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ESSENTIALS OF ALGEBRA
COMPLETE COURSE
(AN ADEQUATE PREPARATION FOR THE COLLEGE OR TECHNICAL SCHOOL)
FOR
SECONDARY. SCHOOLS
BY
JOHN C. STONE, A.M.,
MICHIGAN STATE NORMAL COLLEGE, CO-AUTHOR OF THE
SOUTHWORTH-STONE ARITHMETICS
JAMES F. MILLIS, A.M.,
THE SHORTRIDGE HIGH SCHOOL, INDIANAPOLIS, INDIANA
OV TTOAA' ttA.Aa TToXv
BENJ. H. SANBBrN & CO.
BOSTON NEW YORK CHICAGO
Wv^V J^/^
S7^
Copyright, 1905,
Br JOHN O. STONE and JAMES F. MILLI8.
U '■
PREFACE.
That there is a place for a new Algebra recent corre-
spondence has abundantly proven. This book contains features
which will not only arouse and sustain interest in the subject,
but are now demanded in an elementary course in algebra.
It is believed that the book is modern and progressive, yet
free from fads, and in no sense extreme. It is simple in style
and rigorous in treatment. In order to make this possible, some
of the topics commonly treated in elementary algebra, difficult
for the beginner, and of comparatively little value to him, have
been postponed or omitted. Highest common factors by divis-
ion, and square and cube roots of polynomials and of arith-
metical numbers by formulse, have been put into the Appendix.
These may be read whenever the teacher thinks advisable, or
they may be omitted without detriment to the subsequent
work. The fundamental laws of numbers have been explained
and carefully illustrated when introduced, but rigorous proofs
of these laws have been put into the Appendix, where they may
be read when the pupil has become sufficiently familiar with
algebraic processes. The beginning pupil should not be over-
burdened with the proof of certain simple principles ; yet he
must see, before he leaves the subject, that there is a demon-
stration for every principle.
The following are some of the special features of the book.
iii
1 '4732
iv PREFACE
Extension of number. The notion of a general number, rep-
resented by some letter of the alphabet, is introduced by many
illustrations. The pupil is shown that all of the new kinds of
numbers, — negative numbers, surds, imaginary numbers — like
fractions, arise logicaMy from an attempt to make the funda-
mental processes universal.
Checks. The pupil is taught from the beginning how to
check his work — multiplication, factoring, etc. — by substitut-
ing particular values for the general numbers in the identities
which he obtains. This gives him constant practice in the in-
terpretation of the symbols of algebra, broadens his grasp of
the significance of general number, and trains him to be ac-
curate and self-reliant.
Factors. The subject of rational factors is thoroughly dis-
cussed, and made the basis of subsequent work whenever
possible. The pupil is taught to reduce all expressions to be
factored, when possible, to one of the fundamental type-
forms ; otherwise to try the remainder theorem. Abundant
miscellaneous exercises are given.
The equation. Special care has been taken in the treatment
of equations. The pupil is required to see that every step in
the solution of an equation is the application of a fundamental
principle. " Transpose " and " Clear of fractions " are terms
}vhich the pupil is allowed to use only after he knows the
meanings of the processes which they represent.
Correlation with science. The formula. Any letter is used
as an unknown number and formulae are solved for any
letter in them. For tliis purpose many fundamental formulge
are borrowed from science. Consequently, the algebra which
PREFACE y
the pupil will meet in his subsequent science work will be
familiar to him.
Expression of laws by equations. It is shown that the equa-
tion may be used to express a law of science. Practical appli-
cation is thus made of the theory of variation. Exercise is
given in interpreting the law expressed by an equation, and in
writing the equation which expresses a stated law. The sim-
ple laws and formulae of science are used in the work.
The graph. The graph is used in the treatment of indeter-
minate equations, both linear and quadratic, as a means of
illustration. It is used to illustrate the different kinds of sys-
tems of equations, such as impossible systems, etc. The
graphic solution of a system is shown, and the solution of a
single equation in a single unknown is found by solving a
system.
Exercises. The exercises are numerous, new, and well
graded. They are sufficiently difficult to call for effort on the
part of the pupil.
Reviews. Miscellaneous review exercises are distrilmted
throughout the book. These reviews consist of many exer-
cises and questions intended to bring out in review the funda-
mental principles of algebra.
The Brief Gourse consists of the first twenty-one chapters
of the Complete Course and is intended for classes that go only
through quadratics, inequalities, ratio and proportion, and the
theory of exponents. The Complete Course furnishes an ade-
quate preparation for the College or the Technical School.
The authors take pleasure in acknowledging their indebt-
edness for many valuable suggestions to Professor E. A.
Vi PREFACE
Lyman and to Dr. N. A. Harvey of the Michigan State
Normal College, who have read both the manuscript and the
proof; to Professor J. L. Love of Harvard University, and
John A. Avery, of the English High School, Somerville, Mass-
achusetts, who have carefully read all of the manuscript ; and
to E. Harry English, Supervisor of Mathematics in the Wash-
ington (D. C.) High Schools, who has carefully read the first
proof, and made helpful suggestions.
J. C. S. •
June, 1905. J. F. M.
PUBLISHER'S NOTE
Before accepting the manuscript of this Algebra it was
placed in the hands of Professor James Lee Love of the Law-
rence Scientific School of Harvard University, who gave it his"
cordial approval and endorsement.
The Southworth-Stone Arithmetics, published in January,
1904, have in one year been introduced in more schools, ex-
ceeded in sale any competitors, and received more general
commendation from the leading superintendents and teachers
than any other Arithmetics ever published in this country.
'Many of the features that have made these Arithmetics so
deservedly popular have been introduced in this Algebra, and
the book, with its large clear type and open page, will be cor-
dially welcomed by the educational public.
CONTENTS.
CHAPTER I.
PAGE
Introduction 1
Definitions 1
Signs of Grouping " . . 6
General Number 8
CHAPTER II.
Fundamental Laws of Numbers 16
Fundamental Laws 16
The Equation 19
Axioms 21
CHAPTER III.
Negative Number 30
Definitions 30
Addition of Algebraic Numbers 35
Subtraction of Algebraic Numbers 37
Multiplication of Algebraic Numbers 39
Division of Algebraic Numbers 41
Exercises for Review I 43
CHAPTER IV.
Addition and Subtraction of Literal Expressions 46
Addition of Monomials 46
Addition of Polynomials 47
Subtraction 49
Removing Signs of Grouping , . 53
Inserting Signs of Grouping 58
vii
Viii CONTENTS
CHAPTER V.
PAGE
Multiplication of Literal Expressions 54
Laws of Exponents 54
Multiplication of Monomials 54
Multiplication of Polynomials by Monomials 56
Multiplication of one Polynomial by Another 60
CHAPTER VI.
Division of Literal Expressions 64
Law of Exponents 64
Meaning of a", .... 64
Division of Monomials 64
Division of a Poljaiomial by a Monomial 65
Division of one Polynomial by Another 66
The Fraction 71
CHAPTER VII.
Powers and Roots 75
Powers 75
Roots :.) 83
Roots of Monomials 86
Roots of Polynomials by Inspection 90
CHAPTER VIII.
Special Products and Quotients 92
Products 92
Quotients 96
Exercises for Review II 100
CHAPTER IX.
Factors 103
Factors— Type Forms 103
The Remainder Theorem 123
Synthetic Division 125
CONTENTS ix
CHAPTER X.
PAGE
Common Factors and MuLTiPiiES 130
Definitions 130
Highest Common Factor 131
Lowest Common Multiple 135
CHAPTER XI.
Fractions 139
Definitions, Laws, etc 139
Addition and Subtraction of Fractions 146
Multiplication of Fractions 150
Division of Fractions , 153
Complex Fractions 155
Exercises for Review III 157
CHAPTER XII.
Linear Equations — One Unknown 161
Definitions , 161
The Solution of a Linear Equation 165
Fractional Equations 167
The Formula 170
CHAPTER XIII.
Linear Equations— Systems 180
Definitions 180
Elimination 183
Systems of Fractional Equations 190
The Graph of an Equation 193
Graph of a System 197
Exercises for Review IV 209
CHAPTER XIV.
Surds and Imaginary Numbers 211
Definitions, Reductions, etc 211
Addition and Subtraction of Surds 215
X CONTENTS
PAGH
Rationalization ^ , 220
Powers and Roots of Surds 221
Imaginary Numbers. ... 223
Geometric Representation of Imaginary Numbers 228
CHAPTER XV.
Quadratic Equations— One Unknown 230
Definitions 230
Solution of Pure Quadratics 231
Solution of Complete Quadratics 233
The Discriminant 241
Solving a Quadratic Formula 244
CHAPTER XVI.
Higher Equations— Equations involving Surds— One Unknown 253
Solution of Equations of Higher Degree 253
Solution of Equations involving Surds 256
Introduction of New Solutions 257
CHAPTER XVII.
Systems involving Quadratic and Higher Equations 261
Solution of Systems 261
Special Devices 270
Graphs of Systems of Quadratics 273
Exercises for Review V 285
CHAPTER XVIII.
Inequalities 291
Definitions 291
Principles 292
CHAPTER XIX.
Ratio and Proportion 298
Definitions 298
Principles , 300
CONTENTS Xi
CHAPTER XX.
PAGK
Variation. Algebraic Expression of Law 310
Definitions, etc 310
Law expressed by an Equation 314
Limits of Variables '. 318
Properties of Zero 321
~ CHAPTER XXL
Fractional and Negative Exponents 324
Definitions and Interpretations 324
Exercises for Review VI 334
CHAPTER XXII.
Permutations and Combinations 337
Permutations, — Definitions, etc 337
Combinations, — Definitions, etc 342
CHAPTER XXIII.
The Binomial Theorem 346
Proof — The Exponent a Positive Integer 346
Exponent Negative or Fractional 350
Extraction of Roots by use of Binomial Theorem 352
CHAPTER XXIV.
Progressions. 354
Series, — Definitions, etc 354.
Arithmetical Progression 355
Geometrical Progression ". 362
Harmonical Progression 369
Exercises for Review VII 371
CHAPTER XXV.
Undetermined Coefficients 375
Tlieorem of Undetermined Coefficients 376
Xii CONTENTS
PAGE
Expansion of Fractions into Series 378
Expansion of Surds into Series ... 380
Reversion of Series 382
Partial Fractions 383
CHAPTER XXVI.
Logarithms 388
Exponential Equations, — Definitions, etc 388
Fundamental Principles of Logarithms 390
Exponential Equations 406
Table of Mantissas t 411
Exercises for Review VIII 407
APPENDIX.
Square and Cube root by the formula 1
The Highest Common Factor and Lowest Common Multiple. 16
The proof of the Fundamental Laws : 23
ESSENTIALS OF ALGEBRA
CHAPTER I.
INTRODUCTION.
1. Algebra. No clear line of distinction will be found
between algebra and arithmetic. Algebra deals with number
and with the same fundamental processes which are dealt with
in arithmetic. The processes of arithmetic, however, are
extended in algebra ; and the meaning which is attached to
number in arithmetic is also extended in algebra. Algebra
deals particularly with the equation.
2. Signs. The signs +, — , X, -^, and =, which are used in
arithmetic, are used with the same meanings in algebra.
Other signs used in algebra will be introduced later, when
they are needed.
3. Fundamental processes. The fundamental processes of
algebra, as in arithmetic, are addition, subtraction, multipli-
cation, and division.
a. Addition. The addition of two or more numbers is the
process of uniting them into oiie whole.
As in arithmetic, two numbers to be added are called
addends, and the result is called the sum.
To indicate addition the sign + is placed between the two
numbers to be added.
Thus, 7 + 3 indicates that 3 is to be added to 7, The result,
which is 10, is called the sum.
2 ALGEBRA
The symbol =, placed between two numbers, indicates that
they have the same value. Thus, 7 + 3=10. This expression is
called an equation, and is read "seven plus three equals ten."
We shall have more to do with equations later.
Three or more numbers may be added by adding two at a
time.
Thus, to add 3, 6, 5, and 2; adding 6 to 3 gives 9; adding 5 to
9 gives 14; adding 2 to 14 gives 16, the sum.
From the foregoing it is evident that to find the sum in a
series of additions we may proceed from left to right, uniting
two numbers at a time.
Thus to find the sum of 4 + 6 + 2 + 9^; we have 4 + 6=10;
10*f2=:12; 12 + 9|=21^.
b. Subtraction. Subtraction is the inverse of addition. Given
one of two numbers and their sum, subtraction is the process
of finding the other number.
Subtraction is expressed by placing the sign — between
the two given numbers. When^ so used, it means that the
second number is to be subtracted from the first.
Thus, 9 — 5 indicates that 5 is to be subtracted from 9.
As in arithmetic, the number obtained by subtracting one
number from another is called the difference, or remainder ;
the number to be subtracted is called the subtrahend ; and
the given sum, or the number from which the subtrahend
is to be subtracted, is called the minuend.
From the definition of subtraction, it follows that to sub-
tract the second of two numbers from the first is to find a
third number such that if it be added to the second, the
sum will equal the first. Hence the following important
principle :
subtrahend + remainder = minuend,
INTRODUCTION 3
A series of two or more subtractions may be performed
by proceeding from left to right, subtracting one number at
a time.
Thus, to find the value of 26 - 12^ - 3| ; 26 - 12| = 13^ ; 13^
— 31 = 10, the value.
Likewise, a series of additions and subtractions may be
performed by proceeding from left to right, performing one
operation at a time.
Thus, to find the value of 100 - 8 + 2| + 10 - 25 ; we have 100-8
= 92 ; 92 + 2| = 94| ; 941 + 10 = 104^ ; 104^— 25=79|.
c. Multiplication. Multiplication has been defined in arith-
metic as the process of taking one number, called the
multiplicand, as many times as there are units in the other,
called the multiplier. It is evident that this definition holds
only when the multiplier is a whole number, and fails when it
is a fraction.
Thus, to multiply 7 by 2| would mean to take 7 as many
times as there are units in 2|, that is, 2| times. This is im-
possible. One cannot do a thing 2| times.
Instead of this definition, which is not sufficiently general,
we shall use the following :
To rmdtiply one number^ called the midtiplicand^ by another^
called the nudtiplier^ is to use the multiplicand as we tnust use
unity to obtain the multiplier. The result is called the product.
For example, let us multiply 3 by 4. To obtain 4 we take
1 + 1 + 1 + 1=4.
Hence, to obtain the product, we take
3 + 3 + 3 + 3=12.
Here we have done to 3 what we did to 1 to obtain 4.
^-^2
GEBRA
Again, to multiply 2 by 4| ; we take
Hence the product i^jf^J^'J^^'^' ^ y^ ^-^^ ::- /J'/i^ _
2 + 2 + 2 + 2 + I + |=8 + ^=9|. ^'^
This definition evidently holds when the multiplier is
either an integer or a fraction, hence we see that it is iiion
geoieral than the old one.
The sign commonly used in arithmetic to indicate mul-
tiplication is the oblique cross X? placed between the num-
bers. When an integral multiplier is written first, the sign
is read '• times ; " and when the second number is consi« vired
the multiplier, the sign is read " multiplied by."
Thus, 8x7=56, is read " 8 times 7 equals 56" or " 8 multi-
plied by 7 equals 56." In this book the second number will be
considered the multiplier, thus, 5x2| is read "5 multiplied
by 2|."
In algebra the dot ( ; ) also indicates multiplication when
placed between two numbers, in a position above that which
would be occupied by the decimal point.
Thus, 8x7 may be written 8 • 7.
A series of two or more multiplications may be performed by
proceeding from left to right, performing one multiplication at
a time.
Thus, to find the value of 7 x 8 x | x 2 ; we have 7x8=56 ;
56 x 1= 84 ; 84 X 2 = 168.
The result obtained by a series of two or more multiplica-
tions is called the product of all of the given numbers. In the
above example 168 is the product of 7, 8, | and 2.
d. Division. Division is the inverse of multiplication.
Given one of two numbers -and their product, division is the
process of finding the other number. The given product, or
INTRODUCTION 5
number to be divided, is called the dividend ; the given number,
or number by which the dividend is divided, is called the
divisor ; and the result, or number found, is called the quotient.
Division is indicated by placing the sign -=- between the two
given numbers. When so used, the second number is the
divisor.
Thus, for "divide 10 by 2," we write 10-^2.
In algebra division is often indicated by use of the horizon-
tal line — , oblique line /, or colon : , placed between the dividend
and divisor.
Thus, 10^2 may be written >/, 10/2, or 10 : 2 ; read " 10 divided
by 2."
From the definition of division we have the following impor-
tant principle :
quotient x divisor = dividend.
Thus, since 20^4=5, then 5x4=20.
A series of two or more divisions may be performed by
proceeding from left to right, performing one division at a
time.
To find the value of 256^8^16 ; we have 256^8=32 ; 32^16=2.
Likewise, a series of multiplications and divisions may be
performed by proceeding from left to right, performing one
operation at a time.
To obtain 28-^4x^x2^3; we have 28-=-4=7 ; 7x9=63; 63x2
=126 ; 126^3=42.
4. Expressions. Any combination of number-symbols indica-
ting one or more processes such as addition, multiplication, etc.,
is called an expression.
Thus, 2x6 — 3 + 1/5 is an expression.
6 ALGEBRA
A rational expression is an expression that contains only indi-
cated additions, subtractions, multiplications, divisions, and
powers of numbers. See §13.
5 . Signs of grouping. An expression is often used as a si7igle
number. To indicate that it is to be so used the expression is
usually enclosed within a sign of grouping, or sign of aggrega-
tion.
We shall use four different signs of grouping to enclose ex-
pressions. They are the parentheses ( ), the brackets [ ], the
braces { }, and the vinculum , the last being placed
above the expression. They all have the same meaning, and
the different signs are used in different places merely for con-
venience. They indicate that the expressions enclosed by them
are to be used collectively, i.e., used as one number.
Thus, 9 — (7—2) indicates that 7—2, as one number, is to be
subtracted from 9. We have 7—2=5 ; 9 — 5=4. The student will
note carefully the difference between 9 — (7— 2) and 9—7—2.
The expression 9-(7-2)=9-[7-2]=9- {7-2; =9-Tr2.
Again, 6 x (5 + 7) indicates that 6 is to be multiplied by 5 + 7 as
one number. Thus 5 + 7=12 ; and 6 x 12=72.
24 4
In the expression ^ — 5— the horizontal line serves three pur-
poses at once. It is a vinculum under the 24—4, a vinculum over
the 2 + 8, and a sign of division. Thus ^:ii=?^ =2.
It follows from the foregoing that in expressions containing
signs of grouping., the operations indicated within the signs of
grouping must be perf or n^ed first.
6. Evaluation of expressions. The value of an expression
is the result obtained by performing all of the operations in-
dicated within it.
Tlius, the value of 38-4 + 12-3 -^ (2 + 3 + 4)- ^^ is 34.
INTRODUCTION 7
MathematiciaiTs have agreed that in expressions containing
additions, subtractions, multiplications, and divisions, the
multiplications and divisions shall take precedence over the
additions and subtractions.* Hence the following rule is to be
observed in the evaluation of all expressions :
If the expression contains no signs of grouping : (1) all
operations of multiplication and division should be perfor7ned in
the order in lohich they are written from left to rights before
any of those of addition and subtraction are performed ; (2) in
the resulting expression which vnll contain only additions and
subtractions^ these should then be performed in order from left
to right.
If the expression contains signs of grouping^ all operations
indicated within these must 'first be performed according to
the foregoing rule. - \f^ >2^^ ^ ^ ^
Example 1. Evaluate 98-(10-2 + 7) + (12-2) -^ (1 + 4).
Performing the operations witfiin the parentheses, we have
10—2 + 7=15; 12—2=10; 1 + 4=5. Hence the expression becomes
98—15 + 10^5. Performing the division, we have 98 — 15 + 2.
Then performing the subtraction and addition, we have
98-15 + 2=85.
Example 2. Evaluate ?-±i_-i^ + 7(2 + 3-1).
We have ^_^ + 7 (2 + 3-1) =-|-4 +7^4
=3-2 + 28
=29.
EXERCISE 1.
1. Find, by using the definition of multiplication, the
product 7x5-
2. Find, in the same way, the product 5 X ^^'
* The reason for this will appear later. Members connected by X
or -r form a single term. § 15.
8 ALGEBRA
3. If the subtrahend is 28 and the remainder 15, what is
the minuend ?
4. If the divisor is 8 and the quotient 12, what is the
dividend ?
5. In the product 9x7 what is the multiplier ?
6. What is the difference between 8x5 and 5x8?
7. What conclusion may you draw from Problem 6 as to
the relation between the multiplicand and multiplier in any
problem ?
Evaluate the following expressions :
8. 12-7-3 + 6. 17. (6-2)X(8 + 3).
9. 16 + 4-3-10 + 8. ■ 18. (5-3)- (5 + 3). (8-7).
10. (16-4)-(8 + 2). 19. (9 + l)-(3 + 2)X(ll-8).
11. 18 + 3 X (7-2). ^Q 7 + 19
12. 6-(3 + 9) ^(6-4). ' '^-^'
13. 4x(7 + 8-3). 21. ^5^X3.
14. 5x(3-2) ^ (6 + 4). ^^ 2 + 12^8-(10-4) 3 ^^
15. 8 + 7x9-8-2x7-5. "'"" 7 ' 12
16. 3X7X2---6--7X4. . 23. jl5 + 3]-|15-3
24. 2 + 3^. (6-4)--(l + 5)
2 .- ^ 2.
1 + 2 -^^.J
26. (6 + 5) • (8 -6) -^7^=^. 4
26. 10-4--[13 + 2-7] 15--9r=^
GENERAL NUMBER.
7. In the following paragraphs we shall discuss a new kind
of number.
The student understands that the character or figure 3 is
not a number, but that it is a symbol which represents a num-
ber. All of the numerals 2, 7, 102, |, 5J, etc., and the Roman
characters I, V, X, etc., represent numbers with which the
INTRODUCTION 9
student is familiar. We shall see that there are' other symbols
which may represent number.
8. Definite number. Each of the symbols 1, 2, 3, 4, etc.,
represents a number with one definite value. For example, 3
represents the single number-idea which we have learned to
call three. The numbers which these symbols represent have
the same values in all problems in which they may occur.
Hence they are called definite, particular, or single-valued
numbers.
9. General number. On the other hand, it is sometimes con-
venient to use number symbols that can represent more than
one value. Such a number evidently can not be represented
by one of the symbols 1, 2, 3, 4, 5, etc. It is usually repre-
sented by some letter of the alphabet., as a or x. Note carefully
the following illustrations.
(1) To get the cost of a number of similar articles at a certain
price per article, we always multiply the price of one article by
the number of articles. In other words we always use the rule :
cost of all articles ={pribe of one article) x {number of articles).
Thus, to get the cost of 7 books at $1.25 per book, we have
cost of 7 books =$1.25x7
=$8.75.
A-lso, to get the cost of 12 pencils at 5 cents a pencil, we have
cost of 12 pencils=5c x 12
=60 cents.
Now, if in place of the statement ^'■cost of all articles,^' which
represents a number, we put c, the first letter of "cost" ; in
place of ''price of one article,'' we put p, the first letter of
'' price" ; and in place of ''number of articles,'' we put n, the
first letter in " number " ; the above rule becomes c=p x n.
This rule, c=pxn, in which c, p, and n represent immbers, ap-
10 ALGEBRA
plies to all particular problems of the above sort. The letters c,
p, and w, represent different particular values in different partic-
ular problems. Thus, in the book problem above, c=$8.75, p=
$1.25, n=7 ; in the pencil problem, c=60c, p=:5c, and n=12.
(2) Simple interest on money is always obtained by the rule
interest = principal x rate x time.
Thus, $200 at 6^, for 3 years, would give
interest = $200 x. 06x3
=$36.
Now, if in the rule
interest = principal x rate x time,
we replace each word by its first letter, we get
i = p X r X t^
in which i, p, r, and f, represent numbers. These letters stand
for different particular numbers in different problems.
Note. It is understood, of course, that " time " refers to the ab-
stract number representing the number of years.
(3) The distance an object, as a train, has moved in a given
time is found by the following rule :
distance = rate x time.
In the same manner as above, this may be written
d = r X t.
d, r, and t represent different values in different particular
problems.
7 2 7 + 2
(4) The statement Trr+-=r() ~~iT)" i^ ^ P^^'ti^^^^^ ^^^^ of the
following principle in addition of fractions :
first numerator second numerator _ first num. -\- Sftcmid num.
common denom. common denom. ~ common denom.
This may also be written
which expresses m a way the rule for adding fractions. Here,
c c c
INTRODUCTION H
/, s, c, represent numbers which, in one set of fractions, will
have certain particular values, and in another set of fractions,
will have other particular values. .
