a-b). 24. (a3+a26 + a62+63)(a-6).
22. (a - 6)3(a + 6). 25. (a + * — c) (a+ & + c).
23. (a + 6)2(a-&)\ 26. (3a>-2y-l)(2a> + y).
MONOMIAL FACTORS
177
27.
(5a-3&)(6a2 + 262-l).
39. (i
28.
(3a2-262-3)(4a + 3 63).
40. (i
29.
(1 + a + «2) (1 - a).
41. (:
30.
(1 - a + a2 - a3) (1 + a).
42. (
31.
(a + &)(a-6)(a2 + 62).
43. (
32.
(a + 6) (a2 -aft + 6s).
44. (.
33.
(a - b) (a2 + a& + b2).
45. (i
34.
(x-\-y)(x — y).
46. (i
35.
(100 + 1) (100 - 1).
47. 0
36.
(100 + 2) (100 - 2).
48. (i
37.
(x + 3) (cc + 5).
49. (i
38.
(a? - 3) (a? - 5).
50. (<
(t + $)(t-3).
(x-2y)(x + 3y).
(x* + x + l)(x*-x + l).
(a; + y) (a3— a;2?/ + xy2—f).
(x -y) (tf+xty+xf+f).
(x2 - xy + y2) (x2 + an/ + /).
(ar' + 2 any + */2) (a; - y)2.
(tf-y^^ + xttf + y*).
(x2 + y^(x*-x2y2 + y*).
(x2 + y2) (x + y) (x - y).
(x—y) (xf+xy+y2) (ar'+y5)-
(an- 6") (a2rt-f an6n.+ Z>2").
FACTORS OF NUMBER EXPRESSIONS
130. The factors of numbers are of great importance in
arithmetic. For instance, the multiplication table consists of
pairs of factors whose products are committed to memory for
constant use. Likewise in algebra the factors of certain special
forms of number expressions are so important that they must
be known at sight.
An expression containing no fractions is said to be prime if
it has no integral factors except itself and 1.
Thus 2, 3, x, x + 2, a2 + b2, are prime expressions.
A prime expression may be factored by using fractions or
radicals. See p. 214.
Thus
5 = 2-2^ and 2= V2- V2-
Such factors are not included in what are here called prime
factors.
178 SPECIAL PRODUCTS AND FACTORS
131. Monomial Factors. If the terms of a polynomial con-
tain a common factor, they may be combined with respect
to this factor according to Principles I and II. The ex-
pression is thus changed into a product of a monomial and
a polynomial.
Illustrative Examples.
1. ax + ay = a(x + y), by Principle I.
2. a2 — ab = a(a — b), by Principle II.
3. 9.8 + 3.4.5 = 3.4(3.2+5), by Principle I.
4. 6 a2b - 4 ab2 = 2 ab (3 a - 2 b), by Principle II.
5. 5 xy — 3 x2y + 4 x*y = xy (5 — 3 x + 4 x2), by I and II.
Observe that factoring each term of a polynomial does not
factor the polynomial.
E.g. 330 - 210 - 60 is not factored by writing it 2 • 3 • 5 • 11
— 2> 3.5.7 + 22 .3-5; but by adding the coefficients of the common
factor 2 • 3 • 5, thus,
2.3.5.11 -2- 3- 5- 7 + 22 .3.5 = 2. 3- 5 (11 -7 + 2).
Likewise 10 a2bc — 15 ab2c + 20«6c2 is not factored, although each
term is in the factored form. But if 10 a?bc — 15 ab2c + 20 abc2 is
written in the form 5abc (2 a — 3 b + 4 c), it is then factored.
These examples illustrate the factoring of a polynomial
when it contains a monomial factor common to every
term.
When such a common factor has been found the whole ex-
pression is then written in the form of a product by means of
Principles I and II. The result may be checked by Princi-
ples IV and XV. If the terms of a polynomial which is to be
factored contain a common factor, this should always be re-
moved at the outset.
TRINOMIAL SQUABES 179
EXERCISES
Factor the following polynomials :
1. 8-12-18 + 48. 7. 15 xy* - 20 afy + x>y\
2. 3-4- 5-15-20 + 35. 8. 9 vsw* + 21 v4w3 - 18 v2w2.
3. 3- 11- 4 -22- 2 + 44 -6. 9. 12 a*b3 - 8 cfb4 - 6 a2b2.
4. 34-24-3-23-5.22. 10. 32.34-64.33-16-23.32.
5. 6.54.7+3-53.2-54.9.22. 11. 27 • 23 + 54 • 24-36 • 23.
6. 13 a*b - 16 a3b* - 2 a3b\ 12. 11 aV - 44a3x4 + 33ax.
13. 1.2- 3-4 + 2- 3- 4- 5- 3- 4- 5- 6+4- 5- 6.
14. 72 x4b3a - 36 x*b2a2 - 48 xWa\
15. 84, x9b7y4 + IS x3bsy* + 12 a^by.
16. 19 • 34 • 24 • 53 - 13 • 34 • 23 . 52 - 29 ■ 32 • 22 . 54.
17. 17 a463c2 + 51 aW - 34 aW.
18. 38 a12 614c4 - 76 au&12c3 - 76 aM&uc».
19. 4,x2aySb-^Qx3ay2b-8x'iayib.
20. 3 a4n+263n+4 + 6 a6n+463n+3 - 12 a5n+365n+6.
TRINOMIAL SQUARES
132. In §§ 87 and 88 we found by multiplication :
{a + b)2 = a2 + 2ab + b\ (1)
and (a - b)2 = a2 - 2 ab + 62. (2)
By means of these formulas we may square the sum or dif-
ference of any two number expressions.
E.g. (3x+2y)2=(3xy+2'3x.2y+(2y)2=9x2+12xy+4;y2.
Also[(5 + r)-(s-l)]2=(5 + r)2-2(5 + r)(s-l) + (s-l)2.
The last expression may now be reduced by performing the
indicated operations.
180 SPECIAL PRODUCTS AND FACTORS
In this manner write the squares of the following binomials.
Verify the first ten results by actual multiplication.
1.
t + 8.
11.
7(a-3)-2(6 + c).
2.
r-12.
12.
±(x-ij) + 2(z + tf).
3.
Sx — 5y.
13.
(3a-26) + 5.
4.
6a- 7.
14.
7 x — (4 r — s).
5.
m5 + 3».
15.
7(m2-3)-6(m3 + rc).
6.
Itf + Zx.
16.
3(z + y)-2(3 + x).
7.
3tf-2y\
17.
4(ar5-3y) + (ar! + 4y2).
8.
8 m2n — 7 win2.
18.
2x2y — b (x — y).
9.
5a3-363.
19.
3(5x2-z)-4:x'z2.
10.
4 aWy3 — 3 a?xy*.
20.
7(r-s) + 3(rV-2rs).
133. The binomial a + 6 is one of the two equal factors of
a? + 2ab + b2, and is called the square root of this trinomial.
Likewise a — b is the square root of a2 — 2 ab + &2-
In each case a is the square root of a2 and b of ft2. Hence
2 ab is twice the product of the square roots of a2 and b2.
From the squares obtained in the last article, we learn to
distinguish whether any given trinomial is a perfect square,
as in the following examples :
1. sc2 -}- 4 # + 4 is in the form of (1), since x2 and 4 are squares
each with the sign +, and 4 a; is twice the product of the square
roots of x2 and 4. Hence
a? + 4« + 4 = x2 -f 2 (2a?) + 22 = (a? + 2) (x + 2) = (x + 2)2.
2. cc2 — 4 a; + 4 is in the form of (2), since it differs from (1)
only in the sign of the middle term. Thus
^-4^ + 4 = ar!-2(2a0+22= (as - 2) (x-2)= (x - 2)2.
A trinomial which is the product of two equal factors is
called a trinomial square.
TRINOMIAL SQUARES 181
EXERCISES
Determine whether the following are trinomial squares, and
if so indicate the two equal factors. If any trinomial is not a
square, make it so by modifying one of its terms.
1. x2 + 2xy + y2. 10. 64 + t2 - 16 t.
2. x2 -2xy-\- y2. 11. 16 + x2 — 8 x.
3. x4 + 2 arty2 + y4. 12.9-6?/ + y2.
4. a;4 - 2 arty2 + y4. 13. 25ar! + 16?/2 + 40ar?/.
5. m2 + n2 — 2 mw. 14. 4m2+«2+2 m».
6. r2 + s2 + 2rs. 15. 100 + s2 + 20s.
7. 4a^-8^ + 4?/2. 16. 64 + 49 + 112.
8. a6 + &6 + 2 a363. 17. 16 a2 + 25 62 - 50 a&.
9. a8 + 68-2a464. 18. 16 a2 + 25 b2 + 40 a6.
19. 81- 270& + 225 62.
134. From the foregoing examples we see that a trinomial is
a perfect square if it contains two terms which are squares
each with the sign +, while the third term, whose sign is
either + or — , is twice the product of the square roots of the
other two. Then the square root of the trinomial is the sum
or the difference of these square roots according as the sign
of the third term is + or — .
Since on multiplying we find (a — 6)2 and (b — a)2 give the same
result, we may write the factors of a2 — 2 <xb + b2 either (a — b) (a — b)
or (b — a)(b — a). See page 214.
EXERCISES
Factor the following. If any one of the trinomials is found
not to be a square, make it so by modifying one of its terms.
1. 9 +2-3-4 + 16. 3. 9x2 + I8xy + 9y2.
2. x2 +±y2 +4xy. 4. 4 x2 + 4 xy + y2.
182 SPECIAL PRODUCTS AND FACTORS
5. 4 x2 + 8 xy + 4 y2. 18. 81 a2- 216 a + 144.
6. 25x2 + 12xy + 4:y2. 19. 4 a2 + 8 a62 + 462.
7. 16ar9 + 16a*/ + 4?/2. 20. 9 64 + 18 62c4 + 9 c8.
8. 9 r2 + 36 rs + 25 s2. 21. 4 x2 + 4 y2 - 8 ay.
9. 16x6 + 8x*y + 2/2. 22. 9 a2 - 16a6 + 4 62.
10. 4a8 + 12z4a3 + 9a4. 23. 9z4 - 24^6 + 1662.
11. a10 + 6 a56 + 9 62. 24. 25 + 49 z2 - 70a.
12. (a + l)2 + 2(a + 1)6 + 62. 25. - 30 a62 + 9 a2 + 25 64.
13. (z + 3)2+4(;c + 3)2/ + 4?/2. 26. 16 a2 - 24 a6 + 9 62.
14. ^ + 12 3* + 36. 27. 36 or* - 84 a: + 49.
15. a4 + 18 a2 + 12. 28. 25-90 + 81.
16. 121 + 4 a8 - 44 x4. 29. 64 a?2 - 32 a; + 9.
17. 16x4 + 642/4-64^2. 30. (3 + a)2 + 62-26(3+a).
31. (2 - a)2 - 2(2 - x) (x -1) +0 - l)2.
32. (2 +yf + 2(2 + y)(l +4) + (1 + 4)2.
33. (3a-26)2-10(3a-26) + 25.
34. (6a-6)2+ (2a + l)2-2(6a-6)(2a + l).
35. 25(a + 6)2 + 50(a + 6) (a - 6) + 25(a - 6)2.
36. x? + 12 x(a + 6 + c) + 36 (a + 6 + c)2.
37. 49 (m - 3)4 + 36 (m + I)6 - 84(m - 3)2(m + l)3.
38. W(x - y)2 - 16 (a; - y) (x + y) + 4(a; +y)2.
39. - 30(a + 6) (a - 6)2 +25 (a - 6)4 + 9(a + 6)2.
DIFFERENCE OF TWO SQUARES
183
THE DIFFERENCE OF TWO SQUARES
135. By multiplication,
(a + b) (a - b) = a2 - b\
Translate this formula into words. By means of this formula
the product of the sum and difference of any two number
expressions may be found.
E.g. (3 a + 2 b) (3 a - 2 b) = (3 a)2 - (2 6)2=9 a2 - 4 A2.
Also (a; + y - z) (x + y + z) = [(x + y) - z] [(x + y) + z]
The last expression may now be simplified by performing
the operations indicated.
In this manner form the following products. Verify the
first ten by actual multiplication.
1. (4a + 5ft)(4a-5 6). 9.
2. (24 x + 12 y) (24 x -12 y). 10.
3. (16a2ft3-3c)(16a2ft3+3c). 11.
xm+ yn) (xm — yn).
2n+l_i_£>2n-l\ (tfn+l ffn-\\
4. (5 -6Z2) (5 + 6*2). 12.
5. (3x- 2y) (Sx + 2y). 13.
6. (a8 - f) (a3 + f). 14.
7. (Sa^-S^XSa^ + Sy4). 15.
8. (>+(y-«)]0-(y-«)]. 16.
-(a-6)][c+(«-6)].
«-(# + «)][>+& + «)].
a + ft + c)(a — ft — c).
a — 6 + c) (a— 6 — c).
r-y-z) (r-y + z).
a + ft +c) (a + 6 — c).
17. [a + ft+(c-ef)][a+&-(c — d)].
18. 0 + y + (w+v)][>+?/-(w + v)].
19. [4 x - (a - 2 ft)] [4 x + (a - 2 ft)].
20. [a.+ 2b-(x-y*)][a+2b+(x-y*)l
21. (11 b3x - 3 bx>) (11 b3x + 3 ftz3).
184 SPECIAL PRODUCTS AND FACTORS
From the preceding examples we see that every binomial
which is the difference between two perfect squares is com-
posed of two binomial factors ; namely, the sum and the
difference of the square roots of these squares.
E.g. 16 x2 — 9 y2 is the difference between two squares, (4 x)2 and
(3 y)2. Hence we have
16x2-9#2 = (4x)2- (3y)2 = (4x + 3*/)(4x-3?/).
EXERCISES
Factor each of the following :
1. x'-Ay2. 8. a2-l. 15. 4:-(x-2y)2.
2. 9a?-36y*. 9. l-9x4. 16. 16a-25a&2.
3. x*-b2. 10. 4 -36 a2. 17. 49 a -4 a?/2.
4. 4ar>-9&8. 11. 1-64 a8. 18. 225-64afy4.
5. 16a4-9Z>4. 12. I44z2&4-1. 19. 576 a -144a?/2.
6. 64 -b2. 13. 256 a*b6-c\ 20. 58-38.
7. 1-&2. 14. l-(x + y)2. 21. x*-8ly2.
136. It is important to determine whether a given expres-
sion can be written as the difference of two squares.
E.g. a2 + b2 + 2 ab - c2 = (a + b)2 - c2 = (a + b + c) (a + b - c).
Also, c2 - a2 + 2 ab - b2 = c2 - (a2 - 2 aft + 62) = c2 - (a - 6)*
= (c — a + b)(c + a — b).
EXERCISES
In the following determine whether each is the difference of
two squares, and if so, factor it accordingly ; if not, make it
so by modifying one of its terms.
1. ^-(y-z)2. 5. 4a2&2-(a2 + 62-c2)2.
2. (x-yf-z\ 6. a2-(62+c2 + 2 6c).
3. a2 + &2_2a5_4. 7. (2a-5)2-(3a + l)2.
4. a? + y2-2xy-z2. 8. (3 x2 - yf - (x + y)2.
THE SUM OF TWO CUBES 185
9. (3a-2 6)2-(8a + 5&)2. 18. 9a2 + 6ab + b2 -c2.
10. (3m_4)2-(2m + 3)2. 19. 16aj*-as+4a6 -462.
11. (2r + s)2-(3r-s)2. 20. a2-2a& + 62-c2.
12. 81-(a + 6+c)2. 21. c2-(a2 + 2a6 + 62).
13. x* + ^ + i_4a2. 22. <?-(a2-2ab + b2).
14. a2-0 + 2i/)2. 23. (a + 6)2-(a-6)2.
15. 9 s2 -(a -ft)8. 24. a2 + 4a6 + 62-a^.
16. 25m2- (3 r + 2s)2. 25. a2 + 4a& + 452- x2.
17. 4c2-(4a2 + 12a6-9 62). 26. 9a2 + 16 b2 - 32ab-x*.
27. a2 + ±ab + 4:b2 — (x2 — 2xy + yi).
28. (3a;-2)2-(4^ + 9?/2-12a^).
29. a^ + 4a;y-i/2-(a2 + 2a6 + 62).
30. 16 xAy2- (4 a2 + 9/ + 12 ay).
31. (a + 6)2-(4a2 + 9&2-12 6c).
THE SUM OF TWO CUBES
137. By multiplication we find
(a + 6)(a2-a6+62) = a3+63.
The pupil should perform the multiplication indicated in this
formula and carefully note how the terms cancel in the product.
By means of this formula obtain the following products :
1. (x + y)(x2-xy + y2). 6. (x2 + y2) (x* - x?y2 + y*).
2. (a + 2)(a2-2a + 4). 7. (5 a + b) (25 a2 - 5 ab + b2).
3. (3a + 6)(9a2-3a&+62). 8. (1 + oz2)(l -5x2-\-25xi).
4. (l + 4cc)(l-41r+16arJ). 9. (1 + 2xy)(l-2xy+4:X2y2).
5. (w>2+3a2)04-3a2w;2+9a4). 10. (2x+3y)(4;X2-6xy+9y2).
186 SPECIAL PRODUCTS AND FACTORS
These products show that every binomial which is the sum
of the cubes of two numbers is the product of two factors, one
of which is the sum of the numbers, and the other is the sum
of their squares minus their product.
E.g. (1) x* + y* = (x + y) (x2 - xy + rf).
(2) 8 a3 + 27 bs = (2 a)3 + (3 b)s
= (2a + 3J)(4a2-2a.36+ 9 b2).
(3) X« + y* = (X2)8 + (,,2)3 _. (X2 + ^ (x4 _ xy + ^4).
. Notice the difference between the trinomial x2 — xy + y2 and the
trinomial square x2 — 2 xy + y2.
EXERCISES
Determine whether each of the following is the sum of two
cubes, and if so, find the factors ; if not, make it so by modi-
fying one of its terms. Check the results by multiplication.
1. tf + tf. 9. 27a^ + l. 17. 125 a? + y\
2. a3 + 8 63. 10. 8a3 + 27 63. 18. 1 + rr8.
3. 27a3 + 63. 11. 8a3 + 64&3. 19. 64^ + 27^.
4. 8a3 + l. 12. wV + ar>a3. 20. 83+103.
5. 1 + 64 ar5. 13. 1 + 8 a?bs. 21. 1 + 729 a*.
6. 23 + 33. 14. 64 a-5 + 343. 22. »6 + y12.
7. 125 + 729. 15. 1 + a3. 23. a9 + Z>3.
8. 1 + 125 a6. 16. a3 + 9 ft3. 24. 27^ + 125 s3.
THE DIFFERENCE OF TWO CUBES
138. By multiplication, we find
(a - b)(a2 + ab + b2) = a3 - b5.
In performing this multiplication, note carefully how the
terms cancel in the product.
THE DIFFERENCE OF TWO CUBES 187
By means of this formula obtain the following products.
1. (x-y^tf + xy + y-). 6. (w2-2a2)(w4+2io2a2 + 4a4).
2. (&-3)<7>2 + 3Z>+9). 7. (3a-2&)(9a2+6a&+4&2).
3. (2a-&)(4a2 + 2a&+62). 8. (3a2-l) (9x4 + 3a2 + l).
4. (l-4«)(l+4aj + 16aj2). 9. (1 - 2 a#) (1 + 2 ajy + 4 ofy2).
5. (a2-62)(&4 + a2&2 + &4). 10- (2a?-3y) (4aj! + 6*y + 9jr!).
These products show that the difference of the cubes of two
numbers is the product of two factors, one of which is the
difference of the numbers, and the other the sum of the squares
of the numbers plus their product.
E.g. (1) xs - ya = (x - y)(x2 + xy + if).
(2) 8 a3 - 64 b* = (2 a)« - (4 6)8
= (2a -46)(4a2-2a-45+ 16 b2).
(3) a9 - b9 = (a3)3 - (i3)3 = (a3 - i3)(a6 + «363 + &«).
Notice the difference between the factor x2+xy-\-y2, and the trino-
mial square x2+2 xy+y2.
EXERCISES
Determine whether each of the following is the difference of
two cubes, and if so, find the factors ; if not, make it so by
changing one of its terms. Check the results by multi-
plication.
1. a8 — 6s. 8. 64 or*-?/3.
2. 8a3-&3. 9. 27 -125 a3.
3. a3-8 63. 10. x6-y6.
4. 8 a3 - 8 K, 11. 1 - x6.
5. 33-23. 12. a9 -8.
6. 1-a10. 13. 1-125^.
7.1-8 a3. 14. 8-27 Xs.
22. Also factor 10, 11, 19, and 20 as the difference of two
squares.
15.
27^-64.
16.
2*s"-l.
17.
8 x4 - f.
18.
64 a3- 27 bs.
19.
1-729 a".
20.
x6-y12.
21.
27 r3- 125 s3.
188 SPECIAL PRODUCTS AND FACTORS
TRINOMIALS OF THE FORM x2 +(a + 6) x+ ab.
139. In §§ 82 to 85 were found such products as
(1) (x + 5)(x + 2)= x2 + Ix + 10.
(2) (x - 5)(x - 2)= x2 - 7 x + 10.
(3) (a? + 5)(oj - 2)= a,-2 + Bx - 10.
(4) (a; - B)(x + 2)= ar> - 3a; - 10.
In each case the binomials to be multiplied have one term x
in common, and the other two terms unlike. The trinomial
product in each case is the square of the term common to the
binomials, the algebraic sum of the unlike terms times the
common term, and the product of the unlike terms.
Thus, the coefficient of x in (1) is 5 + 2, in (2) it is (— 5) + (— 2),
in (3) itis(+ 5) + (- 2), and in (4) it is (-5)+(+ 2).
The third term of the product in (1) is (+ 5)(+ 2), in (2) it is
(- 5)(- 2), in (3) it is (+ 5)(- 2), and in (4) it is (- 5)(+ 2).
With these points clearly in mind, such products may be
written out at once, without the formal work of multiplication.
EXERCISES
Find the following products by inspection :
1. (a + 7) (a +9). 9. (a?- l)(a?-+ll).
2. (6-5)(6 + 7). 10. (c-3)(c + 12).
3. (x -7) (x -17). 11. (e»-l)(c»- 5).
4. (y + 8)0 - 3). 12. (x* - 5) {x* - 3).
5- (V + 7)(y - 9). 13. (a2 - 2)(a2 - 4).
6- (y-7)(y-l). 14. (a3-l)(a3 + 4).
7. (*-3)(aj-4). 15. (2a-l)(2a-3).
8. (x-S)(x-9). 16. (2a + 3)(2a-4).
THE TYPE x2 + (a + b)x + ab 189
17. (x-5)(x + 2). 25. (3&c + 4)(3 6c-7).
18. (3 a -7) (3 a + 2). 26. (100 + 3) (100 + 5).
19. (a2-8)(a2-4). 2? (2 a« + &)(2 a« + c).
20. (&3-3)(63-4).
21. (2a? + a)(2a;+6). V' A '
22. (5c-4)(5c + 5). 29' (* + «)<*-»)•
23. O»y+0)(ajy-7). 30- t? r- «)(»-. &)•
24. (l-3a)(l-2a). 31. (3x* + 2 2/)(3^-52/).
140. It is possible to recognize such products at sight, and
thus to find the factors by inspection.
Illustrative Examples. Determine whether the following tri-
nomials are of the kind just considered :
1. x2 + 7 x + 12. The question is whether two numbers can-
be found such that their sum is + 7 and their product 12.
The numbers 3 and 4 answer these conditions. Hence,
x2 + 7 x + 12 = (a; + 3) (x + 4).
2. x2 — 5 x — 14. Since the product of the numbers sought is
— 14, one number must have the sign — and the other -{-; and
since their sum is —5, the one having the greater absolute
value must have the sign — . Hence, the numbers are — 7 and
+ 2, and we have x2 — 5x— 14=(# — 7)(x + 2).
3. x2 -7 x + 12 = (x-3)(x -4). Since (-3)(-4)= + 12
and (-3) + (-4) = -7.
4. x2 + 4:X-12 = (x + 6)(x-2). Since (+6)(-2)=-12
and(+6) + (-2) = + 4.
It is to be noted that it is not always possible to find inte-
gers to fulfill these two conditions.
E.g. given x2 + 5x + 3. By inspection, it is easily seen that there
are no two integers such that their sum is + 5 and their product -f 3.
190 SPECIAL PRODUCTS AND FACTORS
EXERCISES
Determine whether each of the following trinomials can be
factored by inspection, and if so, find the factors ; if not, mod-
ify one term so as to make such factoring possible.
1. ^ + 3^ + 2. 11. b2 + 8b + 15. 21. z4 + 18ar° + 77.
2. tf + x-6. 12. b2-b-56. 22. a4-5a2-104.
3. ^_a;_6. 13. b2 + b-5S. 23. a2+32a + 240.
4. x2-6x+8. 14. c2-3c-15. 24. a4-lla2 + 28.