It must now be evident that numbers may sometimes be
represented by letters of the alphabet ; * and that a number
represented by a letter may have any particular value v^hatever.
Thus, a may equal 1, 2, 3, 4, |, ||, or ayiy other definite or par-
ticular number.
Hence, numbers represented \>j letters are called many-
valued, or general numbers. General numbers are largely em-
ployed in the problems of algebra.
10. The same laws which apply to particular or definite
numbers must evidently apply also to general numbers. Gen-
eral 7iumbers may be added, multiplied, or subjected to any
other operation which may be performed upon definite
numbers.
For example, a + 6, a — 6, a x b^ a-i-b, represent respectively
the sum, difference, product, and quotient of the general num-
bers represented by a and b. When a = 8 and 6 = 4, what are the
values of a + 6, a X 6, and a-^b'i
Hereafter, instead of speaking of the " numbers represented
by a, 5, etc.," we shall speak of " the numbers a, ^, etc."
11. Some special notation in the nmltiplication of general
numbers may now be considered.
In addition to the use of the two signs of multiplication dis-
cussed in § 3, multiplication may sometimes be indicated by
the absence of any sic/n between the given numbers.
* By attaching superscripts and subscripts to the letters of our al-
phabet, other symbols for representing numbers may be obtained.
Thus a', a", a'", etc., and oci, a2, cca, tti, etc. The letters of the Greek
alphabet, a, /3, y, S, etc., are also much used for representing numbers.
12 ALGEBRA
Thus, a X 6, ab, and a&, all indicate the product of a and b.
The product of 4, a, and x may be written 4ax.
The product of a + £c and b + y may be written (a+a?) (b + y)^ as
well as (a + a?) x {b-\-y), or (a+a^)-(6 + 2/).
It is clear that on account of the place value feature of our
notation for definite number by which a figure represents both
an intrinsic value and a local value,, the absence of a sign may
not be used to indicate the product of two definite numbers.
Thus 34 does not mean 3x4, the value of which is 12, but
means 3x10 + 4.
12. Factors. Coefficients. Numbers which are multiplied so
as to form a product are called factors of the product.
Thus, in 5aa?, 5, a, and x are called the factors of ^ax.
Any factor of a product, or the product of any two or more
factors, is called the coefficient of the product of the remaining
factors.
For example, in 5aa?, 5 is the coefficient of ax, 5a of x, a of 5a?,
etc.
If a coefficient is a definite number, it is called a numerical
coefficient.
For example, in 5aa?, 5 is the numerical coefficient.
If no numerical coefficient is written, 1 is understood.
Thus, abc is the same as la6c.
13. Powers. If all of the factors of a product are equal, the
jjroduct is called a power of one of the factors. It is usually
written in an abbreviated form.
Thus, aaaaaa is written a^, and is called a power of a.
Similarly xxxx is written x*.
In the power a^, a is called the base and 6 the exponent. The
exponent is a number written at the right of, and above the
INTRODUCTION 13
base, to show the number of times the base is to be used as a
factor to form tlie power.
Thus, a2, read " a square," or " a second power," denotes aa; a^,
read " a cube," or "a third power," denotes, aaa; a*, read "a
fourth power," denotes aaaa; a^, read "a fifth power," denotes
aaaaa ; a"\ read "a exponent ?i2." or "a mth power", denotes
aaa ... to the product of m factors.-
If no exponent is written, tlie^rs^ power is understood.
Thus, a is the same as a^.
14. Literal expressions. An expression containing one or
more general numbers is called a general or literal expression.
For example, 2a + bG—x is a literal expi'ession.
A literal expression may liave any definite, or particular
value, which depends upon the values of the general numbers
involved in it. A definite value of a literal expression may be
found wlien a definite value is assigned to eacli general number
involved in it.
For example, when
a ='10 and a; = 6, 2a— x = 14 ; 2ax = 120 ; 3a^x = 5.
To find the value of a literal expression use tlie method of
§ 6. Note carefully the following examples.
Example 1. Find the value of a + d—c + b, when a=l, 6=2,
c=3, d=4.
The expression becomes 1 + 4—3 + 2.
Performing additions and subtraction, 1 + 4 — 3 + 2=4.
Hence a+d— c + 6=4.
Example 2. Find the value of 3a^ 6^ when a=4, 6=2.
We have 3a^ ¥ =^aa666=3-4-4-2-2-2=384.
Example 3. Find the value of a + 3S— 2^ + (i-f-2, when a=l,
6=2, c=3, d=4.
14 ALGEBRA
We have a + 36—2c + -3-/a
8. 6.5£c + 3.25 = 15.75-6£c. 16. 5A;-5=X; + 3.
26. The equation may be easily used to solve certain kinds
of arithmetical problems.
To solve such problems, the values of certain unknown nuni-
bers are to be found. If some of these unknown numbers are
represented by letters of the alphabet, the conditions of the
problem will lead to one or more equations containing the un-
known numbers. In some problems there will be but one
THE EQUATION 25
equation containing one tinknoimi number. By solving this
equation the value of the unknown number is found.
It is evident that the unknoton numbers in these problems
become the general numbers in the equations. Consequently,
the general numbers of any equation are sometimes called
unknown numl)ers.
The following examples will show in detail how the equation
may be used to solve arithmetical problems.*
Example 1. Of two unequal numbers the greater is twice the
less, and the two together equal 99. Find the numbers.
Let i»?=the less.
Then 2ic=zthe greater^ for " the greater is twice the less."
Hence, iZJ + 2a?=99, for " the two together equal 99."
Therefore 3x=99. § 21.
a?=33. Axiom 4.
And %x=m. Axiom 3.
Hence the less number is 33, and the greater is ^^.
Observe, that when x was taken to represent one of the num-
bers, the next two statements followed from the conditions stated
in the problem.
Example 2. John, Henry and James have among them 30
marbles. John has twice as many as Henry, and James has as
many as John and Henry together. How many has each ?
Let a= the number of marbles Henry has.
Then 2a = the number of marbles John has. Why ?
And 3a = the number of marbles James has. Why ?
Therefore a + 2a + 3a = 30 . Why ?
Adding hke terms, 6a =30. Law of Dis.
* Some authors prefer to represent unknown numbers by only the
last few letters of the alpliabet. Tliere is no good reason for this. On
tlie other hand the student sliould be able to consider the number
represented by any letter in the equation as the unknown number, and
to solve for its value.
26 ALGEBRA
Dividing by 6, a=5. ^ Axiom 4.
Hence 2a =10. ' Axiom 3.
And Sa=15. Axiom 3.
Therefore Henry has 5, John 10, and James 15.
The student should see that these numbers satisfy the conditions
stated in the problem.
Example 3.
The difference between the ages
of two persons is
10 years,
and the sum of their ages is 60 years.
What are their
ages ?
Let
n= age of younger.
Why?
Then
n + 10= age of other.
Why?
Hence
n-\-n + 10=Q0.
Why?
Then
n+n=50.
Why?
2n=50.
Why?
n=25.
Why?
71 + 10=35.
Why?
What are their ages ?
EXERCISE 6.
1. In a certain algebra class there are 24 pupils, and there
are twice as many girls as boys. How many boys are there ?
2. I paid $7 for a pair of shoes and a hat. The hat cost
three-fourths as much as the shoes. AVhat did the hat costp^
Suggestion. If you let 4ic represent the cost of the shoes, what
will represent the cost of the hat ? If you let x represent the
cost of the shoes, what then must represent the cost of the hat ?
Which is preferable ? Why ?
3. A certain man's age is 8 times that of his son, and in ten
years it will be twice as great as the son's age will then be.
What are their present ages ?
Suggestion. If you let x represent the son's age now, what
must represent the age of the father now ? What will represent
the ages of each ten years hence ? What equation follows ?
7
THE EQUATION 27
4. Divide 125 into two ^arts one of which is 35 greater than
the other.
Suggestion. If a? = the smaller part, what must equal the
larger part ? When x = the larger part, what must equal the
smaller part ?
5. Find two numbers whose difference is 32, and one of which
is 3 times the other.
6. The sum of the ages of two boys is 23 years, and one is
7 years older than the other. What are their ages ? * ^
7. A, B, and C buy a piece of property for $1550. A invests -^
twice as much as B, and C invests $50 more than A and B ^ ^
together. How much does each person invest?
8. A farmer sold a number of cows at $45 each, and three
times as many hogs as cows at $13 each. If all sold for $336,
how many cows and how many hogs were sold ? 'Z ♦ 1 1— .
9. A rectangular field is twice as long as it is wide, and the
distance around it is 672 yards. What are its dimensions ?
10. Two men, starting from points 40 miles apart, walk
toward each other, one at the rate of 2 miles an hour, and the
other at the rate of 3 miles an hour. In how many hours will
they meet?
11. If a man can row 2 miles an hour in still water, what /
must be the speed of the current of a stream, if by its aid he ^
can row down the stream 10 i miles in 3 hours ?
12. What number increased by f of itself will make 112? l
{llint. Let 4£c = the number. Why ? Also solve by letting
x= the number.)
/13. What number must be added to 48 in order that twice
the sum shall be 114 ?
14. If to a certain weight you add its half, its third, and 8
28 ALGEBRA
pounds more, the sum will be twice the weight. What is the
weight ?
15. Divide 63 into two parts whose difference is 15.
16. The difference between two numbers is 7, and their sum
is 137. Find the numbers.
17. The sum of two numbers is 80 and the greater one
exceeds the smaller one by 16. What are the numbers?
J 18. Find three consecutive whole numbers whose sum is
^174.
' 19. Two. men were employed to dig a ditch 630 feet long.
One of them dug an average of 45 feet a day, and the other 60
feet a day. How long was required for them to dig the ditch ?
^;^20. A man started on a bicycle to a town 25 miles away.
He rode at the rate of 6 miles an hour. On the way the bicycle
broke, and he walked the remaining distance at the rate of
^\ 2 miles an hour. It required 6i hours to make the whole
trip. How far was he from home when the bicycle broke ?
—21. A started from a place and traveled at the rate of 3
miles an hour, and 3 hours later B • was sent from the same
place to overtake A. B traveled at the rate of 4 miles an hour.
How long was it from the time that A started until B overtook
him ?
^ 22. A tank holding 300 gallons of water has two pipes.
One lets in 15 gallons a minute, and the otiier draws out 12
gallons a minute. If both pipes are running, how long will it
require to fill the tank ?
23. A chain which contains 60 links is divided into three
segments, whose lengths are as 3, 4, and 5. How many links
in each segment?
Suggestion, Let 3a; represent the number in the shortest
THE EQUATION 29
segment. What then will represent the number in each of
the other two ?
24. One number is twice as large as another. If I take 4
from the smaller and 16 from the greater, the remainders are
equal. What are the numbers ?
26. Four men planned to form a partnership and to buy a
piece of property, but, one man dying, each of the others had
to invest $2000 more than he had planned to invest. What
was the cost of the property.
26. Find the number whose double exceeds 12 by as much
as 9 exceeds the number.
' 27. A man is now seven times as old as his son, and in five
years he will be only four times as old as his son will then be.
W^hat are their present ages ?
28. How far can I drive into the country at the rate of 4
miles an hour, in order that by walking back at the rate of 2
miles an hour, 1 may return in 9 hours from the time that I
started?
' 29. A boy bought equal amounts of two kinds of candy,
one kind at 10 cents a pound, the other at 20 cents a pound.
It all cost him 90 cents. How much of each kind^did he get?
30. A solves a certain number of the problems in this exer-
cise. B solves all of those which A cannot solve and twenty-
two of those which A solves. B solves two more problems
than A. How many does each solve ?
CHAPTER III.
NEGATIVE NUMBER.
27. Let us attempt to solve the equation
Subtracting 2a; from both members,
03+7 = 4. Axiom 2.
Subtracting 7 from both members,
x = 4:—7. Axiom 2.
What is 4 — 7? Evidently this equation has no solution,
unless we may subtract 7 from 4. This leads us to the follow-
ing considerations.
28. Extension of fundamental processes. Counting gives rise
to the series of arithmetical whole numbers
1, 2, 3^ 4, 5, 6, 7, 8, 9, 10, 11, 12, etc.
The sum of any two numbers in this series is another one
of these numbers. Thus 3 + 7== 10.
The product of any two numbers of this series is also another
one of these numbers. Thus 2x6 = 12.
But when we attempt to divide, the quotient of two numbers
of this series is sometimes another one of these numbers, and
sometimes it is not. Thus 12-^-3 = 4; but 10^7 gives no one
of these numbers. However, problems which arise have de-
manded that we be able to divide any number in this series
by any other of the numbers. Hence, since the whole num-
30
NEGATIVE NUMBER 31
bers are insufficient to express all quotients, we indicate the
quotients, and thus obtain a neio kind of 7iimibei\ not found in
the above series of whole numbers. These new numbers in
arithmetic have been called fractions. Thus, 10-r-7=-i^, a
fraction.
The series of numbers in arithmetic has, therefore, been
extended to include fractions as icell as lohoU niimhers^ and this
has followed from the necessity of making division always
Likewise, when we attempt to subtract, the difference be-
tween two numbers of the above series is sometimes another
number of the series, and sometimes it is not. Thus, 8 — 5 = 3 ;
but, 4 — 7 gives no one of these numbers, and we say 7 cannot
he subtracted from 4- In general, we say that a number can
not be subtracted from a smaller number. However, problems
which arise have demanded again that we be able to subtract
any number from any other number. Hence, v^hen the sub-
trahend is greater than the yninuend^ we indicate the remainder,
and thus obtain another new kind of number., not found in the
old series of arithmetical numbers. These new numbers are
called minus numbers, or negative numbers. Thus, 4—7 gives a
negative number.
•
The series of numbers has, therefore, been extended in algebra
to include negative numbers ; and this has followed from the
necessity of making subtraction always possible.
29. Negative and positive numbers. One number may be
subtracted from another by separating the subtrahend into
parts and subtracting the parts one at a time. It follows that
we may write 4— 7 = 4— 4— 3 = 0— 3. Hence it appears that a
negative number may be written to indicate the subtraction of
a, number from. zero.
32 ALGEBRA
Dropping the in — 3, we have — 3=— 3.
Likewise, 4-5 = 4-4-1 = 0-1 = -!;
4-6 = 4-4-2 = 0-2 = -2;
4-8 = 4-4-4 = 0-4= -4;
4-9=4-4-5 = 0-5=-5; etc.
It is clear that a negative number, such as —3, indicates a
reserved subtraction, there being nothing, when it stands alqne,
from wliicli to subtract it. It is in nature always a subtra-
hend.
Negative numbers may be added, subtracted, or used in any-
other operation in which other numbers may be used.
For the sake of distinction, ordinary or arithmetical numbers
in algebra are sometimes called plus or positive numbers. Also
for distinction in writing positive and negative numbers, the
positive or ordinary numbers are often preceded by the sign
+ when standing alone. Thus, 6 is written + 6 ; a is written
+ a. When clearness would not be sacrificed, however, the
sign + may be omitted from the positive numbers. The sign
--^^jnust never be omitted.
When standing alone the positive numbers 1, 2, 3, etc., or
+ 1, +2, +3, etc., are read either « plus 1," " plus 2," " plus 3,"
etc., or " positive 1," " positive 2," " positive 3," etc. And the
negative numbers —1, —2, —3, etc., are read either " minus
1," " minus 2," "minus 3," etc., or " negative 1," " negative 2,"
" negative 3," etc.
The positive and negative numbers of algebra are called
algebraic numbers.
The signs + and — written before numbers are called the
signs of the numbers or signs of quality. They may always be
considered as signs of operation i7i algebra; and when con-
venient, as signs of quality^ or distinction.
Two numbers which differ only in their signs, such as + 8
NEGATIVE NUMBER 33
and —8, are said to have the same arithmetical, or absolute
value.
30. Opposite numbers. Since a negative number, such as
— 5, always implies a reserved subtraction, when it is combined
with an arithmetical or positive number, it tends to destroy of
the positive ntmiber, a part equal to itself.
Thus, when 9 and —5 are combined, —5 destroys 5 of the 9,
leaving 4.
Accordingly, positive and negative numbers are sometimes
called opposite numbers.
In subtraction, when the minuend and subtrahend are equal,
the remainder is zero. This is equivalent to saying that when
two opposite numbers with equal absolute values are combined,
the value of the result is zero.
Thus, 3 and —3 give ; 25 and —25 give ; +a and —a giveO.
31. Opposite concrete magnitudes. Many concrete magni-
tudes are capable of existing in opposite states^ such that one
tends to destroy an equal amount of the other ; and hence their
numerical values may be represented by positive and negative
numbers. A few are here suggested. The student should
study these examples carefully.
(1) Debt and credit. If I have $20 in a bank, and owe the
bank $20, paying the debt will leave me nothing. The indebt-
edness destroys the credit. Hence debt and credit are opposite
magnitudes.
If I have $10 in my pocket, and make a purchase amounting
to $15,1 will have left $10 — $15, or — $5. The — $5 means
that I not only have no money left, but am $5 in debt. In this
sort of problem, then, a negative number means indebtedness.
If we call indebtedness negative^ credit will \)Q positive. Indebt-
edness will destroy credit.
34 ALGEBRA
(2) Forces in opposite directions. If two persons pnll in oppo-
site directions on cords attached to the sanie object, each with
a force of 100 lbs., the object will not move. The two forces
have destroyed each other. If one pulls 100 lbs., and the other
l^ulls only 80 lbs., the -effect upon the object will be the same
as if the first person alone had pulled with a force of 20 lbs.
If the force exerted by one person be represented by a^9ost7iy6
number, the other force will be represented by a negative num-
ber.
(3) Motions and distances in opposite directions. If a person
walk 100 ft. east, then turn about and walk 100 ft. west, he
will arrive at his starting point. The result is the same as if
he had not moved at all. The second motion destroyed the
effect of the first. If the motion in the first direction be called
positive, the second motion will be called negative.
Accordingly, distances moved through, or distances measured
in, opposite directions are called opposite distances. Thus if
distances measured to the right of the point A are called posi-
A
1 .
- +
tive, distances measured to the left of A are called negative.
Hence in representing distances, the signs + and — serve to
indicate the directions in which they are measured.
(4) Positive and negative temperature. If temperature above
zero is called positive, temperature below zero will be negative.
+ 10° means 10° above zero. —10° means 10° below zero.
From the above illustrations it is seen that a negative num-
ber has the property of oppositeness. Thus both positive and
negative numbers may be represented by distances along a
straight line measured from a common starting point. If dis-
tances to the right represent positive numbers, then negative
NEGATIVE NUMBER 35
numbers must be represented by distances in the opposite
direction
—5 ~4 --3 -2 -1 o 1 2 3 4 5
To every point to tlie right there corresponds a similar one
to the left. + 3 is represented by a point 3 units to the right
of some point, as o, on a straight line. —3 is represented
by a point 3 units in the opposite direction. If we measure 3
units in one direction, then from this point, measure 3 units
in the opposite direction, we return to the starting point, ^^e., 3
and -3 = 0. ' .
32. Addition of algebraic numbers. Addition was defined in
§ 3 as the process of combining two or more numbers into one.
We shall indicate the addition of algebraic numbers by writing
them in succession with their signs.
Thus, to indicate the addition of +6, +2, —4, +3, and —5, we
write +6 + 2—4 + 3 — 5. The first term in the expression being
positive, we may omit the sign, giving 6 + 2—4 + 3—5.
Algebraic numbers to be added are sometimes placed in a
column with their signs. Thus,
-3
+ 2
-6
Evidently two positive numbers, since each is an ordinary
arithmetical number, ina\j he combined by adding their absolute
values ; and their sum will be a positive number.
Thus, 5 + 9 = 14.
Since the negative number —5 indicates a reserved subtrac-
tion, then to add — 5 to any positive number means to combine
it by subtraction; that is, to subtract 5 from the number.
(See § 30.) The same reasoning would apply to any other
negative number.
36 ALGEBRA
Thus, the sum of 7 and —5 is 7—5=2. And the sum of 3 and
— 5 is 3 — 5=— 2. Remember that when the subtrahend is greater
than the minuend the remainder is negative.
Also, to subtract each of two numbers in succession is
equivalent to subtracting their sum. Hence two negative
numbers, since each is a subtracted number, inai/ be combined
by adding their absolute values ; and the sum will be negative.
Thus, to add —7 and —5 we have —7—5= — 12.
From the above considerations we evidently have the follow-
ing rules of addition of algebraic numbers :
(i) To add two numbers with like signs, find the sum of their
absolute values., arid %>refix the common sign.
Thus, 7 + 15=22; -7-15=-22.
(^) To add t\no numbers with iinlike signs.^ firul the difference
between their absolute values, and attach the sign of the arithmet-
ically greater.
Thus, -7 + 15 = 8; 7-15 = -8.
{3) To add three or more algebraic numbers, 2^'^oceed from left
to right, performing each step according to (1) or (2).
Thus, to find the sum of 6, —9, 3, and —7, we write 6 — 9 + 3—7;
6-9 = -3; -3 + 3=0; 0-7=-7.
Another rule is sometimes to be preferred to Rule (3).
It is illustrated in the following example :
7-3-9 + 6-5 + 21 = -3-9-5 + 7 + 6 + 21 Law of order.
= ( — 3-9 — 5) + (7 + 6 + 21) Law of grouping.
= _17 + 34. Rule (1).
= 17. Rule (2).
The same method is applicable to any problem. Hence we
have the following rule :
(4) To add three or more algebraic numbers, find the sum of
NEGATIVE NUMBER 37
the positive numbers and the sum of the negative numbers hy
{1); then find the sum of the sums hy {2).
Thusin 21-6-3 + 5 + 1;
21 + 5 + 1=27;
-6-3=-9;
27-9 = 18.
33. Subtraction of algebraic numbers. Since subtraction is
the inverse of addition, the rule for subtraction of algebraic
numbers may be obtained from addition.
In § 3 we established the principle :
subtrahend + remainder = minuend.
Hence if s stand for the subtrahend in any problem, and r
for the remainder, the minuend will be r + s.
Now (r + s) — s represents the sum of the minuend and the
subtrahend with its sign changed.
But {r-\-s) — s = r-\'{s — s) by Law of Order for Addition.
By § 82, s-s = 0.
Hence r+(s— s) = r.
That is, the sum of the minuend and the subtrahend with
its sign changed equals the remainder.
Hence the rule : .
To subtract one algebraic number from another^ change the
sign of the subtrahend^ then proceed as in addition.
Thus, to subtract —5 from 10, we write
10+5 = 15, remainder.
Again, 6 from —7 gives —7—6 = — 13.
The latter might have been written —7
6
-13
38 ALGEBRA
EXERCISE 7.
Find the sum of
1,
7, -2, 3, -5.
9. From 6 take —3.
2.
6, 5, -3, -1.
10. From -6 take 3.
3.
3, -15, -12, 1, -9, 10.
11. From -7 take -5.
4.
-6, -5, -12, -8, 4.
12. From 3 take 8.
5.
12 5 Ql
2'> — ¥» — IT' ^2'
13. 6-(-8)=?
6.
A. -f -h
14. -3-(-7)=?
7.
sV' T6' — 1» — i-
15. 5-(-3)-(-6)=?
8.
5.25,-2.5, -7.34, 4.35.
16. -10-(-4)-(-3)=?
17. Since the sign — always indicates a subtraction, what
name may be given to any expression witliin a sign of group-
ing which is preceded by tlie sign — ?
18. From 7-3-6 + 12 take -6 + 14 + 3-7.
19. From the sum of -3-5 + 2 and 12-6-1 take the sum
of 26 + 111-9 and 3i + 7-l.
20. Is the absolute value of an algebraic number always
diminished by subtraction ? Illustrate.
21. Do you ever add arithmetically when you are subtract-
ing algebraically ? Illustrate.
22. Is the absolute value of an algebraic number ever de-
creased by addition ? Illustrate.
23. A balloon lifts up with a force of 200 lbs. A weight of
150 lbs. is attached to it. What is the effect? If the weight
is represented by a positive number, what will represent the
lifting force of the balloon ? What will represent the result
when the weight is attached ?
24. A freight car is running at the rate of 20 feet per sec-
ond. If a man walks toward the front on the roof of the car
at the rate of 5 feet per second, how fast will he be moving
NEGATIVE NUMBER 39
relative to the ground ? If the speed of the car is called posi-
tive, what algebraic number will represent the speed of his
Avalking ? What algebraic number will represent his speed re-
lative to the ground ? If he walks toward the rear of the roof
of the car at the same rate, what will represent the rate of his
walking? What will be his speed relative to the ground?
What will represent it ?
25. I deposit in the bank at different dates $100, $175, and
$95. I check out at different times $25, $10, $87.25, and $37.75.