5. a2 + 6a+8. 15. rf-lSx + m. 25. a4-lla2-60.
6. 3^-3^ — 8. 16. x2 + 15a;-54. 26. a2 -14 a — 51.
7. a,*2 + 2 a -8. 17. 0^-14 # — 95. 27. a2-3a-54.
8. a2_4a_32. 18. y2 + 21 y + 98. 28. »4-8a^-32.
9. a2 + 4a-32. 19. y2-7y-98. 29. a6-3a3-154.
10. 62 + 15 6 + 56. 20. a2 -19 a; + 78. 30. a?-10x + 25.
31. a266 - 13 a&3 - 30. 41. 9 a2 + 24 a + 16.
32. x2 - 17 xyz + 72 y2z2. 42. 81 a2 -99 a + 30.
33. r8 + 6 r4s - 91 s2. 43. #2 + 26^ + 133.
34. a4c4 + 9a2c2-162. 44. x2 + 5xy-8±y\
35. a2 + lla-210. 45. *+3r— 154.
36. m4 + 4mV + 4w2. 46. u2 + 38 uv + 165 v2.
37. sV-Wst-te. 47. (a + 6)2-19<a + &) + 88.
38. a2b2 - 27 ab + 26. 48. (z-?/)2-14(a;-y) + 40.
39. Z2 + 13Z + 42. 49. (r-s)2-17(r-s) + 60.
40. «Y — 11 xy — 180. 50. x2 + (a + 6) x + a&.
THE TYPE atf + bx + c 191
141. Trinomials of the form ax* + bx + c.
Ex. Find the product of 2 x + 5 and 3 x + 2.
2z + 5
3a + 2
6^ + 15*
4z+10
6a2 + 19a + 10
The products 3 a; • 2# and 2 • 5 are called end products and
2 • 2 a; and 5 • 3 x are called cross products. In this case the
cross products are similar with respect to x and are added.
Hence the final product is a trinomial two of whose terms are
the end products and the third term is the sum of the cross
products.
This fact enables us to write such products at once.
Kg. (3a + 4)(5a-7)=15cr-a-28.
In this case 15 a2 is the first end product and — 28 the second, while
— a is the sum of the two cross products, 20 a and — 21 a.
In this manner obtain the following products :
1. (2a + 3)(a + 3). 11. (3<-5)(* + 4).
2. (4o-l)(3a + 2). 12. (5x-y)(2x + 3y).
3. (2 a + 5) (a -7). ' 13. (3x-2y)(x + 3y).
4. (7r + 8)(3r-6). 14. (4a-3y)(a+jf).
5. (2 x + 8) (9 x -4). 15. (3r-2s)(2r + s).
6. (3 m — l)(4m + 3). 16. (5m-n)(2m + »).
7. (5s-7)(2s-4). 17. (5a + 3x)(3a-±x).
8. (2aj-l)(7* + 4). 18. (4a-56)(a + 3&).
9. (4n-9)(5w-7). 19. (3a + 5 6)(a-6).
10. (&y-l)(5y + ll). 20. (3 c- 7 d)(2 c + 3d).
192 SPECIAL PRODUCTS AND FACTORS
142. Trinomials in the form of the above products may some-
times be factored by inspection.
Ex. 1. Factor : 5 x? + 16 x + 3.
If this is the product of two binomials they must be such that the
end products are 5x2 and 3 and the sum of the cross products 16 x.
One pair of binomials having the required end products is 5x + 3
and x + 1, another is 5 x — 1 and x — 3, and still another 5 x + 1 and
ar + 3.
The sum of the cross products in the first pair is 8 x, in the second
pair — 16x, and in the third pair 16 x. Since the sum of the cross
products in the last pair is the one required, the factors are 5x+l
and x + 3.
Ex. 2. Factor: 6 a2- 5 a -4.
In this case as in the one preceding there are several pairs of bi-
nomials whose end products are 6 a2 and —4, such as 2 a — 2 and 3 a + 2,
6 a— 1 and a + 4, etc. By trial we find that among these 3 a — 4 and
2 a + 1 is the only pair the sum of whose cross products is — 5 a.
Hence 6 a2- 5 a - 4 = (3 a - 4)(2 a + 1).
Obviously not every trinomial of this form can be factored in this
manner. Thus, for example, in 6 a2 + 5 a + 4, no pair of binomials
whose end products are 6 a2 and 4 has the sum of its cross products 5 a.
In this manner factor the following :
1. 3^+5^ + 2. 9. 5r*+18r-&
2. 9a2 + 9a + 2. 10. 14 a-2 - 39 a + 10.
3. 2** +11* +12. 11. 5x2 + 26a-24.
4. 9z2 + 36z + 32. 12. 2a2-5a + 2.
5. 2 a* -a; -28. 13. 2rft2-m-3.
6. 12.s2 + lls + 2. 14. 7c2-3c-4.
7. 6t2 + 7t-3. 15. 5a;4 + 9^-18.
8. 6^-3-2. 16. 7a4 + 123a2-54.
FACTORS FOUND BY GROUPING 193
17. 6c2 -19c + 15. 24. 6-5a;-4iB2.
18. 3a2-21a + 30. 25. 3h2-13h + U.
19. 6^ + 4^-2. 26. lor2-r-2.
20. 20a2-a-99. 27. 2t2 + llt + 5.
21. 12c2 + 25c + 12. 28. 10-5X-15X2.
22. 8 + 6a-5a2. 29. 5 x2 - 33 x + 18.
23. 15-507-10^. 30. 20-9a-20ar!.
FACTORS FOUND BY GROUPING
143. Besides the methods of factoring which have been
applied to the types of expressions thus far considered, there
are various other processes which will be considered in the
Advanced Course. One other method of general application
will suffice here.
Ex. 1. Find the factors of ax + ay -f- bx + by.
By Principle I, the first two terms may be added and also
the last two.
Thus, ax + ay + bx + by = a(x + y) + b(x + y).
The given expression has thus been changed into the sum of
two compound terms which have a common factor, (x + y), and
may be added with respect to this factor by Principle I.
Thus, a{x + y) + b(x + y) = (a + b)(x+y).
Hence, ax -{-ay + bx-\-by = (a-\-b)(x-\-y).
Ex. 2. Factor ax — ay — bx + by.
Combining the first two terms with respect to a and the
second two with respect to — b, we have,
ax — ay — bx + by = a(x — y) — 6 (x—y).
194 SPECIAL PRODUCTS AND FACTORS
Again combining with respect to the factor x — y,
ax—ay — bx-\-by=(a — b)(x —y).
The success of this method depends upon the possibility of
so grouping and combining the terms as to reveal a common
compound term, § 77. The order of the terms in the given
polynomial may be changed as found desirable.
EXERCISES
Factor the following expressions by means of Principles I
and II :
1. aft2 + ac2 — db2 — dc2. 11. 2 n2 — en + 2nd — cd.
2. 6 ms— 15 nt + 9 ns — 10 mt. 12. 5 ax — 15 ay — 3 bx + 9 by.
3. 8 ax — 10 ay + 4 bx — 5 by. 13. 3xa — 12 xc — a + 4 c.
4. 2 a2+ 3 a& — 14 an — 21 wfc. 14. 3 xy — 4 wm -f- 6 my — 2 xn.
5. ac + 6c + ad + 6d*. 15. 1 mn + 1 mr — 2n2 — 2nr.
6. ax2 — ftcc2 — ay2 + 6y2. 16. a — 1 + a3 — a2.
7. 8 ac - 20 ad- 6 6c +15 bd. 17. 3s + 2 + 6s4 +4S3.
8. 2ax — Q>bx + Zby—ay. 18. as2 — 3 bst — ast + 3 bt2.
9. 5 + 4 a — 15 c — 12 ac. 19. 3 m?i + 6m2 — 2 am — an.
10. 156-6-206c + 8c. 20. 2ar + 2as + 26r + 26s.
MISCELLANEOUS EXERCISES
Classify the following expressions according to the fore-
going types for factoring, and indicate the factors :
l.<o? + 5z+6. 5. 4:X2 + 9y2-12xy.
2. 1 — Xs. 6. 5 ar> + 4aa; + 7 xy.
3. a? + 11 ai + 30. 7. 2w2 — 6wc-3wy + 9cy.
4. 4arJ + 9y2 + 12:ry. 8. 4x-2-#2.
MISCELLANEOUS FACTORING 195
9. a3 + 63. 35. 3(a+l)3+4(a+l)2+a + l.
10. 9x2 + yi + Qxy2. 36. (x + a)2 - (x— a)2.
11. 2y2as + 4:ya2-8ya. 37. 15 m2 + 224 m - 15.
12. ^ + 7^+6. 38. 3.^4- 27 a +42.
13. 9 0^ + 36^ + 36 a*/2. 39. cc4 + 49 a2 + 14 aa2.
14. 9y-9z-2xy + 2xz. 40. 27 a6 -aV.
15. a'-l. 41. S^ + aV.
16. a4 + 64 + 2a262. 42. 8M-406e + 3cd-15ce.
17. a4 -25. 43. a2 -11 a + 30.
18. 27 a3 -125. 44. 0-+2)2-4(a-2)2.
19. 4a2 + 4a6 + a&2. 45. x* + 9y2-6yx?.
20. 4 a2 + 9 a4- 12 ax2. 46. 4a2-7ca2- 4d2+ 7cd*2.
21. 1 + ar5. 47. a2 + 15a -16.
22. 2x2 + 5a + 3. 48. 18-27c + 166-24c6.
23. 36 + 4 a6 +24 or5. 49. 4-(a2 + &2-2a&)2.
24. 0_1)2_0 + 1)2. 50. 10r + 36s-66r-5s.
25. 8 + 64 a6. 51. 25 + 64^ + 80a3.
26. a,c — ax — 4&c + 4&c. 52. 1000 — Xs.
27. 27 -216 a3. 53. 103+ ar*.
28. 33 + 63a3. 54. 8a3 + a3&3 + &2a2.
29. 25(x + l)2-4. 55. 100 -49a4.
30. 5cx-10c + 4:dx — 8d. 56. 100 + 625 + 500.
31. 4t{x+2)2+y2 + ±{x + 2)y. 57. a2 -17a + 72.
32. ra + 2 r/t - 5 sa- 10 sA. 58. a2 + 17 a + 72.
33. -2a26 + a4 + 62. 59. a2+1662-8a6.
34. 2/ia — hb+ 6 a- 36. 60. xs — y9.
196 SPECIAL PRODUCTS AND FACTORS
61. 24 a2 + 37 a -72. 87. 24 a4 c4 + a6 + 144 c8 a2.
62. (8* + 15s2 — 100. 88. 9s2 + 4?/4-12s?/2-16.
63. 9« + 83. 89. 81 + 100s8- 180 s4.
64. 9s4 + 16?/2 + 24sy 90. a4 + 27a2 + 180.
65. 1-1000. 91. a4 + 3a2 -180.
66. 16a262+24a& + 3663. 92. a4 -3 a2 -180.
67. 64 + 8. 93. 144-(a4 + fc2-2a26).
68. 16a262 + 9aV + 24a26c. 94. 81 a2 64 + 49 c2- 126 ab2c.
69. a2 + 462 + 4a&-4s2. 95. 12 s2 - 23 st + 10 t2.
70. asb6 + c?. 96. 36 s4 + 12 s2?/4 + ?/8.
71. 5^ + 10s3?/2 + 30 or5?/4. 97. 16s6 + 9y4 + 24sy~49.
72. 16aV + 4c2s2+16ac2s. 98. y2 + 35y + 300.
73. a6y3-z\ 99. 5 ?/2 -80?/ + 300.
74. s4- 7s2- 120. 100. 100 - (16s2 + ?/6- 8s?/3).
75. 9a4&2 — 12a3& + 4a2. 101. ac—bc + ad—bd.
76. 8ao + 27a&7. 102. 6 rd -15 re +22 cd-55ce.
77. s4 + 4s2 + 4 — s6. 103. z3 + ya — ifz3 — ay*.
78. l-125a366. 104. s4 + 2s2 + l-s2.
79. 16 + 16a6+4a262. 105. 60 s2 + 7 s?/ - ?/2.
80. 64a3 + 8a2&3. 106. s2 - 20 s?/ + 75 ?/2.
81. 36a264 + o264 + 12 ab*c. 107. s2- 17s -60.
82. 4a2 + 9&4 + 12a62-16a4. 108. 65^ + 8 r-1.
83. s6 + 17s3 + 30. 109. a2 -13a -140.
84. 25-(a4-2a2Z>3 + &fi). 110. 39s4-16s2 + l.
85. -112a2c3 + 49a4 + 64c6. 111. 625- (31 -4a2)2.
86. a2_a_380. 112. 36a2-29a& + 5 62.
EQUATIONS SOLVED BY FACTORING 197
113. c4- 31c2 + 220. 115. 26 + 39w-22m-33mw.
114. ac + d5a— b4c-b*d5. 116. 12 a2 + 11 a - 56.
117. a2 + 4a&+462-(a2-4a& + 4&2).
118. (3a;-l)2-(a,'2 + 4^-4^)2.
119. (x + 3yy+(x-2yy +2(x + 3y)(x-2y).
120. 16 (a + b)2 - 8 (a -6) (a + 6) + (a - b)\
121. 256a,-2 - (49 x2 + 4?/4 - 28a?/2).
122. (2a -a)2 +100 (a -3a)2 +20 (2a - a) (a - 3a).
123. -48(a-6)(a+6) + 36(a-&)2 + 16(a+6)2.
EQUATIONS SOLVED BY FACTORING
144. Illustrative Problem. There are two consecutive num-
bers the sum of whose squares is 61. What are the numbers ?
Solution. Let x = one of the numbers, then x + 1 is the other.
Hence, x2 + (x + l)2 = 61 (1)
By F, x2 + x* + 2x + 1 = 61 (2)
By F, I, S, 2x2 + 2x =60 (3)
By D, x2 + x = 30 (4)
These equations differ from any which we have studied here-
tofore in that they contain the squares of the unknown num-
ber, which cannot be removed by addition or subtraction.
It is evident on inspection that a = 5 satisfies equation (4).
That is, 52 + 5 = 30.
Hence 5 is one of the numbers sought, and 5 + 1 is the other, and
these numbers satisfy the conditions of the problem,
Since 52 + 62 = 25 + 36 = 61.
145. Definition. Equations which involve the second but
no higher degree of the unknown number are called quadratic
equations.
198 SPECIAL PRODUCTS AND FACTORS
One method of solving quadratic equations is now to be
considered.
By S, equation (4) above may be written
x2 + x - 30 = 0. (5)
Factoring the left member,
(x + 6) (x - 5) = 0. (6)
Equation (6) is satisfied by any value of x which, substituted in
the left member, reduces it to zero ; and, since the product of two fac-
tors is zero if either factor is zero, we seek values of x which make
x — 5 = 0 and also x + 6 = 0. Hence the equation is satisfied by
x = 5 and also by x = — 6. Thus,
(5 + 6) (5 - 5) = 11 • 0 ss 0,
and also (- 6 + 6) (- 6 - 5) = 0 . - 11 =0.
Therefore — 6 and — 6 + 1 = — 5 are two numbers which meet
the condition of the problem.
That is, (- 6)2 + ( - 5)2 = 36 + 25 = 61.
Hence this problem has two solutions, namely, the numbers 5 and 6
and the numbers — 6 and — 5.
Illustrative Problem. A rectangular flower bed 10 feet long
and 6 feet wide is surrounded by a gravel walk whose area is
192 square feet. How wide is the walk ?
Solution. Let the width of the walk be x feet, then
2 • 10 x is the area of the sides,
2 • 6 x is the area of the ends,
4 x2 is the area of the corners.
Hence the total area is
4x2 + 2-10x + 2.6x = 192. (1)
By F, 4x2+32x = 192. (2)
ByZ>, x2 + 8x=48. (3)
EQUATIONS SOLVED BY FACTORING 199
By S, x2 + 8 x - 48 = 0. (4)
Factoring, (x + 12) (x - 4) = 0. (5)
But (5) is satisfied if x + 12 = 0, and also if x — 4 = 0. (6)
Hence x = — 12 and also x = 4. (7)
Check by substituting in equation (3) :
( - 12)2 + 8 (- 12) = 144 - 96 a 48.
Also 42 + 8 • 4 = 16 + 32 = 48.
Hence the quadratic equation to which this problem gives rise has
two solutions, but since the width of a walk cannot be a negative
number, only the number 4 satisfies the conditions of the problem.
Ex. 1. Solve the quadratic equation
5a2 + 30z + 3 = 3-5». (1)
By A, S, 5x2 + 35x = 3-3 = 0. (2)
Factoring, 5 x (x + 7) = 0. (3)
But (3) is satisfied if 5 x = 0, and also if x + 7 = 0. (4)
Hence x = 0, and also x = — 7. (5)
Check. Substitute x = 0 and also x = — 7 in (1).
Ex. 2. Solve the quadratic equation
6x2 + llx = 10. (1)
By S, 6x2+llx-10 = 0. (2)
Factoring, (3 x - 2) (2x + 5) = 0. (3)
But (3) is satisfied if 3 x — 2 = 0 and also if 2 x + 5 = 0.
Hence x = § and x = — |.
Check by substituting each of these values in (1).
200 SPECIAL PRODUCTS AND FACTORS
Ex. 3. Solve the quadratic equation
3z2-5a;-7 = 2a:2-a;-ll. (1)
By .4andS, x2-4x + 4 = 0. (2)
Factoring, (x - 2) (x - 2) = 0. (3)
It follows from the first factor and also from the second that (3) is
satisfied if x = 2.
In this case the two solutions turn out to be identical, while in
Example 1 one solution was zero and the other was — 7.
If we count each of these results as two solutions, then for every
quadratic equation thus far solved we have found two values of the
unknown number.
146. The method of solution above explained consists of
three steps :
(1) Transform the equation so that all terms are collected
in one member, with similar terms united, leaving the other
member zero. This can always be done by Principle VIII.
It is convenient to make the right member zero.
(2) Factor the expression on the left.
(3) Find the value of x which makes each of these factors
zero. This is readily done by setting each factor equal to
zero and solving it for the unknown.
EXERCISES
Find two solutions for each of the following quadratic
equations :
1. ar,-3» + 2 = 0. 6. a2 + 3a = 10 a + 18.
2. x, + 7« = 30. 7. a2 + 10a=-24-4a.
3. a2-lla=-30. 8. 2z2-6:r=-40+12a;.
4. a2 + 13 a = 30. 9. 3x + x2= 20^-72.
5. a2 + 10a + 8 = -3a-34. 10. 17a + 30= -^-40.
EQUATIONS SOLVED BY FACTORING 201
11. 7aj» + 2a; = 30a>-21. 21. x2 + 12 a + 6 = 5a-4.
12. lla + 3x2 = 20. 22. 2a2-7a=60 + 7tf.
13. l§-5x+x2= -2x*-20x-2. 23. 60 a; + 4 ^ + 144 = 8 a.
14. or* -16 = 0. 24. 18 se = 63 - x2.
15. ^-1=0. 25. 24x2 = 12a + 12.
16. x2 — x = 0. 26. 2 a; = 63— x2.
17. x2 + a- = 0. 27. 22 x + x2 = 363.
18. 4 a2 = 25. 28. 3a^ + 7a; = 6.
19. arJ + 4a; + 4 = 0. 29. 2^ = 2- 3».
20. arJ + 8a; + 16 = 0. 30. x-2 = -3x2.
147. It is sometimes possible to solve other equations than
quadratics by the above process.
Ex. 1. Solve the equation :
a8 + 30 x = 11 x2. (1)
By S, x»-llx2+30x = 0. (2)
By §131, x(x2-llx + 30) = 0. (3)
B\ § 139, x (x - 5) (x - 6) = 0. (4)
(4) is satisfied if x = 0, if x — 5 = 0, and if x — 6 = 0.
Hence the solutions are x = 0, x = 5, x = 6.
Ex. 2. a>(» + l)(a;-2)(a;+3) = 0.
Any value of x which makes one of these factors zero re-
duces the product to zero and hence satisfies the equation.
Hence the solutions of the equation are found from
x = 0, a; + 1=0, x — 2 = 0, and # + 3 = 0.
That is x = 0, x= — 1, x = 2, x= — 3 are the values of x
which satisfy the equation.
Notice that this process is applicable only when one member
of the equation is zero and the other member is factored.
202 SPECIAL PRODUCTS AND FACTORS
EXERCISES
Solve the following equations by factoring :
1. x3 — x2 = 6x. 8. x2 — ax — bx + ab = 0.
2. 5a; = 4ar! + arJ. 9. 4 (x _ 2)2 _ (a? + 3)2 = 0.
3. Xs — 25 a = 0. 10. ar'-aic + fo; — ab = 0.
4. Xs — 3x2=— 2 x. 11. x2 + ax — bx — ab = 0.
5. 3 ^ = 15 or* + 42 a. 12. 9(> + 2)2-4 (z-3)2 = 0.
6. 5 a3 + 315 a = 80 a2. 13. ^-aj-3^ + 3 = 0.
7. a* + ax + &a; + a& = 0. 14. or5 -4a;- 8 a2 + 32 = 0.
15. 5 ^ + 120 a2 = 119 a -2a^ + 8a^.
16. (a?+6aj-16)(a>*-7a; + 12) = 0.
17. (aj + 7)(a2-T7a; + 72) = 0.
18. (a?-16)(«8 + llaj8+30aj) = 0.
148. Illustrative Problem. The paving of a square court
costs 40^ per square yard and the fence around it costs $ 1.50
per linear yard. If the total cost of the pavement and the
fence is $ 100, what is the size of the court ?
Solution. Let x — the length of one side in yards.
Then 40 x* = cost in cents of paving the court,
150 • 4 x = 600 x = cost of the fence in cents.
40 x2 + 600 x = 10000. (1)
By A x2+ 15 a; = 250. (2)
By S, x* + 15s - 250 = 0. (3)
Factoring, (x - 10) (x + 25) = 0. (4)
Whence, x = 10, and also x = — 25. (5)
It is clear that the length of a side of the court cannot be — 25
yards. Hence 10 is the only one of these two results which has a
meaning in this problem.
It happens frequently when a quadratic equation is used to
solve a problem that one of the two numbers which satisfy this
equation will not satisfy the conditions of the problem.
EQUATIONS SOLVED BY FACTORING
203
PROBLEMS
In each of the following problems find all the solutions pos-
sible for the equations and then determine whether or not each
solution has a reasonable interpretation in the problem.
1. The dimensions of a picture inside the frame are 12 by 16
inches. What is the width of the frame if its area is 288
square inches ?
2. There are two consecutive integers such that the sum of
their squares is 3961. What are the numbers ?
3. An open box is made from a square piece of tin by cutting
out a 5-inch square from each corner and turning up the sides.
How large is the original square if the box
contains 180 cubic inches ?
If x — length of a side of the tin, then the volume of
the box is : 5(x - 10) (x - 10) = 180. (See the figure.)
4. A rectangular piece of tin is 4 inches
longer than it is wide. An open box contain-
ing 840 cubic inches is made by cutting a 6-inch square from
each corner and turning up the ends and sides. What are
the dimensions of the box ?
5. A farmer has a rectangular wheat field 160 rods long by
80 rods wide. In cutting the grain, he cuts a strip of equal
width around the field.
How many acres has
he cut when the
width of the strip is
8 rods ?
6. How wide is the
strip around the field
of problem 5, if it con-
tains 27^ acres ?
7. In the northwest a farmer using a steam plow starts plowing
around a rectangular field 640 by 320 rods. If the strip plowed
the first day lacks 16 square rods of being 24 acres, how wide is it ?
160
80
X
1
8
X
*(l60-2a:)
160-2*
\x **
X
H
M
8
160-2*
f
8
.so
160
204
SPECIAL PRODUCTS AND FACTORS
8. A rectangular piece of ground 840 by 640 feet is divided
into 4 city blocks by two streets 60 feet wide running through
840 feet it at right angles. How many square
feet are contained in the streets ?
9. A farmer lays out two roads
through the middle of his farm, one
running lengthwise of the farm and the
other crosswise. How wide are the
roads if the farm is 320 by 240 rods, and the area occupied
by the roads is 1671 square rods ?
QUADRATIC AND LINEAR EQUATIONS
149. When two simultaneous equations are given, one quad-
ratic and one linear, they may be solved by the process of sub-
stitution, which was used (§ 116) in the case of two linear
equations.
Illustrative Example. Solve the equations :
r x1 — y'1 = — 16.
I a? — 3 2/ = — 12.
From (2) by S, x = 3 y - 12.
Substituting (3) in (1), (3 y - 12)2 - y2 = - 16.
From (4) by F, 9 y2 - 72 y + 144 -y2=- 16.
From (5) by F, A, 8 y2 - T2 y + 160 = 0.
By A y2 - 9 y + 20 = 0.
Factoring, (y — 5)(y — 4) = 0.
Hence, y = 5, and y = 4.
Substitute y = 5 in (2) and find x = 3.