If my account is then balanced, how much will I have in the
bank ? If the deposits are called positive, what will represent
the checks ? Show how thus to find the balance.
26. On the earth's surface longitude west is called positive
longitude ; longitude east, negative longitude ; latitude north,
positive latitude ; and latitude south, negative latitude. How
could you describe the position of a town whose longitude is
112° east, and latitude 75° north? In what country Avould a
city be whose longitude was +90° and latitude +40° ? Longi-
tude -90° and latitude +40° ?
34. Multiplication of Algebraic Numbers. The laws of signs
for multiplication of algebraic numbers come directly from the
definition of multiplication given in § 3 ; viz, to obtain the prod-
uct of two numbers we must use the multiplicand as we m,ust use
1 (unitf/) to obtain the multiplier.
Suppose the multiplier to be +3.
To obtain +3 we must add three I's.
Thus, +3 = 1 + 1 + 1.
Hence to obtain the product ( + 5) ( + 3), we must add
three +5's.
Hence ( + 5)( + 3) = + 5 + 5 + 5=+15.
To obtain the product ( — 5) ( + 3), we must add three — 5's.
40 ALGEBRA
Thus (-5) ( + 3) = -5-5-5= -15.
Suppose the multiplier to be —3.
To obtain —3 we must subtract three I's from ; that is,
change their signs and add them.
Thus -3=-l-l-l.
Hence to obtain the product ( + 5)( — 3), we must subtract
three +5's from ; that is, change their signs and add them.
Thus ( + 5)(-3)- -5-5-5= -15.
To obtain the product (— 5)( — 3), we must subtract three
— 5's from ; that is, change their signs and add them.
Thus (-5)(-3)= +5 + 5 + 5= + 15.
The preceding reasoning will evidently apply to any multi-
plier and any multiplicand. Hence the following laws :
(1) The product of tvio numbers loith like signs is positive.
Thus ( + 8)( + 9) = + 72, and (-8)(-9)=: + 72.
Note that all of the signs + could have been omitted.
(2) The product of tico numbers vnth unlike signs is negative.
Thus .( + 8)(-9) = -72, and (-8)( + 9) = -72.
35. A product consisting of an odd number of negative
factors is negative ; and a product consisting of an even number
of negative factors is positive.
This follows easily from § 34.
For ( — 1)( — 1)= + 1 ; Even number,
hence (-1)(-1)(-1) = ( + 1)(-1) = -1 ; Odd number,
and (-1)(-1)(-1)(-1) = (-1)(-1) = + 1; Evennumber.
and(-l)(-l)(-l)(-l)(-l) = ( + l)(-l) = -l; Odd number,
and so on.
Thus (-6)(-5)(-2)=-60; (-3)(-4)(-5)(-6)= +360,
NEGATIVE NUMBER 4|
Since the product of two positive factors is positive, it is
easily shown by reasoning similar to the foregoing that a pro-
duct consisting of any number of positive factors is j^ositive.
Thus, ( + 6)( + 5)( + 4) = +120, and ( + 3)( + 7)(+ 1)( + 2)
= +42.
36. Signs of Powers. If the factors of a product are all equal,
the product becomes a power. See § 13.
Thus, (-3)(-3)(-3)(-3)=(-3)*; read "the fourth power
of -3."
Hence from § 35 we get the following laws :
(1) Any power of a positive number is positive.
Thus ( + 2y* =( + 2) ( + 2)( + 2)=+8; ( + 4)-^ =( + 4) ( + 4)=16 ; etc.
{2) Any odd j)ower of a negative number is negative ; any
even power of a iiegative nu^nber is positive.
Thus, (-2)^ =(-2) (-2) (-2) = -8 ; (-3)^ =(-3) (-3) (-3) (-3)
(-3) = -243.
And(-2)*=(-2)(-2)(-2)(-2) = + 16 ; (-5f =(-5) (-5) = + 25.
To find the value of a^6^ when a=3, 6=— 2; we have a^ b^ =
32(-2)^ =9 (-8)^-72.
If a=-3, 6=2, aW={-3yx2^=9x8=72.
37. Division of Algebraic Numbers. The laws of signs in
division of algebraic numbers are easily deduced from the cor-
responding laws in multiplication.
In § 3 it was shown that
divisor x quotient = dividend.
Hence, since- ( + 5) ( + 3)= +15, then ( + 15)^( + 3)= +5 ;
since (-5) ( + 3) = -15, then (-15)--( + 3) = -5 ;
42 ALGEBRA
since ( + 5) (-3)--15, then (-15)--(-3)= +5 ;
since (-5) (-3)= +15, then ( + 15)---(-3) = — 5.
Manifestly the same laws apply here which apply to multi-
plication. The reasoning in the particular cases above will
evidently apply to any dividend and any divisor. Therefore
we have the following laws :
(1) If the dividend and divisor have like signs^ the quotient
will he positive.
Thus,(-18)-(-2)= + 9;^^=+2;^- +4; etc.
{2) If the dividend and divisor have unlike signs^ the quotient
will he negative.
30
Thus, (-21)-(+3) = -7;--g=-5; (- |)- | = - |; etc.
EXERCISE 8.
Find the product of
1. 2 and -7. 7. 9, -2, G, -5.
2. -2 and 7. 8. 5, -7, +4, -2, -1.
3. 2 and 7. 9. -7, -1, -1, -1, -1.
4. -2 and -7. 10. -1, -1, -1, -1, -1, -1.
6. -3, 2 and -5. 11. -4, 3, -2, 0.
6. -4, —3 and -1.
Find the value of
12. ( + 2f. 14. (-4)1 16. (-2)^ (-3)^
13. (-3)*. 15. (-If. 17. (-ly (-2)*(-8)».
If a=2, ^>= — 3, c^ —4, find the value of
18. a'h\ 19. ahc. 20. h'c, 21. a'h'c'. 22. (-2)(-3)i^.
NEGATIVE NUMBER 4.3
Divide :
23. -27 by 3. ^ ^
^ 27. 6.25 by -2.5.
24. -27 by -3. ^ ;
28. 8 by - |.
25. 21 by -7.
29. -2| by 17.
26. - 1 by f .
30. +216 by +12.
If a=— 2, h = Z, c=-
-4, find the value of
31. "^',-5.
32. (v'~-G'yJ)\
34. 3c^--2^. -J^ 4
36. 3 .
33. c^--6--«^
36. d'^&-^h\
EXERCISES FOR REVIEW (I).
1. How does algebra differ from arithmetic ?
2. How is the addition of three or more numbers indi-
cated?
3. What are the signs of multiplication used in algebra ?
Why may ah indicate the product of a and h but 87 not the
product of 3 and 7 ?
4. What is the use of the " signs of grouping " ? Illus-
trate.
6. IIow would you evaluate the following expression?
(l + 2)(9-4)-8-=-(10-6 + 2) + 3.
State a rule for evaluating expressions.
6. What is a general number ? Illustrate. How is it rep-
resented in algebra ?
7. What do we mean by factors of an expression ? What
are the factors of lahc ?
8. What is a ponder of a number? What does a? mean?
What is the 3 called? When a = 2 what is the value of a*? of
a*? oia'^i
44 ALGEBRA
9. What is the coefficient of an expression ? What is the
numerical coefficieyit of lab'^ ? What is the coefficient of b^ in
the expression 7a//? of a? of 7b'?
10. What is a /*7eraZ expression ? Illustrate. ^o
11. When a =10, 6=4, c = 2, find the value of -U '
(a.) {2a-Zb){4a-3c)~{Sa-Sc-b). **
(6.) {a + br-^^^ + c(a + 2b).
12. If a = 4, 6 = 8, c = 3, r?=2, and x = b^ find the value of
(a.) 3 aa5 + 5M-2c'a; + 2cf?^-M\
Aa' + ^ b' Sd'{2G'-^ db' + 2x) d^x
^^■^ a'b + a {b'-c') b '
13. What are similar terms ? Illustrate.
14. State t\iQ ftmdamental laios of nmnbers. Write them in
symbols. Illustrate each law with definite numbers.
15. State the laws used in the following identities:
2a^-56^-7c2 = 2-5-7a^-6^c''=(2-5-7)(a26V) = 70a^6V.
16. What law is involved in
5a; + 2a; + 12£c = (5 + 2 + 12)£c = 19a;?
17. State the axioms given in this book. Of what use are
they?
18. What is an equation? Distinguish between a condi
tional equation and an identity. Illustrate each.
19. What is meant by a root^ or a solution^ of an equation?
20. State the steps in the solution of the following, giving
your authority for each step.
l + 4a; = 2(a! + l)+3.
21. A tree 60 feet high was broken at such a point that the
part broken off was 3 times the length of the part left stand-
ing ; required the length of each part.
^ko^-,
^'^ **' -^ j^ NEGATIVE NUMBER 45
22. The greater of two numbers is 5 times the less, and their
sum is 126 ; required the numbers.
23. What is a negative number ? A positive number ? How
did they originate ? Illustrate.
24. Why are positive and negative numbers sometimes
called opposite numbers ?
25. If a boy weighing 75 pounds, is holding a toy balloon,
pulling upward with a force of 15 pounds, how may these
numbers be represented by positive and negative numbers ?
26. In exercise 25, if 75 is called +75 what is the 15?
What force will be required to lift the boy and balloon ?
27. If I have $1000 and owe 11500, by what number may
my financial condition be expressed ?
28. How else may 50° above zero and 15° below zero be ex-
pressed ?
29. What is the difference between algebraic numbers and
arithmetical numbers f
30. How do we add algebraic numbers ? Find the algebraic
sum of -10, +5, -7, -3, +4, and +8.
31. How do you subtract algebraic numbers? Illustrate.
Upon what fundamental principle is the proof based ?
32. From the sum of 6, 5, —3, —1, take the sum of ~15,
-12, 1, -9, 10.
33. State the laios of signs in multiplication. Upon what
important definition is the proof of these laws based ? By the
use of this definition prove that (—2) (—4)= + 8.
34. Give the values of (-2)^ (-3)^; {\f\ (-i)*-
35. Give the laws of signs in division. Upon what import-
ant principle is the proof of these laws based ?
36. Divide 8 by -2 ; -8 by -2 ; -81 by 3 ; -f by -|.
CHAPTER IV.
ADDITION AND SUBTRACTION OP LITERAL
EXPRESSIONS.
38. Addition of monomials. It was shown in § 21 that
similar terms could be added by use of the distributive law
ax-\'hx-\-cx={a-\-h + c)x.
From this law we have the following rule :
To add similar terms^ add their coefficients and affix to this
sum the common letters with their exponents.
Thus, to add 4x^2/^ — 2a?^2/^ and —z>x^y^^ we have, writing them
in succession with their signs,
A.x'y^ - 23(^1/' - ^x^f = (4 - 2 - 5) x^y"" = - Sx'y^
To add dissimilar terms, write them i/t succession with their
signs.
Thus, to add 2ah, —^ax^ 122/^ and— 32;^ we have
2a6-3aa?4-122/'-32;l
The addition of terms similar with respect to certain letters
may be indicated by grouping the coefficients of the common
letters, and affixing to this group the common letters with
their exponents.
Thus, to add 2ax^y^ —hx^y, and Scoc^y, we write
2ax^y—bx^y + Scx^y={2a—b + 3c)x^y.
46
ADDITION AND SUBTRACTION OF LITERAL EXPRESSIONS 47
EXERCISE 9.
Find the
sum of
1. 2x
2. Za'b
3. 7c'
4. Qabcd
5. ax^
-3a;
4a'b
-be'
— 21abcd
^ax'
5x
- a'b
Qc'
— Sabcd
- ax'
— X
-lOa'b
-12c'
Uahcd
7 ax'
6. bA, 12A, -3.4, -7A. 9. ^pq, 4pq, -V^pq.
7. 16P§, -10P(2, 4P§. 10. 100ylC;-14.4(7, |^(7.
8. 4^"^ 2^^ -B\ -^B\ 11. ccy, -9a;V', 4ajV', -a^y.
12 p5'r, — 10^5'r, ^pqr^ —\pqr.
13. ^mhi^^ 6^iW, — 32/?^^^^ — mW.
Simplify
14. ^x—^x—bx^-^x—x. 16. aWc—2a''¥c^ Q^a^'^c.
15. -12y^-3y^ + 4/-72/^ 17. 4a^»c-2cia-7^>ac.
18. |2/^-3y2;-|2/s + |2/2.
Simplify by adding similar terms
19. 4a + 6a-12a; + 2a + 3a5. ^
Solution.
4a + 6a-12ic + 2a + 3ic=4a + 6a + 2a-12.r + 3£C Why ?
= (4 + 6 + 2) a+ (-12 + 3).T
= 12a-9x.
20. xy — ab + l(iab—\^xij — '^ab.
21. ^abc''\-2a''bc-babc''^7ah^c-\-7a''bc.
22. ^a'' + W-ba'-b\
Indicate the sum of
23. Zax^j bbx\ —7cx^. 24. 2xi/z, axijz, -bxyz.
25. -7cij\ 2y\ Zay\
39. Addition of polynomials. If all of the terms of a poly-
nomial are added to an expression, the polynomial is added to
the expression. This follows from the lavi of grouping.
48 ALGEBRA
Thus, if a, 6, and c are added to a?, we have x-\-a+b-\-c—x+
(a+b+c).
But x+{a+b+c) indicates that the polynomial a+b + c itself is
added to x.
Hence, to add two or more polynomials^ write down all of the
terms in successioii vnth their signs / then combine the similar
terms^ if any.
Example 1. Add 2a + 3&— 4c, —4a— 6 + 5c, and 5a + 6— 2c.
We write 2a+36— 4c— 4a— 6+5c+5a+6— 2c=3a+3?)— c.
The work may often be more conveniently performed by
Avriting the similar terms in vertical columns, then adding the
terms in the resulting columns.
The above example might be written
2a+36-4c
—4a— 6+5c
5a+ b—%c
3a + 36— c
Checks. In much of the work of algebra the student can easily
verify or chech his results ; i. e. , he may perform other operations
that tend to show that the first result is correct. This is called
checking the work.
A short method of checking addition of polynomials consists
of assigning particular values to all of the general numbers in-
volved in the polynomials and in the sum, and seeing if the
sum of the values thus obtained for the polynomials is equal to
the value of the sum of the polynomials. This is illustrated
in the following example.
Example 2. Add 6a— 56-f-3c, 7a+106— 6c, and 8a— 96- 10c.
Work. Check.
6a- 56+ 3c 6- 5+ 3= 4
7a + 106- 6c 7 + 10- 6= 11
8a- 96-lOc 8- 9-10=-ll
21a- 46-13C 21- 4-13=
when
a=l, 6=1, c=l.
ADDITION AND SUBTRACTION OF LITERAL EXPRESSIONS 49
EXERCISE 10.
Add and check :
1. x-{'y-\~z^ x—2i/-\-Sz, —t)x—4ii/ + z.
2. Sa—b + 2G, ba + c—'2b, —a + Sb—4:C.
3. 2P + 4§ + i?-7.S; -6P-§ + 3i? + 2>S'.
4. 7ac-\-Sxi/, 2ac—lxy.
5. 20ji?— <7 + r, 2/) + 5^— 7r, — 7p + 2y + 3r.
6. 2a6 — 3*c + 5ac, 7^c— 2ac + 6a*, — 3ac + 2Z>c.
7. a;=' + i«' + a; + l, x'-x' + aj-l.
8. a^ + 2a^ + ^>^ a'-2ab + h\ 2a' -W.
9. a;^— y^ a;^ + 6£c^y, —^xy'^—y^.
10. |6«-^^-i-c, fa-J^ a + |$-3c.
11.
3£c-'-2£c' + a;-
-4,
x'
+ 4aj^ + l,
3a;^
-2a; + l,
x^-x\
12.
-12a;* + 2a;-
-1,
Zx'
'-2a;^ + 3a^,
, a;^
+ 2a; + 5,
Zx' + A.x\
40. Subtraction. The reasoning in § 33 evidently holds for
algebraic expressions in general, since any expression is itself
a number. Hence, to subtract one expression front miother^
change the sign of the subtrahend and add the result to the
minuend.
The sign of an expression is changed by changing the sign of
each of its terms. This follows directly from Rule 4 in addition,
since an expression is but the algebraic sum of its terms.
Thus, changing signs in 7—3 we have —7+3. Now 7—3 and
—7 + 3 are opposite in sign, but have the same absolute value, 4.
Therefore, for subtraction we have the following rule :
Add to the minuend the subtrahend loith the sign before each of
its terms changed.
Example 1. Subtract — 4a& from — 2a6.
Changing the sign of the subtrahend and adding gives
-2a6+4a6=2a6.
50 ALGEBRA
Example 2. From 2a— 36 + 5c take 3a— 2b— 2c.
Changing the sign before each term in the subtrahend arid
adding, we have
Work. Check.
2a-3h + 5c =4]
-3a + 2b + 2c =1 I when
a=l, 6=1, c=l.
— a— 6 + 7c = 5 J
The change of signs need only be made mentally, the writ-
ten signs of the subtrahend remaining unaltered. This is illus-
trated in the following examples.
Example 3. Subtract —2x^ — 4:y^+15a from 7x^—2y^.
Work. Check.
lx^-2y' = 5 1
-2a;2- 41/2 + 1 5a = 9 I when
x*=l, i/=l, a=l.
9xH2i/2-15a = -4 J
Example 4. From 2a^ + 4^* — 3a7 + 7 take a^ — 3ar* + 2x^ — x.
Work. Check.
2x^ + 4cc* -3a.'+7 = 10]
x^ -Sx'-{-2x-- X ^- 1 lwhena;=l.
af' + 4x' + 3x'-2x'-2x + 7 = 11
EXERCISE 11.
1. From 2 a'b' take —SaW. 4. From —7abc take ^abc.
2. From — £cy take 5a»/. 5. hhy'^—{—%hif)='i
3. From -^aa;Uake -^ax\ 6. 21a;y2-(-3a;y^)=?
7. From Zx—2y-\-lz take x^ 6y — Sz.
8. From 7a + 2a^— 2c take 8a— 12a^ + 5c.
9. From 2x—7y take 3y— 5a;.
10. From x^—x^+x""—! take 2x^—x^^x^—l.
ADDITION AND SUBTRACTION OF LITERAL EXPRESSIONS 5I
n. From a^^-Za'b + ^ab'-^b^ take a'-a^h-\-ah'-h\
12. From lAB^^lxy — ZPQ take 'lxy-V\Pq.
13. 3ic2-5a; + 9-(2ec2 + 6£c-4)=? -
14. The subtrahend is x^^-x'-^x'-^-x-^ the remainder is x^^x^
+ £c"^ + l. Find the minuend.
15. Subtract — 4c^s^ + «*— r^ from 2rt^ + 3i^'^— cV.
16. Subtract \—x^^x^ from a?^ ; from a\ from 0.
17. From the sum of cv' + a'b'—d'b'-\'If and ^aW—2b'—a'
take a^— 5^ + 2£c.
18. From the sum of ^x'—^^x' + l and ^x' + p' + ^x take ia;^'
-x' + 2.
19. From 1.5a-7.2ic'-3.25m=' take the sum of .4a;^-7.5a + 5m'
and l-.125a + 3m^
20. From the sum of a^^— 1, ce^+aj + l, and x^—1 take the sum
of x^ + a?^ + 2 and x^ — 1 .
21. What operations are indicated by 3«^ + 2«— (a^— a^ + «— 1) ?
22. Simplify ^x^ -\-Sx^+1- (2x^ -3 + 0^).
23. From 7a^ + 3^2-2 take a' + Sa + b.
If ^ = 2a;^ + a;-3, J?=a;='-a;'^ + 2a! + 2, C=Sx'-6x + l, find
the value of
24. A-B+O. 26. -A + B+O.
2b. A + B-a 27:^ A-B- a
If a= -2, b=~l, c=3, f?=2, find the value of
28. 2abc + ^a'cl 33. 3a -2/; + 4c.
29. b'c + adc. 34. -2ac-2bcl
30. a^-5^ . 35. a' + b' + c' + dK
31. 2^>c + 3«(?. 36. 4a'^' + 2aiV-^>cW.
32. a + Sd-2c + b. 37. after?- 2a^ + cr.
52 ALGEBRA
41. Removal of signs of grouping. The negative sign (— )
always indicates that the number following it is a subtracted
number. Hence, an expression inclosed within a sign of
grouping which is preceded by the negative sign is a subtrahend.
Thus, in 3a — (5a— 26), the expression (5a— 26) is a subtrahend.
But subtraction is performed by changing the sign of each
term of the subtrahend and adding the resulting expression to
tlie min'uend.
Hence, a sign of grouping ^yreceded by the negative sign may
be rernooed if the sign before each tertn inclosed is changed.
Thus, 3a-(5a-26)=3a-5a + 26 ; -{-Qx-{-5x') = 6x—5x\
The positive sign ( + ), preceding an expression inclosed
within a sign of grouping, either indicates an addition or serves
as a sign of distinction.
Hence, a sign of grouping preceded by the jyositive sign
may be removed vnthout changing the sign of any of the terms
inclosed.
Thus, 4x + 7.r+(3a?— 2j?+1)=4:c + 7x + 3x— 2j?+1.
Sometimes signs of grouping are inclosed within other signs
of grouping. In such cases the use of different kinds of signs
is advantageous.
Thus, a-{2a-(a— 26)} ; x-^y—{2x—y) + \;Zx-{^y—x)].
In expressions of this kind, it is best for the beginner to
remove the innermost sign first ,' then the innermost remaining
sign ; etc.
Thus, a-h-{-a-{-h-a-b))
=a— 6— .{a— (— 6— a + 6)}, removing vinculum,
=a — 6— {a + 6 + a— 6}, removing parentheses,
' =»a— 6— a— 6— a4-6, removing braces,
= — a— 6, adding like terms.
Again, Ix-Zy- {(4a— 6)— [5«— 6— (3a:— 2^)— 36] j
ADDITION AND SUBTRACTION OF LITERAL EXPRESSIONS 5^
=7x—:^^J—{4:a—b—[5a—b—3x + 2y—Sb]}
=7x—3t/—{4.a—b—5a + b + Sx—2y + Sb}
=7x—3i/—4a-\-b + 5a—b—Sx-h2y—3b
=4x — y + a—3b.
Some Avork may be saved by adding the like terms as each
sign is removed.
EXERCISE 12.
Simplify by removing signs of grouping and adding like
terms :
1. 2a + Sb + (a-4:b.) 3. 2B'-{B'—4.AC).
2. {lx + 2y) + (Zy-2x). 4. « + [2a-(26-r0].
5. x' + x'-{x-'lx^)+{x'-{%x''-l'Mx'-x)\,
6. a;V-(2.^'^-3.^//)-(2.^^V- {Zxy^-^^f-l^-f^}).
7. —(a— {« — [«— a— 26]})
8. -(^^ + 2iC"-a5)+(3a;-a;3+l)_(2cc2_^8.T + 5).
9. 6a5-- (2y2-4aj^) -7y' + (3a;y-2y^) -(3a3'^-4/).
10. -[-{-(-,7=^)}].
11. 10rt-(3^»-4a)-{2rt-(35 + «)}-{3/>-(2« + 6)}.
12. 2.«-{a?-(-y-a^^^)}. 13. 7- {8-[3 + (6-2^=a;)]}. .
14. —\_x— {x+ {a—x) — {x— a) —x) —a~\—x.
15. ^(-.(^(^(^(-.1))))).
16. 10-[16-(14-{12-2}-4)-10]-2.
17. Solve the equation
8i«-(5-2^+{3 + 4)=«-(2a;-10).
42. Insertion of signs of grouping. It follows immediately
from § 41 that terms of a pohpioniial may he inclosed within a
sign of (frouping^ ichen this sign of grouping is preceded by the
positive sign, icithout changing the signs of the terms ; and.
54 ' ALGEBRA
when preceded by the negative sign., by changing the signs of the
terms.
Thus, to inclose the last three terms of 3a?-'' + 2a:^— 4a; +1 in brack-
ets preceded by the sign +, we have 3a?^ + [2x^—40? + 1]. If pre-
ceded by the sign — , it becomes 3^— [— 2a?2 + 4a?— IJ.
This principle is of use In combining the terms of a poly-
nomial which are similar with respect to some general num-
ber.
Thus, combining the terms having the same powers of x,
a£C* + &ir^ + 3Ce»+5x*-3iC=*-x + 4=(a + 5)x* + (6-3)^+(3c-l)x + 4.
Again, 4-5^ + 3ca^-ai;c-|-6aa^ + 3.r-7ii;2^4-(a-3)x-|-(3c-7)ar'
-(5-6a)ar\
EXERCISE 13.