Substitute y = 4 in (2) and find x = 0.
Therefore (1) and (2) are satisfied by the two pairs of values,
x = 3, y — 5, and x — 0, y — 4.
Check by substituting these pairs of values in (1) and (2).
(1)
(2)
(3)
(4)
(5)
(6)
(7)
(8)
(9)
QUADRATIC AND LINEAR EQUATIONS 205
EXERCISES
In the manner just illustrated solve the following :
\x+2y = 8, \x-y=-l,
2.
3.
5«« + 12?/2 = 128. Ua;2 + 3i/2 = 147.
1^ + ^ = 1. ' Ix2— 52/2 = 4.
f 2 a? — y = 6, 13 fas-yssi,
U^ + 5?/2 = 36. ' 1 3 as2 — 2 2/2 = — 5.
fa; + 3y = 6, J5z-7y=-28,
la2 + 3/ =12. ' 115 a2 + 49 1/2 = 784.
a;_22/=-2, j6z-7*/ = 18,
x2-6?/2 = 10. " 1 36 ^-7^ = 324.
.8* -16?/ = -120, 16 Ja-9y = 2
7.
\x-\
7 x* + 2 y2 = 585. ' Ick2- 45^ = 4.
j7a + 9?/ = 88, f« + y=8,
17^4-9^ = 736. ' ll3»2 + 32/2 = 160.
8 ix-y = 6, \2x-5y=-16,
U2-7?/2=36. ' 14^ + 15/ =256.
|3a;4-22/ = 7, J7a + 4y = 7,
l3^ + 8y2=35. ' l49a2-8?/2 = 49.
f»-8y«-ll, fa»-3y^-12,
* 1 3 cc2- 16/ = 11. ' lx2-y2 = -16.
150. Illustrative Problem. The fence around a rectangular
field is 280 rods long. What are the dimensions of the field,
if its area is 30 acres ?
Solution. Let w = the width of the field in rods,
and I sr length of field in rods. (1)
Then 2 1 + 2 w = 280,
and lw — 4800 (one acre = 160 square rods).
(2)
206 SPECIAL PRODUCTS AND FACTORS
From (1), I + w = 140, and w = 140 - I.
Substituting this value of w in (2),
Z(140 - 0= 4800,
or, Z2 -U01 + 4800 = 0.
Then, (I - 60) (/ - 80) = 0.
Whence 60 and 80 both satisfy the equation.
If I = 60, then from equation (1) w = 80, and if I = 80, then w = 60.
We group these pairs of numbers as follows :
| / = 60>andM = 80,
I w = 80, I w = 60.
Substituting these pairs of values in both (1) and (2), we have
| 2 • 60 + 2 • 80 = 280, j 2 • 80 + 2 • 60 = 280,
I 60 • 80 = 4800, an 1 80 • 60 = 4800.
Hence, we obtain two pairs of numbers which satisfy both of these
equations. The solution I = 60 and w = 80 is applicable to this prob-
lem only by calling the greater side the width instead of the length.
AVe have seen before that solving a quadratic often results in one
solution which is without meaning in the problem that gives rise to
the equation.
In any case each solution should be carefully examined to ascertain
whether under any reasonable interpretation they are both applicable
to the problem.
151. If squares are constructed on the two sides, and also on
the hypotenuse of a right-angled triangle, then the sum of the
squares on the sides is equal to the square on the hypotenuse.
This is proved in geometry, but may be verified by counting
squares in the accompanying figure. This proposition was first
discovered by the great philosopher and mathematician Py-
thagoras, who lived about 550 b.c. Hence it is called the
Pythagorean proposition.
We now proceed to solve some problems by this proposition.
QUADRATIC AND LINEAR EQUATIONS
207
^
0
s
t
1
11
rf
6
s
1-
*k
f<-
0
s
1-
ft
i
PROBLEMS
1. The sum of the sides about the right angle of a right tri-
angle is 35 inches, and the hypotenuse is 25 inches. Find the
sides of the triangle.
Solution. Let a = the length of one side in inches,
and b = the length of the other.
Then a + b = 35, (1)
and a2 + b2 = 252 = 625 (Pythagorean proposition). (2)
208 SPECIAL PRODUCTS AND FACTORS
From (1), a = 35 - ft.
Substituting in (2), (35 - ft)2 + ft2 = 625,
or 1225 - 70 b + ft2 + ft2 = 625,
2 ft2 - 70 ft + 600 = 0,
ft2 -35ft + 300 = 0,
(ft- 20) (ft- 15)= 0.
Whence ft = 20, and ft = 15.
From (1), if ft = 20, a = 15, and if ft = 15, a = 20; that is, the
sides of the triangle are 15 and 20.
2. The difference between the two sides of a right triangle
is 2 feet, and the length of the hypotenuse is 10 feet. Find
the two sides.
3. The sum of the length and width of a rectangle is
17 rods, and the diagonal is 13 rods. Find the dimensions of
the rectangle.
4. A room is 3 feet longer than it is wide, and the
length of the diagonal is 15 feet. Find the dimensions of
the room.
5. The length of the molding around a rectangular room
is 46 feet, and the diagonal of the room is 17 feet. Find its
dimensions.
6. The longest rod that can be placed flat on the bottom
of a certain trunk is 45 inches. The trunk is 9 inches
longer than it is wide. What are the dimensions of the
bottom ?
7. The floor space of a rectangular room is 180 square feet,
and the length of the molding around the room is 56 feet.
What are the dimensions of the room ?
8. A rectangular field is 20 rods longer than it is wide, and
its area is 2400 square rods. What are its dimensions ?
9. A ceiling requires 24 square yards of paper, and the
border is 20 yards long. What are the dimensions of the
ceiling ?
QUADRATIC AND LINEAR EQUATIONS 209
10. The area of a certain triangle is 18 square inches, and
the sum of the base and altitude is 12. Find the base and
altitude.
11. The altitude of a certain triangle is 7 inches less than
the base, and the area is 130 inches. Find the base and
altitude.
12. The sum of two numbers is 17, and the sum of their
squares is 145. Find the numbers.
13. The difference of two numbers is 8, and the sum of their
squares is 274. Find the numbers.
14. The difference of two numbers is 13, and the difference
of their squares is 481. Find the numbers.
15. The sum of two numbers is 40, and the difference of
their squares is 320. Find the numbers.
16. The sum of two numbers is 45, and their product is
450. Find the numbers.
17. The difference of two numbers is 32, and their product
is 833. What are the numbers ?
CHAPTER VII
QUOTIENTS AND SQUARE ROOTS
QUOTIENT OF TWO POWERS OF THE SAME BASE
152. Illustrative Problem. To divide x6 by x\
Since by § 66 the quotient times the divisor equals the dividend,
we seek an expression which multiplied by x* equals x6.
Since by Principle XIV two powers of the same base are multiplied
by adding their exponents, the expression sought must be that power
of x whose exponent added to 4 equals 6. Hence the exponent of the
quotient is 6 - 4 = 2. That is, x* ■*■ a;4 = x6-4 = x2.
EXERCISES
Perform the following indicated divisions :
1. 24-=-22. 8. 513-=-512. 15. x* + x*.
2. 23--22. 9. T-4--!22. 16. tu + t*.
3. 24-=-2. 10. 83-=-8. 17. m3-*-ra.
4. 33-=-32. 11. 64-=-62. 18. n6-=-n2.
5. 34h-3. 12. as + a2. 19. (20)4h-(20).
6. 34-5-32. 13. a*--- a3. 20. (101)14--(101)13.
7. 9u-*-9*. 14. m4-t-m*. 21. 417-;-416.
153. The process of division by subtracting exponents
leads in certain cases to strange results.
Thus, according to this process, x4 + x4 = a;4-4 = x°, which is as yet
without meaning, since an exponent has been defined only when it
is a positive integer. It cannot indicate, as in the case of a positive
integral exponent, how many times the base is used as a factor. We
know, however, that x4 + x4 — 1, since any number divided by itself
210
QUOTIENTS OF POWERS OF THE SAME BASE 211
equals unity. Hence if we use the symbol x° it must be interpreted to
mean 1, no matter what number x represents. It is sometimes con-
venient in algebraic work to use it in this way.
Again by this process x2 + x* = x2-4 = x-2, which is as yet without
meaning, since negative exponents have not been defined. Cases of
this kind are considered in the Advanced Coui-se.
The preceding exercises illustrate the following principle :
154. Principle XVI. The quotient of two powers of the
same base is a power of that base whose exponent is the
exponent of the dividend minus that of tJie divisor.
For the present only those cases are considered in which the expo-
nent of the dividend is greater than or equal to that of the divisor.
Notice that Principle XVI does not apply to powers of different
bases.
E.g. 37 + 24 does not equal any integral base to the power, 7 — 4.
This division can be performed only by first multiplying out both
dividend and divisor.
EXERCISES
Perform the following indicated divisions by means of Prin-
ciple XVI :
1. 27-r-23. 6. xin + x2n. 11. a?*+h -+- x°+b.
2. a7 + as. 7. 32"-1 -f- 3a"2. 12. ic^ + w'.
3. 34-=-32. 8. 5"+5--5"+2. 13. (17)14h- (17)13.
4. x* + x*. 9. xa+i-^xa+2. 14. 43-i-4.
5. 33"-*-32n. 10. t4a + ta. 15. (12)4--(12)3.
In the following use Principles V, VI, and XVI :
16. (24 + 23) -=- 23. 21. (a2ra4-&2m3)-r-m3.
17. (3 • 24 + 5 • 23) -=- 22. 22. (4 • 32 - 33 • 5 • 7) -f- 32.
18. (3.43-5-44)-=-42. 23. (23 • 3 + 24 • 33 - 23) -23.
19. (a3b - a462) -*- a2. 24. (12afy- llaV+5a;4) + X2.
20. (4 ^ + 3 a4)-*- a?2. 25. (x3m+4+x2m+3-5xm+2)-i-xm+\
212 QUOTIENTS AND SQUARE BOOTS
DIVISION OF MONOMIALS
155. In finding the quotient of two numbers each in the
factored form, if the factors are represented by Arabic figures,
the operation may be carried out in either of two ways.
E.g. 28 • 3» . 5 - 2 • 3 = 1080 - 12 = 90.
Also 23 • 33 • 5 - 22 • 3 = 2 • 32 • 5 = 90.
In the second process we divide by one of the factors, 22 or 3,
and divide this result by the other. 23 • 33 • 5 divided by 22
gives, according to Principle V, 2 • 33 • 5, and this result di-
vided by 3 gives by the same principle 2 • 32 • 5. In practice
such operations may readily be performed simultaneously.
In the case of literal factors the second process only is
available.
E.g. 5 aWc + a262 = 5 a4-268~2c = 5 a'Wc = 5 a2bc.
EXERCISES
Perform the following indicated divisions in two ways when
possible :
1. 58-79-h5-72. 5. 15asb* + 5ab.
2. 23-33-53-^2.32-5. 6. 12 x*y ~ 4 x.
3. 44.5<h_4.53. 7. 18 ^--3 J2.
4. 34-52-5-3-5. 8. 28m3n-=-7m.
The preceding examples illustrate the following principle :
156. Principle XVII. TJie quotient of two monomials is
found by dividing the dividend by each factor of the divi-
sor in succession.
Each factor of the divisor is associated with any desired
factor of the dividend according to Principle V, and when the
bases are the same the exponents are subtracted according to
Principle XVI.
DIVISION OF MONOMIALS 213
157. If there are factors in the divisor not found in the
dividend, this process terminates before the operation is com-
pleted. The remaining steps of the division must then be indi-
cated, which is usually done in the form of a fraction.
E.g. x2-r-3 xsy = — — — = — .
3 x1 • xy 3 xy
Again, 15 a%2c + 3 a2bx2y = 15^>% = £«*? .
3 a-bx2y x'2y
By this process all factors common to dividend and divisor
have been canceled. Notice that in the first example the fac-
tor 1 remains when x2 of the dividend is divided by x2 of the
divisor.
_L. . , EXERCISES
Divide :
1. 4 • 7 • 9 by 2 . 3. 6. 5 a4bnc by ab*<*.
2. 12 • 8 • 20 by 2 • 4 • 5. 7. 10 aWV by 2xb4c.
3. 6 x?y2z by 2 xyz. 8. 36 x4f by 6 xSf .
4. 64 • 34 • x3 by 62 • 52 • x2. 9. 35 a*- y+i by 5 xa~hf+\
5. 12 x12y13 by 4 xtfz. 10. 2 m20*4™3—2 by ma+2na-2.
In each of the following exercises state which of the Prin-
ciples I-XVII are used :
Divide :
11. 23.32-24.33by 23 . 32.
12. 5-27.38 + 7-25.34by 2s • 34.
13. 4 aV- 3 x?y2 by x2y2.
14. 18 xy - 12 xAf + 6 x2y2 by 6 afy2.
15. 49 a4 -f 21 a3- 7 a by la.
16. 12 ax4}/3 — 16 a2ary -f 8 a3xy by 4 axy.
17. 2 ar3* + 4 x4" — 8 x2" by 2 a8.
18. 6 x2" f J -f 12 ar3n+] - 10 iC+1 by 2 an+1.
19. 4 a;13 - 6 «n6 — 10 x4c by 2 x4.
20. 10 a362 - a2b3 + 15 u464 by 5 a2b2.
214 QUOTIENTS AND SQUARE ROOTS
SQUARE ROOTS OF MONOMIALS
158. Definition. The radical sign, Vj indicates that we are to
find one of the two equal factors of the number expression
which follows it, and the vinculum is attached to it, V , to
show how far its effect is to extend.
E.g. V9 is read the square root of 9.
Similarly, Va2 -f 2 ab + b2 is read the square root of a2 + 2 ab + b2.
The square root of any number is at once evident if we can
resolve it into two equal groups of factors.
E-g.
V576 = V2- 2- 2- 2- 2 -2. 3. 3= V(23. 3)(28 . 3) = V24 . 24 = 24.
It should be noted that every square has two square roots.
E.g. V9 = - 3 as well as + 3, since (- 3)2 = 9 and 32 = 9.
In obtaining a square root the two results should always
be indicated. This is usually done by attaching the double
sign ± to the square root, e.g. V9 = ± 3.
EXERCISES
Find by i
inspection the following
; square roots :
i. VI.
8.
VI21.
15. V324.
22.
V24".
2. V9.
9.
VI69.
16. V289.
23.
V51"2.
3. VI6.
10.
V225.
17. V625.
24.
V71".
4. V25.
11.
12.
V196.
V256.
18. V900.
25.
26.
Va12.
5. V36.
19. Vioooo.
V31"4.
6. V49.
13.
V576.
20. Va4.
27.
Va14.
7. V8T.
14.
VIoo.
21. Vi?.
28.
V3^.
SQUARE BOOTS OF MONOMIALS 215
159. The square root of the product of several factors, each
of which is a square, may be found in two ways if the factors
are expressed in Arabic figures.
E.g. V4 • 16 • 25 = V1600 = V40 . 40 = ± 40,
or V4 • 16 • 25 = V2'2 . 42 • 52 = ± 2 • 4 • 5 = ± 40.
But with literal factors, the second process only is available.
E.g. Vl6 a*b4c* = V42 a%*c2 = ± 4 ab2c.
EXERCISES
Find the following indicated square roots, keeping the re-
sults in the factored form :
1. V22-32. 7. V312-514. 13. V9zy2.
2. V81 • 121. 8. V222 • 312. 14. V121 a¥.
3. V49-25-169. 9. Vl6aW. 15. V74a462.
4. V82.52-32. 10. V64aV. 16. V625 x*y\
5. V54.32-44. 11. V44a264. 17. Vl225a2r.
6. V25 • 36. 12. ^W¥y\ 18. V36r&4™c2".
Notice that the square root of a sum is not obtained by taking
the square roots of the terms separately. Thus, V9 + 16 is
not equal to V9 + Vl6.
The preceding exercises illustrate the following principle :
160. Principle XVIII. The square root of a product is
obtained by finding the square root of each factor sepa-
rately and then taking the product of these roots.
In order that a factor may be a perfect square it must be
a power whose exponent is even. Its square root is then a
216 QUOTIENTS AND SQUARE ROOTS
power of the same base whose exponent is equal to one-half
the given exponent.
Thus, Vx^= Vx3 • xs = xs. The exponent 3 of the root is one-half
the exponent 6 of the power. Hence to find the square root of a mo-
nomial we divide the exponent of each factor by 2.
EXERCISES
Find the following square roots :
1. V4a%«. 6. VlO4 a4b\ 11. V81 x*y8cw.
2. V3VV4. 7. Vo^m". 12. V729 a«y™zH.
3. V5- • 3- fM. 8. V54^38^72. 13. V64 • 625 a2b\
4. V121 xy\ 9. V314 - 712 a4. 14. V216 arV".
5. V576 a2b*. 10. V2^W2. 15. V32* • 52".
DIVISION BY A POLYNOMIAL
161. The simplest case of division by a polynomial is that
in which the dividend can be resolved into two factors, one
being the given polynomial divisor and the other a monomial.
E.g. To divide 4 x3 + 4 x2y by x + y, factor the dividend and we
have 4 x2 (x+y) h- (x + y) = 4 x2.
In case the dividend cannot be factored in this manner,
then, if the division is possible, the quotient must be a polyno-
mial. The process of finding the quotient under such circum-
stances is best shown by studying a particular case.
Illustrative Example 1. Consider the product
(x2 + 2xy + y2)(x+y) = x2(x + y) + 2 xy (x +y) + y2 (x + y).
The products, x2(x + y), 2xy(x + y), and y2(x + y) are called partial
products, and their sum, x3 + 3 x2y + 3 xy2 + y3, the complete product.
In dividing x3 + 3 x2y + 3 xy2 + y3 by x + y the quotient must be
such a polynomial that when its terms are multiplied by x -f y the
DIVISION BY A POLYNOMIAL 217
results are these partial products, which in the solution are called 1st,
2d, and 3d products.
The work may be arranged as follows :
Dividend or product : x3 + 3 x2y + 3 xy2 + y3
1st product, x2(x+y) : x3 + x-y
x + y, divisor.
x2 + 2 xy + y2,
Dividend minus 1st product: 2 x2y + 3 xy2 -f y3 [quotient.
2d product, 2xy(x + y): 2 x'2y + 2 xy2
Dividend minus 1st and 2d products : xy2 + y3
3d product, y\x+y): xy2 + y3
Dividend minus 1st, 2d, and 3d products : 0
Explanation. Since the dividend or product contains the
term x3, and since one of the factors, the divisor, contains the
term x, the other factor, the quotient, must contain the term
x2. Multiplying this term of the quotient by the divisor, we
obtain the first partial product, x3 + x*y.
Subtracting the first partial product from the whole product
Xs + 3 x2y + 3 xy2 + y3, the remainder is 2 x2y + 3xy2 + y3, which
is the product of the divisor and that part of the quotient
which has not yet been found. Since this product contains the
term 2 x?y and the divisor contains the term x, the quotient
must contain the term 2 xy. The product of 2 xy and x + y is
the second partial product.
Subtracting this second partial product from 2x?y + 3 xy2
-t-y3, we have xy^ + y3 still remaining after the first and second
partial products have been subtracted from the whole product.
This remainder is the product of the divisor and the part of the
quotient not yet found. Since the product contains the term
xy2 and the divisor contains the term x, the quotient must contain
the term y2 ; hence, the third partial product is xy2 + y3.
Subtracting the third partial product the remainder is zero.
Hence the sum of the three partial products thus obtained is
equal to the whole product, and it follows that x2 + 2 xy -f- y2 is
the required quotient.
218 QUOTIENTS AND SQUARE ROOTS
162. Problems in division may be checked by substituting
any convenient values for the letters. For example, in this
case, x = l, y = l, reduces the dividend to 8, the divisor to 2,
and the quotient to 4, which verifies the correctness of the
result.
Since division by zero is impossible (see Advanced Course),
care must be taken not to select such values for the letters
as will reduce the divisor to zero.
Illustrative Example 2. Divide 2xi-\-xz — 7 a? + 5 x — 1 by
Solution. [divisor.
x2 + 2x-l,
Dividend or product: 2 x*+xi — 7 x2+5 x—
1st product, 2x2(x2+2a;-l): 2x*+ix8-2x2 12 x2- 3 x + 1,
Dividend minus 1st product : — 3 x8 — 5 x2+ 5 x— 1 [quotient.
2d product, -3 x(x2 + 2x-l): -3x8-6x2+3x
Dividend minus 1st and 2d products : +i2+2i-1
3d product, 1 • (x2 + 2 x- 1): x2+2x-l
Dividend minus 1st, 2d, and 3d products : 0
Check. Substitute x = 2 in dividend, divisor, and quotient.
Illustrative Example 3. Divide 20 a? - 8 + 18 a* + 22 a - 19 a3
by 2u2-3a + 4.
Solution. Arranging dividend and divisor according to the de-
scending powers of a, we have Tdivisor.
Dividend or product: 18a4-19a8 + 20a2+22q-8[2 a2-3 a + 4,
1st product : 18a4-27a8+36a2 9a2+4a-2,
Dividend minus 1st product : 8 a3— 16 «2 + 22 a— 8 [quotient.
2d product : 8a8-12a2 + 16a "
Dividend minus 1st and 2d products : — 4a2+ 6 a — 8
3d product : - 4a2+ 6a-8
Dividend minus all products : 0
Check. Substitute a = 1 in dividend, divisor, and quotient.
DIVISION BY A POLYNOMIAL 219
163. From a consideration of the preceding examples the
process of dividing by a polynomial is described as follows :
1. Arrange the terms of dividend and divisor according to
descending (or ascending) powers of some common letter.
2. Divide the first term of the dividend by the first term
of the divisor. This quotient is the first term of the quotient.
3. Multiply the first term of the quotient by the divisor
and subtract the product from the dividend.
4. Divide the first term of this remainder by the first term
of the divisor, obtaining the second term of the quotient.
Multiply the divisor by the second term of the quotient and
subtract, obtaining a second remainder.
5. Continue in this manner until the last remainder is zero, or
until a remainder is found whose first term does not contain
as a factor the first term of the divisor. In case no remainder
is zero, the division is not exact.
EXERCISES
Check the result in each case, being careful to substitute
such numbers for the letters as do not make the divisor zero.
Divide the following :
1. a2 -f- 2 ab + b2 by a + b.
2. a2 -2 ab + b2 by a-b.
3. a3- 3 a2b + 3 ab2 -W by a- b.
4. 2 a^ -f 2 x*y — 4: x2 — x — 4= xy — y by x -\- y.
5. x3 -f- %y2 — x?y — y3 by x — y.
6. tf + lrf + x — 6bycc + 3.
7. a^ + 4 xP + x — 6 by x — 1.
8. :B4-6ar, + 2ar!-3:K + 6by x — 1.
220 QUOTIENTS AND SQUARE ROOTS
9. x3 + 3 x?y -f- 3 xy2 + y3 by x2 + 2 an/ + y2.
10. a? — 8 3? + 75 by a? — 5.
11. 2 a? + 19 a2b + 9 ab2 by 2a+b.
12. a;4 — 4 ar3?/ + 6 a%2 — 4 a$3 + ?/4 by x — y.
13. x4 + 4 arfy -f 6 ar2/ + 4a#J-j-?/4byar! + 2ar?/ + 2/2.
14. a;4 -f- a^y + a;?/3 + y* by a; + ?/.
15. x* + a%2 + y* by x* — xy + 1/2.
16. x* — y4 by x — y.
17. a3 + 63 + c3 — 3 a&c by a + 6 + c.
18. 2a4+ll33-26a-2 + 16a;-3by a^+7»-3.
19. x5 + 5 afy + 10 x'y2 + 10 a,-2?/3 + 5 xy*+ y5 by x2 + 2 xy + y2.
20. a,-5 - x* - 27 ar3 + 10 x2 - 30 x - 200 by a;2 - 4 x - 10.
21. 3a^ — 4a# + 8a;z — 4y2 + 8y« — 3z2 by x — 2y + 3z.
22. 9rV-4r¥ + 4rs£2-s¥by3rs-2rt + s*.
23. 9 a2b2 + 16 x2 - 4 a2 - 36 Vx2 by 3 a& + 6 bx - 2a - 4 a;.
24. a^ + a?2?/ + a^z — a;?/2 — ?/2z — yz2 by x2 — ?/z.
25. a5 + a4& + a3 - a362 - 2 a&2 + &3 by a2 + ab- b2.
26. a3 -{- 2Z3 — 2s + 3 xyz by a; + y — z.
27. a3 + &3 + 3 ab - 1 by a + b — 1.
28. 6 x5*- 11 a;4* + 23 0^ + 13^ -3 a;* + 2 by 3o* + 2.
29. a3* - 3 a2kbk + 3 a*62* - ft3* by a* - 6*.
30. 32 sia - 9 s^P + 12 sH2h - 18 saP - 17 £46 by -«•-*».
164. Since the dividend is the product of the divisor and quo-
tient, it follows that if one factor of an expression is given, the
other factor may be found by division.
SQUARE ROOTS OF POLYNOMIALS 221
EXERCISES
Obtain the factors of each of the following products, one
factor being given in each case :
Product Given factor
1. x^ — y8. x — y.
2. ar3 + y*. x + y.
3. a5— b5. a — b.
4. a5 + b5. a + b.
5. a6 + &6. a? + b2.
6. a9 + 69. a3 + 63.
7. r3 — s3. r2 + rs -{- s?.
8. r4-s4. r3 + r2s + rs2+s3.
9. r5 4- s5. r4 — ^s -f rV — rs3 -|- s4.
10. a3-12a2 + 27a + 40. a-5.
11. ar*- 5 a;4?/ + 11 arty2 a2 - 3 xy + 2 y2.
— 14 a:2?/3 — 51 xy* + 54 f/5.
12. a;4 + xPy2 + y*. aP — xy + y2.
13. a3 + 5a2-2a-24. a2 + 7a + 12.
14. a5-5a4& + 10a362 a2-2a6 + 62.
- 10 a2b3 + 5ab*- bs.
15. x5 — 5 x?y2 — 5 x2^3 + y5. x2 — 3 xy + y2.
SQUARE ROOTS OF POLYNOMIALS
165. In §§ 87, 88, we found certain trinomials which were
perfect squares, namely,
a2 + 2a& + 62 = (a + &)2, (1)
a2-2ab + b2=(a-b)2. (2)
Hence we know the square roots of all trinomials which are
in either of these forms. These trinomial squares may be used
222 QUOTIENTS AND SQUARE BOOTS
to discover a process for finding the square root of any poly-
nomial which is a perfect square.
Finding a square root may be regarded as a process of
division in which divisor and quotient are equal and both are
to be found simultaneously.
Illustrative Example. Find the square root of 4arJ+12sc?/+9?/2.
Considering the formula (1) we are to pass from the square
a2 + 2 ab + b2 to the square root a + b, and for this purpose
we write a2 + 2 ab -f b2 in the form a2 + 6 (2 a + b) and arrange
the work as follows :
Square or product, 4 x2 + 12 xy + 9 y2\2 x -f 3 y, square root.
4 x2 1st par'l product.
1st par'l divisor, 4 x
1st compl. divisor, 4 a: + Sy
\2 xy + 9 y% square minus 1st par'l prod.
12 xy + 9 y 2, 2d par'l product.
0
Supposing that 4 x2 is the a2 of the formula, a is then 2 a,
which is the first term of the root. Squaring 2 x gives 4 x2, the
first partial product. Subtracting 4 cc2 from the total product
leaves 12 #?/ + 9 y2, which is the b (2 a + b) of the formula.
Since b is not yet known, we cannot find completely either
of the factors of b (2 a + b) ; but since a has been found, we can
get the first term of the factor 2 a -f- b, viz. 2aor2-2sc = 4a;,
which is the first partial divisor. Dividing 12 xy by 4 a; we
have 3y, which is the b of the formula. Then 2a + b = 4:X + 3y
the first complete divisor.
To obtain the second partial product, b (2 a + b) or 12 xy + 9y2,
we multiply 4 x + 3 ?/ by 3 y. On subtracting, the remainder is
zero and the process ends, whence the required root is 2 x + 3 y.
It should be clearly understood that the sum of the first and
second partial products is a square, viz., the square of 2 x + 3 y,
because it has been constructed just as a? + b(2 a + b) was
formed from a + &.
SQUARE ROOTS OF POLYNOMIALS 223
EXERCISES
Find in the manner just described the square roots of the
following trinomials :
1. 36<c2-84a^ + 49?/2. 4. 81 m4+ 144 m?n2 + 64 n*.
2. 16 a2 - 40 ab + 25 b2. 5. 49 s6 - 84 s3 + 36.
3. 121 t2 + 264 ta + 144 a2. 6. 1 - 20 a4 + 100 a?.
166. The process just applied to trinomials is applicable to poly-
nomial squares having a larger number of terms, by regarding the
a of the formula at each step as representing the part of the root
already found and b as the term of the root about to be found.
Illustrative Example. Find the square root of x2 + 2 xy + y2
+ 6 sc+ 6 y + 9. The work may be arranged as follows :
Square or product, x2 + 2 xy + y2+Gx + Qy + 9\x + y + 3, square root.
x2 1st par'l product.
2xy + y2+6x+Qy + 9, 1st remainder.
2xy + y2, 2d par'l product.
6 x + 6 y + 9, 2d remainder.
6 x+ 6 y+ 9, 3d par'l product.
1st par'l div'r, 2x
1st compl. div'r, 2x + y
2d par'l div'r, 2 x+ 2 y
2d compl. div'r, 2x + 2y + 3
~ 0"
After the first two terms of the root have been found, namely,
x and y, then we consider x + y as the a of the formula and
call it a'. The new b, which we call b', is then found to be 3.
Subtracting the first and second partial products is the same
as subtracting (x-\-y)2, that is, the square of a'. Hence, the
second partial divisor, which is twice a', is 2 (x-\-y).
In case there are four terms in the root, then the sum of the first
three, when found as above, is regarded as the new a, called a".
The remaining term of the root is the new b, and is called b".
The formula (a — b)2 = a2 — 2 ab + 62 is not needed, since this may
be written (a + (- ft))2 = a2 + 2 a (- b) + (- by, which is in the
form of (a + b)2 = a2 + 2 ab + 62.
224 QUOTIENTS AND SQUARE ROOTS
EXERCISES
Find the square roots of the following polynomials:
1. x2 + y2 + z2-2xy + 2xz-2yz.
2. 9arJ + 4?/2 + z2 — 12xy + 6xz — Ayz.
3. a?-8ab + 16b2-2ac + c2 + 8bc.
4. a4 + 4a3 + 6a2 + 4a + l.
5. c4_4c3 + 6c2-4c + l.
6. ?/4 - 4 ?/2- 8 a-?/2 + 16 a + 163? +4.
7. x4 — 2 a8 + 3 a2 — 4* + 4.
8. a2 + a262 - 2 a26 + 2 ode - 2 a&2c + &2c2.
9. a4 + 4a36 + 6a2&2 + 4a&3 + &4.
10. a;4 — 4 a??/ + 6 x^y2 — 4 xy3 -f 2/4.
11. a;6 - 4 as" + 4 x4 + 6 a;3 - 12 cc2 + 9.
12. a4 + 53 a2 + 14 a3 + 28 a + 4.
13. a-2 + 16afy2 + 289 + 8a:2?/ + 34a: + 136a:?/.
14. 9 a4 + 4 a2 + 256 -12 a3 -96 a2 + 64 a.
15. a6 + 6a5 + 15a4 + 20a3 + 15a2 + 6a + l.
16. a6-6a5 + 15a4-20a3 + 15a2-6a + l.
17. 16x6+4,y2-[-l-16xiy + 8xi-4:y.
18. 25 + 49 a2 + 4 a4 -70 a -20 a2 + 28 or5.
19. 64 x6 + 192 x5 + 240 a4 + 160 ar5 + 60 x2 + 12 x + 1.
20. 4a6-12a,-5 + 13a;4-14a*! + 13av!-4a; + 4.
21. «4J4-2aV + a2-2fl26+2a6+52.
22. 16a6 + 24a5 + 25a4 + 20a3 + 10a2 + 4a + l.
23. xHf + 2 ary + 3 a;4?/4 + 4 ary + 3 tfy2 -\-2xy-\-l.
24. l + 2a; + 3ar2 + 4ar3 + 5ar4 + 4ar5 + 3a;6 + 2a;7 + ar8.
25. 9 a2 - Gab + 30 ac + 6 ad + 62- 10 6c -2 M + 25C2
+ 10cd + d2.
SQUARE ROOTS OF ARABIC NUMBERS 225
SQUARE ROOTS OF NUMBERS EXPRESSED IN ARABIC FIGURES
167. The square root of a number expressed in Arabic
figures may be found by the process just used for polynomials.
Illustrative Example. Find the square root of 5329.
In order to decide how many digits there are in the root we
observe that 102 = 1000 and 1002 = 10000 ; hence the root lies
between 10 and 100, i.e. it contains two digits. Since 802 =
6400 and 702 = 4900, we see that 7 is the largest number pos-
sible in tens' place of the root. The work is arranged as
follows :
The given square,
5329 |70 + 3, square root.
a2 = 702
4900, 1st partial product.
2 a = 2 x 70 = 140
ft = 3
2 a + b = 143
429, 1st remainder.
429 = 6 (2 a + b).
0
Having decided as above that the a of the formula is 7 tens,
we square this and subtract, obtaining 429 as the remaining
part of the power.
The first partial divisor, 2 a = 140, is divided into 429 giving
a quotient 3, which is the b of the formula. Hence the first
complete divisor, 2 a + b, is 143, and the second partial prod-
uct, b (2 a + b), is 429. Since the remainder is zero, the pro-
cess is exact and 73 is the square root sought.
EXERCISES
Find the square roots of the following :
1. 3249. 5. 6241. 9. 1849.
2. 8836. 6. 7056. 10. 7225.
3. 7569. 7. 9409 11. 3025.
4. 8281. 8. 9801. 12. 9216.
226 QUOTIENTS AND SQUARE ROOTS
168. The square of any integer from 1 to 9 contains one or
two digits ; the square of any integer from 10 to 99 contains
three or four digits ; the square of any integer from 100 to 999
contains five or six digits ; etc.
Hence it is evident that, if the digits of a number which is
a perfect square be separated into groups of two each, count-
ing from units' place toward the left, the number of groups
thus formed is the same as the number of digits in the square
root.
Illustrative Example. Find the square root of 120,409.
Since this number is divided into three groups, the first digit is in
hundreds' place. The work is arranged as follows :
Square, 12 04 09 1300 + 40 + 7 = 347, square root.
a2 = 3002 9 00 00, 1st partial product.
2a = 2 x 300=600
b= 40
2 a + b = 640
2 a' = 2 x 340 = 680
3 04 09, 1st remainder.
2 56 00 = b (2 a + b).
V= 7
2 a' + V = 687
48 09, 2d remainder.
48 09 = V (2 a' +b').
0
The first partial divisor, 2 x 300, is completed by adding the second
term of the root, 40, that is, 2 a + b — 600 + 40. At first glance it
might be supposed that the second digit is 5 instead of 4, since 600 is
contained in 30,409 50 times, but account must be taken of the addi-
tion to be made to the partial divisor, and when this is done the
quotient is 40, not 50.
In the second partial divisor, 2 a' stands for 2 times (300 + 40) = 680,
and V stands for the third digit of the root.
In case a square contains an odd number of digits the last
group at the left will have one instead of two digits.
E.g. 3 47 21 has three places in its square root, of which the first,
hundreds' digit, is the largest square in 3, namely, 1.
SQUARE BOOTS OF ARABIC NUMBEBS
227
EXERCISES
Find the square root of each of the following :
1. 294,849. 5. 3481. 9. 100,489. 13. 35,721.
2. 37,636. 6. 7569. 10. 26,569. 14. 16,641.
3. 872,356. 7. 1849. 11. 874,225. 15. 32,761.
4. 599,076. 8. 73,441. 12. 170,569. 16. 223,729.
169. Since the square of any decimal fraction has twice as
many places as the given decimal, it is evident that the square
root of a decimal fraction contains one decimal place for every
two in the square.
E.g. (.15)2 = .0225 ; (.012)2 = .000144.
Hence, for the purpose of determining the decimal places in
the root, the decimal part of a square should be divided into
groups of two digits each, counting from the decimal point
toward the right.
Illustrative Example. Find the square root of 4.6225. Ac-
cording to §§ 168, 169 the root contains one digit in the integral
part and two in the decimal part. The work is as follows :
a2 = 22
2a = 2 x 2
b =
2a + b =
4.62 25 1 2 + .1 + .05 = 2.15
4
4.0
4.1
2 a' = 2 x 2.1 = 4.2
b'= _J;05
2 a' + b'= 4.25
.62 25, first remainder.
.41 = b (2 a + b).
.21 25, second remainder.
.21 25 = b' (2 a' + b').
0
This process is also applicable for the purpose of approxi-
mating the square root of a number which is not a perfect
square
228
QUOTIENTS AND SQUARE ROOTS
Illustrative Example. Find the approximate square root of
582 to three decimal places. The solution below shows all the
steps of the work.
5 82 1 20 + 4 + .1 + .02 + .004 = 24.124
4 00
a2 = 202
2 a = 2 x 20
= 40.
b =
2a + b
' 4.
= 44.
2 a' = 2 x 24
= 48.
b' =
.1
2 a' + V =
48.1
2 a" = 2 x 24.1
= 48.2
b" =
.02
2 a" + b" =
48.22
182,
176
6.00,
first remainder.
= b (2 a + b).
second remainder.
4.81 =b' (2 a' + 6').
1.1900, third remainder.
.9644 = b" (2 a" + b").
2 a"' = 2 x 24.12 = 48.24
V" = .004
2 a'" + V" = 48.244
.225600, fourth remainder.
.192976 = V" (2 a'" + 6'").
.032624
The decimal points are handled exactly as in division of deci-
mals in arithmetic, the chief care being needed in forming the
divisors.
170. Evidently the process in this example may be carried
on indefinitely. 24.124 is an approximation to the square root
of 582. In fact, the square of 24.124 differs from 582 by the
small fraction .032624. 24.12 is the nearest approximation
using two decimal places. If the third figure were 5 or any
digit greater than 5, then 24.13 would be the nearest approxi-
mation using two decimal places. Hence three places must be
found in order to be sure of the nearest approximation to two
places.
SIMPLIFYING RADICALS 229
EXERCISES
Find the square
roots of the folio
wing,
correct to two deci
mal places :
1. 387.
7. 2.
13. .02.
2. 5276.
8. 3.
14. .003.
3. 2.92.
9. 5.
15. .5.
4. 27.29.
10. 7.
16. .005.
5. 51.
11. 8.
17. .307.
6. 3.824.
12. 11.
18. 200.002.
SQUARE ROOTS OF FRACTIONS EXPRESSED IN ARABIC FIGURES
171. Since in arithmetic the product of two fractions is
found by multiplying their numerators and their denominators,
a fraction is squared by squaring its numerator and its denom-
inator separately.
Hence, to extract the square root of a fraction, we find the
square root of its numerator and its denominator separately.
E.g. Vif = |, since f x | = if.
However, in approximating the square root of a fraction whose
denominator is not a perfect square, if the final result is to be
reduced to a decimal, the direct use of this method is cumber-
some since it usually involves the approximation of two square
roots and always necessitates dividing by a long decimal fraction.
E.g. Vf = a/2 ■+■ V3 = 1.4142 + 1.7321, in which we are now
obliged to divide by 1.7321.
This difficulty may be avoided in either of two ways :
1. The fraction may be reduced to a decimal before the
root is approximated.
E.g. Vy = V.666 • • • = .8165.
230 QUOTIENTS AND SQUARE BOOTS
2. The denominator of the fraction may be made a perfect
square before approximating the root.
y *3 *9 3 3
By either of these two processes only one square root is
approximated, and there is only a short division by 3 instead
of a long division by 1.7321.
It is clear that any fraction can be changed into an equal fraction
whose denominator is a perfect square by multiplying numerator and
denominator by the proper number.
E-ff- J = h I = M> j = il> etc. If a fraction is given in the form
— , it may be written Vf = vf = |V3. In like manner,
V3
J__ = 7. _J_ = 7 JT=7 JH=7 ^
vn vn vn Xi-Ji ii
EXERCISES
Find approximately correct to two decimal places the follow-
ing square roots.
In the first ten obtain the results in three different ways :
(a) Find the root of each numerator and denominator separately ;
(6) reduce each fraction to a decimal ; (c) reduce each fraction
so as to make its denominator a perfect square.
•In the remaining exercises use methods (6) and (c) only.
In each case compare the results obtained.
5.
v|.
6. VJ.
ii. V3?.
16.
Vf.
Vf.
Vf
7. V*f.
8. vjf
12. V^.
13. V|.
17.
1
V5*
9. V|.
io. Vf
14. Vf.
15. Vf
18.
5
V13
19.
3. V*. 8. V*f. 13. Vf. V5 20.
21.
3
V7*
7
y/Tf'
Vf
SIMPLIFYING RADICALS 231
172. Principle XVIII may be used to advantage in approxi-
mating the square roots of certain integral numbers.
E.g. suppose V2 has been computed, and VS is desired. It is
unnecessary to compute the V8 directly, for by XVIII,
VS= vTT2 = y/i- V2 = 2V2.
This sort of simplification is possible whenever the number
under the radical sign can be resolved into two factors, one of
which is a perfect square.
E.g. suppose the V5 to have been computed, then
VT25 a V25V5 = V25 • V5 = 5 V5.
In like manner, Va568 may be written
Va462 • ab = Va*P ■ Vab = a?bVab.
Definition. A radical expression is said to be simplified
when the number under the radical sign is in the integral form
and contains no factor which is a perfect square.
E.g. the simplified forms of Vl^5> VaW VT — ~> are respectively,
5 V5, a% Vab, i V5, i >/3.
EXERCISES
Given V2 = 1.4142, V3 = 1.7321, V5 = 2.2361, compute
the following, correct to three places of decimals, without
further extraction of roots :
1.
V80.
6.
V2-3.
11.
V27 + V|.
2.
vj.
7.
V72.
12.
V45 + VJ.
3.
VJ-
8.
V98.
13.
V50-VJ + V8.
4.
V48.
9.
V363.
14.
V48+V75-V3.
5.
V75.
10.
VI25.
15.
V32+V72-V18.
232 QUOTIENTS AND SQUARE ROOTS
Simplify the following :
16. V32a26. 19. VioVy^. 22. V500 x7 asb.
17. VSlxW2. 20. V63 b(fd\ 23. -yJ'Stf + Kxy + dy2.
18. VoO aW. 21. V900 o6V. 24. V8«2-12y2.
25. V32a2-64a6 + 32 62. 26. VF25 x2 + 250 ajy + 125 y\
27. Find approximately to four decimal places the sides of
a square whose area is 120.
28. Approximate to four decimals the side of a square having
an area equal to that of a rectangle whose sides are 15 and 20.
29. How many rods of fence are required to fence a square
piece of land containing 50 acres, each acre containing 160
square rods?
30. A square checkerboard has an area of 324 square inches.
What are its dimensions ?
173. In adding or subtracting expressions containing radi-
cals it is always best to first reduce each radical expression to
its simplest form, since this often gives opportunity to com-
bine terms which are similar with respect to some radical
expression.
Ex.1. V32+ V72- Vl8 = 4V2 + 6V2-3V2 = 7V2by
Principles XIII, I, and II.
Ex.2. VJ+Vi2-V| = |V3 + 2v/3-|V3
= a + 2-i)V3=l|V3.
EXERCISES
Simplify each of the following as far as possible without
approximating roots.
1. V27 + 2V48-3V75.
2. V20 + V125- VT80.
3. 3V432-4V3 + V147.
APPLICATIONS OF SQUARE BOOT 233
4. 3V2450-25V2 + 4V13122
5. 3y2^xJz + 2^xJz?-yziyJ-2
6. V4 x*y + V2o a-?/3 — x -\/xy
7. Vaa-2 — bar + V4 cwV — 4
8. 4A/|-fV^-3V27.
9. 2 V| + V60 + Vf.
10. 5 V3- 2 V48 + 7 VIM.
11. Va3 - a2b - Vai* - b3 - V(a + 6)(a2 - 62).
12. A/a + 3 V2~cT- 2 V3a + V4~a- V8a+ Vl2a.
13. Va-3 -\-2xry + xy2 — Va? — 2 x2y + a;?/2 — V4 xy\
14. Vr - s + Vl6 r - 16 s + Vrt2 - st2 - V9(r— *).
15. V(m — w)2a + V(m + n)2a — -\/am2 + Va (1 — ra)2 —
16. V32 x2y* + V162 arY - V512 arfy4 + V1250 tfy*.
APPLICATIONS OF SQUARE ROOT
174. Some of the most interesting and useful applications of
the square root process are concerned with the sides and areas
of triangles.
The fact that the sum of the squares on the two sides of a
right triangle equals the square on the hypotenuse was used
in Chapter VI. (Pythagorean Proposition, page 206.)
If a and b are the lengths of the sides, and c the length of
the hypotenuse, all measured in the same unit, this propo-
sitionsays: <? = a2 + b\ (1)
Hence, by S, a2 = <? - b\ (2)
and b2 = c2 - a2. (3)
234 QUOTIENTS AND SQUARE EOOTS
Taking the square root of both sides in each of these
equations,
c=Va2 + 62. (4)
a = Vc2^^62. (5)
b = V^^a2. (6)
The negative square root is omitted here, as a negative
length cannot apply to the side of a triangle. By these formu-
las, if any two sides of a right triangle are given, the other
may be found.
E.g. if a = 4, b = 3, then, by (4),
c = V42 + 32 = Vl6 + 9 = V25 = 5.
If c = 5, b = 3, then, by (5),
a = V52 - 32 = V25 - 9 = Vl6 = 4.
If c = 5, a = 4, then, by (6),
b = Vo2-42 = V25 - 16 = \/9 = 3.
Illustrative Problem. If the two sides of a right triangle are
8 and 12, find the hypotenuse correct to two decimal places.
Solution. We have c - Va2 + b2 = V64 + 144 = V208,
V208 = VI6TT3 = VIS • V13 = 4VI3 = 4(3.605)= 14.420.
PROBLEMS
In solving the following problems, simplify each expression
under the radical sign before extracting the root. Find all
results correct to two decimal places. In each case construct
a figure.
1. The sides about the right angle of a right triangle are
each 15 inches. Find the hypotenuse.
2. The hypotenuse of a right triangle is 9 inches and one
of the sides is 6 inches. Find the other side.
APPLICATIONS OF SQUARE ROOT 235
3. The hypotenuse of a right triangle is 25 feet and one
of the sides is 15 feet. Find the other side.
4. The hypotenuse of a right triangle is 7 rods and one of
the sides is 5 rods. Find the other side.
5. The hypotenuse of a right triangle is 12 inches and the
two sides are equal. Find their length.
Let s equal the length of one of the equal sides.
Then s2+s2=144.
2 s2 = 144.
s2 = 72.
s = V72 = 6V2= 6 x 1.414 = 8.484.
6. The hypotenuse of a right triangle is 30 feet and the
sides are equal. Find their length.
7. The hypotenuse of a right triangle is h and the sides
are equal. Find their length. Solve 5 and 6 by means of
the formula here obtained.
8. The diagonal of a square is 8 feet. Find its area.
9. The diagonal of a square is d. Find an expression in
terms of d representing its area.
10. The side of an equilateral triangle is 6 inches. Find
the altitude.
A line drawn from a vertex of an equi-
lateral triangle perpendicular to the base meets /
the base at its middle point. Hence this /
problem becomes : the hypotenuse of a right /
triangle is 6 and one side is 3. Find the /
remaining side.
11. The side of an equilateral triangle is 10. Find the alti-
tude.
236 QUOTIENTS AND SQUARE BOOTS
12. The side of an equilateral triangle is s. Find the
altitude.
This is equivalent to finding a side of a right triangle whose hypote-
nuse is s, the other side being -. Let a equal altitude.
Then • = V"-(§)'-V^
;V4-^=V¥!=V^
-V
j-VS=fv^.
This formula gives the altitude of any equilateral triangle in
terms of the side. By means of this formula solve 11 and 12.
It is interesting to notice that the square root of 3 is the only
root required in finding the altitude of any equilateral triangle
whatever.
13. Find the altitude of an equilateral triangle whose side
is 41. Substitute in the formula under 12.
14. Find the area of an equilateral triangle whose side is 5.
Since the area of a triangle is \ the product of the base and alti-
tude, we first find the altitude by means of the formula under 12, and
then multiply by J the base.
15. Find the area of the equilateral triangle whose side is s.
s2 —
Show the result to be — V3.
4
16. If the area of an equilateral triangle is 16 square
inches, find the length of the side.
Let s equal the length of the side. Then bv the formula derived
s2
under 15, 16 = — V3.
Hence (§§ 171, 172), s2 = -^4 = — V3 = 21.33 x 1.732.
V3 3
APPLICATIONS OF SQUABE BOOT 237
17. The area of an equilateral triangle is 50 square inches.
Find its side and altitude.
18. The area of an equilateral triangle is a square inches.
Find the side.
Solve the equation a = — V3 f or s, and simplify the expression,
finding s2 = -J , and s = V ~V~ = - V3aV3.
V3 0 6
19. The area of an equilateral triangle is 240 square inches.
Find its side. (Substitute in the formula obtained under 18).
20. Find the area of a regular hexagon whose
side is 7.
A regular hexagon is composed of 6 equal equi-
lateral triangles, whose sides are each equal to the
side of the hexagon (see figure). Hence this prob-
lem may be solved by finding the area of an equi-
lateral triangle whose side is 7, and multiplying the result by 6.
21. Find the area of a regular hexagon whose side is s.
(Solve 20 by substituting in the formula obtained here.)