Without changing the values of the expressions, inclose the
last three terms of the following expressions in signs of group-
ing preceded by the sign — :
1. ^—x'-^x^-x'+x-l. 3. a + 2^>-3c + 4(^.
2. ax—by — cz-Vdw. f L(/^»Aj^. Sd-lOe^bf-g.
In the following expressions combine the terms having the
same powers of x, so as to have the sign + before each group :
5 . 2x^ — Sx^ + aoi? + b^ + hx — ex.
6. 7 + dx'-^x'—2ax—4ax' + Qbx^ + ^x.
7. ax*—l + 2x^ — dx* + x'^ + ax^ — cx + dx — bx^.
In the following expressions combine the terms having the
same powers of y, so as to have the sign — before each group :
8. -y + b + 2y' + ay-by\
9. py-qy + ry- sy* + 2py' - Sqy\
10. -Qxif + xY-dx'y-2y'-Sy' + by.
ADDITION AND SUBTRACTION OF LITERAL EXPRESSIONS 55
In the following expressions remove all signs of grouping,
and then combine the terms having like powers of x :
11. 2x-(ax'-bx)+{cx-(2x'-10)\.
12. ax'-('2x-6x'-^x).
13. (x'-x+l)-{2x-(Sx'-2)-x'}.
Add the following expressions, combining like powers of x :
14. a^—1, ax^ + bx, ax^ — bx^ + cx, x-\-b.
15. ax^ + a^x\ 2ax—Sx\ ax^ — %^.
16. x^'^—ax^^b, bx' + c, 2x>—d.
17. From a^ + bx + c take bd^ + gx— d.
18. From 2£c^ — 3a; + 5 take a—bx^-\^cx.
1 9. From 2^^^ ~ 2^ take px^ -{-rx—q.
CHAPTER Y.
MULTIPLICATION OF LITERAL EXPRESSIONS.
43. Law of exponents. The Lxav of exponents in multiplica-
tion is derived immediately from the meaning of an exponent.
It is understood here that Ave are dealing only with exponents
whidi are positive integers. The law expressed in symbols is
flr™-a" = a"'+".
WehnYG a"' = aaaa torn factors;*
and . a" — aaaaa to n factors.
IIe«ice a"'-a" — aaaaa to 7n-j-n factors.
m+u
= a
Thus, d'^a*=aaaaaaa=a' ;
tA/ tA/ — eA-/«A/eAytA-/«A/ tAytAy tAj >
and y^-y^=i/+^ = y^.
By similar reasoning this law may be extended to any num-
ber of factors. Hence,
fl'"-a"-a;'=fl'«+«+/-; etc.
That is, the product of two or more poicers of the same base
is equal to that base tcith an exponent equal to the sum of the
exponents of the given poicers.
* Tlie sign • is called the sign of continuation, and means
" and so on," or " and so forth."
Thus, 1, 3, 3, 4 is read " 1, 3, 3, 4, and so on."
And aaaa is read "aaaa and so on."
56
MULTIPLICATION OF LITERAL EXPRESSIONS 57
44. Multiplication of monomials. By use of the law of order,
the law of grouping, the law of signs, and the law of exponents,
the product of two or more monomials may now be found.
Thus, to find the product of 2i)c^y, — 3ic^2/^ ^"d 7xy^^ we have
{2oc^y){—3x^y%7xy^) = —2-3-7x^a^xyy^y^, law of order and signs,
= — {2-S-7){oc^3cr^x){yy^y^)^ law of grouping,
=—42x®2/*', law of exponents.
By these laws we have the following rule for the multi-
plication of monomials :
Find the j^roduct of the numerical coefficients^ using the law
of signs ; and affix, to this the products of the literal factors^
using the law of exponents.
Example. The product of — 3a&^ 7a^x^, and — 2&V is
( - 3a62) (Ta^x^) ( - 2h'x^) = ^2aWx\
Note. — The student must be careful not to combine the exponents of
different kinds of bases, as of a and 5. It must be remembered also
that if no exponent is written above a base, the exponent 1 is under-
stood.
EXERCISE 14.
Find the product of :
1. aW, and -a'b'\
2. Qa'b, -baWand -2a'b\
3. Sx% — 2a;y, —bxg and 7x\ .
4. —iJfy^g'^z^, —^axg^ and ^a^.
5. \d'h'c'd^ and -Ua''hc\
6. -'Ia-b\ -4a=^^^c and l^c.
7. 1^, -lA'B and ^A'B\
8. F\ -P'QsLudP'Q'.
9. — ^mW and J^mn^.
68 ALGEBRA
10. 2.bx\ S.2bxy and l.lbxi/.
11. 12pqr, — 4/9^$'r^, jo^^V and —r^.
12. B''^, -lOi^/S'^ and -^BS.
13. (a + by and 2(a + b)\
14. 3«2(^> + c) and -ba{b + cy.
16. What is the meaning of x-? Of »;'»+«? Of j^"-*? Of
16. What is the product of x"^", £c"+»' and x*"-'"?
17. What is the product of a^+S —2a^" and --3«?
18. What is the meaning of {a'f? Of (ay? Of (a;'')''?
19. Find the product of (aj\ (a'f and {a')\
45. Multiplication of a polynomial by a monomial. The rule
for multiplying a polynomial by a monomial is obtained
directly from the law of distribution. Stated in symbols the
law is
(fl + 6 + cH- ) x = ax-\-bx-\-cx-\-
If a + ^ + c+ represents the polynomial, and x the
monomial, then ax-^bx-\-cx-^ is the product.
Hence, the rule :
The product of a polynomial midtiplied by a monomial is the
si(,m of the products obtained by multiplyi?iy each term of the
polynom,ial by the monom^ial.
Example 1. Multiply 3ic*— 2a^+6a?— 5 by 4a^.
(3x*— 2x2 + 6x— 5)4.r^=3x*-4£e— 2x2-4aT^ + 6x-4a^— 5-4a?*
= 12x' - 8^ + 24a?* - 20x^
The work is conveniently arranged thus :
3ic*-2^2 + 6ic-5
4ic^
12x' - 8x^ + 24x* - 20itr»
MULTIPLICATION OF LITERAL EXPRESSIONS 59
Check. When ir=l, multiplicand=3, multiplier =4, product =
8, as it should.
Observe that since any power of 1 is 1, substituting 1 for x does
not check the exponents 7, 5, 4 and 3, but merely the coefficients.
It is always more convenient to perform the multiplication
from left to right.
EXERCISE 16.
Multiply and check :
1. a'-'lab + h'^hy a'h\
2. cc^— i// + a;— 1 by £c*.
3. 6«'-5«2^>+25^by -«^6^
4. - 2a;* - ZxY + 5y * by - Ix^y"-,
5. x^y^—m^z^ + x^w by Sxyzw.
^6. 2pq—Sqr-\~4:rphjbpqr.
7. A'-B' by AB.
8. 20x'-dx' + 2hy—4x\
^- .9. i a^y + 1 xyz-^ £cV by -xh/z'.
10. —ha^h'^&^'^abc—ax—hy—cz by —Zabcx^y^.
11. I a;y^— i «icy + J f^ — f aV by —'la^xyK
12. Is there any difference between 4(a— ^) and («— ^)4?
Why?
Remove the signs of grouping and simplify :
13. x{a—h) — (a^h)x.
Note. — In these expressions, products preceded by the sign — are sub-
trahends. Hence, when the signs of grouping in products preceded
by the sign — are removed by multiplication, the signs of all terms
arising from such products must be changed. Thus, — (a +6) a? is a sub-
trahend. Hence x (a — 5) — (a + h) x =^ax — hx — ax — bx 2bx.
14. A(2x-Sy)-2(4:X + y). 16. 10{x-2y)-(2x-y)S.
16. {a-b)c-(a + b)c. —17. -Sy(xy-x') + 2x(x'-y').
60 ALGEBRA
18. -^cJ?h-a¥)-{ah^'-a'b).
19. -2(— 2j«* + 3a;V-y') + 3(^* + 2£cV-2y*)-
20. 3[2a;y-4fc{y-2(a^-2/)}].
46. Product of polynomials. The product of two polyno-
mials is also obtained from the law of distribution. Thus,
(ct + 6 + c)(£c+y + ^) = «(^+y + 2!) + i(£c+y + 2;) + c(£c + 2/ + s)
= ax-\' ay -}- az-{-hx-{-by -\'hz-\- cx-\- cy -{- cz.
Here the product shows that each term of the multiplicand
has been multiplied by each term of the multiplier ; and the
product is the sum of all the products thus formed. The same
method will evidently hold for any two polynomials. - Hence
the rule :
To obtain the product of ttoo polynomials^ multiply each term
of the multiplicand by each term of the multiplier^ and take the
sum of the residting products.
Example 1. Multiply 2x'^—^xy-\-y'^ by ^x^—xy.
Write the multiplier below the multiphcand ; then multiply
each term of the multiplicand by 3^"^ ; then multiply each term
of the multiplicand hj —xy. The similar terms obtained by
multiplying should be arranged in columns and added. Thus
2aj2- Sxy + y^
Sx^— xy
(2a^-Sxy + y^){Sx^)= 6x'- dx^'y + Sx'y'
{2x^—3xy + y^){—xy)= — 2x^y + 3x''y^—xy^
6x* — 1 IxTf + 6x^2/'' — ^if
Check. When x=2 ; i/=l ; multiplicand =3 ; multiplier =10 ;
product =30.
Note. — A polynomial is said to be arranged according to the powei^s
of some letter when the exponents of that letter either increase or
decrease in the successive terms as we pass from left to right.
MULTIPLICATION OF LITERAL EXPRESSIONS. 61
Thus, x*—2oiy^+x^+x~5, is arranged according to the descending
powers of x ; while ^if+xy^+x^y'^+a^y+x^ is arranged according to
the ascending powers of x.
The student will find it an advantage in multiplication of poly-
nomials to arrange, if possible, both multiplicand and multiplier
according to the powers of some letter.
Example 2. Multiply x^—2-\-2xhjx — ^ + x^.
Arranging both trinomials according to the descending powers
of x^ we have
x''-\-2x -2
x^-^ X —6
a?* + 2ar^-
2x'
ie +
2x'-
- 2x
—
ex'-
-12;r+12
x'* + 3x^- 6x'—Ux+12
Check. When x=3 ; multiplicand =13 ; multiplier =6 ; pro-
duct=78, as it should.
The product of three or more polynomials may evidently be
obtained by multiplying the product of any two by a third, and
so on.
EXERCISE 16.
Multiply and check :
1. a + b by a + b. 9. 2a;-5 by 2£c + 7.
2. a + b by a—b. 10. 4a+7 by 4«-7.
3. 2x-ij by Sx + 2i/. 11. -^-10 by -a^ + lO.
4. 2x' + by' by x—Sij. 12. Scd+Qx^/ by 2xy—cd.
5. x^-\ by 3£c^ + 4. 13. \x^-\ by fcc^-i
6. 4m^-5/^^ by 3?^^ + 2m^ 14. |a + |^ by \a-\b.
7. X^pq^^pr by 2pq—'lpr. 15. |£c'4-|y' by \x—\xj.
8. a; + 3 by £c + 2. 16, 2.25a; + 7.5y by -1.5y + 2.5a;.
62 ALGEBRA
17. .2a-lM by 4.7a + 2M. 20. a + b + c hy x + y-\-z.
18. ax + b}/ + czhj ax-hy. 21. x^—'lx-^^ by 2a;^+5ic-4.
19. pq-Vqr—pr hj pq + qr. 22. £c^ — 1 by a;^ + ic + l.
23. 4£c=^-2a;-^ + 3ix;-5by a;' + aj^ + l.
24. £c* + a;^ + l by £c*-a;^ + l.
25. a^-}-a^x—ax'^—x^ by «^— aa; + £c^
/ 26. 2a'^ ba'b' - 36* by a' - ^a'b' + 26*.
27. a'-^ab + b'hj a'-ab + b\
28. x' + x'-i-x' + x + lhy x-1.
29. J?^-4^C^by J^^-4i?(7.
30. pc+qd—rehy Sqd—2pc-^re.
31. iK^ + 2aa;^-£c* by 2x'-Sa + ax\
32. «-^-3a'6 + 3a6^-6=^ by a— 6.
33. a'x'-2ax' + Sa'x-x' by aa;-a;^ + 6e^
34. ab + bc—cd—bdhjab — bc + cd+bd.
35. a''—aW + a'b^ by a*-6l 39. ic^'^+^ + y"-^ by a;^-2/^
36. aj'^ + y'^by aj + y. 40. x''—x''-' + x''-''hjx''-i-l.
37. X"— y" by x^ + y"". 41. |t«'— ^ 6«6 + i6'' by ^a—^b.
38. ^=^'' + 6^'' by a^^' + J^^ 42. ^x'-^a' by ia^^ + ^a^
43. ^x'-^x^^ by 2£t— 1.
44. i«=^-^a;2 + ia;— ^ by 3.c^ + 9£c-27.
45. f^^ + |^-iby|a.^-|;^-i.
46. 1.4i«^-2.7a; + 3.2 by 2»5''- 1.4a; -3.2.
47. x''—y^\)yx^—y'. 48. £c»+y'^ by a;™ + y'".
Find the product of
49. x-1, iB + 2, aj + 1. 52. x' + x-^l, x-1, a;M-l.
50. x^-1, a;2 + l, aj* + l. 53. a^-6', a^-\-b\ a^ + b\
51. 2a-36, 3a + 26, a + b. 54. ic + 1, aj + 2, £c + 3, aj+4.
MULTIPLICATION OF LITERAL EXPRESSIONS 63
56. ia^ + iy, ix + y, i^ + iy-
6 6. £c» — y % ic« + y ", x"' + y "'.
57. f/ -h ab -}- b'^, a — b, a^ — ab + b'^, a-\-b.
Remove signs of grouping and simplify :
58. {a-{'b){a + b)-(a-b){a-b).
- 59. 2(Qi'-Sx+l)-(x + A)(x-l).
60. (x + y)(x-2ij) + 2(x'-y') — {x—y)(x+2i/).
61. 2Xx+l)(x-l)(x + 2)-4x(l-x)(x + S).
62. -2a^{a^-3a(a-6)} +(a' + a^>-^>')(«'-6' + a^»')-
CHAPTER VI.
DIVISION OP LITERAL EXPRESSIONS.
47. Law of exponents. The law of exponents for division is
obtained directly from the corresponding law for multiplica-
tion, by means of the principle
quotient x divisor = dividend.
Expressed in symbols the law is :
This follows from the preceding principle, for the quotient
«'"-", multiplied by the divisor a" gives «"*-"«'", or a% the
dividend.
Thus, a^-i-a^=a^-^=a\ This agrees with the above principle,
for a'^-a^=a^.
48. Meaning of a". By § 47 we have
But «"^a'' = l, for any quantity divided by itself gives 1.
Hence fl°=l. Axiom 7.
That is, am/ base loith the exponent zero equals 1 .
Thus, a^=l ; 2"=1 ; 10''=1 ; 45"=1; x^-^x'=3(P=\.
It is therefore evident that if a base appears to the same
power in both dividend and divisor, it will have the exponent
zero, and hence gives the value 1, in the quotient.
49. Division of monomials. Since division is the inverse
of multiplication, i. e^ since quotient X (Umsor = dividend^ then
from § 44, the rule for multiplying monomials, we obtain the
following rule :
64
DIVISION OF LITERAL EXPRESSIONS
(;5
To divide one monomial by another^ divide the numerical co-
efficient of the dividend by that of the divisor^ using the law of
signs ; then divide the literal ^factors by subtracting the ex-
ponents of the bases in the divisor from the exponents of the like
bases in the dividend to obtain the exponents of these bases in
the quotient.
Thus, since (4a'&V)(-3a*60=:-12a«6V,
( - 1 2a«65c*) ^ ( _ 3a*6c'0 = 4.a'b*c\
which may evidently be obtained by the above rule.
Likewise, ( - IGx'y'z') -- {Sx^y^z') = - 2o^z'.
EXERCISE 17.
Divide :
1. a^ by a^.
2. a'b' by a'b.
3. -QaW by 3a'
4. lSx^y''z* by
6. 4«^6''c^ by a^bcK
6. 2Sa*b'c' by -7ab\
7. -lOOyyby -2bp'r/.
8. -bOx'a' by 4x'a\
9. 42aWc by QaWc.
10. \m^ii^p^ by — \rnn^p^.
11. r'sH' by — 3rV^.
13. -5a;™+i by -ic».
13.
by a"
14. — 12£C'»+"2/™+"by Sy'^aj"
15. x^'-^'y by of-'^y.
16. -12s¥by ^st\
17. 18y^"-^ by 2y"-l
18. IZz'^-^y"' by 62"-"^'".
19. 7i5"'5'" by 3^"'5".
20. lls^+V-^ by Qs-'-^r'-^'
50. Division of a polynomial by a monomial. If a polynomial
be divided by a monomial, the quotient multiplied by the
divisor must equal the dividend. Hence, the quotient must
be such an expression that the product of its terms by the
divisor will give the terms of the dividend. Therefore, the
terms of the quotient must be obtained by dividing the terms
of the dividend by the divisor. Hence the rule :
5
QQ ALGEBRA
To climde a polynomial hy a tnonomial^ divide each term
of the dividend hy the divisor, and take the sum of the residting
quotients.
Example. Divide a?^ 4-40?^— 5x* by ic^
{x^ + 4:Qd' — ^x^) -^x^=x^ + 4a^ — ^x^.
The work might be written
x^ 4- 4x'-5x' ^^, 4x'-5x;\
x^
Another form often used is
x^ )a^ + 4x^-5x^
x* + 4:X^ — 5x^ '
EXERCISE 18.
Divide :
1. x' + x' by x\ 5. -S0a'b'-27a'b' by -SaW.
2. xy-xy + 4:S(^y by xy. 6. Sa'-6a*b + 9d'b' by Sd\
3. x''—bx*^Sx' by —x\ 7. a'-d'b-d'c by -a'
4. ^Ix^-l^x' by -9x\ 8. 1x'''-\^x'y^ by ^x\
9. 4a;y + 8a;y-12j«yby2£cy.
10. 3£c2— |a;y + | a^y by fa^.
11. — «+5— cby— 1.
12. «=^ + a6 + ^'by aW.
1 3. ^ficy + 5£cV' - ^a^y by — liaj^'.
14. 3.25,73' -5.2a;« + 9.75x^ by .25i«l
15. 10a;'»+='-4a;»+3 by 2a!l 17. y»+V"+^ + y'"+V'+^ by a;»+y+».
16. a^'^—a"^^ by a». 18. x"- + x''+'' -V x'^^" by a;'.
5 1 . Division of one polynomial by another. The process of
dividing one polynomial by another is based upon the principle
that the quotient multiplied by the divisor gives the dividend.
The process is best explained by taking an example as follows,
DIVISION OF LITERAL EXPRESSIONS 67
First arrange both dividend and divisor according to the descend-
ing powers of x. See § 46. The work may be indicated as
below.
Dividend x*+ a^ + 7x^—&x + 8 a^ + 2x-\-8 Divisor.
{x'^-\-2x + 8)x^= x^ + 2x^ + 8x^ of— x+1 Quotient.
—x'— af—Qx + S
{x^-\-2x + 8)-{—x)= —af'—2od'—8x
x^ + 2x + 8
(x^4-2x + 8)-l= ' a;^ + 2a?+8
Now the product of the term of highest power in x in the
quotient and the highest term in the divisor must give the highest
term in the dividend. Hence, the highest term in the quotient is
the quotient obtained by dividing the highest term of the dividend,
X*, by the highest term of the divisor, x^. This gives x^^ the first
term of the quotient.
Multiply the whole divisor by the term of the quotient just
found. This gives x* + 2x^ + 8x'^, which is placed below the
dividend.
The dividend is the product of the divisor by the whole quo-
tient. And X* + 2x^ 4- 8x^ is the product of the divisor by the term
of the quotient found. Hence, subtracting this from the dividend,
the remainder —x^—x^—Qx-\-8 must bo the product of the divisor
and the part of the quotient to be found.
Therefore the product of the next highest term of the quotient
by the highest term of the divisor must equal the highest term of
the remainder. Hence, dividing —x^ of the remainder by x'^ of
the divisor gives — x, the second term of the quotient.
Multiply the whole divisor by the new term, —x ; subtract the
product from the remainder. This leaves oc^ + 2x+8.
Evidently the third term of the quotient will be obtained from
this second remainder just as the second term was obtained from
the first remainder.
By continuing this process, all of the terms of the quotient may
be found.
68 ALGEBRA
The above reasoning will evidently apply to any dividend,
divisor, and their quotient.
If the divisor is an exact divisor of the dividend, the work
may be carried on until a remainder zero is found. Otherwise,
the process may be continued until a remainder is obtained in
which the highest term is of lower power than the highest term
of the divisor. This is a true remainder.
It is evident that, in the latter case, the dividend is com-
posed of the remainder and the product of divisor and quo-
tient. That is,
dividend = quotient x divisor + remainder.
If, in the preceding example, we had arranged both dividend and
divisor according to the ascending powers of x^ we would have
obtained the same result, except that the order of the terms
would have been reversed.
We have, therefore, the following rule for dividing one poly-
nomial by another :
Arrange both the dividend and divisor according to the de-
scending or ascending powers of some letter.
Divide the first term of the dividend by the first term of the
divisor to. obtain the first term of the quotient.
Multiply the lohole divisor by this term of the quotient^ and
subtract the result from the dividend.
Treat the remainder as a neio dividend, {being careful to
arrange the terms as before) and repeat the process, continuing
until either the ronainder zero, or a true remainder, is found.
Example 1. Divide a^— 11a + 30 by a— 5.
a^- 11a + 30
a^— 5a
— 6a + 30
— 6a + 30
a — 5
a—^ Quotient.
DIVISION OF LITERAL EXPRESSIONS 09
Check. When a=l ; dividend =20; divisor =—4; quotient =—5;
as it should.
Example 2. Divide 789ify' + 45xy^ + Uy* + Ux* + 45a^y by 2x' +
7y^ + 5xy.
Uoc* + 45a?*2/ + 78ic22/2 + 4:5xy^ + Uy*
14x* + 35^ + 49.rV
2xM- 5xy + 72/^
7^^;=^ + 5xy + 2i/''
lO.r'2/ + 29xY^ + 45it?2/"^ + 14y*
10x'y + 25x'y' + S5xy'
4:X^y'^ + 10xy^-{-14y^
4;rV + 10a^y' + 14*
Here the terms are arranged according to the powers of x
without reference to the formation in y.
Check. When a?=l, i/=l; dividend=196; divisor=14; quotient
=14, as it should.
Note. Where the error may be in the exponents, other values than 1
sliould be used for the letters or general numbers. See " observation"
§45.
Example 3 . Divide 2x'' — 7a? + a?* + 1 — 7x'^ by x'' + 2— Sx.
a?* + 2a?^— 7a?2— 7£C + 10 |a?2— 3x + 2
oc*—33(^ + 2x'^ \x^-^5x + Q
IX'
\x'
5x'— 9x^— 7.r + 10
5x^-15x'' + 10x
6ir*-17a^4-10
6ar^-18;y+12
X— 2^ True Eemainder
Check. When x=S; dividend=61; divisor=2; quotient=30;
rem. =1.
Observe, that either a?=l or a?=2 would reduce the divisor to
zero, hence, these values could not be used. See § 219.
Example 4. Divide x^ + y'^—z^-\-3xyz by x + y—z.
Here each remainder should be arranged with the term or
terms containing the highest power of x preceding all others.
70
ALGEBRA
ar* +Sxyz + y^—z^
x + y-z
\ sc^ + x'^y-x^z
x'^—xy + xz + y'^ + yz + z^
—x^y + x^z Sxyz T ^
^ -x'y -xy'+ xyz
x'^z + xy^ + 2xyz
> x'^z + xyz—xz^
xy^+ xyz + xz^-Vy^
^ xy"" -vy'-y^z
xyz + xz"^ +y''z—z^
. xyz -vy^z-yz^
xz^ +yz'^—2^
[^ 3dz^ +yz'^—z^
EXERCISE 19.
Divide and check :
1. Grt^-7a-3 by 2a-3. 3. 17.y + 2y^ + 21 by 3 + 2y.
2. lQx + bx' + ^ by £c + 3. 4. 5a-^ + lla + 2 by a + 2.
5. 3a^-7a-2-2a^ by 1 + a.
6. a;*-2a;y + y*by ic'-2a;y + 2/l
7. c^-lOc + 24 by c-6. 11. aj^' + y^^ by x + y.
8. a' — (^' by a + b. 1 2. ic' + y ^ by aj + y.
9. a'-h' by a-^>. 13. ««-^»« by ci'-h\
10. a;*-.y* by x-y. 14. a« + ^»« by a=' + 5^
15. 144x^-1 by 12a; + l.