22. The area of a regular hexagon is 108 square inches.
Find its side.
If the area of the hexagon is 108 square inches, the area of one of
the equilateral triangles is 18 square inches. Hence this problem can
be solved like 18.
23. The area of a regular hexagon is a square inches. Find
its side. (Solve 22 by substituting in the formula obtained
here.)
24. Find the radius of a circle whose area is 9 square inches.
The area of a circle is found by squaring the radius and multiply-
ing by 3.1416. The number 3.1416 is approximately the quotient
obtained by dividing the length of the circumference by the diameter
of the circle. This quotient is represented by the Greek letter tt
238
QUOTIENTS AND SQUARE BOOTS
(pronounced pi). In this chapter we use 3^ as an approximation to ir.
This differs from the real value of it by less than .0013, and hence is
accurate enough for most purposes. If a represents the area of a
circle, the above rule may be written
a
irr'.
Hence if a = 9,
and
9 = 9^ _ 63_ _ 2 g63
j 3f 22 '
25. Find the radius of a circle whose area is 68 square feet.
26. Find the radius of a circle whose area is a square feet.
We have
Hence
a = irri, or rz =
1 7T \ 7H IT
In problems stated in terms of letters, the results, of course, cannot
be reduced to a decimal. In such formulas it is best not to replace
the letter ir by any of its approximations.
27. Find the sum of the areas of a circle
of radius 6 and the square circumscribed
about the circle.
The area of the circle is 627r = 36 ir, and the
area of the square is 4 • 62 = 4 • 36 ; i.e. the
square contains 4 squares whose sides are 6.
The sum of the areas is
4 • 36 + 36 7T = (4 + tt) 36 = (4 + 3f) 36.
28. Find an expression for the sum of the areas of a circle
of radius r and the circumscribed square. (Solve 27 by sub-
stituting in the formula here obtained.)
29. If the sum of the areas of a circle and the circumscribed
square is 64, find the radius of the circle.
By the formula obtained under 28,
64= (4 + 7r)r2 = ^r2.
Hence,
r2 = 64^ = 8.96,
r = V8.96 = 2.99.
APPLICATIONS OF SQUARE ROOT
239
30. If the sum of the areas of a circle and the circumscribed
square is 640 square feet, find the radius of the circle.
31. The sum of the areas of a circle and the circum-
scribed square is a. Find an expression representing the
radius of the circle. (Replace w by 3| before simplifying.)
32. If the radius of a circle is 12, find
the difference between the area of a circle
and the circumscribed square.
33. If the radius of a circle is r, find the
difference between the area of the circle and
the circumscribed square. (Solve 32 by
substituting in the formula obtained here.)
34. If the radius of a circle is 16, find the
area of the inscribed square. (This is the same problem as finding
the area of a square whose diagonal is 32. See problems 8 and 9.)
35. If the radius of a circle is r, find an expression repre-
senting the area of the inscribed square.
(This is problem 9, the hypotenuse being
2r.)
36. If the radius of a circle is 12, find
the difference between the area of the circle
and the area of the inscribed square.
37. If the radius of a circle is r, find
an expression representing the differences
between the areas of the circle and the inscribed square.
38. The radius of a circle is 10. Find the area of an
inscribed hexagon. See note, problem 20.
39. The radius of a circle is 6. Find
the difference between the areas of the
circle and the inscribed hexagon.
40. Find an expression representing the
difference between the areas of a circle with
radius r and the inscribed regular hexagon.
240 QUOTIENTS AND SQUARE ROOTS
SOLUTION OF QUADRATIC EQUATIONS BY MEANS OF SQUARE ROOT
175. Illustrative Problem. A rectangular field is 18 rods
longer than it is wide and its area is 50 acres. What are its
dimensions ?
Let x = the width of the field.
Then, x + 18 = the length of the field.
And x (x + 18) = the number of square rods in the field.
Hence, rr2 + 18 x = 8000, (1)
or x2+ 18 x- 8000 = 0. (2)
We are not able to factor the left member of the equation
by any method thus far known, and hence we cannot solve the
equation as in § 145. We therefore proceed to study a general
method of solving quadratic equations.
Consider the equation in the form (1) above. We seek a
number k2 such that x2 + 18 x + k2 shall be a perfect square.
By § 134 the middle term of a trinomial square is twice the
product of the square roots of the two square terms. Hence,
18 x=2 kx, that is, k must equal 9.
Hence, adding 92(=k2) to each member of (1),
a^+18.T + 92 = 8000 + 92, (3)
or, x2 + 18 x + 81 = 8081. (4)
Since the left member is now a perfect square, we may
extract the square root of both sides, approximating the root
on the right.
Hence, x + 9 = ± V8081 = ± 89.89,
giving x = - 9 + 89.89 = 80. 89,
and also, x = - 9 - 89.89 = - 98.98.
In this case the negative result is not applicable to the
problem. Hence the width of the field is 80.89 rods, which
is correct to two decimal places.
QUADRATIC EQUATIONS 241
Illustrative Example. Solve the equation :
x2 - 12 x + 42 = 56. (1)
By S, x2-12x = 14. (2)
By A , x2 - 12 x + k2 = 14 + F. (3)
By § 175, - 12z = 2 fee or *= - 6.
Hence, x2 - 12 x +(- 6)2 = 14 + 36 = 50. (4)
Taking square roots, x — 6 = ± V50 = ± 5 V2. (5)
By ,4, * as 6± 7.071. (6)
Hence x = 6 + 7.071 = 13.071,
and also x = 6 - 7.071 = - 1.071.
176. This process is called solving the quadratic equation by
completing the square, since in each case a number is added to
both sides which makes the left member a trinomial square.
Since the process always involves extracting the square root
in order to find the value of the unknown, the two solutions
of a quadratic equation are commonly called the roots of the
equation. By analogy the solutions of any equation are some-
times called its roots.
EXERCISES
In solving the following quadratic equations the result may
in each case be reduced so that the number remaining under
the radical sign shall be 2, 3, or 5. (§ 172.) Use these
square roots only.
1. x2-4a; = 8. 9. a2- 4a; = 16.
2. JB»=3-6aJ. 10. 2x =14 + 4 x.
3. 4 a; = 16 -a8. 11. 24 = 3 x2 + 12 x.
4. a;2 + 6 a; = 9. 12. 69 - 18 x = 3 a;2.
5. a;2 + 6a; = ll. 13. 84 + 24 x = 12 x>.
6. af-^x = 12. 14. 25-3^ = 5*;.
7. a?-8a;=-14. 15. x2 + ix = 2.
8. aj8 = 2a;+l. 16. a^-fa; = -2^.
242 QUOTIENTS AND SQUARE ROOTS
177. In case the coefficient of x2 is not unity, both members
may be divided by this coefficient.
Example. Solve 3 x2 + 8 x = 4. (1)
By A i»+J* = f (2)
By A, *P + |* +*» = ! + *■. (3)
By § 175, §x = 2£xor&=f.
Hence, **+ t* + (*)? = ♦ + V = ¥• (4)
Taking square roots, * + f = ± V-^- (5)
Hence, x= -|±|V7^ (6)
and the two roots are x = 0.43,
and x = — 3.10.
The preceding example may also be solved as follows :
Multiplying each member of (1) by 4 • 3 = 12,
then, 36 x2 + 96 x = 48.
By A, 36 x2 + 96 x + t* = 48 + k2,
where - 12 kx = 96 x or & = 8.
Hence, 36 xa + 96 x + (8)2 = 48 + 64 = 112,
and 6 x + 8 = ± VTl2 = ± 4 VT.
Therefore x = - f ± ] VT.
178. The advantage of this solution is that fractions are
avoided until the last step, and the value of k is found to be
the same as the coefficient of x in the given equation. This
may always be accomplished by multiplying the members of
the given equation by four times the coefficient ofx2.
In the solution of the following equations only the square
roots V2, V3, V5, V6, V7, need be used. In all cases where
the roots are integers or exact fractions the solution may also
be obtained by factoring as in § 145.
1. 2x* + Zx=2. 4. 6x + l = -Sx2. 7. 2x2-3x = U.
2. Sx2 + 5x = 2. 5. 2ar2 = 5z + 3. 8. 3^ = 9 + 2^.
3. 3z = 9-2ar!. 6. ±x = 2x2-l. 9. 4x2 = 2x + l-
QUADRATIC EQUATIONS 243
10. §x-l = 3x2. 23. 2z-l = -4 a2. 36. 6^-12^ = 2.
11. 2^ + 4^ = 23. 24. 5x»+16:r=-2. 37. 3x2 + 2x = 5.
12. 3^-7 = 40;. 25. 4a* + l = 8ar. 38. 2 + 3x = 2x>.
13. 2x*-5 = 3x. 26. 2^-3^ = 20. 39. 8« + l = -4arJ.
14. 4ar> = 6a;-l. 27. 2arJ-3 = -5z. 40. 8 + 4a; = 3a;2.
15. 2x = l-5x2. 28. 3ar + 4a; = 8. 41. 10 + 4^ = 5^.
16. 3*-20«-2a£ 29. 10-4:r = 5a;*. 42. 2 + 5^ = 3^.
17. 2x + 3x2 = 9. 30. l+4x2 = -6x. 43. 3x-l± = 2x2.
18. 4a^-l = 3ic. 31. 5-3a = 2z2. 44. 3x2-2x = 5.
19. 4z = 7-2;r!. 32. 7 + 4x = 2xi. 45. 2a^+4*»l.
20. 2x+l = bx2. 33. 6 a2 + 12 a = 2. 46. 3x-l = -4ors.
21. 3x2 + 4x=7. 34. 6x2-12x=-2. 47. 23 + 4 a = 2 a2.
22. 3z + 9 = 2x2. 35. &x2+12x=-2. 48. Zx-l = 2x2.
179. Solution by Formula. Solve the equation
ax2 + bx + c = 0. (1)
By S, M,
4 a2x2 + 4 aftx = — 4 ac.
(2)
By ,4,
4 a2x2 + 4 aftx + £2 = - 4 ac + k2,
(3)
where
4 a£x B 4 aix or k = b.
Hence
4 a2x2 + 4 a&x + b2 = b2 - 4 ac.
.(4)
Taking square roots, 2 ax + b = ± \/&2 — 4 crc.
(8)
By 5, A
_ - 6 ± Vb2 - 4 ac
(6)
Calling the two values of x in the result xx and x2 we have,
_-6+V62-4ac _-6-V62-4gC
ri~~ ~2«r~ ~; Jr2~~ ~2^~ "7
244 QUOTIENTS AND SQUARE ROOTS
Any quadratic equation may be reduced to the form of (1)
by simplifying and collecting the coefficients of x2 and x. Hence
any quadratic equation may be solved by substituting in the
formulas just obtained.
Ex. 1. Solve x2-4x + l = 0.
In this case a = 1, b = —4, c = 1.
Hence
, _-(-4)±V(-4)2_4.1.1
2-1
x1 = 2 + VS
From which
and
x2 = 2 - V3.
Ex. 2. Solve
3a:2 + 16 x - 12 = 0.
Here
a = 3, b = 16, c = - 12.
Here
_ _ 16 ± V(16)2-4-3(- 12)
2-3
From which
Zj = §, and x2 — — 6.
180. A quadratic equation may be proposed for solution
which has no roots expressible in terms of the numbers of
arithmetic or algebra so far as yet studied.
Example. Solve x2 + 4 x = - 8. (1)
By .4, a:2 + 4x + 4=-4. (2)
Taking square roots, x + 2 = ± V— 4. (3)
V— 4 is unknown to us as a number symbol, since there is no num-
ber thus far considered whose square equals — 4. (See Principle XI.)
Such symbols are defined and used in the Advanced Course, and are
called imaginary numbers. For us any quadratic equation which gives
rise to such a solution is to be interpreted as stating some impossible
condition.
EXERCISES
1. 7-3^ = 5 x2. 4. 12 -51 a = 36 + 6^.
2. 51^-33 = 3^. 5. 5aj + a? + 8 = 0.
3. Ux + 8-x*=52-3x2. 6. 5x*-31x=-§.
QUADRATIC EQUATIONS 245
7. x2-8 + 3a;=-15a;. 10. 37 - 4 x2- 12x = 79 -5a*
8. llar9-49a: + 57 = 0. 11. 10 a? + 41 + 7 a: = 44.
9. 3a^ + 18-16x=5. 12. 45 + 3^-85-2^ = 0.
MISCELLANEOUS QUADRATICS
Solve as many as possible of the following equations by fac-
toring. When this is not convenient, use the formula of § 179,
or complete the square independently in each case. Show
which equations state impossible conditions. Approximate all
square roots to two decimal places.
1. x2 + 11 x = 210. 21. 2a:2 + 3a;-3 = 12a;+2.
2. 5^-3^ = 4. 22. 3^-7^ = 10.
3. 7x + 3 x2 -18 = 0. 23. 17a + 31 + 2ar° = 0.
4. 2= 5 a; + 7 a* 24. 18 - 41 a = 3 + a*
5. 6z-lla:2 = -7. 25. 10 x + 25 = 5 - 2 x -x2.
6. -51+42a;-3ar! = 0. 26. 3a-59 + ar> = 0.
7. 3ar! + 3a: = 2x- + 4. 27. 5ar+7z-6 = 0.
8. 13-8a + 3ar = 0. 28. x>+A2=lx.
9. 2^ + 11 a=32 a; -a?- 27. 29. 8 a; -5 a?2 = 2.
10. 176 + 3a;-a;2 = 2x. 30. 5^ + 3^-22 = 0.
11. x2 + 6x-54 = 0. 31. 50 +20 ^ + ^ = 5 x.
12. 5 x2 + 9 x + 12 = 4 x2 + a;. 32. x2 + x + 4 = 0.
13. 2a*-4x-25 = 0. 33. 20 05+2 3? + 42=33 a> + a*
14. 7.^ + 11 z = 6. 34. 17x-3x-2=-6.
15. 2z2-ll£ + 5 = 0. 35. 8a? + 5a?2=-2.
16. 2ar!-lla; = 6. 36. 10 + 15 x + a;2 = 26 x.
17. 25 x - 95 = x2. 37. 3 x2 - 2x -7 = 0.
18. 11 x2 - 42z = 2. 38. 5x2-9aj-18 = 0.
19. a;2- 8 a;- 4 = a?-22. 39. 7 a; -7 a;2 + 24 = 0.
20. 8a;2 + 5a; = -8. 40. 31 +2 a? + a;2 = 0.
41. 7z2 + 7a:-5a;2 + 20 = ar2-2a; + 2.
246 QUOTIENTS AND SQUARE BOOTS
181. The solution of two equations in two variables, one of
which is linear and the other quadratic, can be reduced to the
solution of a quadratic equation in one variable. (See § 149.)
Example. Solve f x + y = 3, (1)
(3x2-y2 = U. (2)
From (1), y = 3-x. (3)
Substituting in (2) and reducing,
2x2+6x-23 = 0. (4)
Substituting in formula § 179, * = - 6 ± V36 - 4 -2(- 23)^ _
4
Hence xl = 2.21 and x2 = — 5.21.
Substituting these values of x in (1) we have as the approximate
roots,
= 2.21 1
= 0.79 J
yi = 0.79 J amS2 = 8.21 |
yx and y2 are here used to designate the values of y which correspond
to xx and x2 respectively.
In this manner solve the following equations simultaneously,
finding in each case two pairs of roots. In the case of roots
which are neither integers nor exact fractions, find the approxi-
mate results to two places of decimals.
f*-y=l. K_2/! = 3
' lar, + w8 = 13.
#2 = 13. 6- 9 4
[x — y = 4.
x + y=9. J
^ + 2/2 = 41. [x-y = l.
3- W = 42. l» 16
. j3x-y = 5. o \2x+y = 5.
fa: + 4y = 26. f*-y»a
' laj,-«' = ll. ' 1^-3^ = 13.
!:
QUADRATIC EQUATIONS 247
10. J*-3^1- 16. J* + 2/=9-
f3a-4</ = l. ly-2x = 5.
* 1 ic2 — y2 = 24. ' l?/2-3a;y = 16.
fz + 2/ = 4. f22/-3a,- = 0.
' l2^-3^4-2/2 = 8. ' l?/2 + a2 = 52.
13 J*-*-*- 19 \y-2x = 5.
\±xt + 2xy-y2 = 19. ' larJ + y2 = 40.
f5* + y = 12. 2Q f*-4y = 12.
12^-3^ + ^ = 0. I3a2 + 2a:y-6y = 44.
|x-22/ = 3. Jy-3a = f.
' l2y2-a2 = 4. " 12^-3^ = 4.
PROBLEMS
In each problem find the two roots of the quadratic equation
and determine whether both are applicable to the problem :
1. The area of a window is 2016 square inches and the length
of the frame is 180 inches. Find the dimensions of the window.
2. The area of a rectangular city block, including the side-
walk, is 19,200 square yards. The length of the sidewalk when
measured on the side next the street is 560 yards. Find the
dimensions of the block.
3. A farmer starts to plow around a rectangular field which
contains 48 acres. The length of the first furrow is 376 rods.
Find the dimensions of the field.
4. A rectangular blackboard contains 38 square feet and
the length of the molding is 27 feet. Find the dimensions of
the board.
5. A park is 120 rods long and 80 rods wide. It is decided
to double the area of the park, still keeping it rectangular, by
adding strips of equal width to one end and one side. Find
the width of the strips.
248 QUOTIENTS AND SQUARE ROOTS
6. A fancy quilt is 72 inches long and 56 inches wide.
It is decided to increase its area 10 square feet by adding a
border. Find the width of the border.
7. A city block is 400 by 480 feet when measured to the outer
edge of the sidewalk. At 4 cents per square foot it costs $ 416.64
to lay a sidewalk around the block. Find the width of the walk.
8. A farmer starts cutting grain around a field 120 rods
long and 70 rods wide. How wide a strip must he cut to
make 12 acres ?
9. The sides of a right triangle are 6 and 8 inches respec-
tively. How much must be added to each side so as to increase
the hypotenuse 10 inches ?
10. A rectangular lot is 16 by 12 rods. How wide a strip
must be added to one end and one side to obtain a rectangular
lot whose diagonal is 1 rod greater ?
11. A picture is 15 inches by 20 inches. How wide a frame
must be added to increase the diagonal 3 inches ?
12. An athletic field is 800 feet long and 600 feet wide.
The field is to be extended by the same amount in length and
width so that the longest possible straight course (the diagonal)
shall be increased by 100 feet. What will be the dimensions
of the new field ?
13. A starts north from a certain place going 4 miles per
hour and B starts east from the same place at the same time
going 3 miles per hour. In how many hours will they be
16 miles apart, the earth's surface being considered as a plane?
Let t equal the required number of hours.
Then (4 02 + (3/)2 = 162 = 256.
16 P + 9 fi = 256,
25 fi = 256.
5t= ±16.
t = ± 3£.
The solution t = — 3£ is not applicable to this
problem.
QUADRATIC EQUATIONS
249
14. In the preceding problem if A goes 5 miles per hour and
B 4 miles per hour, in how many hours will they be 24 miles
apart ?
15. A rectangle is 12 inches wide and 16 inches long. How
much must be added to the length to increase the diagonal
4 inches ?
Let x = number of inches to be added to the length. The diagonal
of the original rectangle is Vl22 + 162 = 20. Hence the diagonal of
the required rectangle is 24.
Then 122 + (16 + x)2 = 242,
or x2 + 32 x - 176 = 0.
Solving, x, = - 16 + 12 V3 = 4.78,
and x2= - 16-12 V3 = -36.78.
The negative solution obtained here may be taken to mean that if
the rectangle is extended in the opposite direction from the fixed
corner, we shall get a rectangle which has the required diagonal. See
the figure.
16. How much must the width of the rectangle in problem
15 be extended so as to increase the diagonal by 4 ?
17. A trunk 30 inches long is just large enough to permit an
umbrella 36 inches long to lie diagonally on the bottom. How
much must the length of the trunk be increased if it is to
accommodate a gun 4 inches longer than the umbrella?
18. A rectangle is 21 inches long and 20 inches wide. The
length of the rectangle is decreased twice as much as the
width, thereby decreasing the length of the diagonal 4 inches.
Find the dimensions of the new rectangle.
250 QUOTIENTS AND SQUARE ROOTS
19. In a rectangular table cover 24 by 30 inches there are two
strips of drawn work of equal width running at right angles
through the center of the piece. What is the width of these
strips if the drawn work covers one-tenth of the whole piece ?
20. A certain university campus is 100 rods long and 80
rods wide. There are two driveways running through the
center of the campus at right angles to each other and parallel
to the sides. What is the width of these driveways if their
combined area is 356 square rods ?
21. A farm is 320 rods long and 280 rods wide. There is a
road 2 rods wide running around the boundary of the farm and
lying entirely within it. There is also a road 2 rods wide
running across the farm parallel to the ends. What is the area
of the farm exclusive of the roads ?
22. A rectangular park is 480 rods long and 360 rods wide.
A walk is laid out completely around the park and a drive
through the length of the park parallel to the sides. What is
the width of the walk if the drive is three times as wide as
the walk and the combined area of the walk and the drive is
3110 square rods ?
23. The sum of the sides of a right triangle is 18 and the
length of the hypotenuse is 16. Find the length of each side.
24. The length of a fence around a rectangular athletic field
is 1400 feet, and the longest straight track possible on the field
is 500 feet. Find the dimensions of the field.
Using 100 feet for the unit of measure the equations are
ix + y = 7,
\x* + y* = 25.
25. The difference between the sides of a right triangle is 8
and the hypotenuse is 42. Find the lengths of the sides.
26. A. room is 5 feet longer than it is wide and the distance
between two opposite corners is 25 feet. Find the length and
width of the room.
QUADRATIC EQUATIONS 251
27. One side of a right triangle is 8 feet, and the hypotenuse
is 2 feet more than twice the other side. Find the length of its
hypotenuse and of the remaining side.
28. A vacant corner lot has a 50-foot frontage on one street.
What is the frontage on the other street if the distance between
opposite corners along the diagonal is 110 feet less than twice
this frontage.
In an old Chinese arithmetic said to have been written about 2600
B.C., we find the following two problems in each of which the square
of the unknown occurs but cancels.
29. In the middle of a square pond whose sides are 10 feet
there grows a reed which reaches 1 foot above the water.
When the reed is bent over to the side of the pond, it just
reaches the top of the water. How deep is the water ?
30. A bamboo reed 10 feet high is broken off so that the top
reaches the ground just three feet from its base. How far
from the ground is the reed broken off ?
31. The sum of the squares of two consecutive integers is
13,945. Find the numbers.
32. The product of two consecutive integers is 4422. Find
the numbers.
33. A square piece of tin is made into an open box, contain-
ing 864 cubic inches, by cutting out a 6-inch square from each
corner of the tin and then turning up the sides. Find the
dimensions of the original piece of tin.
34. A rectangular piece of tin is 8 inches longer than it is
wide. By cutting out a 7-inch square from each corner and
turning up the sides, an open box containing 1260 cubic inches
is formed. Find the dimensions of the original piece of tin.
35. By cutting out a square 8 inches on a side from each
corner of a sheet of metal and turning up the sides, we obtain
an open box such that the area of the sides and ends is 4 times
the area of the bottom. Find the dimensions of the original
sheet if it is twice as long as it is wide.
252
QUOTIENTS AND SQUARE BOOTS
36. An open box whose bottom is a square has a lateral area
which is 400 square inches more than the area of the bottom.
Find the other dimensions of the box if it is 10 inches high.
(By lateral area is meant the sum of the areas of the four
sides.)
37. A box whose bottom is 4 times as long as it is wide
has a lateral area 600 square inches less than 4 times the area
of the bottom. Find the dimensions of the bottom if the box
is 6 inches high.
38. A train approaching Chicago from the south at the
rate of 50 miles per hour is 75 miles away when a train
starts west from Chicago at the rate of 25 miles per hour.
How long after the second train starts will they be 50 miles
apart measured diagonally across the country ?
If t is the number of hours required, then (75 - 50 1)2 + (25 <)2= (50)2.
This may be written : (25)2 • (3 -2 02+ (25)2 • P = 22 - (25)2.
Hence dividing both members by (25) 2, we have (3 — 2 t)2 + t2 = 4.
39. An automobile run-
ning northward at the rate
of 15 miles per hour is 20
miles south of the inter-
section with an east and
west road. At the same
time another automobile
running westward on the
cross-road at the rate of 20
miles per hour is 15 miles
east of the crossing. How
far apart (diagonally) will
they be 15 minutes later ?
One hour later ?
40. Under the condi-
tions of problem 89 how
long after the time speci-
4/ C
/ CO
/ <±
/ 15-3 n
15
s J
15-r2 /
QUADRATIC EQUATIONS 253
fied will the automobiles be 10 miles apart? Is there more
than one such position?
41. What are the rates of motion of the automobiles in
problem 39 if one hour later they are 5 miles apart and 3
hours later they are 35 miles apart? (See the figure.)
If the rates of the automobiles are rx and r2, then after 1 hour we
have (20-r1)2 + (15-r2)2 = 52, (1)
and after 3 hours we have (20 - 3 r,) 2 + (15 - 3 r2)2 = 352. (2)
Simplifying (1) and (2), r* + r22 - 40 rx - 30 r2 + 600 = 0, (3)
and 9 rt2 + 9 r22 - 120 rx - 90 r2 -600 = 0. (4)
Multiplying (3) by 9, subtracting from (4), and simplifying,
4 rt + 3 r2 = 100. (5)
The solution of this problem may now be completed by solving (5)
and (1) simultaneously.
Note. In equation (2), 20 — 3 r± and 15 — 3 r2 are both negative
numbers. This means that in this problem the distances north and
west of the crossing are negative, while those south and east are
positive.
REVIEW QUESTIONS
1. Define exponent. Explain the difference between an
exponent and a coefficient.
2. Explain why and under what circumstances exponents
are added in multiplication.
Show that (a2)3 = a8, (a3)4 = a12.
3. Explain the method of multiplying two monomials.
How is Principle III involved in Principle XV ?
4. What is meant by factoring ? Is the following expres-
sion factored ? x (a + b) + y (a + b). Why ?
5. What are the characteristics of a trinomial square ?
Are the following trinomials squares ? If not, change one
term in each so as to make it a square, x2 + xy + y2 ; x* + x*y2
+ y*. a2-2ab-b2; 4a2 + 4a6 + 462.
254 QUOTIENTS AND SQUARE ROOTS
6. What are the factors of the difference of two squares ?
Factor x6 — ?/6 as the difference of two squares.
7. What are factors of the difference of two cubes ?
Factor x6 — y6 as the difference of two cubes.
8. What are the factors of the sum of two cubes ?
Factor x6 + y6 as the sum of two cubes.
9. Explain how to factor a trinomial by inspecting the
end products and cross-products of two binomials.
By this method factor,
3^-7 a-10; a2-9a: + 18; 3^ + 5^-12.
10. By means of the following examples explain the pro-
cess of factoring by grouping.
x2 + ax + bx + ab ; x3 — x — 3 x2 + 3.
11. How may a quadratic equation be solved by factoring ?
Why is it essential that one member of the equation be zero
when the other member is factored ?
12. Explain the method of solving the quadratic equation
by completing the square.
13. How many roots has a quadratic equation ? When does
the solution of a quadratic equation indicate that the condi-
tions of a problem are impossible ?
14. Explain why and under what circumstances exponents
are subtracted in division.
15. Explain the method of finding the quotient of two mo-
nomials. Show how Principle V is involved in Principle XVII.
16. State Principle XVIII. Given V7 = 2.645, find V28 by
means of this principle.
17. Explain how a number in Arabic figures is divided into
groups for the purpose of finding its square root.
18. Show how the value of the following may be approxi-
mated by finding only one square root.
5 V20 + 2 V45 - 3 V80 + 2 VJ.
CHAPTER VIII
LITERAL FRACTIONS
COMMON FACTORS
182. If a number is a factor of each of two or more numbers,
it is said to be a common factor of these numbers.
Thus, 8 is a common factor of 16 and 48, and 12 is a common
factor of 12, 36, and 48.
If each of a given set of numbers is factored into prime factors,
any common factor which they may have is at once apparent.
Illustrative Example. Find the common factors of 72, 96,
120, and 288.
Factoring, we have
72 = 2
. o
. 2
3
• 3 = 2s • 32.
96 = 2
■ 2.
, 2 .
■ 2
• 2 • 3 = 26 . 3.
120 = 2
• 2
• 2
• 2
• 3 . 5 = 24 • 3
•5.
288 = 2
■ 2
• 2
• 2
• 2 • 3 • 3 = 26
8*.
The common prime factors are 2, 2, 2, and 3. The other
common factors are the various combinations of these, namely,
2.2 = 4,2.2.2 = 8,2.3 = 6,4.3 = 12, and 8-3 = 24. 24
is, therefore, the greatest common factor of these numbers.
To find common factors of literal expressions we proceed
in the same manner.
Illustrative Example. Find the common factors of x + y,
x2 — y2, and x2 + 2 xy + y2.
Factoring, we have x + y = x + y.
x2 -y2 = (x + y)(x - y).
x2 + 2 xy + y2 = (x + y) {x + y) .
Hence, x -f y is the only common factor of these expressions.
255
256 LITERAL FRACTIONS
Illustrative Example. Find the common factors of
10 x2 + 20 xy + 10 y2, 5(x + y)(x> - y2), and 15{x + y)(x* + f).
Factoring, 10 x2 + 20 xy + 10#2 = 2 • 5(x +y ) (x + y~).
5(x + #)(x2 - #2) = 5(x + y)(* + #)(x - y).
15(x + ^)(x8 + ys) = 3 • 5 (x + y)(x + ?/)(x2 - xy + #2).
The common prime factors are 5, x + y, and a; + ?/. The
other factors, obtained by combining these, are 5(cc + y), (x + y)2,
and 5(x + y)2. The last factor, 5(sc2 + 2a$ + 2/2), is called the
highest common factor.
The name highest instead of greatest is used in algebra referring
to the degree of the factor. Thus x2 is of higher degree than x, al-
though if x = \, x2 is not greater than x.
183. Definition. In general that common factor which con-
tains the greatest number of prime factors is the highest common
factor. This is usually abbreviated to H. C. F. (See § 130.)
EXERCISES
Find the H. C. F. of the following sets of expressions :
1. x-y, x?-y2, x2 — 2xy + tf.
2. x2 + 2x + l,3x + 6x? + 3xi.
3. x2 + 4 x + 4, x2 - 6 x — 16.
4. x2-8x + 16,x2 + 10x-56.
5. as-b3, a2-2ab + b2.
6. xP+y3, x2 — y2, x2 + 2xy-\-y2.
7. x2 — 7 x + 12, ax - 3 a — bx + 3 b.
8. a2-13a + 42, a3-216, a2-a-30.
9. 27 + f,y2 + 9y + 18, f-9.
10. b2 + 7b-S0, 62+H6-42, 62-&-6.
COMMON MULTIPLES 257
11. a3 + 2a2 + a, a2 + a, a3 + 5 a2 + 4 a.
12. Xs + y3, Xs + afy + #?/2 + S/3.
13. a4 + 3 ar5 + 2 a2, & + x2, x* + 7 a?+ 6 x2.
14. ar2 — 11 x -f 30, xz - 5 2 + x2 — 5 x.
15. ra3 — w3, 2 xfm2 4- 2 tfmn + 2 a^w2.
. 16. x>-l,a?- 1,3^-13 x + 12.
17. 1 - 64 a,-3, 1 - 16 ar2, 5 - 2 2 - 20 x + 8 a*.
18. 1 + 125 a3, 1 + 10 a + 25 a2, 1-25 a2.
19. ac — ax + 3 be — 3bx, a3+27&3.
20. 5 c — 2, 5 ac + 20 c — 2 a - 8.
21. 4x* -x2,2xi + xi-x2,2x4-3xi + x2.
22. 3a3-3a, 3a3-6 a2+-3a, 6a3 + 12a2-15a.
23. 6 a; — 10 xy + 4 xy2, 18 a; — 8 xy2, 54 a; — 16 xy3.
24. 3x5 + 9x4-3x3,5x2y2-{- 15xy2-5y% 7 ax2 + 21ax-7a.
25. 18 3^-57 3^ + 30 a:, 9 a3 - 15 a:2 + 6 a;, 18 ar» - 39 x2 + 18 a;.
COMMON MULTIPLES
184. A number is said to be a multiple of any of its factors.
In particular any number is a multiple of itself and of one.
Thus, 18 is a multiple of 1, 2, 3, 6, 9, and 18, but not of 12. 3 a2x2
is a multiple of 3, 3 x, 3 x2, etc.
Since a multiple of a number is divisible by that number, it
must contain as a factor every factor of that number.
E.g. 108 is a multiple of 54 and contains as factors all the factors
of 54, namely 3, 3, 3, and 2, and also 2, 6, 9, 18, and 54.
Definition. A number is a common multiple of two or more
numbers if it is a multiple of each of the numbers.
Thus, 18 is a common multiple of 6, 9 and 18. Evidently 3 • 18,
4 • 18, 5 • 18, etc. are also common multiples of 6, 9 and 18. Of all
these common multiples 18 is the smallest and is called the least or
lowest common multiple.
258 LITERAL FRACTIONS
185. The process of finding the least common multiple of a
set of numbers in Arabic figures is shown as follows :
Illustrative Example. Find the least common multiple of
16, 24, and 98.
Finding the prime factors of each number,
16 = 2 • 2 • 2 • 2 = 2*
24 = 2- 2. 2- 3 = 28- 3
98 = 2 • 7 • 7 = 2 • 72
Any multiple of 16 contains all its prime factors, namely,
2, 2, 2, and 2. Any common multiple of 16 and 24 contains in
addition to the prime factors of 16 any prime factors of 24 not
in 16, namely 3. Any common multiple of 16, 24, and 98 contains
in addition to 2, 2, 2, 2 and 3 those prime factors of 98 not in 16
or 24, namely 7 and 7. Hence 2- 2- 2- 2- 3-7-7 = 1176 is
a common multiple of 16, 24, and 98. Moreover, it is the least
common multiple because no unnecessary factor has been included.
We proceed with a set of literal expressions in the same
manner as above.
Illustrative Example. Find the L. C. M. of
ar2 — 2Z2; x2 + 2 xy + y2 ; x?-2xy + y\
Factoring, x2 — y* = (x — y) (x + y). (1)
x* + 2xy + y*= (x + y)(x + y). (2)
x*-2xy + y* = (x-y)(x-y). (3)
In order that an expression may be a multiple of (1) it must con-
tain the factors x — y and x + y. To be a common multiple of (1)
and (2) it must contain a second factor x + y, giving (x — y), (x+y),
(x + y). To be a common multiple of (1), (2) and (3) it must
contain a second factor x — y, giving (x — y), (x + y) , (x + y), (x — y).
The product thus found contains the fewest prime factors possible
in order to be a common multiple of (1), (2), and (3). Hence
(x — y)(x + y) (x + y)(x — y) = (x — y)2 (x + y)'2 is called the lowest
common multiple of (1), (2), and (3), since it is the common
multiple of lowest degree.
COMMON MULTIPLES 259
In general, the process may be described as follows : to
obtain the lowest common multiple of a set of expressions,
factor each expression into prime factors; use all factors of
the first expression together with those factors of the second
which are not in the first, those of the third which are not in
the first and second, etc. It is evident that in this manner we
obtain a product which is a common multiple of the given
expressions, but such that if any one of these factors is omitted,
it will cease to be a multiple of some one of the expressions ;
that is, it will no longer be a common multiple of them all.
Thus, if in the example above either of the factors x — y is omitted,
the product will no longer be a multiple of x2 — 2 xy + y2.
186. Definition. We now define the lowest common multiple
of a set of expressions as that common multiple which contains
the smallest number of prime factors. The lowest common
multiple is usually abbreviated to L. C. M.
EXERCISES
Find the L. C. M. of the following expressions :
1. 2-3.4; 3-7-8; 2s- 3 -4. 9. 25^-1, 125^-1.
2. 5«Y, lOafy, 25ofy. 10. 2^-7 a+6, 4^-11 a>+6.
3. 2 ab, 6 a2, 4 b2c. 11. x^ — y^jX — y^ + xy-^y2.
4. x2 — y2, x2 — 2xy + y2. 12. x3 — y3, Xs + y3, x2 — y2.
5. x-y^ + y^tf — y2. 13. 5 ^+7^-6, x2— 15a;— 34.
6. 4 — x2, 2 — x, 2 + x. 14. cc3 + 2/3, x2 — y2, (x— y)2.
7. a2+2a&+&2,a2-2ao + o2. 15.' 3a6c,a2-4ac+4c2, a-2c.
8. x2+ 3 x + 2, x2- 4,^-1. 16. x2-l, x + 1, a2 + 8a + 7.
17. 4 x3y - 44 x2y + 120 xy, 3 a?x2 - 22 a?x + 35 a3.
18. x2 + 2xy+y2, 2aa^— 10ax + 12a.
260 LITERAL FRACTIONS
19. 3 6a^-216a; + 36 6, ar*- 5a; + 4.
20. 5a262-5aV,62 + 26c + 6 + c + c2.
21. 15 (fax2 + 16 <?ax + (?a, 2 cax2 + 10 cax + 8 ca.
22. ac — 4 a + be — 4 6, aa: -f ac + 6a; + 6c.
23. a^ + 5a; + 4, a;2 + 7a;-f 12,a;2 + 5a; + 6.
24. x2 + 2x+l,x2-2x + l,x2-l.
REDUCTION OF FRACTIONS TO LOWEST TERMS
187. In arithmetic the denominator of a fraction is usually-
regarded as indicating the number of equal parts into which a
unit is divided, while the numerator designates a certain num-
ber of these parts.
Thus, | means 2 of the 3 equal parts of a unit.
5
However, a fraction such as — cannot be regarded in this
5
way, since a unit cannot be divided into 3^ equal parts. —
5
indicates that 5 is to be divided by 3£. I.e. — = 5 -*- 3\>
In algebra any fraction is usually regarded as an indicated
division in which the numerator is the dividend and the de-
nominator is the divisor.
Thus, - is understood to mean a ■+■ 6.
6
The numerator and denominator are together called the terms
of the fraction.
In case the numerator and denominator have common factors,
these may be removed, without changing the value of the frac-
tion, by means of Principle XVII, as applied in § 157.
Thus, — - — '——— = — ■ — — , where the common factor 3 is can-
, ' 3 • 7 • 11 7 • 11 '
celed.
REDUCTION TO LOWEST TERMS 261
Similarly — - — ■ — - = - — -, the factors 23 3, and 4 being can-
24 • 3 • 42 2 • 4
celed; and <*" 7* + 12> = (»-3)(«-*) = fez^.
' (^-Sa-t-e) (a? -2) (a? -3) (a;-2)
If the terms of a fraction have no common factor, the frac-
tion is said to be in its lowest terms. Evidently the process of
canceling common factors in the numerator and denominator
may always be continued until the fraction is reduced to its
lowest terms.
EXERCISES
Reduce the following fractions to lowest terms :
t 3 . 92 . 2G aW x*-f
' 24.53-94' ' a262' ' 2x2-3xy + y2'
4 a W x* + 2xy + y2 64 -ft3
8aW* " tf-tf ' ' 16-86 + &2'
. afyV a a:2 + 7a;-30 _ ar3 + 27z3
a^V a?2 — 7 a; + 12 xy — 5 x+ 3 yz — 15 z
10 1-216C3 27 • 3s . 54 - 26 . 34 • 57
a;-4y-6ca; + 24c2/' ' 24-32 • 55-25 • 32- 52'
,j 14 6z — 2 6a; + ax — 7 az 5 x?y* — 12a*Y + 7 a3?/2
ar!-4922 ' 6 afy2 + 3 x?y2
.„ 3a2-29a + 56 nQ 5c + 106- 6c-262
63 -9 a- 7m + ma 8c3 + 64 63
13 a(x - Vf 19 4 a;4 -28 3^ + 48 0^
(a?-tf)(x-y) ' 2x*-$a? + 6x2 '
14 a;3 +27 20 9a264+ 18a263c+9a262c2
4a2 + 24a; + 36' " 3a63-3a6c2
15 a2-3a-36 + a6 7 a?y2 - 133 xy + 126 a;
(a2-b2)(a-3) ' 15xy2-36xy + 21x '
262 LITERAL FRACTIONS
22 20xi + 20x2y + 5xy2 (a?-l)(s-2)(a?-3)(s-4)
. Wtf-Wafy2 ' (a>-l)(a>-3)(a-3)(a;-4)'
23 3a^-3a52c2 (s8-yg)(as» + 2sy + y8)
27 a362 + 27 a36c * (a2 - 2 ^ + ?/2) (a? + y) '
24 4a3-42a2 + 20a (s»-l)(a3 + l)(3s? + 3)
2a466-20a3&6 3(»4-l)
28 (3 a* ~ 3 a2b2) (a2 + 13 a + 42)
(a2 + 3a + 2)(a2 + 5a-6) '
29 23 • 43 • 54 (x2 - b2) (x2 + 19 x + 90)
' 22.4-53(«2 + 9a;-10)(ari + 10a; + 9)'
REDUCTION OP FRACTIONS TO A COMMON DENOMINATOR
188. Since we have just seen that a common factor may be
removed from the numerator and denominator of a fraction
without changing the value of the fraction, it is evident that a
factor may be introduced into both terms without changing
the value of the fraction.
Thus since — ^— is reduced to - by removing the factor 3 from both
3 • 5 5
4 3-4
numerator and denominator, so - is changed to by introducing
o o • o
the factor 3 in both the terms. Likewise any other factor may be
introduced. E.g. * = ^ , ^ = («-»)(« + ») = «^J*# '
y 5 5-7 a + b (a + b)(a + b) (a + 6)a
In this manner any fraction may be changed into an equal
fraction whose denominator is any given multiple of the de-
nominator of the given fraction.
Thus, f can be changed into a fraction whose denominator is
q o _ -i o
72 (a multiple of 4) by multiplying both terms by 18, i.e. '-. — j— ttt •
REDUCTION TO COMMON DENOMINATOR 263
and can be changed into a fraction whose denominator is
x- y
x2 — y2 by multiplying both of its terms by x + y, i.e.,
Sa -2b=(Sa-2b)(x + y)
x — y x2 —y2
Any two or more fractions may therefore be changed into
respectively equal fractions which shall have a common de-
nominator, namely, a common multiple of the denominators of
the given fractions.
Illustrative Example. Keduce ~ , . „ , to frac-
as+1 x — 1 x l — 1
tions having a common denominator.
The L.C.M. of the denominators is (x — l)(x + 1). Multiply the
numerator and denominator of each fraction by an expression which
will make the denominator of each new fraction (x — l)(x + 1).
x-1 __ (x- l)(x- 1) _ x2 - 2 x + 1
x + 1 (x + l)(x-l) (z + l)(x-l)'
X+1=(X+1)(X+1) = (X + 1)2 .
x-l (x-l)(x + l) (x + l)(x-l) '
2x + 3_ 2x + 3
x2-l (x + l)(x-l)"
It is best to indicate the multiplication in the common denomina-
tor, since this makes it more easily apparent by what expression the
numerator and denominator of a fraction must be multiplied in
order to reduce it to a fraction with the required denominator.
It should be noticed that there are three signs in connection
with a fraction, the sign of the fraction itself, the sign of the
numerator, and the sign of the denominator. It follows from
the law of signs in division (Principle XII) that any two of
these signs may be changed simultaneously without changing
the value of the fraction.
-p a _ — a — a _ _ a
'9' b~ -b~ ' b -b
Thus,
264 LITERAL FRACTIONS
This is useful in cases like the following:
x + 1 x 1
Reduce , — — -, and to fractions having a com-
1 — x x'- — 1 x-\-l
mon denominator.
x + 1 _—x-l _(x + l)(—x — l) _—x* — 2x — l
1-x x-1 ' ' (x + l)(x — l) '' x*—l
X X -, 1 X — 1 X— 1
, and
tf-\ tf-l' x + \ (aj + l)(a?-l) x*-l
EXERCISES
Reduce each of the following sets of fractions to equivalent
fractions having a common denominator.
1 ^+3 4 4 2 g - 1 a? + l_
x—y' x* — 2xy + y'2 3 — x' x + l' x — 3
. a — 1 a+1 c a + & a— 6 a
a2-&2' a2 + 2a6+62 ' b-a (a + &) a2-&2
3-a a;+4 .1 1
^-9^+20' 7<e*-26a;--8 ' a2 — 3<e — 4' a;2 + 3a; + 2'
a+1 a— 1
7.
8.
a2-2a& + &2' a2 + 2a&+62
1 1 1
a3_&3> 6_a' a2+a& + &2*
a &
10.
5a2-4a-12' a2 + 4a-12' a-2
a; — 2 a + 2 cc-1
^-Sa-G' rf + lZx-lOtf x* + VZx + 18
In each of the following exercises reduce the given fractions
to a common denominator and check by substituting a con-
venient number for each letter, taking care that no denomina-
tor becomes zero :
REDUCTION TO COMMON DENOMINATOR 265
11 mr d 16 *{T-t) W(T-Q)
m_l' i + n' ' w(Q-T)' ~w(Q-t)
12 Rr 1 17 P d(V-w)
(B + r)(m-l)' B + r ' W(p-2t+l)' W-w
CS Br + Sr + SB t. V V 1
B + BS' BSs ' ' V-v' V+v' V'-v1
. . a b , « BA 1 1
14. , • 19.
15.
n—an—b a — A'B-\-r'A — a
W(T-Q) V 9ft B r 1 Bx
w((J — t) "jo x y x — y x-\-y
189. Since any number may be written as a fraction with
the denominator 1, the above process may be used to reduce
an integral expression to the form of a fraction having any
desired denominator.
Thus, 3=^5; x-y=(x-Mxi-V,ete.
5 x2 — 1
It is sometimes convenient to reduce expressions, some of
which are not fractions, to the form of fractions having a com-
mon denominator.
Illustrative Example. Reduce 5 x, ~ ', —r^, to
ar — 1 x — 1
fractions having a common denominator. The lowest common
denominator is ar2 — 1.
„,. 5 x (x2 — 1) 5 x8 — 5 x
Thus, 5x=- — \ — , - — • — T~>
x2 — 1 x2 — 1
5x — 1 _ 5x — 1
x2 - 1 " x2 - 1 '
2 x — y _ (2 x — ?/)(x + 1) _2x2 + 2x — yx — y
x-1 " (x-l)(x+l) ~ x2-l
266 LITERAL FRACTIONS
EXERCISES
Reduce the following mixed expressions to fractions having
a common denominator :
- 5x— 3 oil . 2 «+ 1
1. 5, a>-y,. • 3. 1 + a + a2, — ^rr-
ar + 2 #?/ + ?/J a — 1
2. 3a~c, !•=•, 2c + 2> 4 *+«*+*•, *±*.
x—y x+y x — y
5. rf — xy + y2, x* — y2,
7. 3 a — 2& — c,
8. x*-l, a?-l,
9. x2 + 2z?/ + 2/2,
10. x + y, x-y, - -*-, :
x? + y2 x-y
In each of the following reduce the expressions to the
form of fractions having a common denominator and check
the results by substituting convenient numbers for the
letters :
„. ** ,. „. SA.,B-r. 13. r, «±-(-
a — A a — A 2
14. TP(g-Q sT} 8(t-q)m u H} hd,
w 2 Z>
# >
x + y
X
y .
x—y
x + y
5
a — &'
2
b-c
x + 1
x-1
1
x + y'
1
1-x
x-y
x + 1
ADDITION AND SUBTRACTION 267
ADDITION AND SUBTRACTION OF FRACTIONS
190. Fractions having a common denominator may be added
or subtracted, exactly as in arithmetic, by adding or subtracting
the numerators and dividing the result by the common
denominator.
For we have, by Principle VI, — - — = — + — = 4 + 5. Like-
o | e o k
wise — ■ — = t + t , the division being indicated in this case. Hence,
4 4 4 °
3 5 3 4-5
reading this identity from right to left, we have ■? + r = — -z — .
ti • 6 4 _6-4 2 T
Likewise — = . In general,
x — y x — y x — y x — y
a b _a + b , a b_a — b
c c~ c ' c c~ c
If fractions which are to be added or subtracted do not have
a common denominator, they must be reduced to this form
Example. Add ^ and a2 + 2ah + **.
a + b a2 - 2 ab + b2
Reducing the fractions to the common denominator
(a-b)(a - b)(a + V),
we have
a-b __ (a-b)(a-b)(a-b) _ a8- 3a2& + 3 ab2- b* }
a + b (a + b)(a - 6)(a -b) (a + b) (a-b) (a-b)
&nd a2+2ab + b* _ (a + b) (a2 + 2 ab + b2) _ a8+ 3 a2b + 3 ab2 + bs
a2-2ab + b2 (a + b)(a - b)(a - b) ~ (a+b)(a - b)(a- b)'
Adding the numerators, we have 2 a8 + 6 ab2 ; whence the sum of
the fractions is
2 a8 + 6 ab2
(a + b)(a- b)(a-b)'
268 LITERAL FRACTIONS
EXERCISES
Perform the following additions and subtractions :
1. 3 | 2 I a + b 8 2x* y y
4 7 3 a— b x? — y2 x — y x + y
x-y x-y a + 1 q-1
(x + y)2 tf — y2 tf + a + l a2 — a + 1
3 ^-9^ + 18 | x 1Q _&_, & b2
or2 -13 a; + 36 4 -a l_&'i + &2 1_&2
4 2 I CT I & 11 a^ + 1 ■ a + 1 , 3a+2
3 a + 6 a-& * x-2 x + 2 x2 — !