16. a;^ + 3a;2 + 3a;+lbya;^ + 2ic + l.
17. 03*— 2ic^y + 2a!y— y* by a;^— 2/1
18. a;^ + a!* + l by aj^-a^^ + l.
19. aj^-1 by a?-l.
20. 7aV-3a*-5aV + 3aaj«-2a;« by a;? + 2aaj^— aj*.
21. 2a*-9a^ + 17a^-14abya'-2a.
22. x^-y^hy ^-y\
DIVISION OF LITERAL EXPRESSIONS 71
23. iz;*-12£c=^ + 54£c^-108a;+81 by x^-Qx+9.
24. x'+x' + l by x* + x' + l.
25. 15/ + 13y-17/-3 by 6/ + 3-4y.
26. x'-a^hj x' + 2x'a + 2xa' + a;\
27. l-x-Sx'-x' by l + 2a; + £fl
28. x' + 2a;=^y2 + 9y* by x' - 2xi/ + 3yl
29. a3* + 81 + 9£c^ by Sx-x'-9.
ZO. a'-Sb'-l-Qabhya-l-2b.
31. a^"^— 41a^-120 by x'-h^x + b.
32. ^—\f^2yz—z^ by x V y—z.
33. x^—x^y—xy^-\-y^ by x- Ty^ — 2xy.
34. a-' -243 by a- 3.
35. 13«^^ + 71t«-70a^-20 + 6(«'by 3a'^ + 4-7a.
36. i6''-'-y' by cc^'— y.
37. a* + 2«^-8a + 12-7a^ by a^ + 2-3«.
38. £c^ + y^ + 2=* — 3£cys by £c + y + ^.
39. ic=^ + 3xV + ^xy^-{' y' + 2' by a^ + y + s.
40. \a''\-^^ab' + ^^h' by ia + i^.
41. ^i^a^»-32^^ by \c(}-2b.
' ^% ^-^x'-\-\xY + y'hy ^x^ + ^xy^f.
43. \xY-\-Th^\^l\x^ + \xy.
^4. 6«"'"* + 3a^"*6*" + a'"^"''*— ^=^'" by (r + ^'".
45. £c*"— y*" by x'+y".
46. 12,x''*+^ + 8£c'^— 45iK"-^ + 25£c»-^ by 6a;- 5.
• 47. -^\x^^ + ly^'' by ^aj^u _|_ i.yu^
48. 4a2 + 4a^ + <^^-12ac-6^>c + 9c2 by 2a + ^-3c.
52. The fraction. An indicated quotient is called a
fraction. The terms used in arithmetic are also applied to
72 ALGEBRA
algebraic fractions. In a fraction, the dividend is called tlie
numerator, and the divisor, the denominator. The numerator
and denominator are called the terms of the fraction.
A fraction may be expressed by any of the signs used to ex-
press division.
a , x^ + 1
Thus, ,— a/6, a-v-6, a : 6, -' are fractions.
' h, ' ' x+1
Any laws that apply to quotients must evidently apply to
fractions.
In the following sections a few principles are established
concerning fractions that will be needed in the subsequent
work. For the full treatment of fractions see Chapter XI.
53. Since quotient X divisor = dividend, it follows from
the preceding definitions that
fraction X denominator = numerator ;
a
that is, J-- b = a.
This is a useful principle.
54. The product of tioo or more fractions equals the product
of the 7iumerators divided by the product of the deno)ninators ;
a b c abc
that is, , — = .
X y z xyz
To establish this, call the product p ; i. e., let
abc
-.-.-=p.
xyz
Multiply both members of this assumed equation by x, y
and z in turn.
DIVISION OF LITERAL EXPRESSIONS
Multiplying both members by x we have
Then
or
Then
a
X- •
X
b
y
c
z
=px.
' a
b
y
c
z ~
=px
h
y
c
z
• a =
=px.
bers of this
equa
h
c
z
a =
=pxi/.
z
• a =
=2)xi/
or
a b=pxy.
^3
Axiom 3.
§ 63.
Law of order,
y, we have
Why?
Why?
Why?
Then multiplying both members by z^ we have
cab=pxyz^ or abc=pxyz.
Now dividing both members by xyz^ gives
abc
xyz
abc abc
X y z xyz
since each member equals ^x
This reasoning can be extended to any number of fractions
P-
Therefore,
Axiom 4.
Axiom 7.
Example 1. — — k-^ • -^ = ^ ,. ., — ■.
5 2a^ a* 5-2a^ a
Example 2.
2^2
3x^
3z/
42/' -2/'
32/-42/^-(-2/3)
60^
12/-
55. To dlolde any number by a is equwaUnt to muUiplying
the number by the fraction — ; that is,
rV
^4: ALGEBliA
n 1
~ = n. —
a a
For, since quotient X divisor = dividend^ to multiply -
by a gives ~a = n. And to multiply n- by a gives n—a =
n- { --a] =nl^n, for the the same reason. Therefore - and
?i-must each be a quotient obtained by dividing n by «,
and hence must be equal.
Thus, 184=J/=3.
56. The law of distribution holds also for di vision ; that is,
a-\-b + c _a be
X XXX
For, " + ^' + '' = (« + j + c)i, by § 55,
= a- + /'>>- + c-,by law of distribution,
XXX
a h c
= - + - + -, by §55.
XXX
CHAPTER VII.
POWERS AND ROOTS.
57. Involution, The definition of a power of a number was
given in § 18, and the laios of signs ofpoioers were established
in § 36. Tlie student sliould now reread those two sections.
The process of raising a base to any power is called involution.
The following laws of exponents will now be established for
involution, wliere the exponents are assumed, of course, to be
positive integers.
58. Power of a power.
(a")"' = fl"™.
That is, the mth povter of the nth poioer of any number equals
the nmth povner of that number.
For, by definition of an exponent,
(«»)'" = a" • a" -te" to m factors,
:^^«+n+n+ m terms, ^^^ ^f CXpOUeUtS, § 43.
Thus, {a^)^=a^-a^-a^-a^=a''^ ; (x^)^=aj^-'=a;2^
59. Power of a product.
{aby = a'b'.
That is, the nth poioer of the product of tioo numbers equals
the product of the nth powers of those numbers.
For, by the meaning of an exponent, *
{abY = ababab to n factors,
75
76 ALGEBRA
= {aaa to /i factors) (^^^ to ?* factors).
Laws of order and grouping.
By similar reasoning the law can be shown to hold for any
number of factors.
Thus, {xyzwy=xYz'iv'; {2aby=2^aW=8aW;
{-3aWy={-3yia'y{b'y=81aW\ by § 58.
Combining the laws of §58 and §59, we get the following
rule :
To raise a monomial to a required poioer^ raise the numerical
coefficient to the required poicer^ using the laws of si(/?is; then
inultiply the exponent of each literal factor h\j the exponent in-
dicating the required power y then indicate the product of the
Example. Raise — Sa^V^^^ to the third power.
We have (-5a.-V^2)'=(-5)'^*V'^'''=-125^^2^V.
60. Power of a fraction.
\b] "F
That is, the nth power of a fraction equals the nth poicer of
the numerator dioided by the nth poioer of the denominator.
For, It) = ----- to n factors, by def. of a power,
\oJ h b h
a aa • • to 7i factors e c^
§ 54,
Thus,
-hbh
to n factors'
a^
-3xVV
2d'b' J
(-3^V)' 81^1/1^
POWERS AND ROOTS 77
EXERCISE 20.
Raise to the indicated powers :
1. (ay. 13. (-ba'xyy
2. (ay. 14. (Smhy.
3. (-aWy. 15. (-2d'x'y.
4. (-xyy. .a^
21
22. u-^^:
5. (xYzy. *"• V^V 23. (ay.
6. (axy)«. ^^ /^'Y 24. (a^hy\
». ( ^^2/). 18. ^— ^^). 26. (-x^Y'^^K
9. (7a^yni
10. (2.y)l 19. (-1^). 27. (^).
12. (-xyy. 2«- ^7-wV- 28. (^) .
6 1 . Square of a binomial.
(a + 6)^ = a^ + 2a6 + 6^
That is, the square of a blnotnial. equals the square of the first
term^ plus two times the product of the tico terms^ plus the square
of the second term.
For, (a-^hy means (a+^)(« + ^). By actual multiplication
this becomes a'^ + 2ai + ^^
Note. — It is understood here that the symbols a and h represent any
terms whatever. Either term may be positive or negative.
Thus, Zx^—2y is of the form a + &, where a stands for 3a?^ and
6 stands for —2y.
Example 1. Square 2x-{-3y.
{2x+Syy={2xy + 2{2x){3y) + {Syy=4x'-{-12xy + Qy\
78 ALGEBRA
Check. When x=2 and y=l; base =7; po\Yer =49, as it should.
Example 2. Square 5x*— 2?/^
{5o(f-2y'y={5x'y + 2(5sc'){-2y')-h{-2yY=25af-20x'y' + 4y'.
Check. When x=2 and y=S ; base =22 ; power =484, as it
should.
Examples. Square —2aH 66.
(-2aH66)2=(-2a=^)2 + 2(-2a'')(66) + (66)'
=4a'-24a^b + 3Qb\
Check. When a=l, h=2; base =10; power =100, as it should.
Example 4. {(a + 6) + l}'=(a + 6y' + 2(a + 6) + l
=a' + 2a6 + ^'-^ + 2a + 26 4- 1.
It is observed that since the square of any number is positive,
the terms obtained by squaring the terms of the given binomial
are always positive. The other term is positive^ if the terms of
the given binomial have like sigtis, and negative., if they have
unlike signs,
EXERCISE 21.
Write out the following squares by the above rule :
1. {x + y)\ 11. {^x'-^y. 21. {m'~^)\
2. (2a + J)^ 12. {x-Viy. 22. {mn-2y)\
3. i'Za^Uy. 13. (2i« + 3)l 23. (46^-5)1
4. {2x^-\-yy. 14. (3ic=^ + 4)l 24. {x'-l)\
5. {a-bf. 15. (5«^' + l)l 26. {d'-l)\
6. {x-y)\ 16. (a* + 10)l 26. {x'-yy.
7. {^x-Zy)\ 17. {la^-^-lxf. 27. {x' + xy.
8. {a'-by. 18. (Sa^' + Sa^^)^ 28. (^y-7)^
9. (2a;^-2/'/. 19. (a;y + 2a)l 29. (£c'-10)^
10. (aj*-5/)l 20. (m-3)l 30. {x'-by.
POWERS AND ROOTS 79
50. U +
36. (2x-Shjy.
37. (2ab-{-4:bcy.
31. (a;*-a;^)l 41. (x^'-iy.
32. (2a;*-' + 0^)^ 42. (x" + iy. """ V* ' ^/
33. {6x'-xy. 43. (cc"-l)l 51^ (-,-~X
34. (7a + 2^.)l 44. (^._2^»)2. ' \*' 2/7-
36. (3aV + ^T- 46. (2.» + 3a»)l ^2. (^l + l^.
46. (a^+'-n'^-^y. 53. {(2a + ^) + 6-} I
38. (-2^xyy. ^7- i^^^^' + ^-^^y- 54. {3a + (5-c)}-
39. (-a^i/^ + a^V)^ ^^' («"*"-!)'• 56. {4-(2a + ^.)}^
40. (.t'"+1)1 49. (a"5»-^-a"-'^«)'- 56. {(4-3/>)-3c}^-
62. Square of a polynomial.
(a + 6 + c)^=fl^ + 6' + c'+2fl6 + 2flc + 26c.
By actual multiplication, it will be found that
(a + 5 + c)' = a^ + ^' + c\+ 2ab + 2ac + 2bc ; also
{a + b-\-c + dy = a' + b' + c' + d' + 2ab + 2ac + 2acl
-^2bc + 2bd+2Gd;
and so on, for any number of terms.
That is, the square of any polynomial equals the sum of the
squares of all of its terms, plus two times the product of each
term into all of the terms following it.
Example 1 . Square 2x^ + 3a? + 5. '"
(2a?^ + 3.^+ 5)2= (2ic'0=^ + (3a^)2 + (5)2 + ^{^x'^Zx) + 2{2x^){S>) + 2(3if)(5).
= 4a?^ + 9a;2 ^ 25 + 12x^ + %W + 30^.
=4x^12x^ + 29x^ + 30.^ + 25.
Check. When x=l;* base=10; power=100. To check ex-
ponents also let X equal some other number than 1. Check when
x=2.
go ALGEBRA
Example 2. Square a*—2a^ + 3a^—^a.
{a*-2a' + 3a'-4ay={ay + {-2a'r + {Sa'y + (-4.ay + 2{a')(-2a^)
+ 2{a'){Sa')+2(a*){-4a) + 2{-2d'){3a')-{-
2(-2a^)(-4a) + 2(3a'-^)(-4a)
=aH4a«4-9a*4-16a2-4aH6a«-8a5-12a^ + 16a*-24a3
=a«-4a^ + 10a«-20a5 + 25a*-24a^ + 16a^ '
Check. When a=l; base=— 2; power=4, as it should. Let
a=2 and check.
Note. — The student should learn to write out these values without
indicating the work as in the first step. He should always check his
work.
Example 3. Write out (S—2x + x''y.
{S-2x + x''y=9-\-4x^ + x*-12x + 6x''-4x^
=9-12x+10x^-4x^-{-x\
Check. Whena?=2; base=3; power=9.
EXERCISE 22.
Write out the squares of the following polynomials :
1. l+x + x\ 10. 4x'-a' + Sax.
2. a* + a' + 2. 11. Sx'-^y"-] bb-a\
3. ^x + Sx' + 4x\ 12. b'-4ac+10.
4. x'+x' + x + l. 13. ab-hc\cd-ad.
5. 2iK*-3a3^ + 4. 14. a'-^¥-&-d\
6. a-2J + 3c-4f7 H5e. 15. 2.^•^ - 5a; + 7 - 3al
7. l_a;-|-a;2_|_a;3 4-a;4. 16. \-x-Vx^—x^-^x'-x\
8. 5a;•'-2 + 7«^ 17. ic" + y« + ^".
'^,"x-\-y—z-\-w. 18. a?" — y"+^4-2«+^
19. Show that the rule of § 61 is a special case of § 62.
20. Show also how § 62 could have been obtained from § 61
by so grouping as to form a binomial.
POWERS AND ROOTS 81
63. Any power of a binomial. By actual multiplication it is
found that
(a-{-by = a'+3a'b + Sab' + b';
(a + by=a' + ^a'b + Qa'b' + ^ab'-}-b';
(a-^b)'=a'-^5a'b + 10a'b' + 10a'b'-{-5ab'-hb';
(a-^by=a'-}-Qalb + 15a'b' + 20a'b'-\-16a'b'-\-Qab'-\-b';
and so on.
Now by comparing these few values of the different powers
of a+b, it is found that they all may be written out by the
following laws :
(i) The first term in each case is a vnth an exponent equal to
the exponent of the binomial ; the last term is b with the same
exponent.
(^) The expo7ient of a in each term after the first is less by 1
than its exponent in the preceding term, b appears to the first
poy>er in the second ter^n., and its exponent in any term, after the
second is greater by 1 than its exponent in the preceding term.
The sum of the exponents of a and b in any term is the same for
all terms., and equals the exponent of the binomial.
{S) The coefficient in the second term equals the exponent of
the first term, 'jind the coefficient of any term is obtained from
the preceding term by multiplying the coefficient of term by the
exponent of a and dividing the product by a number greater by 1
than the exponent of b in the term.
{4) The number of terms is always greater by 1 than the ex-
ponent of the binomial.
Note.— These laws constitute what is known as the Binomial
Theorem. This theorem was first estabhshed about the year 1665 by
the great mathematician, Sir Isaac Newton.
6
82 ALGEBRA
In chapter XXITI this theorem will be proved to hold for
any positwe integral exponent, and its application to fractional
and negative exponents will be shown.
Example 1. Write out the value of {x-\-yf.
By (1) and (2), the terms without the coefficients will be
ixf" x'y ixfy'^ scr'if x^y^ x^y^ x^y^ xy^ y^
By (3), the coefficients will be
1 8 28 56 70 56 28 8 1
Hence, {x + yf=j(^ + ^x^y + 2%x^y'' + mx>y^ + H)x'y^ + 56xV'^ + 28ic'7/«
+ ^xy'' + if.
Check. When x=\, y=l ; base=2 ; power=:256.
Example 2. Write out the vahie of {x—yf.
The exponents and coefficients may be calculated at once.
{x-yf=x' + ^x\-y) + ^x{-yy + {-yf
= x^ — 3x^y + 3xy^ — y^ .
Check. When x=3, y=l ; base=2 ; power=8.
Example 3. Write out the value of (2x^—3y-^y.
{2x'-3fy=(2xy + 4{2xy{-:iy') + ()(2xy(-3yy + 4(2x')(-:iy'y +
(-3yy =16x^-96x^y^ + 216xY-21Qx'y^ + SU/\
Check. When x=2, y=l ; base=5 ; power=625.
EXERCISE 23.
Write.out the values of the following powers :
1. (x + yy. 8. (Ax'-Syy. 15. (f x— fy^.
2. (x~yy, 9. (x + iy. 16. (ix' + Wy-
3. (x-ay. 10. (1 + ay. ^^ /a c
4. (x + yY\ 11. (x-iy. ' \^ ^^
5. (^x + 2yy. 12. {x-2y. 18. (^-2
6. (^x^ + yr- 13. (1+yy. /I^ , M'
7. (x'-yy. 14. (x + ^yy, ^' \a'^ bj'
POWERS AND ROOTS 83
20. (2x-yy. 22. (2a-^y. 24. (f«-f^»)«.
21. {x + i^y. 23. (i-2a)«. " 25,{ix-yy
ROOTS.
64. If all of the factors in a product are equal, one of the
factors is called a root of the product.
Thus, a is a root of aaaa, or a\
The nth root of an expression is one of its 7i equal factors.
Thus, the square root of an expression is one of its two equal
factors.
The cube root of an expression is one of its three equal factors.
The fourth root of an expression is one of its four equal factors.
The fifth root of an expression is one of its five equal factors ;
and so on.
To indicate a root of an expression the radical sign (|/ ) is
used, ^/x represents the fourth root of x. Here 4, the num-
ber placed above the radical sign, is called the index of the
root. The index of a root of an expression indicates what root
it is, or the number of equal factors in the expression.
The index of a square root is usually not written.
Thus, f/x represents the cube root of x; i.e., one of the three
equal factors of x. ya represents the fifth root of a. ya is
the same as j/d. y 81 represents one of the four equal factors of
81 ; i.e., y 81=3.
A root is called an even root if its index is an even number,
and an odd root if its index is an odd number.
Thus, yd represents an even root ; yd an odd root.
It follows from the above definition that to find the nth root
cf (i gimn number is to find a second number whose nth power
equals the given number ; that is,
{y^ay = a.
84: ALGEBRA.
Thus, since 5^^=25, therefore |/25=5 ; since {a}f=a^, therefore
The process of obtaining a root of an expression is called
evolution.
6 5 . Laws of Signs. The laws of signs of roots are obtained
from the laws of signs of powers.
The following principles are true :
(J?) A 2^ositwe number has at least two even roots which differ
only in signs. For, if two numbers have the same absolute value,
but differ in sign, their like even powers are equal and positive.
Thus, since ( + 3)^=9 and (-3)^=9, therefore v'^9=+3 or -3.
Also, since ( + 2a^)*=16a^2 and (— 2a=')*=:16a'^ therefore y'T^'^
+ 2a^ or -2a^
The two even roots of a positive number are often written
together, by use of the double sign ± . Thus ]/25a*= ± 5a^, means
]/25a*= + 5a^ or — 5a^
{2) Any jjositive nnmher has at least one odd root., which is
also positive / and any negative nnmher has at least one odd root,
which is also negative. For any number has the same sign as
any odd power of itself.
Thus, since ( + 3)=^ = + 27 and (-3)=^= -27, therefore ^27= + 3
and ^^^=-3.
Since ( + 2)^= + 32 and (-2)'^=-32, therefore |/32=+2 and
5/
V -32=-2.
(3) A negative number has no even root that can be expressed
as a positive or negative number. For any even power of a
positive immber, or of a negative number, is positive ; i. 6., no
positive or negative number, raised to an even power, can give
a negative number.
POWERS AND ROOTS " 85
Thus, y/ —9 is neither +3 nor -3, for ( + 3)'= + 9 and (-3)"
= + 9.
The indicated even root of a negative number is called an
imaginary number, and does not belong to the series of num-
bers with which we are now acquainted. Imaginary numbers
will be discussed in Chapter XIV.
66. Root of a power. A root of a j^otrer of a base equals that
poioer of the base vnhose exponent ^.9 tJie quotieiit obtaiued by
dioidlng the given exponent by the index of the root. That is,
n — 77
ya"' = a"'^".
For, by § 58,
(a'"-5-") " = a"'-5-''x» = a"'.
m
Hence, i/o^^^"'^'* or <^"-
Thus, j'/^o :^ ct2o - 4 ^ ^5 . ^^ = x^ = x' ; f{x-yf=(x-yy^-=
{x-yf.
67. Root of a product. The nth root of a product equals the
product of the nth roots of its factors ; that is,
y ab — yay b.
For, by § 59,
^yay'br = {{raY-{yby=-ah.
Hence, ^} ^ == ^^ ~ . ^« ^.
Thus, ^^^ = -,^^l>^; -i>^Y^= |/^|/^=a^Y; y" - 32a?^ =
y^^y'¥^=-2-x' or -2x\
68. Root of a fraction. The nth root of a fraction equals the
nth root of the numerator divided by the nth root of the denomi-
nator ; that is,
V\
S6
ALGEBRA
For, by § 60,
Hence,
^ * yb
Thus, |/g =
^'' y 2436^^ ^'2436^'^
2a«
36=^'
69. If the value of the mdicated root of a rational expres-
sion can not be exactly obtained, the indicated root is called a
surd. An expression which contains one or more surds is called
a surd expression, or an irrational expression.
Thus, |/3, ]/ c?, -y/a + b, are sia^ds ; Vx—\/y'^ 24-i/5, ixvQ irra-
tional expressions, yl + y^ is not a surd since 1 + ^/3 is not a
rational expression.
A perfect ntla. power is an expression whose ni\\ root can be
obtained ; i. e., the ?^th root of which is a rational expression.
Thus, since (it^— 4)2=x*-8a^' + lG, then yiJc^—%x'-\-U=x'-4..
Hence, a?*— 8a?^ + 16 is Si perfect square.
70. Roots of monomials. Roots of monomials may be extracted
by means of § 65, § 66, § 67, § 68.
Example 1. Find the square root of 25a^y*.
We have |/25aV= V25\/a^i/y* § 67.
= ±.5aV. §65, §6G.
Example 2. Find the fifth root of ~B2x'Y^.
We have ^-32x'Y'= \/~^^ y'x^^Vy^ % 67.
= -2icy. § 65, § 66.
si 125a;V
Example 3. Find the value of \~ 21 6a'>&' >'
./ I2r»y_ rm£i- ^^ j„
4
216a»&8 f 216 a«6«
POWERS AND ROOTS
87
§67.
-"""^2* §65, §66.
J^Y5-
Example 4. Find the value of i/f+
Adding tr.c«on.. l/ff^.y/S
= /
16^2
25
, =±i|/2, surds.
EXERCISE 24.
Find the value of
Al ^9Z^5^ 11. tM6^5«. J^^ i/J^,
2. |/T6-^y. ^12. ^>8k,Tvy^. ' /^l^l
4. v/225m'»«'. /^U. ^"=«Tpy?^. 23. r5?^«.
^24. j;/i?5^
^
^5. if 27^^«. 15. i:^32^
6. f-^xY\ '16. ^/-243«^^y. *^^"'** ^^ "^
7. r^^^. n. i^-,^V^a^6-A26. V^T^.
8. f/-125m^Vi«. ,18. i^e^^^s. \ 26. i/TT^T.
9. iTGlSy: 19. ^-Vl^aH'^ \ 27. ^/fT].
10. i^ip^l /20. v'afy^ \28. Vf^.
71. Square roots of trinomials by inspection. In §61 it was
shown that the square of a binomial was a triuoraial. I'lie
square root of a trinomial that is the square of a binomial can
be found by inspection.
88 ALGEBRA
Note. — It is not always possible to extract any root of any expres-
sion. See § 69.
Since the square root of the trinomial is to be a binomial, it
must take the form a + h. Plence, the trinomial must be of the
form {a-^b)\ov ce \'lah-\^h\ From the form of a' + 2ah-^b\
we have the following :
A trinomial is a perfect square if tioo of its terms are perfect
squares (a^ and 6^), and the other term is tvnce the product of
t?wir square roots (2a 6).