3 5 2 .. »-l aj+1 , ar>-5
23-32 22-34 24-33 a + 1 a?-l a2-l
6 a2 — 9 &2 a2 — 6 q& y2 y y
a2+6& + 9&2 a2-962' ' y2-l y + 1 l-y'
7. ^±1 + ^^-2. 14. i-I L. + ^l.
»— y » + y « y x—y x+y
15 I ^ . 2 a2 — a
2 6_a"l"&2_a2 + & + a-
16 ^ + 4^
a^ + y3 » + y tf — xy + y
W. .-JL-. + : * *
1 — ar* 1 — a; 1 + aj + a^
18. a~3 CT~1
q2-3a + 2 a2_5a + 6 o2-4a + 3
ADDITION AND SUBTRACTION 269
19- . * ..+ 2
20.
a^-5a;-14 tc-7 a,-2 -9 a; + 14
a &
ac + ad—bc — bd a? — 2 a& + &2
,!. » +.,39 x. 3
a-5 a2 + 3a-40 a + 8
22. ; " 1 " „-- ^,+ 4
2/2 + 82/+16 y(y + 4) y2(y+4)
23. = — +
a* + 4:X-60 v*-±x-12
n. a—c , c — a 2
24. -= „ +
25.
a2 — c2 a2 + 2 ac + c2 a — c
_9 8
! + 7a;-18 ^ + 6^-16'
26. a + 2 | «~4 « + 2
a2-a-6 a2-7a + 12 a2-2a-8
27. *^_ * + 1
^-11*5 + 30 ^-36 a^-25
(z-l)(a + 2) + (a; + 2)(a-3)+(a;-l)(3-z)'
29 1 1 , 1
(«-&)(&-c) (&-a)(c-d) (b-c)(c-d)
30 4 a-1 a2-38a-3
'a-3 a2 + 3a + 9 a3 -27
270 LITERAL FRACTIONS
MULTIPLICATION AND DIVISION OF FRACTIONS
191. The product of a fraction and an integer. In arithmetic
the product of a fraction and an integer is obtained by multi-
plying the numerator of the fraction by the integer.
Thus, 2 x 4 = — and 7 x - = — •
3 3 5 5
Since in algebra a fraction is an indicated quotient, and since
multiplying the dividend multiplies the quotient, it follows that
in algebra also the product of a fraction and an integral expres-
sion is obtained by multiplying the numerator by the integral
expression.
That is, in general, a • - = = —
c c c
It is best to factor completely the expressions to be multi-
plied and keep them in the factored form until all possible
cancellations have been made.
EXERCISES
Find the following indicated products and reduce the frac-
tions to the simplest form.
1. (l-o) x1 + a + a\ 6. <V+9a;+18)x X~
a — 1 x2—2x — 15
2. (a3-?/3) X <^ty. 7. (1 _^) x - 1~X •
x-y v J 1+x + a?
3. (a»-2«a + a!)x^ 8. (27 a3 - 1) x — ~^~. r-
x — a 9or+3a + l
5- Jfrax(o!-5a+4>- 10- *-****t& .
MULTIPLICATION AND DIVISION 271
192. In arithmetic a fraction is divided by an integer by
multiplying its denominator or dividing its numerator by the
integer.
— ***-& ¥^fH-
Since multiplying the divisor or dividing the dividend divides
the quotient, it follows that an algebraic fraction is divided by
an integral expression by dividing the numerator or multiply-
ing the denominator by the integral expression.
That is, in general, - ■+■ c = — = — j- —
b be b
EXERCISES
Find the following indicated quotients and reduce the frac-
tions to their lowest terms :
, x3 — w3 / o , . 2\ n #24-4# + 4 , j, ..
1. ?--i- (ar + xy + y2). 3. „„ — — — 5-(ar — 4).
z + y v * * } x2-2x-\-l v ;
2. t±t+(x>-y*). 4. ^^^(1 + 3^+9^.
a — ?/ 1 — 9ar*
2^-35 ^_^_4a._5)>
a? + 10 x + 21
6. ^7 ^ X +,3r9 -*- (s2 - x - 156).
or — 8 a; -f- 15
7. ^ + 6c-ad-M c_d
xy — 4:X—3y + 12 x '
x2 + ax + bx + a6
a2 -fax — 3#— 3 6
(cc2 + aa? — 5 x — 5 a).
9. : -s- (3 r — xr + 3 s — xs).
mx — m—nx + n
10- ^~Q3;~99^(a:2+9a;+2)-
ar — 9 x — 22
272 LITERAL FRACTIONS
TO MULTIPLY A FRACTION BY A FRACTION
193. In arithmetic, to multiply a number by the quotient of
two numbers is the same as to multiply by the dividend and
then divide the product by the divisor.
E.g. 10 • — = 10 • 6 = 60; or, 10 • — = (10 • 18) + 3 =60.
2 5_/2.*\ . „_»«6 - 2-5
Likewise, | • - = U -5J--7
3 3-7
Since in algebra a fraction is an indicated quotient, to mul-
tiply a number by an algebraic fraction we multiply by the
numerator and divide the product by the denominator.
rru fl o fa \ . ac . . ac
Thus- r^-fr-v* *" w
Hence, the product of two algebraic fractions is a fraction
whose numerator is the product of the given numerators and
whose denominator is the product of the given denominators.
Illustrative Example. Multiply ^^ by £=§*±f
and reduce the resulting fraction to its lowest terms.
x2- 1 ^-3x+2^(g-l)(ic + l)(x-2)(x- 1)
x2-7x + 10 x2 + 2x+l (x - 2)(x - 5)(x + l)(* + 1)
_(x- l)(x - 1) _ x2 - 2 x + 1
(x-5)(x+l) x2-4x-5*
It is desirable to resolve each numerator and denominator
into prime factors, and then cancel all common factors before
performing any multiplication.
MULTIPLICATION AND DIVISION 273
EXERCISES
Find the following indicated products and reduce each frac-
tion to its lowest terms :
1 3 arfy2 6 az 32 . 43 10-2
2yz2 9x*' 52 . 24 34
5o(q-6) 9(a + &)2 6 32 . 23 54. 72 6 • 32
3 c (a + 6) 15 (a2- 62) ' 52 34 • 23 54 • 73
12<*b 35 (c2+c5 + ft2) . 7 fl2-^ 2^ + 43; + 2
' 5(c3-63) 14c*&» ' ' x2-! 3a? + 6x
4 y» + 3y + 2 y»-7y + 12 g a2-10q+16 a + 3
' y«_5y + 6 f + Sy + 7 ' a2+6a + 9 a2-4'
9# (x + yf-z2 x a; x (a; - y)2 - z\
x2 + xy — xz (x — zf—y* xy — y2 — yz
10 a2 + 7a + 12 3a3 + 27a2 + 42a
a3 + 5a2 + 6a 6a2 + 66a + 168 '
3<(a - 5)(a + 2) 22(a - 3)(a - 5)
' 23(a-3)(a-2) 33(a-5)2
12.
3(g + 4)2 Q-7)2
4(« + 4)(as-7) 3(a> + 4)(a>-7)'
13 a3+5a2-36a (q-16)(a-3)
a2-7a-144 a(a-4)(a + 2)"
14 3a(a-7)(a-5) 6(a-3)(a+10)
7 6(a-3)(a-7) a(a-5)(a-10)
15 42(6-3)0-4) 6(6 + 7)0-5)
3(6-4)(6+7) 14(6-3)(6-6)
a(»-<*) (6»-q»)» (6 + a)2
6(6 + a) b2 + ba + a2 (6 - a)2
274 LITERAL FRACTIONS
17 a2-4g + 3 a2- 9a + 20 a2 -la
a2 - 5 a + 4 a2 - 10 a + 21 X a2 - 5 a'
18 3 a(a - 7)(o - 5) 6(a - 3)(o + 10)
' 7 6(a-3)(a-7) a(a-5)(a- 10)'
19 a2- 12a + 35 a2 -8a + 15
' c(a2- 10a + 21) a2 + 3a-70'
20 c*-10c + 21 c^-3c-88
c2- 18 c + 77 c2-8c + 15'
ar^ar8 - 16) y2 - 12 y + 35 y«(g + 5)
' y2(y2 -3 y- 28) a?-9x+2Q a?(x-4)
22.
3 a262(c2-14 c+33) 2(c2 + c - 56) c2-2c-15
4 a* (c2 - 10 c + 21) a2(c2 - 16 c + 55) 62(c2 + 12 c + 32) '
St2-2t-l 2t2 + 5t-S 4t2 + 10t + 4,
2t2 + t-l 3t2 + 7t + 2 4:t2-2t-2'
m. 6^-7^ + 2 „6a?-5x-l 10a^ + 3a;-l
' 10a^-7a; + l 6^+a-l Ztf-kx-l
452-175+4 1052-216 + 9 362-5 5 + 2
• 662-76 + 2 562-236 + 12 462-56 + l"
TO DIVIDE A FRACTION BY A FRACTION
194. In arithmetic, to divide a number by the quotient of
two numbers is the same as to divide by the dividend and
multiply the result by the divisor.
E.g. 72 + — = 72 + 6 = 12 ;
if 3
or, 72 - ^ = (72 - 18) . 3 = 4 . 3 = 12.
Likewise, jUf = (§ + B) . 7 = (^) . 7
MULTIPLICATION AND DIVISION 275
Since in algebra a fraction is an indicated quotient, to divide
a number by an algebraic fraction we divide by the numerator
and multiply this result by the denominator.
y b d \b J \b-cl
ad
be'
Hence, in algebra, as in arithmetic, a number is divided by a
fraction by inverting the fraction and multiplying by the new-
fraction thus obtained.
Thus, *^*S5Sfx^»«f.
b d b c be
EXERCISES
Perform the following indicated divisions, and reduce the
resulting fractions to their lowest terms :
q3-[-53 a + b g^-6o;-16 . x*+9x+U
d2-9b2 ' a + 3b' ' x2+4a;-21 : tf-Sx+W
x2 + x-2 . a3 4-2^ x2-! . a? + 2x-3
x>-3x ' x* + 9x-36' ' ^-4^-5 ' x>-25
a2 -11 a -26 , a2- 18 a + 65
a2_3a_18 * a2-9a + 18 '
e x2+9xy + lSy2 . x2 + 6xy + 9y2
x2 — 9 xy + 20 y2 xy2 — ky*
x2 4- mx + nx + mn a? — n2
x2 — mx — nx + mn x2 — m?
3a4-9a8-54a» . a34-8a24-15a
9 a3 -117 a2 + 378 a ' 3a2-33a + 84#
ft a2- 11 a + 30 a2- 3 a . a2 -9
a3-6a24-9a a2-25 ' a24-2a-15
276 LITERAL FRACTIONS
10.
x* + x-56 x-2 + 4x-32
a2-3a + 2 a2-8a + 15 . a2-9a + 14
a2-7a + 12 «2-6a + 5 ' a2-12a + 32*
12 n2-ll?i + 18x n2-8n-9 , 6n-12
m2 — 7 ?« — 18 m2 — 5 m — 14 an + a
13 a2-62 &(a-5) . b(a + b)
' ab2x a2 + 2a6 + 62 ' a2-2a& + &2
14 a + 5 a2-&2 . (a- &)2(a + fr)2
ab 3(a2+62) ' 3 a3?/ + 3 a?/3
5a54-5ar> ^-9^ + 8;
15.
7a2-56x-63 14 a;2 + 14 a; -1260
8y2(,y + 4)(y + 5) , y(y + 4)(y + 8) .
' 2*(y + 6)(y-7) ' 23G,-7)(j/ + ll)
17 (c + 4)(c-4) (c + 2)(c-5) . (c + 4)(c-13)
• (C_3)(c_2) (c-5)(c-6) ' (c-6)(c + ll)
2a26(6-4)(& + 4) (& + 6)(5+8) , a(& + 8)(&-4)
• 5(& + 4)(6 + 6) (6 + 8)(6-9) ' (&-9)(6-5)
19 ^(s-?)2 (s + 2)(a-3) , x\x-3)(x- 5)
' (x + 2)2 (a>-2)(a>-7) ' (s-5)(a>-7)
a^2(c + 5)(c-4) (C-8)(c + 9) , o6(c+-9)(c-l)
• (c_4)(c-8) (c + 4)(c + 7) " (c + 7)(c + l)
21^ + 23a;- 20 6 x2- 11 »- 10 . 7a2 + 17a;-12
" 10^-27^ + 5 3a»+2o;-5 "" 5x> + 9x-2 '
COMPLEX FRACTIONS 277
s* + 6a; + 9 9a^ + 9a;-10 , 3a?-7a;-20
' 9^-4 7a2 + 20a;-3 ' 35a?-12a> + l"
23 (a + 4)2-62x 3as-4 . ab* + 4:b2 + bs
12 as3 — 16 x* ay? 4- 4 a2 — 6X2 jc3
12 ya^2- 18 ya; 3 6s8 4- 7 6a; . 3&x3 + 76a;2
15a;2 + 32aj-7 6a^4-15y ' 10a;24-23a;-5'
ac + bc — ar — br 6 a2 + 23 a 4- 7 . 3 a2 4- a 4- & + 3 afr
3 a2 — 8 a — 3 c& — r& + c2 — re ' a& — 3 6 4- ac — 3 c
COMPLEX FRACTIONS
195. Sometimes fractions occur whose numerators or de-
nominators, or both, contain fractions.
1+1 1 +. 1
E.g. 2 and x + 1 x~1.
* 1 1 _1 1_
a x — 1 x + 1
Such fractions are called complex fractions. A complex frac-
tion is said to be simplified when it is reduced to an equal
fraction whose numerator and denominator are in the integral
form.
1+1 «+l
a a a a+1 a — 1
Ex.1.
■a 1 a _ 1 a a
a a a
__<x + l a _a4-l
a a—1 a—1
This result may also be obtained directly by multiplying both
I 1
terms of the given fraction by a, finding at once "*" .
a — 1
278 LITERAL FRACTIONS
1 + x
Ex.2.
x + 1 x — 1
x — 1 x+1
X — 1 +
x + 1
X2-
1
x + 1 —
x + 1
X2-
2x
>
■ 1
. x2-!
x2-!
By multiplying the terms of the given fraction by (x + V)(x — 1)
■r 1 4- X + 1
we may also get directly — — — :!— = x.
x — j~ X — X ~r x
Such fractions seldom occur in the problems of elementary
algebra. A more extended discussion is given in the Advanced
Course.
EXERCISES
Reduce each of the following complex fractions to its
simplest form :
m + n + 1 a? — 8 b3
1. 1±£. 4. 3 7.
1 m — n — 1 3a — 2b
x
i_l! i i a + ft Jf
- ■ S 2 fl.
1 + a' . a — b 1 + d2
4-
2 Z>2
, x tf-tf H hd
3. —1. 6. 4 9. c cD
a g-f-y 1 + £
*~2 2 c.
RATIO AND PROPORTION 279
RATIO AND PROPORTION
196. Definitions. A fraction is often called a ratio. Thus -
may be read the ratio of a to b, and is also written a : b.
The numerator is called the antecedent of the ratio, and the
denominator the consequent. The antecedent and consequent
are called the terms of the ratio.
An equation, each of whose members is a ratio, is called
a proportion.
Thus, - = - is a proportion, and is also written a : b = c : d.
b d
It is read the ratio of a tob equals the ratio of c to d, or briefly,
a is to b as c is to d.
The four numbers a, b, c, and d, are said to be in proportion.
a and d are called the extremes of the proportion, and 6 and c
the means.
IMPORTANT PROPERTIES OF A PROPORTION
1. If, in the proportion - = -, both members of the equation
b d
be multiplied by bd, we have, ad — be.
That is : If four numbers are in proportion, the product of the
means equals the product of the extremes.
2. Show that if - = c , then - = -.
b d a c
Hint. Divide 1 = 1 by the members of the given equation.
This process is called taking the proportion by inversion.
3. Show that if -=-, then - = -•
b d c d
Hint. Multiply both members of the given equation by - .
c
This process is called taking the proportion by alternation.
280 LITERAL FRACTIONS
4. Show that if a- = c-, then ?±± = c_±*.
b d b d
Hint. Add 1 to both members of the given equation.
This process is called taking the proportion by composition.
5. Show that if - = -, then *I2*»£i±i!.
b d b d
Hint. Subtract 1 from each member of the given equation.
This process is called taking the proportion by division.
6. Show that if ?«*, then ^±& = ^M.
b d a— b c — d
Hint. Divide the members of the equation obtained under 4 by
the members of the one obtained under 5.
7. Show that if g = ^ = g, then a + c + e = g.
b d f b+d+f b
Let - = - = - = k : then a = bk, c = dk. e = fk.
b d f ■ J
Hence, a + c + e = bk + dk +fk =(b + d +/) k,
a + c + e 7, a c e
and
b+d+f b d f
That is, If several ratios are equal, the sum of the antecedents
is to the sum of the consequents as any antecedent is to its
consequent.
EXERCISES
a c
1. If ad = be, show that- = -• Hint. Divide by bd.
b d
2. If ad = be, show that - = - •
c d
3. If ad = be, show that - = - •
c a
RATIO AND PROPORTION 281
4. If ad = be, show that - = - ■
b a
5. If - = -, show that — - — =— — -
b d a c
m T£ a c i ,i , o- b c — d
6. If - = - , show that =
b d a c
m T£ Cb C 1 i/U i. <* — b C — d
7. If -=-, show that =
b d a+b c+d
8. If ^ = C-, show that 1+b^a^dL
b d c+d c—d
9. If - = -, show that — ! — = -•
b d c+d c
10. If - = -, show that — - — =-•
b d b+d b
11. If- = -, show that
a — b a
b d c—d c
12. If - = -, show that ^=^ = --
b d' b-d b
13. Solve the equation - = - for each letter in terms of all
6 d
the others. If a = 3, 6 = 5, c = 8, find d. If b = 7, c = 9,
d = 3, find a. If c = 13, d = 2, a = 5, find 6. Ifd = 50, a = 3,
6= -7, find c.
14. If - = -, then x is said to be a fourth proportional to a, b,
, b x
and c.
Find a fourth proportional to 3, 5, and 7 ; also to 9, 5, and 1,
and to 3, — 2, and — 5.
282 LITERAL FRACTIONS
15. If - = -, then x is called a mean proportional between
a and b.
Solve the equation ~ = ~ for * in terms of a and &- Snow
that there are two solutions, each of which is a mean propor-
tional between a and b.
Find two mean proportionals between 4 and 9 ; also between
5 and 125, and between — 4 and — 36.
16. Which is the greater ratio „ , — or — t— —
5 + 4d 5+5d
Hint. Reduce the fractions to a common denominator and com-
pare numerators, (d is a positive number.)
17. Which is the greater ratio, a "** , or a ~*~ ,t ♦
a + 86 a + 10 6
18. Which is the greater ratio, - or ~*~ , if b and c are posi-
b b + c
tive, and a less than b ? a equal to b ? a greater than b ?
19. Find two numbers in the ratio of 3 to 5 whose sum is 160.
20. Find two numbers in the ratio of 2 to 7 whose sum is
-108.
21. Find two numbers in the ratio of 3 to — 4 whose sum
is -15.
22. What number added to each of the terms of the ratio
^ makes it equal to f| ?
23. What number must be added to each term of the ratio
Jy to make it equal to the ratio f ?
24. What number added to each of the numbers 3, 5, 7, 10,
will make the sums in proportion, when taken in the given
order?
25. Two numbers are in the ratio of 2 to 3, and the sum of
their squares is 325. Find the numbers.
EQUATIONS INVOLVING FRACTIONS 283
EQUATIONS INVOLVING FRACTIONS
197. We have already seen, § 40, that in solving an equa-
tion involving fractions the first step is to multiply both
members of the equation by a number which will cancel all
the denominators. Evidently, any common multiple of the
denominators is such a multiplier. For the sake of sim-
plicity, and for reasons which are fully discussed in the
Advanced Course, the lowest common multiple is always used
for this purpose.
Illustrative Example. Solve the equation :
x .x — 6_1— 2 a; _ 55 ,^.
4+_6 8 "~8" U
Solution. The L.C.M. of 4, 6, and 8 is 24. Multiplying both
members of the equation by 24,
24* , 24(z-6) 24(1 - 2 x) _ 24 • 55 /0.
~T+ 6 ~ 8 ~~ 8" {2)
ByF,V, 6x + 4(:r-6)-3(l - 2x)= 3 • 55 = 165. (3)
ByF, 6z + 4x-24-3 + 6x = 165. (4)
ByF, A, 16 a; = 192. (5)
By A x = 12. (6)
As in this solution, so in general, each denominator is can-
celed by Principle V, after multiplying by the L.C.M. In
practice, however, equation (2) may be omitted, the canceling
being done mentally, and equation (3) may be written down
at once.
The dividing line of the fraction acts like a parenthesis with
respect to the sign preceding the fraction.
284 LITERAL FRACTIONS
EXERCISES
Solve the following equations, and check each solution by
substituting in the original equation :
4cc — 5 2x + 5__x — 3 . 7a; + 5 ,
3 5 = 2 + 8
_ x + 3 5x — 3 5z — 3_ ,
2a? + 5 = 5a; + 20 7a: + 13 15fl + 3
3 15 2 4
3s + 10 7a: + 15 5 a:-14=3 a;-12 2
5 3 7 4
,3a + 5,7-3a; 15 — a; , 3 — 11 x
5. X-\ -—- + — 7> = — o 1 a
4a: 2x 2x 4
. 7aj-2 3a;-24 3z + 4 5-4a?
"• — ~ 7i "
2^_5» 7_a_^ = 45
3 12 8 2
2_x _ Zx 5x _ 11a; _ n*
3 4 7 " 12 =
„ s-3 . 2s+3 5s-3 3s-l Q
9- -2^ + "^ 6~- = ~^ 3-
,- 2t + l £-1 , 4*-8 3*-l,„
10- "1 8_ + "l5~ = ^0" + 3-
,„ z + 3 4» + 5 . 3 a-5_5a;-7 , .
"' ~2 7~~ + ~4 ^2~ + 4,
EQUATIONS INVOLVING FRACTIONS 285
8fc-6 13fc 21k-12 = k-2 Uk-3 d
' 5 2 5 10k 5
5a; — 1 2a; + 3 5a; + l_13a: + 5 1
' 2 3 4 11
14 o ft-15 ft-20 4fe + 2 = Q
2^3 11
17 + ra 2m-7 . 5m— 3 . Q 3
15. . _ 1 _ — __f-ii__.
4 3 2 m
3r + 4 5r + l 8r+4=o 6r
5 12 10 7 *
17 9 + 2*/ 4y + 3_27 + 3y s
17>"^ 3» 6~ "
1Q 2 a; + 5 5a; + 2 , 4a;-5_3a;-f 4
z+3 2z-15 5z-ll 11 =Q
'2 « 8 2'
OA3a5 + 20.5a;-3 4a>-l Q
20. __ + — — =3.
198. Illustrative Example. Solve the following equation :
2a; — 1 4 _ 3a; „ m
x-1 x + 1 a*-l ' K '
Solution. The L. C. M. of the denominators is x2 — 1. In multi-
plying both members of the equation by x2— 1, x — 1 is canceled in
the first fraction, x + 1 in the second, and x2 — 1 in the third, giving
(2x-l)(x+ l)+4(x-l)- 3x = 2(x2- 1) (2)
Solving, 2x2 + x- l + 4x-4-3x = 2x2- 2, (3)
2 x = 3, (4)
and x = §. (5)
Check by substituting x = § in (1).
286 LITERAL FRACTIONS
EXERCISES
Solve the following equations and check each solution by
substituting in the original equation :
- Sx-1 4s + 3 x2 27 1
2 3a; + 5 2x + l_ x — 1
x-9 x + 2 a?-7x-18
3 3a;-4 4cc-l ar>+44 _Q[
aj + 5 a?+4a? + 9aj + 20-
4 4-s2 3a; + 6 = 2a: + l
ic2 — 5ic — 14 a:+2 ~ a- — 7
a;_4 3a-15 3^-114
5.
2z-10 2z-6 4a2-32a:+60
6 6(a;+4) 3(2a?-l) = 7[
a + 5 a + 1 2*
„ 3z-4 . 5a-7 9a^-38
7- r + ;
a;-4 2a;-2 2^-10^ + 8
g a; + 17 2(a; + 6) = a;-l
a; + 5 ar + 3 ' a:-f-3
9 a; + 2 3a;-15 = 3a:-21
ic — 5 a; — 3 x — 3
2a-3 . 3a + l 4z + 17
10.
-4a; <c — 2 a; — 2
EQUATIONS INVOLVING FRACTIONS 287
11 3x-2 = 2xi + 15x + 2$ 2x-\
" 23 + 3 2^ + 53 + 3 3 + 1 '
23-3 3-8 _ x + 2
23 + 2 53 + 2 23 + 2*
12.