Thus, a*— 6a2 + 9 is a perfect square ; for |/c? is either a^ or —a^,
|/ 9 is either 3 or —3, and twice the product of two of these
values, a^ and —3, or —a^ and 3, gives the other term — 6a^ In
fact a*— 6a''' + 9 is the square of a^— 3, or of — a^ + 3. Why not
select a'^^- 3 or -a='-3 ?
From the type form,
fl^^-2a6^-6^ = (a^-6)^
we have the following rule for obtaining the square root of
the trinomial :
Write the sum of the square roots of the terms that are p>er-
feet squares^ using such signs that twice the product of the result-
ing terms will gim the other term of the trinomial.
Example 1. Find the square root of a;^— 14a? + 49.
We have j/p=±ic,|/49=±7'. ^ Since the product — 14.:r^ is
negative, the terms must have unlike signs. Hence we use ^x,
and —7 or —x and +7. Therefore y x' — \^.x\ ^^—x—1 or— ^ + 7.
See §65, (1).
Check. When x=\ ; base=36 ; root= — 6, or 6.
Example 2. Find the value of |/9x* + 30x'-^?/ + 25 z/^.
Here |/ 9x*= ± 3a?', |/25p= ± 5?/ Since + ^^x^y is positive, we
must use like signs. Hence |/9^*T30^pT252/^=3a?H5?/ or
-3ic2-52/.
Check, When a?=l, 2/=l ; base^^Oi ; root=8 or —8.
POWERS AND ROOTS g^
EXERCISE 25.
Determine which of the following expressions are perfect
squares :
1. 4a' + 4ab + b\ 8. a'b'-2a'bc' + c\
2. a'-^ab + 9b\ 9. {x' + x' + l.
3. lQx' + 24x}/ + 9if. 10. l--ia + ^i-a2
4. x + 2xij-\ri/, 11. 25xY- + 20axij + 4a\
5. 9r«^ + 24a&-16^>l 12. 121aj«-20ic^ + l.
6. m^-10m?2 + 25?zl 13. «* + 50a"' + 625.
7. // + 16^yV + 64c\ 14. a^-4a/y^ + 4^>l
15. IQx'-Ux + SQ.
Find by inspection the square roots of the following trino-
mials :
16. ii;' + 10a; + 25. 30. 25a'' + 10a'x + x\
17. x' + 12x + SQ. 31. a'x' + 2axy + 7/\
18. a;^ + 16a; + 64. 32. 9.^y-24«^Z.% + 16rt*6«.
19. CC2 + 18.T + 81. 33. l-6a; + 9a;l
20. a3--20i^ + 100. 34. 1^.2 + ^^^^ + ^^2^
21. ic^-30ic + 225.
22. a^ + 50« + 625.
23. 4a;^ + 28a!-f49.
24. 9^^ -30a; + 25.
25. l6a2-48« + 36.
26. 81a;^ + 36£c^ + 4.
27. 121aj«-22a;M-l.
28. lQ9a' + U2ab + 4db\
29. 9a;*-12ajy + 4y.
— 40. -^
144
35.
9^
-v^v
'+¥2/*-
36.
a'
%•-
2/^'
37.
2- + 1.
2/
38.
16a;^
49
-2 +
49
39.
4 +* + «*
2 + ^^^.
^ 1
iC«
90 ALGEBRA
72. Roots of polynomials by inspection. Since the process of
finding the ?^th root of an expression consists of finding a second
expression whose nth power is the first expression, the roots of
some polynomials may be found by the aid of § 63.
Note. — A general process of finding the square roots and cube roots
of polynomials, and of arithmetical numbers, will be found in the
Appendix. These methods by inspection will suffice for the present.
Example 1. Find the cube root of x"* — 3x'^y + 3xy^ — y^.
Here the first and last terms are perfect cubes, and there are
four terms. This suggests that the given expression may be the
cube of a binomial. See § 63, (1) and (4). Taking the cube roots of
the two terms which are perfect cubes, we get x and —y. Their
sum, x—y, is the cube root required; for if we cube x—y by
§ 63, we get xr'-Sx'y + Sxy^—yK
Example 2. Find the fourth roots of 16x^—96x^y + 21QxY—
216xV + 8l2/^
This expression has five terms, and the first and last terms are
perfect fourth powers. Taking the fourth roots of these two
terms, we get ^^320; and J^Sz/, respectively.
Hence the fourth root will be either 2x^-{-^y^ 2x^—3y, —2x^
+ 3^, or —2x^—3y. Of these, the fourth powers of 2x^—3y or
-2x^ + Sy will give 16x«-96^«i/ + 216xY-216xV' + 8l2/*.
Hence, the fourth roots are 2x^—3^ and —2x^ + 3y.
From these examples it is seen that
If a perfect cube contains just four tenns^ arranged according
to the poicers of some letter^ the cube root is the sum of the cube
roots of its first and last terms y and if a perfect fourth poicer
contains just five terms^ arranged according to the powers of
some letter^ its fourth root is the sum of the fourth roots of its
first and last terms ; and so on for other poicers.
The terms of the nth root of an expression must always be
given such signs that when the root is raised to the iith power
by the method of § 63, the result will be the given expression.
POWERS AND ROOTS 91
Tlie work should be checked by seeing if the root will produce
the given power.
EXERCISE 26.
Find the cube roots of
1. x' + V2x' + 4Sx + Q'^. 3. 8a;^ + 36a;- + 54.^ + 27.
2, x'-Ux' + 7bx-12^. 4. 27cf;'-10Sa'b + lUab'-Ub\
5. 64^«-144e^y + 108a^y-27/.
Find the fourth roots of
6. a'-4:'ab^Qa'b'-4ab' + b\
7. x' + 20x'-\ 150^^ + 500a; + 625.
8. Sla''-4^2a'P'}'SQ4a'b'-lQSc(;'b' + 2^Qb'\
Find the fifth roots of
9. aj5-10a;* + 40£c-^-80a;^ + 80a;-32.
10. S2x' + 240x\i/ + 720£cy + 1080£cy + 810a^y* + 243y^
Find the sixth roots of
11. 729a«-1458tr^^^ + 1215«V-540a^5« I Uba'b'-Uab'' + b'\
12. x'' + 12x''a + QOx'W + 160icV + 240icV + 192^3^^ _|_ q^^b
CHAPTER VIII.
SPECIAL PRODUCTS AND QUOTIENTS.
73. There are some especially important products and quo-
tients which it is essential that the student should master before
proceeding. They are fundamental forms that are often met
in algebra. The student should learn to write out the pro-
ducts or quotients that come under these forms by using the
rules or formulae without performing the actual multiplications
and divisions.
PRODUCTS.
74. Product of the sum and the difference of two terms.
By actual multiplication,
{a + b){a-b)=a'-b\
when the symbols a and h stand for any terms ichatever.
That is, the p)roduct of the sum and the difference of the same
two expressions equals the diff^erence between their squares.
Example 1. Find the product of x-\-2 and x—2.
{x + 2){x-2)=x'-2''
=x^-4..
Check. When x=4: ; factors are 6 and 2 ; prodnct=12.
Example 2. Find the value of (2^^+ 6a') {2x'-5a').
{2x' + 5a^) (2x'' - 5a^) = {2x'f - {^a^
=4x*—2oa\
Check. When x=l, a=l ; factors are 7 and— 3 ; product= —21.
Example 3. Find the product otx + y + 5 and ,r + ?/ — 5.
92
SPECIAL PRODUCTS AND QUOTIENTS 93
By grouping terms, these trinomials can be written in the type
form of the binomials a-^h and a—h.
We have x-\-y->t-^= (-r + i/)+ 5 ; x-\-y — ^= (x^y) — 5.
Hence ix + y + 5){x + y—5)=[(x + y) + 5][(x+y) — b]
= {x + i/Y-5'
=x'^ + 2xy + y^—2o.
Check. When x=l, y=l\ factors are 7 and— 3; product =—21.
Example 4. Write out the product {a + h-\-c) (a—b—c).
Grouping terms, a-\-b'\-c=a + {b-\-c) ; a—b—c=a—{b-\-c).
Hence {a-\-b + c)(a—b—c) = {a + b-\-c){a—b-\-c).
=a^-{b-\-cy
=a'-(jb^ + 2bc + c')
=a^-b''-2bc-c\
Check. Whena=4, 6=2, c=l; factors are 7 and 1; product ;=7.
EXERCISE 27.
Write out the following products without performing the
actual multiplication :
1. {x + y){x-y). ^^ 10. 0«^ + y^)(a3^-y^).
2. {m-n){m-Vn). 11. {a''-b'){d' + J)').
3. (a— 5) (a-f-5). 12. (m«— /i'')(m*^ + n*').
4. (a + 10)(a-10). 13. {xy-\-ab){xy-ah).
5. (y-3)(y + 3). 14. (xY-z>){xY-Vz').
6. {a + x){a-x). 15. {'lx'-hy'){^x'-\-by^).
7. (2a + 3)(2a-3). 16. (3a^-75^)(3a^ + 7^0-
8. (3a;-2y)(3a5 + 2y). 17. (4aic-^ + 5%)(4aar'-5%).
9. (5m-4?z)(5m + 47z). 18. (\x'^-ly'){\x'-lf).
19. (2.5a^-1.7^)(2.5a^ + 1.76).
20. {^a'x' - 88iy) (3|a V + 33 ly*) .
21. (a!H-y + 2)(a! + y-2). 22. (a^-y-8)(a!-y + 8).
94: ALGEBRA
23. (x + a + b)(x-a-b). 25. (r'-r + l)(r' + r-^l).
24. f2a;-3y + 4)(2a; + 3y-4). 26. (x-l)(x^l)(x' + l).
27. (s' + s + l){s'-s + l)(s'-s'-{-l).
28. (x*-4:)(x' + A)(x' + lQ).
29. {x + y-i-z)(x'{-y—z)(x—i/-{-z)(—x+y+z).
30. (10a;" + a») (a"-10«").
31. (af+' + y-')(x-+'-y--'),
33. ^-i + J-VJi-l-Y
\4x' 2i/y\4x' 2yV
75. Product of two binomials having a common term bs (x + a)
(jr + b).
By multiplication,
(x + a)(x + b) =x'^ + ax + bx + ab.
Adding like terms, this becomes x^-^{a-^b)x-hab.
Hence, (x-^a)(x-{-b) = x~ + (a-^b)x-\-ab.
That is, the j^^'odiict of tvno binomials having a common term
equals the square of the common term^ plus the product of the
common, term atid the sum of the other terms^ plus the product of
the other terms.
Example 1. (x' + 2)(e;i; + 3)=x2 + (2 + 3)x + 2-3
Check. When a?=l ; factors are 3 and 4 ; product =12.
Example 2. {x-4) {x + 2)=x'+ (-4 + 2)a? + (-4-2).
=a?'-2x-8.
Check- Let x=l \ then (-3-3) = l-2-8, or-9=-9.
SPECIAL PRODUCTS AND QUOTIENTS 95
Example 3. {5x'-{-2y''){5x''-7y') = {5a^Y-\-{2y'-7y'')5x'-\-2y'{-7if)
=2^3C^—2^xhf-l^y\
Check. Let x=2^ and 2/=3. (Left to the pupil).
Example 4. (a + fo + 5)(a + 6— 2) = (a + 6 + 5)(a + 6— 2)
= (a + 6)' + (5-2)(tt + 5)-(5-2)
=a^ + 2a& + 62 + 3a + 36-10.
Check. Let a=l and h=2. (Left to the pupil).
Note. — In many of the examples worked out in this book hereafter
the process of checking will he omitted, in order to save space. .But
the student is advised to always check his work. The liabit of check-
ing cultivates the indispensable habit of accuracy.
EXERCISE 28.
Write out the following products ;
1. (^ + 3)(^ + 4). 15. {A^-^){A'-1),
2. (a; + 7)(i»-3). 16. (2+i>)(jt>-5).
3. {h-Q){h-b). 17. (mV + 8)(6 + mW).
4 (ic-10)(i« + 2). 18. (6-a;)(12-a;).
5. (a-8)(a + 6). 19. (3-a)(10-a).
6. (m-ll)(m-2). 20. (-c + 5)(-c-7).
7. {x-b){x-^). 21. (a;» + 3)(£c'' + 7).
8. (s-10)(«-3). 22. (a;" -3) (a;" -5).
9. (jt)^ + 12)(jt>^ + 10). 23. (««+^-6)(«"+^ + 5).
10. (a;=^ + 6)(a^^ + 8). 24. {B'-4.AC){B'-QACy
11. (r^-5)(r^-3). 25. (« + 5 + 3)(a + ^>-2).
12. (a;=' + 12)(a;2-4). 26. (aj-y + 7)(«-y+ 3).
13. (a;^H-12)(a;^-3). 27. (i? + ^ + 10)(jt> + ^-16).
14. («;y^-4)(l+a;2/-^). 28. (a;^ + a; + 6)(a;^ + a;-3).
96 ALGEBRA
29. (««-«« + 10)(a«-«^-20). 32. (x'-l + x)(x'-2-^x),
30. (xy + xy + xi/) (xy + xy 33. (x' - a') (x' - 3a^).
— 3a;y). 34. (aj« + 3y»)(a;" — Ty"). .
31. {z + b-b)(z^7-b). 35. (iC«+^-22/"-^)(3y»-^ + a;"+0.
QUOTIENTS.
Since division is the inverse of multiplication, from the type
forms of multiplication, certain type forms of division follow.
.76. Difference of two squares. Since, by § 65,
(a + b)(a—b)=a~ — b% then
a'-b
a-^b; _ . L =a — b-
fl-6"~""' a + b
That is, the quotient obtained by dimding the difference be-
tween the squares of tioo expressio?is by the diff^'erence between
those expressions equals the su7n of the expressio9is. And the
quotient obtained by dividing the difference betioeen the squares
of two expressions by the su7n of those expressio7is equals the
difference between the expressions.
EXAMPLE 1. ^^ = (£!rr:(5)^=^. + 5.
XT— 5 X*— 5
Check. Whena?=l: divisor= — 4; dividend =—24; quotient=6.
E^^^^^^- 2- 9a^a^ + 2y^ ^ 9a-.^+2^- "^^ ^^"^^ •
EXERCISE 29.
Perform the indicated divisions :
1 x'-9 o «*— 25 K 4ic'
3 'a'- 5 2x-l
a; +3* ' ^=^ + 4 3a +1*
1
— lax
1-
-IQx^
1-
-4.x* '
r-
-1
SPECIAL PRODUCTS AND QUOTIENTS 97
y 2^a'^l .. 64x*-Sly' 2ba'-22^b'
' ^ab+1 ' ' Sx'-Qf ' ^^' ba-lbb' '
' ^ " • lla^«+10a * a«-6«*
1 — Kymrw
10 ^'"-^ 14 169a;y- 144a^ - (« + ^)2_^2
• 1 + ^^* ■ V6xy'^Vla' ' -^^5-— •
a + h—x—y ' ' {a + by-\-{x—yf
77. Sum and difference of like powers of two expressions.
By actual division,
^■^—^ = a' + ab + b';
a —0 '
^lz^' = c(?-\-ceb-Va¥-VW',
CI/
d'^ — b^ is not divisible by a + J;
^^J^ = d' - cr5 + «6^ - h' ;
d^^-b^ is wo^ diinsible by a— ^ ;
a* + ^* is 7/^0^ diuisible by a—b\
d^ + b^ 2 7 I Z.9
z=a' — ab + b^\
a -{-b
a^ + 5Ms ?io^ divisible by 6« + 5.
These examples illustrate the following principles.*
(a) a" — b* is always divisible by a — b.
The quotient is a''-^ + a''-264-a"~^5^+ +6"~^
Thus, a'^—b^, a^—¥, a*—b*, etc., are divisible by a—b.
* The general proof of the divisibility in these various cases is left
for the student in Exercise 43.
7
98 ALGEBRA
(b) a" — b" is divisible by a-\-b only 'when n is even.
The quotient is a"-^ — W'-'^b + a''-W— — 6«-i.
Thus, a^—b^, a*—b\ a^—lf^ etc., are divisible by a + 6 ; whil^
a^— 6^*, a^—b^,a'—b\ etc., are not divisible by a + b.
(c) a''-^b'' is never divisible by a — b.
Thus, a^ + 6^ a^ + 6% a* + b\ are not divisible by a—b.
(d) fl" + 6" is divisible by a^b only when n is odd.
The quotient is a"-'^—a>'-^b + a''-W— +6""^
Thus, a^4-6^ a^ + 6^, a^ + 6^ etc., are divisible by a + &, while
a^ + b^, a* + b\ a^ + b^^ are not divisible by a + b.
From the preceding, observe that :
W7ien the divisor is a — b, the signs of all terms of the quo-
tient are +.
When the divisor is a + 6, the signs of the terms of the quo-
tient are alternately -\- and —.
The exponent of a in the first term of the quotient is less by 1
than the exponent of a in the dividend., and decreases by 1 in the
successive terms.
The exponent of b is 1 in the second term of the quotient ., and
increases by 1 in the successive terms.
Example 1. Divide a^—Whj a—b.
^^ =a' + a'b + a'¥ + a'b'-\-ab* + b\
Check. When a=2, b=l ; dividend=63 ; divisor=l ; quo-
tient =63.
Example 2. Divide 64^2/' + i25 by 4xy + 5.
Uoc^y^ + 125 ^ (4xyf + 5^
4:Xy-^5 ~ Axy + 5
= (4xyy-(4:xy) (5) + 5'
=^lQx'y'-20xy + 25.
SPECIAL PRODUCTS AND QUOTIENTS 99
Example 3. Divide 32a^-2436^" by 2a-Sb\
32a^ - 2436^ ^ (2af - {Sb'f
2a-3b' 2a-W
= {2ay + (2a)^(36*0 + {2ay{^by + {2a) {Wf + {W)'
= 16a* + 24a=' h' + 36a' b' + 54a6« + 816«.
EXERCISE 30.
1. Can a*— 1 be divided by a-\-l ?
2. Can cc'—af be divided by a—x?
3- Can a'—x^ be divided hj a + x?
4. Can a' + i' be divided by a + ^* ?
By Avhat expressions can the following be divided ?
6. ay' + S. 6. l-£c\ 7. a;^-32. 8. a= + ^>l 9. xy-l.
Determine which of the following indicated divisions are
possible, and write out all possible quotients :
10.
m—n
11.
1-/
l+y-
12.
a^ + 1
a-1*
13.
aM-1
a + 1*
14.
x'-l
x-1'
15.
a'-b'
IB
16a«-81J*
18.
19.
20.
21.
x'^'y"
x^-\-y^
'xTy"
a* + b*
a—b'
y
x-y
22.^'
x+y
23.^
'Id' + Zb' xy'-a' «• + !
100 ALGEBRA
EXERCISES FOR REVIEW (II).
1. What is the rule for adding similar terms ? From what
law does it come? Add ^xihj^ —\Qi'}y^ |a?^y, — 6a;-y, —\xhj.
2. How do you add dissimilar terms ? Add 2a^, — 3a;, — 2c,
^x\
3. Simplify 3.T^- 22/ -7y + 6£c^ + 2«.
4. In what letter are the terms 3icy^, 2aa?, Zcxy similar ? Add
them.
6. How do you add polynomials ? Add 6a;^— 2£c + 5, —^x^^
4i«-l, 7a;^-5£c-2.
6. What is meant by checking work in algebra ? How would
you check the result of exercise 5 ? Check the work.
7. Add d^—y^^ ^a^y—xy'^^ x^ + ^xy'^—y^. Check the work.
8. How do you subtract polynomials ? From 3c«"' — 2«^ + a— 4
take a? + 4a^ + 1. Check the result.
9. What are the laws of signs to be observed in removing
signs of grouping ?
10. What is indicated by 6£c-4y — (3a?— 2y) + (5£c + ?/)?
From what fundamental processes do the laws of grouping
follow?
11. Simplify a-[3^>+ {3c-(c-5) + «} -2a].
12. What laws must be observed when inserting signs of
grouping ?
13. Group like terms in x so as to have the sign + before
each sign of grouping : Ix^ — 3c^£c — ax^ + 5ic + Ix^ — ahx".
14. Group like terms in x so as to have the sign— before each
sign of grouping : lyx — ax^ — hx" + 3a;^ — ^ax — ^bx? + 2ic^ — ex?.
15. Add by combining like powers of x\ «^— 2a;, aa;^ + 5,
ax'-Zx^^-hx^^
EXERCISES FOR REViEW 101
16. What is the law of exponents in multiplication ? Find
the value of a^a^ ; a-a}^-(]^\ x^-x'-x^.
17. How do you multiply monomials ? Find the product of
— 2£cy, Za^xif^ —ax^, and — |^.y.
18. What is the meaning of x' ? Of x" ? Of (xy ?
19. From what law do we obtain the rule for multiplying a
polynomial by a monomial? State the rule. Multiply
x'-'2x' + Sx-bhj 2x\
20. How can you check your work in multiplication?
Check the work in the preceding multiplication.
21. From what do we obtain the rule for multiplying a poly-
nomial by a polynomial? Multiply 2a^^ — 3a'' + 26' by d^~ab
-\-b\ and check the work.
22. Simplify ^[^ab-'2a{b-4(a-b)}'].
23. What is the law of exponents in division ? Find the
values of «'"^-a^ ; a^-^a^ ; a^~a\ a}-^a^.
24. What is the meaning of a^ ? How is it shown ?
25. How do you divide a polynomial by a monomial ?
Divide a'-'2a;'b' + b' by a'-2ab + b\
26. What is the relation between the dividend, divisor, quo-
tient and remainder ?
27. What is a fraction ? -
X
28. Prove that ~y=x.
29. How do you find the product of two or more fractions ?
TV ^ ^^ ^
y 21a; 1
30. {ay='i («»)"•=? {by{my='i (3^)*=? State the law.
31. {xyy='i {xYy=? {4d'b'y=? State the law that you
used.
102 ALGEBRA
32. What are the laws of signs in involution? { — 2x^yh^y^ = '>
%7 •
33. What is the square of a-\-b? Show that your result
gives a rule for squaring any binomial. (2x^ — Sy*y-=?
{la' + Shy = ? {x'-2y=?
34. Square « + ^ + c and show that the result gives a rule
for squaring trinomials. Square ^x—Sy + zhy the formula you
have just derived.
35. What is a root of a number ? What is the index f
36. What are the laws of signs in evolution? ]/4iK^=?
f/27a^^-? f'-32a'"^>'^=? fl^hf=^. |/'=4=?
37. What is a perfect nth power ? Illustrate.
38. i/16£c*-8£c^a + a^ = ? |/81 + 25/iM-90??;^ = ?
f/8a;'^-36ajy + 54£cy-27y'* = ?
39. Find the product 0,1 a— h and a-\-h and show that this
gives a rule for finding the product of two binomials.
(4£c' + 3y-)(4a;^-32/=^)-=? (i-6cc-^)fi + 6£c^)-?
40. Find the product of a^ + a and x^h and show this gives
a rule for finding the product of two binomials which have a
common term. (a;^ + 6)(a;^-4) =? i^lah + 3)(2a^> + 7) =?
(4c.^-3)(4c£c + ll)=?
41
^'-y'_^ ^.B_y6_^ a^«+y«_,
X —y ' x^-y^ ' sc^+y'
d' + l
42. AVhen is a" — b" divisible by a + b?
43. When is a'^ + b" divisible by a + 6?
CHAPTER IX.
FACTORS.
Definitions and type forms.
78. Factors were defined in § 12. It follows from the prin-
ciple quotient X diinsor = dimdend^ that a factor of an expres-
sion is an exact divisor of it. The process of obtaining the
factors of a given expression is called factoring. Hence factor-
ing, like division, is the inverse of multiplication and depends
upon certain type forms established by multiplication.
Thus, since (a + 6) (a— 6)=a^— 6^ the factors of a"-'— 6'' are a-\-h
and a—h.
A common factor of two or more expressions is an exact dioi-
sor of each of those expressions.
Thus, a is a common factor of a&, ac, and ax+ay.
It is understood that in this chapter only rational factors
of an expression will be considered.
79. Any monomial expression can be factored.
Thus, 6x^y^=2Sxxyyy. Hence its factors are 2, 3, x, x,
2/, 2/, y-
80. Not all polynomials can be factored into rational factors.
But there are certain types of polynomials which can be factored.
These types will be discussed in this chapter.
8 1 . Monomial factors. A polynomial containing a monomial
factor may be factored by aid of the distributive law :
ffjr+6jr+cjr+ =(a + 6 + c+ )x
103
104 ALGEBRA
This identity shows that x, which is a factor of ever i/ term of
the polytiomial ax + bx + cx-\- • • • • , is a factor of the poly-
7iomial itself And the other factor^ « + 6 + c+ ^may
he obtained by dimding the given polynomial by the monomial
factor.
Hence the rule :
Find., by inspection,, a monomial vihich ivill divide every term
of the polynomial. Divide the given p)olynomial by this mono-
mial. The divisor and quotient are the monomial and polyno-
mial factors., respectively., of the given jyolynomial.
Example 1 . Factor 2ax'^ —4ay'^ + 6az\
By inspection, 2a is seen to be a factor of each term. Hence 2a
is the monomial factor. Dividing by 2a, we get x'^—2y^-\-oz^, the
polynomial factor.
The factor 2a may itself be factored into 2 and a. Therefore
all of the factors of 2ax^ — 'iay^ + ^az^ are 2, a, and x^—2y^-{-2>z^.
The given polynomial may be written 2a{x^—2y^ + Sz^).
Note. — The factors 2, a, and x^ — 2y^+^z'^, no one of which can be
factored, are sometimes called the prime factors. 2a is called a
composite factor.
Example 2. Factor 4:xy^—2x'^y^+x^y.
Each term may be divided by xy. Dividing by xy^ gives
4y^—2xy-{-x\
Hence the required factors are a?, y^ and 4:y^—2xy + x^ ; and
4xy^ — 2x'^y^ -\- x^y = xy{'iy'^ — 2xy -f- ^0-
EXERCISE 31.
Factor :
1. xy'^—xy-\-x.
5. xY-\-xY-
2. a5='y + 3£cy-5a;y^
6. bd'-l^a'b.
3. a;'— 3a;.
7. 2^x' + lxY,
4. a;' + 5a;^
8. 18a;^-9aJ^
FACTORS X05
9. a''-^cv>h + 2a*b\ 15. 24a;y-12ar'y + 42a!y.
0^, 6a;* + 9icy + 3a;y ^6. 27wVi + 36mW + 81wi/il
11. 'Ua'b''^ZMb\ ^7. 56ay-14ay + 28ay.
12. 4x' + 4:x\ 18. ic" + a£c«.
^3. 8a'* + 4a^^> + 2a«^>^ l9. a^?r'-a\
14. «^^>V4-«'^>'c''' + «'^^V. />^0. 5a«y«-^ + 10a"-V.
82. Polynomials that are powers of binomials. Polynomials
that are poioers of binomials may be factored by aid of § 71 and
§ 72. In any polynomial the monomial factor, if one exists,
should first be discovered and divided out.
Example 1. Factor a?*— 2ay*2/ + ^*V-
By inspection x'^ is seen to be a mono7nial factor. Dividing by
a?^, we get x^—2xy + y'^. This is the square oix—y.
Hence, x*—23c'y + x'^y'^=x^{x'^—2xy-\-y'^)
=x~{x—yY.
The factors are a?, x^ x—y^ ^c—y.
Check. Whena?=2, i/.=l; polynomial=4; factors are 2, 2, 1, 1.
Example 2. Factor ^^xHj-2Ux^y' + ^2^x'y^-1^2xyK
Qxy is a monomial factor. Dividing by Qxy^ we have
8x^ — 3Qx^y + 54:Xy^—27y^. This quotient is the cube of 2x—3y.
Hence, /"T^
. 4Sx*y—21Qafy'' + S24x^y''-162xy*=6xy{8x'-(^^y + Mxii^-27y") •
=6xy{2x-3tjf. '
The factors are 6, a?, ?/, 2x—Sy, 2x—3y, 2x—3y.
Check, When x=l, y=^\ polynomial =—768; factors are 6, 1,
2, -4, -4, -4.
Example 3. Factor —x^ + 2xy^ — y*.
Taking out the factor —1, we have a perfect square.
Hence, —x' + 2xy'-y*=-l{x^-2xy^-\-y*)
= -l{x-yy
= -{x-yy.
106 ALGEBRA
EXERCISE 32.
Factor :
1. bx'-40x + S0. My. -a' + Sa'-16.
2. 4:a'-Sab + 4:b\ 8. 2ba'b'-10ab'x'y-{-b'xY.
)^3. ax' + Qax' + 9a. \9. Sx'(/-Sxy + ^xf.
4. 4x'f-^Sbxij + 4b\ 10. 2SSa' + 4S0a'b + 200b\
J>(p. 20a'-60a' + 4ba. >L1. 7x' + 21xh/ + 21xi/' + 7i/\
6. Qx-9x'-l. Vl2. ba'-lbd'b + lba'b'-^ab\
M^. 2x'^-{-QxY + QxY + 2xy\
>(i4. -^a'x-\-4:^a'bx-lSba'b'x+U^a¥x.
VI 5. 32a* - 64a=^^ + 48a^^>^ - IQab' + 2/A
Vl6. Sa'-na'b + SOa*b'-nOaW + na'b'-Sab\
8 3 . Trinomials of the form x'-{^ax + b. By § 75,
{x + m){x-i-n) = x^ + (m-\-n)x-\-mn.
Now a;^ + (^ + ^?')£c + m/i is of the form x'^ + ax + b, where
m-\-7i = a and 7n?i = b.
But the factors of x'^-{- (m + n)x + mn are cc + m and x + ii.
Hence, if x^-hax + b has rational factors, they vnll consist of
two binomials, like a^ + m andx-\-n, having the common term x,
and the other terms such that their sum is a and iwoduct b.
Thus, since (a? + 3) (a? + 4)=x' + 7a7+12, the factors oix^-\-lx-\-\2
are a? + 3 and x-\-4. Here 7=3 + 4, and 12=3x4.
Example 1 . Factor qc}^^x^ 18 .
Here the factors must have the common term x, and the other
terms of the two factors must be such that their sum is 9 and
product 18. Two numbers whose su7n is 9 and product 18 are
3 and 6.
Hence x'' + 9x + 18={x + 3)(x + Q).
Example 2. Factor a?2—2x— 35.
FACTORS 107
Evidently, here we seek two numbers whose product is a nega-
tive number^ —35, hence the numbers must have unlike signs.
And since their sum is —2, the one having the greater absolute
value must be negative. Hence, they are 5 and —7.
Therefore, x^—2x—i^5 = {x—7){x+5).
Example 3. Factor f -25^ + 150.
The common term here is t. Since the product of the other
two terms is +150, they must have like signs. And since their
sum is —25, they must both be negative. Hence these terms
are —10 and —15.
Therefore, t'-25t + loO=(t-10){t-15).
Example 4. Factor 3—x^—2x.
This can be thrown into the type form, x'^-\-ax+b, by taking
out the factor —1.
Then 3-x'-2x=-l {x' + 2x-S).
But x'' + 2x—3=(x+S)(x-l).
Hence, 3—x'—2x= — lix-\-S){x—l),
or multiplying first and last factors, this may be written
{x + S)(l-x).
Example 5 . Factor a'^x* + 5a V + 6 .
This is of the form x^ + ax-\-b, which may be more easily seen
by writing a*x^ + 5a^x^ + 6 = (a'^x'^y + 5 (a V) + 6 .
The X of the standard form is a^x^ in this exercise, hence the
common term of the factors will be a^x"^.
Therefore, {a'xy + 5 (a^x^) + 6 = {a'x'' + 3) (a V + 2) .
Example 6, Factor x^ + ^xy + 14^/^
If we write this x^ + {^y)x-\-lAy'^, it is seen to be of the given
type form.
The common term is x and we have now to find two expressions
whose sum is 9^/ and whose product i^ 142/^.
It is easily seen that these two expressions are 2y and 7y.
Hence, x^->r^xy-irl4.y^={y-\-2y){x + 7y).
108
ALGEBRA
Factor :
1. x' + lSx + 42. 8.
2. x' + 2x-4S. 9.
3. £c^-9a^ + 20. >\^0.
4. x'-Sx-2S. 11.
5. a;^ + 17^ + 72. ^12.
6. x' + bx-bO. 13.
/,.-^2. x'-^bx' + Q.
23. aj*-7£c^ + 10.
26. fc« + 3£cM-2.
27. a^^'-Sa^^+e.
28. a;^« + lla;-'-26.
29. a;'^-2a;«-224.
30. a'x' + 9ax+-U,
31. bhf-lhy-2,0.
32. a^^^ + 30a^> + 200.
33. £cy-28a!y + 160.
34. mV + 4m?i-60.
Y35. «^^V + 13r^^6— 30.
^36. a;ys^-19£cy^ + 90.
37. xy-^xY + 2.
38. a;y + 14£cy + 33.
39. a;y-5ajy-126.
^40. «'"^>^-2«-^^>-35.
EXERCISE 33.
a;^-12a; + 32. 15. c^417c-84.
a^ + 3a-180. ^\l6. cP-M-bb.
m'-m- 240. 17. 2-{-Sr + r\
f-t-420. M8. 24:-2s-s\
a2+3« + ^. 19. 15 + 2y-yl
3c-70.
^JO.
2-rt-a^
5''' + 20Z>+84. 21. z'-z'-2.
41 a;y + 7£cy^— 44.
]s>^42. pY-Sp'q' + 2.
43. a;=^ + 3a!y + 2yl
* 44. £c^ — 5^cy + 6y^
^5. a3^ + 17a;y + 70y^
^"^6. a'^10ab-S9b\
47. a'-lSab-40b\
\s, -^a^-\-2ab + b\
i)*9. -«^-5a^» + 104^1
lO. a^y — babxy + 6a^^^
>/.
^61. xY^^Uibx}f-V^ceb\
62. £cy-3c.^?/-10cl
53. a!y + 9axy + 14al
\
64. x^if — la^bx^y^ -f 1 2«:«'^6l
65. c«^m^ + llac^w' + 30c^
C^6. l-3a + 2a^
57. \^-Qx-21x\
58. (a + ^)^ + 7(a + ^») + 10.
V59. {x-yy-^x-y)-40.
FACTORS 109
60. (x + yy + U(x+>/y + SS. 66. 2£c^-34a;-400.
61. (a + by-S(a + b)(x+i/)+ ,66. ax' + bax-Ua.
2{x-{-yy. 67. a'x' + 2a''x-Sba\
62. 2ic^-10x-168. 68. x' + ax'-42a'x.
(First remove tlie monomial factor. ) 69. dx^ — QOx^i/ — 288£cy^
63. Sx'^Sx-lS. ^70. xy + 9xy + UxY.
64. 5ic^ + 45ic+100. 71. 260a; + 62«V + 2a;y.
72. 220-2a;-2ajl
84. Trinomials of the type form ax'^-\-bx+c.
There are different methods of factoring the general quad-
ratic trinomial of the type form ax'^ + bx-\-c when it has rational
factors. Such a form may be factored by first changing it to
the form discussed in § 83. Thus, multiplying by a, and at the
same time indicating the division by a, in order not to change
the value of the expression, we have
ax^^bx +c=
aV-
abx
a
(axy + b(ax) + ac
~ a
This numerator is now a quadratic trinomial in («.x'), the form,
discussed in § 83 v^hen x is replaced by ax. Hence, we have the
common ter7n ax^ and the other terms such that their product
equals ac and their sum equals b ; then finally dividing by a,
we get the required factors of ax^-{-bx-\-c.
Example 1. Factor 2xH8a?+l.
4x^ + 6^^ + 2
2ic-^ + 3x+l =
2
{2xf + 2>{2x)-^2
2
(2.r + 2)(2a!+l)
2
110 ALGEBRA
Dividing the first factor of the numerator by 2, we have (x + 1]
(2ic+l).
Check. When x=l ; trinomial=6; factors are 2 and 3.
Example. 2. Factor Qx^ + llx + S.
6
_ i6x + 9)iQx + 2)
= {2x + 3)(3x + l).
Here, to divide the product of the two factors by 6, divide the
first one by 3 and the second by 2.
Example 3. Fsictor ax'^+{a-\-b)x + b.
a
= (ax + a){ax + b)
a
=(a? + l)(ax + 6).
Example 4. Factor 1 2a?'— 23a?// + 10?/^
1/*
_ {12x-8y){n x-15y)
12
= {3x—2y){4x-5y).
A second method may be obtained as follows :
Since the factors are buiomlals of the form (mx + ri) and
{rx \-s) whose product is rnix^^{rn + sm)x + s7i^
therefore, ax^ -\-hx + c= rmx^ + {rn + sm)x -[- sn = (mx + ?i) (rx + s).
From this it is observed that if a general trinomial of this
form can be factored, the ^ first and last terms of the trinomial
must he so factored as to give the terms of the hinomial factors^
and the sum of the products of the first term of each hinomial
factor by the second term of the other factor must give the middle
term of the trinomiO/h
FACTORS 111
Example 1. Factor 2x^ + 5x+2.
Now the first terms of the binomial factors are factors of 2x^,
and the second terms are factors of 2. The sum of the products
of the first term of each by the second term of the other, called
cross-products, is ^x.
The possible pairs of factors may be conveniently arranged as
follows :
2x+2. 2a?+l
x+1 _ x^-2
from which we may easily select the pair that gives the proper
cross-product. The factors then are 2a? +1 and x-\-2.
Example 2. Factor 3ir^ + £c— 10.
Four of the possible pairs of factors are
x—2 x + 2 x—5 x-\-5
3x + 5 3j?— 5 3x+2 3a;— 2
from which it is seen that the second set is the correct one.
Each set should he tested as vnritten. If this is done., it will
usually be unnecessary to write all possible sets. The simpler ex-
pressions of this type can be factored by inspection.
Note. — The student should always look for monomial factors first,
and remove any that are found. iTse the method that seems best.
EXERCISE 34.
Factor :
1. 2£c^ + 5a; + 2. 7. 12x' + bx-2.
]2< Sx' + 7x-\-2. 8. 6x'-llx + b.
3. 2x'-h9x-\-10. 9. 2t'-bt-l.
4. 2x' + bx-S. 10. 8a;' + 15a;-2.
5. 2x'-^x + ^. 11. 12c^-25c+12.
6. bx'-9x-2. 12. bb'-79b-U.
112
ALGEBRA
13.
15.
16.
Vl7.
tA 21.
23.
fv 24.
26.
7«- + 36a + 5.
10/ + 21y-27.
2r*
6.
Ua' + ba'-l.
6aV + 13aV + 6.
5a'^>V + 19a^>c— 4.
2icy — x^i/ — 15.
Qa'b'-2Sa'b' + 20.
12x' + 21x-Q.
Qax'^ + lbax + 9a.
8Qx'i/ + U2xj/-Ui/.
Qa'b-4:Qab-72b.
2x' + Sxi/ + y\
26. 12x'-7xy+f.
S 27. 2x'-Sxi/-2i/\
K28. bx'-19xi/-'^tf.
\29. 12ni'-lQa77i-Sa\
30. 2£cy + «y-15.
'Nl. 3a;^ + 10a;y-8/.
V 32. 21ax^—hlaxy^^ay^.
• 33. 6a^^»^ + 2«Z>c—4cl
34. ax^-\-{b-a)x-b.
(X35. 2y^ + (4a + % + 2«^.
36. 2;2^-(2a + ^)2; + a*.
' 37. ax'-^{ab^-\)x-^b.
38. 2a£c'''+(2a^-2«^»)a;-2«^^>.
85. Binomials of the type form a^ — b^. Binomials of the type
form a^—V^ may be factored by aid of § 74. Since
(a + 6)(fl-6)-a^-6%
the factors of a^ — b- are a-^b and a—b.
Example 1. Factor 4x^—25.
= (2ic+5)(2a!-5).
Note. — We might also write
= (-2a^-5)(-2if+5).
Usually, however, we have no use for this second set of factors.
It is easily seen in general tiiat if an expression has two factors, tlie
expressions obtained by changing the signs of those factors will also be
factors. Thus, the factors of ah are a and 6, or — a and — 6.
FACTORS 113
Example 2. Factor 27a?^— 12a?.
Kemoving the monomial factor 3a;,leaves 9a?^— 4.
= (3x-2)(3.r + 2).
Hence 27x'-12^=3x'(3ic-2)(3x + 2).
Note. — We need not write down the second step of the above solu-
tion, but write merely
9^2-4=(3j;-2)(3a;+~2).
'
EXERCISE 35.
Factor :
1, x'-m.
8.
x^-n\ ^^62bx'-
-2256^
2. x'-l^jd.
9.
4x'—a\ 16. 9a V -
1.
3. 0^—49.
4. x'-U.
10.
11.
9a3^-a^ 17. i6a^y-
-9.
■25.
5. a^^-144.
6. 03^-196.
12.
13.
lQx'-2ba\ 19. 25£cV-
lOOy^-495^ "^20. lOOx'y
-16.
^-815^,
7. x'-a\
14.
Sla' — 64b\ 21. IG^cV-
-b'c\
^ 22. imxYz'-^ha'h
y%Z. \x'-\y\
24. 1 a2_ 1 52
^cl
36a.V^ -,
2^^- 25a^5-^ 1-
33 1 fiu2
2 5 16
K26. T-V^«'6'-^V«^.
„. 4x' 81?/
V^- Sly^ 4x^ '
27. /T-IKy'-
\28. 1— Vra'*'«'-
35. 1 ^, .
" 29. Ti^-T-v*y.
36. (a + by-1.
30. 2-:_«i
„, a' 25(8'
^7, {a+by-(c+dy.
38. (x + yy-(a + by.
. 39. (a^-y)'^-(«-^')^
114 ALGEBRA
40. 4(x+(/y-9{a + by. -o (a + by (c + d y
{a-by {c-dy
(2a+by 2Hx+j/y_ w(x-j/y
^ ib{x Zy). to. ^Q^^^j^^y 8i(^^_^)2-
86. Polynomials which can be written in the type form a^—b^.
Some polynomials, by grouping terms, can be put into the
tjpe form a^—b^^ and can then be factored by the method of
■§ 85.' .•
Example 1. Factor aH2a& + 6='- c^
Grouping terms,
o? + 2a6 + 6' - & = {a' + 'itah + h^) - &
=(a + 6 + c)(a + 6— c).
Example 2. Factor a^—Jy"— 2bc — c\
Grouping terms,
=a^-{b + cy
= {a + (b + c)} {a-{b + c)}
= {a + b + c) (a—b—c).
Learn to omit the second and third steps and to lorite out the
icork as follows :
Example 3. Factor x" + 6x— a' + 4a&— 46' + 9.
Grouping terms,
x^* + 6a? - a' + 4a6 — 46H 9 = (x^ + 6.r + 9) - (a' - 4a6 + 4&')
= (a; + 3 + a-26)(a74-3-a + 26).
EXERCISE 36.
\Factor :
1. a^—lab^^—x^. 4. a' + 6a5+9^>'— 4c^
2. \-\-'lx-Vx^-if. ^b. l-x'-Sxy-Uf.
^, x^—1xy\y''—\, 6. a;'+4a£c-y' + 4a^
I^ ACTORS 115
\y, a'-l + 10ab + 2^b\ 10. l + Qax-a'-9x\
8. -4-2ab + a' + b\ "^ 9x'-ia'-9c' + 12ac.
K^ a'-^c'-Sab+lQb'. 12. a' + 2ab + b^ + 2cd-c'-d\
14. a' + 12m7i—10ab-4m' + 2bb'-9n^.
15. ic2-12aa!-4^•y + 36a'-4?/2-6^
X XFactor, and simplify the factors :
Al6. (x + 2i/y-(2x-^yy 19.- (5a-3^)^ + 12«5-^>2-36a2.
17. (2x-Si/y-(^j-xy. \20. a^^-8a;?/ + 16/-(£c + 4y)^
3^8. (.^-5y)^ + 24£cy-9£c--16/.ai. W0a' + 20ab + b'-(a-2by,
87. Special trinomials of the type form x*+axy^+y.
It often happens that trinomials ol the form x*-\ ax'^y'^+y*
may be written in the form a^—¥ by the addition and subtrac-
tion of a term. The addition and subtraction will not, of course,
change the value of the given expression.
Example. 1. Factor aj^ + o^y + ^/^-
Adding and subtracting ar^i/^, we have
x^ + x^y^ + y*=x^ + 2x^y^ + y*—x^y^
= (x'' + yy-x'y'
= {x'' + y^ + xy) {x'' + y^—xy).
Example 2 . Factor a?* + 2x'^y'' + 9y\
Adding and subtracting 4x^y^, we have
x* + 2xY + ^y*=x' + Qx'y' + ^y*-4xY
= {x' + 3yy-4xY
= (s(f + 3y' + 2xy){o(f + Sy''-2xy),
Examples. Factor a?*— a? V + 2/*-
Adding and subtracting Sx^y^,
x*-x^y^ + t/=x* + 2xY + y'-Sx^y'
={x' + yy-BxY
={a^ + y'+xyy'S){oc' + y'-xyi/S).
116 ALGEBRA
Note. — These factors are rational in x and y, but not in the coeffi-
cients. For this reason the expression is not considered factorable in
the sense in which the term is usually taken.
Example 4. Factor 4ic^—16a?*^* + 92/^
Adding and subtracting 4:X*y*,
4a?«— 16ic*2/* + ^y^=^o(f-12xY + 9i/«-4^Y
= (2x*-3yy-4xY
= {2x' - By' + 2xY) {2x* - 3y' - 2x'y') .
EXERCISE 37. 4 ^
Factor :
1. x'-^-x'-i-l. ^9. 36a*-76ay + 25i/.
^^. l + Sx' + 4x*. 10. 49a'-^19a'b'-\^4b\
3. x' + x* + l. J^^- 4iz;'-16i«y + 25yi
/ ^ 4. a* + 2a'b' + 9b*. 12. 04.Ty + 12a;y+l.
5. a'-Sa' + 9. ^ }^- «'-22aVy + 9a.'y-
\«. ,4£c* + 16a!y + 25y*. ^4. 4m*n«-45a^^^mV + 25a*i«.
" lyi^ 4a' + 4a'b' + 2^b\ />\^ 16. x''-^x' + U.
V 8. a;*-15a!y + 9/. vV 16. 25a* + 16a'^»V + 4^»V
88. Binomials of the type form flf'4 6", when /7 is odd.
By § 77, a" + b" is divisible by a^ b mhen n is odd. Hence,
the factors of a" + b'\ when n is odd^ are a + b and the quotient
obtained by dividing a" + 6" by « I ^.
By division,
a'+b'={a^-b){a'-ab -\-b')',
a'-\-b'=(a-^b){a'-d'b + a'b'-ab^ +6*) ;
a''\rb''=:{a^b){a'-a'b + a'b'-a'b'-\^a'b'-ab' + b%
Example 1. Factor x? + Sy^.
X? + ^y^ may be written a?^ + {2yf^ and therefore may be divided
by x-[-2y.
FACTORS
11'
Hence, o(?-^Sy^={x-\-2y){3(?—2xy + 4ty^),
Example 2. Factor 2>2w' + 2436^
32a5 + 24365=(2a)H (36)5
= (2a + 36) j (2a)* - {2a)\Zb) + (2a)'^(36)^ - (2a) (36)» + (36)*
= (2a + 36)(16a*-24a^6 + 36a'^62-54a6H816*).
Example 3. Factor a^ + 6^
a» + 6»=(a»)3 + (63)»
= (a« + 6^) {{ay-{a^){h^) + (6^)^}
=(a3 + 6^)(a«-a='6H6«).
But aH6=^=(a + 6)(a2-a6 + 62).
Hence, a^ + lf={a + h){o}-ah + h''){a^-aW + W).
<, Factor :
1. x^-^y\
2. Q^-\-yK
/y^Z, x' + y\
4. a?+y\
^b. x'' + y'\
6. 27x^+a\
/X7. Ux' + 27y'.
8. 125a^ + 216^>=^
9. 40a' + Ubb\
10. 432 + 2a;^
EXERCISE 38.
0^11. 81a;^ + 3.
^2. ^' + 1.
13. l + x\
h4:. S2 + x\
15. l + 3125a^
\6. «W + 1.
,'\17. «V + 32.
18.
i25a;y.
19. J + 1.
1 7,5
/^20. ?" - 1
2/' •
21. ^i^^' + ^V.
^^2. 1024a''
23.- 12Sx^+^i^y\
24. i^ + 2/.
26. ic^ + l.
58. l + a'^».
89. Binomials of the type form a'*— 6", when n is odd.
By § 77, fl" — 6" *s divisible by a — b when n is odd. Hence,
the factors of w—b'\ when n is odd^ are a—b and the quotient
obtained by dividing a"—b" by a—b.
By actual division,
X18 ALGEBRA
a^-b^={a-b){a'-\-ab^-b') ;
a'-b' = {a-b){a'^-a'b^-a'b'^-ab'^-b')\
a''-b'=^{a-b){a' + a'b^a*b' + d'b'+a'b'^ab' + b').