20^ + 73-3 3a; + l=1
9a^-l 33-1
732 + ll3 + 4 3 + 3 ^73 + 11
6^ + 13a; + 5 23 + 1 33 + 5
33 + 1 a;-3 _= 2ar2-10a; + 12
5o;-7 23-7 10.-^-493 + 49'
199. Sometimes it is best to add fractions before multiplying
by the L. C. M. and in other cases to multiply by the L. C. M.
of part of the denominators first, and, after simplifying, multiply
by the L. C. M. of the remaining denominators.
Ex. 1. Solve the equation
111
3—2 3 — 1 3 — 4 3 — 3
Adding fractions on the right and left,
1 1
(3-2)(3-l) (3-4)(3-3)
(1)
(2)
Multiplying by L. CM., (3-4)(3-3)=(3-2)(3-l). (3)
Hence 4 3 = 10, (4)
and 3 = 2\. (5)
Check by substituting 3 = 2£ in equation (1).
288 LITERAL FRACTIONS
Ex. 2. Solve the equation :
4t-3 t-2=2t-2
16 4 5t + 2
3.
8 dx + 2
7* + 3 21t + 9 = 17<-3 2
' 5 15 3* + ll
11^-15 33^ + 15^5^ + 5
10 30 v-5 '
x — 1 x — 2 _x — 3 a — 4
#— 2 3 — a; a; — 4 a; — 5
(1)
Multiplying by 16, 4 i - 3 -4 « + 8 = ^|f=^- (2)
of-)-*
Hence, ^W" <3>
Multiplying by 5 1 + 2, 25 £ + 10 = 32 £-32. (4)
Hence, 7* = 42, (5)
and I-.6L (6)
Check by substituting £ = 6 in equation (1).
EXERCISES
3x + 6 9a; + 3_ x + 1 *
5 15 6 x -8
7a:+l 14s-22 = lla; + 5
' 12 24 8 a- 28*
3z + 4 12a + l 5ic-l
EQUATIONS INVOLVING FRACTIONS 289
? 1 2 1 4
X
-1
2
X + l
a;
-2
4a: + l
X
X
-2
-3
a;
X
-3
-4
X —
X —
i*
x — 5
6 — x
X
9
-7
X
9
-2
5
X —
8
5
x + l
X
-1
+x
-2
X —
3
x — 5
9.
10.
x— 2 3— # #— 4 a— 6
SIMULTANEOUS FRACTIONAL EQUATIONS
200. When pairs of fractional equations are given, each
should be reduced to the integral form before eliminating, ex-
cept in special cases like those in the second following illustra-
tive example.
Ex. 1. Solve the equations :
x — y x + y tf — y*
3 2 -18
(2)
\2x-y x—3y (2x — y)(x-3y)
From (1) by M, 4 (x + y) + 6(a> - y) = 36. (3)
BjF,D, 5x-y = l8. (4)
From (2) by M, 3(x - Sy) - 2 (2 x -y) = - 18. (5)
VyF, D, x + ly = l%. (6)
From (4) by M, 35x-ly = 125. (7)
Adding (6) and (7), 36 x = 144. (8)
ByZ>, a = 4. (9)
Substitute x = 4 in (6), y « 2. (10)
Check by substituting x = 4, y = 2 in (1) and (2).
290
LITERAL FRACTIONS
Ex. 2. Solve the equations
x y
20_21
Las y
= 3.
(1)
(2)
In this case it is best to solve the equations for - and - in-
stead of for x and y.
From (1) by M, — +
x
?i = 14.
y
(3)
Adding (2) and (3),
21 = 17.
X
(4)
Hence by D,
1_1
x 2
(5)
Substituting - = - in (1),
1 1
y 3*
(6)
From (5) and (6) by M,
x = 2,y = 3.
(7)
EXERCISES
Solve the following equations :
1. {
2a?-l 3y-l_
— «y
oj + 1 y+1 (aj + lXy + l)'
cc -f- 2 2x-l_ 5xy
I2y-1 y + 1 (2y-l)(y+l)
2. 4
3a5 + 2_:B + l
3^-5 f — 1*
3a;-2 3z-l
2
y + 1 y-1 (2,_l)(y + l)
3.
PROBLEMS INVOLVING FRACTIONS
(3 2
291
5-6- = 2,
t v
V t
f- + - = 21,
x y
I-S— la
a;
y
a o
3
ft
+ 1-? = 24.
la &
6. <
!l
= -4,
-6 + Ll = 52.
7.4
ri2_io_1
a; ?/
a 6
a 2/
c , d
U y
h
1.
9. <
10.^
r1.1 ia
- + - = 16,
a; j/
1 + 1=14,
y z
i+i=i2.
2 X
1 ■ 1
- + -=«,
x y
- + -=&,
y z
- + - = c
Iz a;
<>
In 9 first add all three equations, and from half the sum subtract
each equation separately. Likewise in 10.
PROBLEMS LEADING TO FRACTIONAL EQUATIONS
In solving the following problems use one or two unknowns
as may be found most convenient.
1. There are two numbers whose sum is 51 such that if the
greater is divided by their difference, the quotient is 3^. Find
the numbers.
2. There are two numbers whose sum is 91 such that if the
greater is divided by their difference, the quotient is 7. Find
the numbers.
3. There are two numbers whose sum is s such that if the
greater is divided by their difference, the quotient is q. Find
an expression in terms of s and q representing each number.
Solve 1 and 2 by substituting in the formula just obtained.
4. What number must be subtracted from each term of the
fraction \\ so that the result shall be equal to ^ ?
292 LITERAL FRACTIONS
5. What number must be subtracted from each term of the
fraction f ^ so that the result shall be equal to f ?
6. What number must be subtracted from each term of the
fraction" - so that the result shall be equal to - ? Solve 4 and
o d
5 by substituting in the formula obtained under 6.
7. What number must be added to each term of the fraction
\ to obtain a fraction equal to \$ ?
8. What number must be added to each term of the fraction
- to obtain a fraction equal to — ?
b H d
By means of the formula thus obtained, solve problem 7, and also
4, 5, and 6. (See work under Problem 23, p. 114.)
9. There are two numbers whose difference is 153. If their
sum is divided by the smaller, the quotient is equal to f£.
Find the numbers.
10. There are two numbers whose difference is b. If their
sum is divided by the smaller, the quotient is q. Find the
numbers. Solve 9 by substituting in this formula.
11. Divide 548 into 2 parts, such that 7 times the first
shall exceed 3 times the second by 474.
12. There are two numbers whose sum is 48 such that 3
times the first is 8 more than 5 times the second. Find both
numbers.
13. There are two numbers whose sum is s such that a times
the first is b more than c times the second. Find both numbers.
14. What number must be subtracted from each of the num-
bers 12, 15, 19, and 25 in order that the remainders may form a
proportion when taken in the given order ?
15. What number must be added to each of the numbers 13,
21, 3, and 8 so that the sums shall be in proportion when taken
in the order given ?
PROBLEMS INVOLVING FRACTIONS 293
16. What number must be added to each of the numbers' a,
b, c, d so that the sums shall be in proportion when taken in
the given order ?
17. What number must be subtracted from each of the
numbers a, b, c, d so that the remainders shall be in proportion
when taken in the given order ?
Compare the results in 16 and 17 and explain the relation between
them. (See remark under Problem 23, page 114.)
Solve 14 and 15 by substituting in the formulas obtained in 16
and 17.
18. There is a number composed of two digits whose sum is
11. If the number is divided by the difference between the
digits, the quotient is 16f. Find the number, the tens' digit
being the larger.
19. There is a number composed of two digits whose sum
is s. If the number is divided by the difference between the
digits, the quotient is q. Find the number, the tens' digit being
the larger.
20. Illustrative Problem. A can do a piece of work in 8 days,
B can do it in 10 days. In how many days can they do it
together ?
Since A can do the work in 8 days, in one day he can do \ of it,
and since B can do it in 10 days, in one day he can do T^ of it. If x
is the number of days required when both work together, in one day
they can do - of it. Hence we have the equation,
x
1 + 1 = 1
8 10 a;
21. A can do a piece of work in 12 days and B can do
it in 9 days. How long will it take both working together
to do it ?
22. A pipe can fill a cistern in 11 hours and another in 13
hours. How long will it require both pipes to fill it ?
294 LITERAL FB ACTIONS
23. A can do a piece of work in a days and B can do it in b
days. How long will it take both together to do it ?
24. A cistern can be filled by one pipe in 20 minutes and
emptied by another in 30 minutes. How long will it take to fill
the cistern when both are running together ?
25. A pipe can fill a cistern in 12 hours, another in 10 hours,
and a third can empty it in 8 hours. How long will it require
to fill the cistern when they are all running ?
26. A man can do a piece of work in 18 days, another in 21
days, a third in 24 days, and a fourth in 10 days. How long
will it require them when all are working together ?
27. A and B working together can do a piece of work in 12
days. B and C working together can do it in 13 days, and A
and C working together can do it in 10 days. How long will
it require each to do it when working alone ?
Suggestion : Let a = the fraction of the work A can do in one day,
b = the fraction of the work B can do in one day, and c = the fraction
of the work C can do in one day.
Then a + b-^, b + c = ^, c + a = ^.
28. A and B working together can do a piece of work in I
days. B and C can do it in m days and C and A can do it in
n days. How long will it require each working alone ?
29. The circumference of the rear wheel of a carriage is 1.8
feet more than that of the front wheel. In running one mile
the front wheel makes 48 revolutions more than the rear wheel.
Find the circumference of each wheel.
If x is the number of feet in the circumference of the front wheel,
then — — is the number of revolutions in going one mile.
x
30. The circumference of the rear wheel of a carriage is 1
foot more than that of the front wheel. In going one mile the
two wheels together make 920 revolutions. Find the circum-
ference of each.
MISCELLANEOUS PROBLEMS 295
31. The distance from Chicago to Minneapolis is 420 miles.
By increasing the speed of a certain train 7 miles per hour the
running time is decreased by 2 hours. Find the speed of the
train.
4°0
If r is the original rate of the train, then — — is the running time.
r
32. The distance from New York to Buffalo is 442 miles.
By decreasing speed of a fast freight 8 miles per hour the
running time is increased 4 hours. Find the speed of the
freight.
33. A motor boat goes 10 miles per hour in still water. In
10 hours the boat goes 42 miles up a river and back again.
What is rate of the current ?
34. A train leaving New York over the Pennsylvania Boad
requires 9 hours to overtake a train leaving Philadelphia west-
ward at the same time. If the Philadelphia train had started
toward New York, they would have met in one hour. Find the
rate of each train, the distance from New York to Philadelphia
being 90 miles.
35. The average of the numbers x, 34, 0, — 58, — 19, 0, — 20,
and y is 12; while the average of 2x,3y, — 18, 50, and — 30 is
— 4. Find x and y.
36. The length of a rectangle is 8 feet greater, and its width
is 4 feet greater, than the side of a certain square. The sum
of the areas of the square and rectangle is 736 square feet.
Find the dimensions of each.
37. The difference between the areas of a circle and its cir-
cumscribed square is 12 square inches. Find the radius of
the circle. (See problem 33, p. 239.)
38. The difference between the areas of a circle ard its
inscribed square is 12 square inches. Find the radius of the
circle.
296 LITERAL FRACTIONS
39. The difference between the areas of a circle and the
regular inscribed hexagon is 12 square inches. Find the radius
of the circle.
40. The altitude of an equilateral triangle is 6. Find its
side and also its area. Find the side and area, if the altitude
is h.
41. The radius of a circle is 3 feet. Find the area of the
regular circumscribed hexagon. Find the area if the radius
is r feet.
42. The radius of a circle is r. Find the difference between
the areas of the circle and the regular circumscribed hexagon.
43. The difference between the areas of a circle and the
regular circumscribed hexagon is 9 square inches. Find the
radius of the circle.
44. A circle is inscribed in a square and another circum-
scribed about it. The area of the ring formed by the two cir-
cles is 25 square inches. How long is the side of each square?
45. A square is inscribed in a circle and another circum-
scribed about it. The area of the strip inclosed by the two
squares is 25 square inches. Find the radius of the circle.
In the following five problems solve the resulting literal equations
for each letter in terms of the others, and for each solution state a
corresponding problem.
46. A hound pursuing a deer gains 400 yards in 25 minutes.
If the deer runs 1300 yards a minute, how fast does the hound
run ? If the hound gains vx yards in t minutes and the deer
runs v2 yards per minute, find the speed of the hound.
47. A disabled steamer 240 knots from port is making only
4 knots an hour. By wireless telegraphy she signals a tug,
which comes out to meet her at 17 knots an hour. In how
long a time will they meet ? If the steamer is s knots from
port and making i>j knots per hour, and if the tug makes v2
knots per hour, find how long before they will meet.
MISCELLANEOUS PROBLEMS 297
48. A motor boat starts 7| miles behind a sailboat and runs
11 miles per hour while the sailboat makes 6^ miles per hour.
How far apart will they be after sailing 1^ hours ? If the
motor boat starts s miles behind the sailboat and runs vy miles
per hour, while the sailboat runs v2 miles per hour, how far
apart will they be in t hours ?
49. An ocean liner making 21 knots an hour leaves port
when a freight boat making 8 knots an hour is already 1240
knots out. In how long a time will the two boats be 280 knots
apart ? Is there more than one such position ? If the liner
makes v^ knots per hour and the freight boat, which is st knots
out, makes v2 knots per hour, how long before they will be s2
knots apart?
50. A passenger train running 45 miles per hour leaves one
terminal of a railroad at the same time that a freight running
18 miles per hour leaves the other. If the distance is 500 miles,
in how many hours will they meet ? If they meet in 8 hours,
how long is the road ? If the rates of the trains are vx and v2
and the road is s miles long, find the time.
51. In going 1200 yards the rear wheel of a carriage makes
60 revolutions less than the front wheel. If the circumference
of each wheel be increased by 3 feet, the rear wheel will make
only 40 revolutions less than the front wheel. Find the cir-
cumference of each wheel.
SCIENCE.
High School Physics.
By Professor Henry S. Carhart, of the University of Michigan, and
H. N. CHUTE, of the Ann Arbor High School. New edition, thor-
oughly revised. i2rao, cloth, 440 pages. Price, $ 1.25.
NO other text-book on Physics, published in this country, has
ever enjoyed the popularity or the success that has, from
the first year of its publication, attended Carhart and Chute's
High School Physics. Throughout the country the demand for
the book has been far in excess of that for any other manual
covering the same field, while in many states it has been used
more widely than all its competitors combined. The new edition
of the book is a distinct improvement on its predecessor. A
comparison of the two books will show numerous changes in
details and a smoothing down of the rougher spots. Physics is
not an easy subject ; but the authors have made an honest effort
to relieve the difficulties for immature students without such an
emasculation of the subject as to diminish its value for either
discipline or scientific information.
The problems have all been replaced by entirely new ones, in
which the purpose kept in view is the concrete illustration of
principles, with a minimum of arithmetical computation. Atten-
tion has been given also to the careful grading of the problems.
None of them have been inserted as puzzles to test the student's
intellectual skill, while, as a whole, they are distinctly easier than
were those of the preceding edition.
The book remains, as it was before, the most attractive manual
for the study of Physics that has been published. Its method
cannot be improved. The principles of the science are stated in
a simple, clear, and direct manner ; their application is illustrated
by apt experiments ; finally, numerous practical problems test the
pupil's mastery of the subject.
Its arrangement of material, its completeness along the lines of
the most recent developments of physical science, its excellent
woodcuts, its wealth of problems, and the ease and precision of
its style deserve and have received unstinted praise.
62
SCIENCE.
First Principles of Chemistry.
By Raymond B. Brownlee, Far Rockaway High School; Robert
W. Fuller, Stuyvesant High School ; William J. Hancock, Eras-
mus Hall High School; Michael D. Sohon, Morris High School;
and Jesse E. Whitsit, DeWitt Clinton High School; all of New
York City.
THIS manual includes a treatment of the common elements
and their important compounds, together with a full dis-
cussion of the fundamental theories of chemistry. These theo-
ries are simply, clearly, and accurately stated without being buried
in a confusing mass of detail. The recent developments in chem-
ical theory are given due recognition.
In the development of laws and hypotheses, the historical order
is followed, the experimental facts in each case being described
before stating the discovery that resulted from them. Theoret-
ical topics are introduced as soon as the pupil is able to take in
their full significance ; this, on the ground that to delay their
presentation is to deprive the pupil of a very useful tool in his
acquisition of chemical ideas.
This book marks a step in advance by its omission of much
material that has found a place in other elementary manuals
through tradition rather than for its real value. On the other
hand, a number of the metallic elements, sometimes neglected, are
given the thorough treatment that their industrial importance
deserves. One of the chief aims of the descriptive matter is to
show the student the many points of contact between the life
about him and the science he is studying.
An important feature of the book is the brief summary and the
test exercises given at the end of each chapter. The summary
is a series of pithy statements emphasizing the essentials and
affording systematic review.
This book has been prepared by five teachers of long experi-
ence in both college and secondary school work. Three years of
careful discussion and revision by the authors have produced a
unified text-book especially planned for the beginner in chemistry.
56
SCIENCE.
Practical Physiography.
By Dr. Harold Wellman Fairbanks, of Berkeley, California.
8vo, cloth, 570 pages. 403 Illustrations, Price, $1.60.
IN this volume the author has tried to work out a practical, con-
crete treatment of the subject of Physical Geography. The
book is intended as an aid to study, not as a compendium of
information ; consequently a description of the world as a whole
is omitted. Attention is devoted specifically to the region of the
United States, and the typical examples afforded by it are studied
as representatives of universal processes.
The purely descriptive method has been discarded as far as
practicable, the object being to lead the student to investigate
and find out for himself. Instead of being told everything, he is
asked to use his observing and reasoning powers.
No separate chapters are devoted to the relation between
physical nature and life, but, instead, this relation is brought
out in its appropriate place in connection with each topic through-
out the book. Such an arrangement, it is believed, will make the
whole matter much more vital.
Another feature which the author trusts will meet with favor
from the practical teacher is the distribution of the questions and
exercises throughout each chapter, in close connection with the
descriptive portions to which they refer. The placing of ques-
tions and exercises by themselves at the close of each chapter, as
is done in many text-books on the subject, puts a premium upon
mere memorizing of the text and the omission of all practical
work. It is not expected, however, that the pupils will be able
to answer all the questions without aid and direction from the
teacher.
The illustrations are a marked feature of the book. Photo-
graphs have been used wherever possible, as they appeal with
much more force to pupils of high school age than do diagrams
or sketches. Most of the views are from the author's own
negatives.
* 68
SCIENCE.
Lessons in Elementary Botany for Secondary
Schools.
By Professor Thomas H. Macbride, Iowa State University. i6mo,
cloth, 244 pages. Price, 80 cents.
THIS is a new edition of Professor Macbride's well-known
text-book, enlarged and revised. This edition contains
ten new full-page plates, and while it calls for no material
but such as is easily accessible to every teacher, its method
is nevertheless in perfect harmony with the best instruc-
tion of the time. The student is not asked to study illustra-
tions or text ; he is sent directly to the plants themselves
and shown how to study these and observe for himself the vari-
ous problems of vegetable life. The plants passed in review are
those which are more or less familiar to every one, and the life
history of which every child should know.
Austin C. Apgar, State Normal School, Trenton, N.J. : There are many
points in the book which please me. Not the least of these is that the
author has not been misled by the craze for that " logical order " which
begins with protoplasm and, some time or other, if the subject be pursued
long enough, reaches such things as can easily be found and examined.
He begins with the known and gradually advances to the unknown, —
the only order in which successful teaching can be accomplished.
A Key to Native Plants.
To accompany Macbride's Elementary Botany.
IN response to a desire expressed by many teachers, Professor
Macbride has prepared a Key to the More Common Species
of Native and Cultivated Plants occurring in the Northern United
States. The material thus furnished has never before been
offered in such convenient form for class-room work. Its use will
in no way change the original plan of the Elementary Botany,
but it offers many advantages for additional study.
The Key will be furnished without charge to those ordering
the author's Botany. Copies ordered without the Botany will be
furnished at twenty-five cents each.
60
MA THE AT A TICS.
A Practical Course in Bookkeeping.
By Carlos B. Ellis, of the Central High School, Springfield, Mass.
8vo, cloth, coo pages. Price, $0.00.
THIS book is the outgrowth of long experience in teaching
and in actual business. It is designed to give pupils the
knowledge of the principles of bookkeeping and of the customs
of business that every man needs, together with sufficient practice
in the application of these principles and customs to meet the
requirements of the best business offices.
A feature of the book is its logical development of principles
whereby the pupil makes a careful study of the essential accounts
instead of merely memorizing rules. The pupil who understands
the uses of the principal accounts and the relations which each
account sustains to the others, will not need to remember rules
for journalizing. In a word, the book is designed to teach the
pupil to think.
A large amount of work is provided for supplementary drill,
and there is abundant practice in the use of business forms and
of the variations in method adapted to special requirements.
Introduction to Geometry.
By William Schoch, of the Crane Manual Training High School,
Chicago. i2mo, cloth, 142 pages. Price, 60 cents.
THE aim of the author is to furnish a series of exercises and
problems that will enable the pupil to pass easily from the
individual notions of his old experience to a firm knowledge of
the elementary facts and concepts of Geometry. The pupil is
to convince himself of geometric truths through measuring and
drawing, and thus gradually to come into possession of the mate-
rial with which formal Geometry deals. Along with this, the
practical side of the subject is taken into account, the pupil's
knowledge being tested constantly by his power to apply it.
61
MA THE MA TICS.
Plane and Spherical Trigonometry.
By President ELMER A. Lyman, Michigan State Normal College, and
Professor EDWIN C. GODDARD, University of Michigan. Cloth, 146
pages. Price, 90 cents.
THIS text-book aims to furnish sufficient material in analytical
trigonometry, and also in the solution of the triangle, both
of which have been adequately met heretofore by no single book.
Particular attention has also been given to the proofs of the
formulae for the functions of a ± y8. Nearly all other text-books
treat the same line as both positive and negative in the same
discussion, thus vitiating the proof, and in many cases proofs
are given for acute angles and are then supposed to be estab-
lished without further discussion for all angles. These diffi-
culties have been avoided by so stating the proofs that the
language applies to figures involving any angles and proving
the general case algebraically to avoid drawing an indefinite
number of such figures.
Computation Tables.
By Lyman and Goddard. Price, 50 cents.
Plane and Spherical Trigonometry, with Com-
putation Tables.
By Lyman and Goddard. Complete edition. Cloth, 214 pages.
Price, $ 1.20.
Principles of Plane Geometry.
By J. W. MacDonald, Agent of the Massachusetts Board of Educa-
tion. i6mo, paper, 70 pages. Price, 30 cents.
A Primary Algebra.
By J. W. MacDonald, Agent of the Massachusetts Board of Educa-
tion. i6mo, cloth. The Complete Edition, 218 pages (containing the
Teacher's Guide, the Student's Manual, and Answers to Problems).
Price, 75 cents. The Student's Manual, cloth, 92 pages. Price, 30 cents.
64
MATHEMATICS.
Plane Trigonometry.
By Professor R. D. BOHANNAN, of the State University, Columbus,
Ohio. Cloth, 379 pages. Price, $2.50.
SOME of the features of this book are the use of triangles hav-
ing variety of form, but always such form as they should
have if they had been actually measured in the field ; the sugges-
tion of simple laboratory exercises ; problems from related sub-
jects, such as Physics, Analytical Geometry, and Surveying.
With a view to minimize the difficulties of the subject, the
author has ventured to present the subjects in the following
order: the sine, its inverse and reciprocal; the cosine, its in-
verse and reciprocal ; the sine and the cosine family in union ; the
tangent, its inverse and reciprocal ; all the functions in union.
A pamphlet containing four-place mathematical tables is fur-
nished free with the book.
Calculus with Applications
By Ellen Hayes, Professor of Mathematics at Wellesley College.
i2mo, cloth, 170 pages. Price, $1.20.
THIS book is a reading lesson in applied mathematics, intended
for persons who wish, without taking long courses in mathe-
matics, to know what the calculus is and how to use it, either as
applied to other sciences, or for purposes of general culture.
Nothing is included in the book that is not a means to this
end. All fancy exercises are avoided, and the problems are for
the most part real ones from mechanics or astronomy.
Logarithmic and Other Mathematical Tables.
By William J. Hussey, Professor of Astronomy in the Leland Stan-
ford Junior University, California. 8vo, cloth, 148 pages. Price, $1.00.
VARIOUS mechanical devices make this work specially easy to
consult ; and the large, clear, open page enables one readily
to find the numbers sought. It commends itself at once to the
eye, as a piece of careful and successful book-making.
65
THIS BOOK IS DUE ON THE LAST DATE
STAMPED BELOW
AN INITIAL FINE OF 25 CENTS
WILL BE ASSESSED FOR FAILURE TO RETURN
THIS BOOK ON THE DATE DUE. THE PENALTY
WILL INCREASE TO SO CENTS ON THE FOURTH
DAY AND TO $1.00 ON THE SEVENTH DAY
OVERDUE.
OCT 24 IMS
MAR 12 101[,BRARV USE
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THE UNIVERSITY OF CALIFORNIA LIBRARY
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