Example 1. Factor 27x^—8?/^
27jL^—^y^ may be written {^xf—i^yf, and consequently may be
divided by Zx—2y. Hence,
27x'-8y'={Sx-2y) {{'^x)^ + {3x){2y) + {2yY}
= (Sx-2y){9x' + Qxy + ^y').
Example 2. Factor x^^—y^.
x^^—y^={x^)^—y^
={x'-y){ix'y + {x'fy + {xYy' + {oc')y' + y*}
= {x^ — y) {x^ + x^y + x*y'^ + x^y^ + y*) .
EXERCISE 39.
Factor :
1. x^—y^. 11. 1— i^c^ \ gi ^^ V^
>"2. a^-h\ \'\2. x'-\. ' ^~^'
V3. x'-y\ \ 13. 32-y\ \ 22. «;^_'i'.
^ 5. .-2/^ 15. .-32.^ )^^^ -^,_
N^ 7. 8a^-2/^ Vl7. l-a\ , ^^^,
'^ 8. 27a;^-64al 18. 128-a3y. ^^6. 1^ — i.
^
9. 125-34aal > 19. 8l£c^-3. i
27 ' iL
10. 8a«-^»\ 20. 648-3^1 ' «'*
28. x^-\y\ 29. (a-*)•'^-(c-<7)^
30. (a;-2y)5-(2a + ^>)l
90. Special Binomials of the type form a" + 6", when n is even.
When n is divisible hy Jf.^ hinomials of the form, a" + b'' may
be factored as in § 87.
FACTORS 119
Example 1 . Factor x^ + a".
Adding and subtracting 2a*x\
ix^ + a^=3(^ + 2a* X* + a^— 2a*x*
={x* + a*y-2a*x*
= {x* + a* + a^a?^/ 2)(.r* + a*-a''x'\/2).
Note. — These factors are rational with respect ioX\\e general numbers
X and a but not with respect to their coefficients. Factors of this kind
are sometimes useful.
Example 2. Factor 81a'Hl.
= (9a«)2 + 2(9a«) + l-2(9a«)
= {9a^ + iy-2{9a^)
= (9a« + l + 3a"^|/2)(9aHl-3a=^y'2).
When the exponent n has an odd factor^ binomials of the form
a" + b''mai/ be loritten as the sum of two odd powers^ and fac-
tored as in § 88.
Example 1. Factor if-\-lf.
={y'^h') {{yy-{y') {h') + m'}
:={if+h') {y'-h'y'+h*).
Example 2. Factor x^" + 1024.
£c" + 1024=(ir=')^ + 45
={x^+4){(xy-{xy-4.+{xy-^^-{x:'ye+4*}
= (a?2 + 4)(.T«-4a?«4-16iC*-64a72 + 256).
\ EXERCISE
40.
Factor :
*^1. a*-^b\
\
■'x 7. x'*-\-l. \
l\12.
l + 81a;^
Y2. a' + b\
8. x* + l.
13.
64a^« + 729/.
A 3. a'-^b\
4. a'^'^b'^
9. 03^°+ 1.
14.
X* y*
y' ^''
^ 5. a^^-\-b'\
'Via l+ar^
,„ 1
6. a«+l.
11. 16a;« + l.
15.
-'"+5r»-
120 ALGEBRA
91. Binomials of the form a"—b'\ when n is even.
Binomials of the form w—b''^ when n is even, should always
be written in the form a}—h\ and factored as in § 85.
Note. — Factors should themselves be factored when possible.
Example 1 . Factor a?* — ?/*•
Writing in the form a^—lf^ we have
Factoring x^—if, =('«^ + 2/^)(<^ + 2/)('^— 2/)'
Example 2. Factor 729a«-64.
Writing in the form a^—b^, we have
729a*^-64=(27a=')^-8"''
= (27aH8)(27a=^-8). '.
But 27aH8=(3a)H2^
= (3a + 2){(3a)2-(3a)-2 + 2=}
=(3a + 2)(9a2-6a + 4).
And 27a3-8=(3a)^-2='
= (3a -2) {(3a)- -f (3a) -2 + 2--^}
= (3a-2)(9a2 + 6a + 4).
Hence 729a«-64=(3a + 2)(3a-2)(9a2-6a4-4)(9a2 + 6a + 4).
EXERCISE 41.
V)
Factor :
1. a*-b\
8. a'-b\
15. l-x'\
2. a'-b\
9. a''-b\
16. 4-a;*.
3. a'-b\
10. a''-b\
17. 16-a^^
4. a''-b'\
11. x'-l.
18. 81-a;«.
5. a''-b'\
12. x'-l.
19. 100-a^«.
6. a''-b'\
13. l-x\
20. 256a«-l.
7. a'-b\
14. l-a;^".
21. 81a'^-16a3'^
FACTORS l^l
22. l-256a^^ 26. 1024-a'». 29. x^'-a"'.
X
/2«+4
23. 6V-64a;«. 27. -,-1. ^0. x'^'-^-if
24. xY-a'b\ ^ ^^4^, 31. (a + by-c\
26. 16-81«^^>*. ^^- ^'~T6* 32. (x-yy-(a-b)\
33. a^ + 4a=^i + 6a-^»^ + 4«^»^ + ^»*-c\
34. {2x-i/y-{a + Uy.
92. Polynomials made to show a common factor by grouping
terms.
The terms of a polynomial may often be so grouped as to
show a common factor^ then factored as in § 81.
Example 1 . Factor ax +ay + hx + by.
Grouping those terms which show a common monomial factor,
we have
ax+ay + bx+by=a{x + y) + h(x + y). .
This is a binoinial whose terms are a{x-\-y}^nd b{x+y), and
shows a common factor x + y.
Hence a{xA-y) + b{x+y) = {x+y){a + b).
Therefore ax + ay + bx+by=(x-\- y) {a-{-b).
Example 2 . Factor x^ + x'^—4x—4.
x^+x'—4:X—x^x'^(x + l)—4ix + l)
= {x + l){x'-4:)
={x + l){x + 2)(:x-2).
Example 3. Factor a^—¥—(a— bf.
a-^-b'-{a-by=(a''-¥)-ia-bf
= {a-b){{a + b)-{a-b)}
= ia—b){a + b—a + b)
=2b{a-b).
Example 4. Factor a^—b^ + ax— bx.
a^—b^-\-ax—bx={a'—b^) + (ax—bx)
=(a + 6)(a-^&) + x(a— 5)
={a—b){a + b + x).
ALGEBRA
EXERCISE 42.
\
%
^'
/ l^
Factor :
1. ac-\-bc—ad—bd.
2. ax—bx—ay + bj/.
X3. ax-^bx + 2a-^2b.
4. ax-\-4:X—ay—4:y.
>\5. a^b-^b-Va'c-^^c.
6. ax^'^-ax-\-bx^-^bx-
k
14. £c— «+(»;— a)'.
15. a^x-\-a^y-^x + y.
vVl6. x^a-Vx'b-a-b.
17. j?/ + 3ic*-a;-3.
18. x^—x—y'^^y.
19. ax^—a—bx^ + b.
7. 2a-2^• + aJ-^>6^+2c + c(7.^^ 20. aW-a*-** + l.
8. ax—bx—a + b.
Jl\9. xy—x—y^\.
10. a^>-3a-25 + 6.
^11. a;^ + aa; + ^i« + «5.
12. a;^ + aa;-2a;-2a.
13. a;^— a^ + C£c— «c.
27. aV
>^^1. ax^ + bx—ax—b.
K23. a'-b'-{a'-by.
24. mnpq + 2+pq-^2mn.
'^' 25. 4:ax—2ay—2bx + by.
26. a^^>^-6^c^^-aW^ + cW^
1-a^-a;^
— V^8. aaj,?f 3air+5^2 + 15a; + 2a + 10.
29. a;^4-4a!^ + 4ic— aj^y— 4ajy— 4?/.
ioO. TTi^n*^ — my — q'^n^ + (^y .
31. a'x'-ay-b'x'-i-by, 36. a(a + c)-^>(^> + c).
32. x'a*-y'a*-b*x^ + by. 37. (a + ^>)2 + 5c(a + ^>) +6c'.
33. «V-a2 + a^>i«=^-«7>. 38. b'-c' + a(a-2b).
34. a;^
■2/^
■2^?^
2(xz-yw). 39. aj*+£c^ + y'-y*.
35. 4a;y-(a;' + 2/'-2')'. 40. x' + x'z + xyz + y'z-y\
41. 2(l-a!)(l+a!)^+(l + a3').
42. a^x + abx + ac + b'^y + aby + bc.
93. The methods of the foregoing sections will enable the
student to factor almost any expression. Another good
method, however, based upon what is known as the remainder
FACTORS
123
theorem is discussed in the following sections. This method is
very convenient when the given expression cannot readily be
thrown into one of the familiar type forms.
94. The remainder theorem. The remainder obtained by
dividing a polynomial in x by x — a is the expression that would
be obtained by replacing x by a in the dividend.
Thus, dividing x^-^-Zx—^ by x—a,
ic^' + Sr-S
x^—ax
x—a (divisor)
a? + (3-|-a ) (quotient)
(3 + «)a?— 5
(3-|ra),r— 3a— g-^
a'^' + Sa- 5, the remainder, which is the same
as the dividend when x is replaced by a.
Dividing i)(f—x^ + dx+2 by x—2^ we have
x^- x^ + 3x+ 2
x-2
x^-2x'
x' + x + 5
x^ + Bx+ 2
x'-2x
5x+ 2
5£C-10
12, the remainder
Now if, in the dividend, x is replaced by 2, the dividend be-
comes i^, which is the reinainder.
To show that this principle is true in general, let any divi-
dend be represented by the general expression,
Ax- + Bx'^-'~\'Cx^^-^+ -VMx + JSr.
Let the quotient and remainder, when this is divided by
.c— a, be represented by Q and B^ respectively, k
Then, by § 51, M^ « \m3){x^-2x+4).
Note. — In attempting to find a value for x that will reduce the given
expression to zero, it is wise to begin by trying 1 or — 1 ; then try larger
possible numbers until the required number is found.
Example 4. Factor o(?-\-2x^y—xy'^—2y^.
This expression becomes zero when x=y. Hence, x—y is a
factor. Similarly a? + 2/ and a? + 22/ are factors.
Hence, a^ + 2x''y-xy^-2y^={x-y)(x + y){x+2y).
126 ALGEBRA
Example 5. Show that a + 6 is a factor of a" + 6" only when n
is odd, and thus prove {d) of § 77.
Putting —b for a in a" + 6", we get (—6)" + 6". If n is odd,
(— 6)" + 6"=— 6" + 6"=0. If w is even, (— 6)" + 6''=6'^H-6''=:26».
Hence, a + & is a factor of a" + 6" when n is odd and not when m
is even.
When a factor of the form x—a has been found, the work
of dividing it out may be abridged. The method may be
derived from the following example :
Divide ar*— 5a;^ + 5x— 7 by £C— 3.
ar»— Sar' + Sic—
7
x-S
a^-3x^
a;'— 2a;— 1
-2o(f
-2x' + 6x
— X
—x-\-
3
10, the remainder.
Observe : (l) that the coefficient of the first term of the quotient is
the coefficient of the first term of the dividend, and the succeeding
coefficients are the coefficients of the first terms of the succeeding
remainders.
{2) That since the first term of each partial product caficels the
term of the dividend immediately above it, the first term of each
partial product need not have been written.
(3) That if the sign of each term of the partial products had
been changed, this term micfht have been added to the corre-
sponding term of the divide^id. (This can be done by changing
the sign of each term of the divisor.)
(^) That if the dividend is written in descending powers of the
first term of the divisor, this term of the divisor need not be written.
Omitting all work that is unnecessary, writing the coefficients
FACTORS X27
only, and raising each number of the several remainders into the
same line, the work may be written as follows :
1-5 + 5-7 |3_
+3-6-3
1-2-1,-10 where 1, —2, and —1 are
the coefficients of the quotient, and — 10 is the remainder.
This abreviated form of division is called synthetic division.
EX4.MPLE 1. Divide 2x^—4x^ + 7x+23hy x + S.
2- 4+ 7 + 23 1-3
- 6 + 30 -111
2-10 + 37,- 88
Hence the quotient is 2xf—10x+S7 and the remainder is —88.
Observe that the first coefficient of the dividend is brought doivn
for the first coefficient of the quotient.
This number is multiplied by the number in the divisor and
added to the next term of the dividend for the second coefficient
of the quotient, and so on to the last term.
Example 2. Divide 2a?*— 10x^-13 by x—2.
2 + 0-10 + 0-13 |2
+ 4+ 8-4- 8
"^ 2 + 4-2-4, -21
Hence, the quotient is 2xr^-\-4iXi^—2x—4: with a remainder of— 21.
Observe that where any power of x is ivanting its place must
be supplied by a zero.
Formulate a rule for the process.
Perform the following divisions by the synthetic process :
1. x^ + Zx'-lx-Z by a;-2. 6. x'-la?-\-2x-12 by x-\-2.
2. 3a;^ + 7a;^-3a; + 15 by aj+3. 7. x' + Zx' + x—l by ic-3.
3. x'-^x'^lx-l by,£c-2. 8. 2a;''+jc=^-7 by a; + 3.
4. 2iK*— 9a;' + 3a;— 5 by cc— 4. 9. x'-i-hx'^-x—^ by a; + 4.
5. 3a;*-7a;^ + 13 by cc + 2. 10. a;* + 3aj^-9 by x-2.
The pupil should now be able to factor any factorable ex-
pression that is likely to occur in elementary algebra.
128 ALGEBRA
The following general suggestions may be useful :
1. First remove all monomial factors.
2. Try to bring the resulting polynomial under some one of
the binomial or trinomial types.
3. If imsuccessful^ try the remainder theorem.
Jf.. Be sure that all factors are prime. ^ s
\^'
EXEBCISE 43.
vi By the use of the remainder theorem, factor the following :
1.
Ix^-x-l.
6.
'M-
-9.^ + 9.
9. x'-^x'-Sx-2.
2.
3^^ + 5cc + 2.
6.
3a3^ + 14a!+15.
10. x'-{-7x'i-7x-16.
3.
2£c2-5ic + 2.
7.
Zx'-
-13^-10.
11. x* + x'-2x-2.
4.
^x^-^lx-1.
8.
x'-
-Qx' + Ux-
-6. 12. x^ + Sx'-i.
13.
2a;-"' + 3i«'-50i« + 24.
19. a'-
-a' + a'-l.
14.
^x'-1x^'\-x-
-2.
20. 2x'
-Sxy + y\
Jl5.
a' + 3a' + 3a-
+ 2.
21. ^x'
+ 2xy-y\
16.
rt^ + 2a-'-4a-
-3.
22. x'--
h 2x'^i/ —xii^ — 2?/.
17. a^-21a-20. 23. x^-2x^y-^xy''-^\%y\
18. a* + 7a=» + 13a^ + 7a + 12. 24. a^^^a^b^ab'-W.
25. Show that w—b" is akoays divisible by a—b, and thus
prove (a) of § 77.
26. Show that «"—?>>" is divisible by « + ^ only when oi is
even^ and thus prove {b) of § 77-
27. Show that a" + ^" is never divisible by a—b, and thus
prove (c) of § 77.
By any method of this chapter, factor the following :
28. ax'-Q>^a\ 31. 4a^ + 32«^. + 39Z>l
29. x'+--2. 32. {x-\-iy-bx-2^.
30. ^ahx^ + 2abx-ab. 33. ax'-Wx^-haK
FACTORS 129
34 ^ + ?? + l ^^* 4«* + 3a;y + 9y*.
' a' a ' ,60. a'b-bc' + a'c-c\
35. xy^-z'-xz-yz. ' ^^ a.3_^2_4^_g^
36. x'+^x'^-Qx+n. 62. a'-Va?-la-^.
37. 4(a-6y-(y + 2)l 63. x^+x'-VJx^U,
38. 9a;^-27a;y + 20y^ 64. a;^-ll£c^ + 31a;-21.
39. a' + aW-{-b\ 65. £c=^-6£c'' + lla;-6.
40. cc*— ISicy + y*. 66. a;=' + 2a3'-9a)-18.
41. a;6 + a;='-42. 67. 2/='-92^ + 25a;'^-10a;y.
42. ^*+2xy-35/. 68. Qx^'^ + x^r-lbf^
1 9 69. 3(a-^)'-14(a-^>) + 8.
^^* ^'"^a*"!* 70. 7(a + ^)'-llc(a + ^>)-6c2.
44. x'^-^x'-x'-^x. 71. £c*-aj^^-17x^ + 5a; + 60.
45. l + ^»y-(a;' + Hi/^ 72. a^-a'b-a^h.
46. (aj^'+y'-s'O'-^a^y- 73. x'-%x-ax-V^a.
47. x'-2xy'-y'-\-^x^y. 74. aa;^-3aV + 2a'a;.
48. a;/ + 7a;y-30a;. sy75. 2£c^-3x^-9a; + 14.
49. ac-\-cd—ab—bd. 76. 35a;'— 74a; + 35.
50. iK*-(a;-6)l 77. a;^ + a;*-56a-^
51. 72«' + 41^-45. 78. x'-^bx-^^-W.
52. a'b'-a'-b'-\-l. 79. a;^ + 7x^-5a;'-35. t^
53. 2s' + 5s«!-12^^ 80. a;^ + 4£c'4-a;-6.
54. 6a='a;-aV-9a*. 81. £c=^-3a;' + 7£c-21.
55. bcx-\-acx^-\-bx^-\-ax'^. 82. a;3 4-25a;2_[-8a;-16.
56. a*-2(^»' + c')a' + (52 + c')l 83. 10 {x-\-yy + lz{x^y)-^z\
57. a;'-(4a=' + ^>')a'£c + 4a''6l 84. 24(a+&)2-5c(a + 6)— SGc^.
58. a;' + (a-%-2a(a+5). 85. a;'+4ic+4-4a'^+4ay-yl
9
CHAPTER X.
COMMON FACTORS AND MULTIPLES.
96. Integral and fractional terms. A fractional term is a term
which contains one or more general numbers in the divisor.
Otherwise, the term is integral.
Thus, r, ^^7 2a ^ ^^^ ct-^{^—y), ^Y^f^CLctional terms, for
there are general numbers in the divisor. And a^, :^x^y^ ^^- ,
and — |a&, are integral terms, for the divisors are not general
numbers. A fractional term may become a whole number, and
an integral term a numerical fraction, when definite values are
assigned to the general numbers involved. Hence, the classifica-
tion of terms into integral and fractional terms has no reference
to arithmetical whole numbers and fractions.
97. An integral expression is an expression whose terms are
all integral.
Thus, ^a^—§ab + b^ and Soc^ + 2x'^ + x + 1 are both integral ex-
pressions.
Observe that the ^ and | do not make the terms fractional.
The nature of a term depends upon its general numbers.
A fractional expression is an expression whose terms are all
fractional.
2«]t^ 6?y^ z^ abc x^
Thus, — — -J- + — and n^- 4- rr are fractional expressions.
An expression may be integral with respect to certain general
numbers and fractional with respect to others. That is, certain
130
COMMON FACTORS AND MULTIPLES 131
general numbers appear in the dividend only, while others
appear in the divisor.
Thus, ^—2^ + -^ is integral with respect to a, hwi fractional
with respect to y.
A mixed expression is an expression some of whose terms are
integral and some fractional.
3c^ a^ a
Thus, — + a and 2ab + ^ "^ 3x ^^^ wi«a?ed expressions. The sec-
ond, however, is integral with respect to a.
98. The degree of a rational integral term is the number of
literal factors in it. This amounts to the sum of the exponents
of all letters in the term.
Thus, the degree of Qa^b is 3. The degree of loc^y^z^ is 6. And
4abcde is of the fifth degree. The term 5a^a^ is of the third
degree in x. And ab^x*y^ is of the seventh degree in x and y.
The degn^-ee of a rational integral expression is the same as
the degree of its ternj of higliest degree.
Thus, 5x^—x-{-4: is an expression of the third degree. And
2x^y^—Sxy-{-l is of the fourth degree. This expression is of the
second degree in x. The expression is of what degree in y ?
An expression whose terms are all of the same degree is
called a homogeneous expression.
Thus, 5x'^—2xy + y^ is a homogeneous expression.
99. The highest common factor of two or more expressions
is the expression with greatest numerical coefficient and of
highest degree that will exactly divide each of them.
Thus, the highest common factor of a^6V and aWc^ is a^b^c^.
And the highest common factor of lOor^T/z* and Sx^y^z^ is 2o(?yz^,
It follows from the definition above that the numerical co-
efficient of the highest common factor is the greatest number
132 ALGEBRA
that will exactly divide the numerical coefficient of each ex-
pression ; i. e., the greatest common divisor of the numerical
coefficients. By the coefficient of a polynomial is meant its
numerical factor. Thus, 2 is the numerical coefficient of
2a;2-6i« + 4 or of 2(£c^-3a^ + 2).
Also the highest power of any literal factor that will exactly
divide each expression is the lowest power of that factor found
in any one of the expressions.
A factor which does not appear in every one of the expressions
can not appear in the highest common factor. Hence the rule :*
To form the highest common factor of ttoo or more m^onomials^
take the product of the greatest common divisor of their numerical
coefficients^ and each letter raised to the lowest power to which it
appears in any of the expressions.
In polynomials^ factor each expressio7i qnd proceed as with
monomials^ treating each factor as yon woidd treat a single
letter.
Example 1. Find the highest common factor of 12a?^i/^ and
The greatest number that will exactly divide 12 and 16 is 4,
their greatest common divisor. The highest power of x that will
divide both a^ and x^ is a?% the lowest power present. And the
highest power of y that will divide both y^ and y^ is ?/^, the lowest
power present. Since z does not appear in both expressions, it
can not appear in the highest common factor. Hence, the highest
common factor is the product Ax^y^.
* There is a method of obtaining the liighest common factor of two
expressions without resolving them into their factors. It is similar to
the EucUdean, or long division, process sometimes employed in arith-
metic of finding the greatest common divisor of two numbers. The
discussion of this method is found in the Appendix. This method
is seldom used in practice. The factoring process here discussed is
sufficient for our purpose.
COMMON FACTORS AND MULTIPLES 133
Example 2. Find the highest common factor of SOa'b^c,
12a^¥c\ and 18d'b*c\
The greatest number that will divide 30, 12, and 18, is 6. The
lowest powers present of a, 6, and c, are a^, 6^ and c, respectively.
Hence, the highest common factor is 6a'*6^c.
Example 3. Find the highest common factor of iK^— 6^— 27
andx' + 6a?+9.
x^-ex-27={x + 3){x-9); x'' + 6x + 9={x-\-3Y. The lowest power
of ic + 3 present is the first power, x + S. And x—9 is not a com-
mon factor. Hence, the highest common factor is a? + 3.
Example 4. Find the highest common factor of
a^x—a^bx—Qab^x^ a^bx^—4ab^x'^ + S¥x^, and a^x'^—2a^bx^—Sab'^x^,
We have a^x—(i%x—Qa¥x=ax{a—3b){a + 2b);
a''bx'-4ab^x'' + 3b^x''=bx-{a-Sb)ia -b);
a^x^-2a%x^-Sab''x^=ax\a-3b){a + b).
Hence, highest common factor =x{a — ^b)
=ax—Sbx.
In finding the If. C.F. of two or more expressions when hut
one of the expressions is easily factored by inspection, we may
use the most likely factors of the factored expression as divisors
of the other expressions.
Example 5. Find the highest common factor oix^—Zx-\r2 and
ar* — 4x^ + 4a?— 1.
Qi?—Zx-^2—{x—\){x—2). Now since —2 of the second factor is
not a factor of —1 of the second expression, the "most likely
factor" is x—\ which by trial is found to be a factor of
x'-^x'^^x-X. Hence, x-1 is the H.C.F.
Example 6. Find the highest common factor of 2x!^+Sx—2
Sind4x^-\-lQx'-19x + 5.
2x^ + 3x—2={x + 2){2x—l). Herethemost likely factor is 2a?— 1.
Why ? By trial 2x—l is found to be a factor of the second ex-
pression and hence is the H.C.F.
134: ALGEBRA
EXERCISE 44
Find the highest common factor of :
' 1. ^a'b\ Mb\ 13. SGa^Sys, 72xyz, lSOxy'z\
2. 9a*!}'c\ 12a'b'c\ ^14. 17baWc\ 70a'b% 10ba*bcP.
3. UxY, SOxyz\ 15. 24m'na\ 42ma\ ISm'a'b.
4. Sa^yV, 9Qxy\ ^16. 15aV?i^ 40«Vwi, 35aV//2^
/>t. a'b\ 10a'bc\ 1 17. ^ba'b\ bQa'b\ 9SaW.
6. 2a; V^', y^^- ^18. -98^;^', 21a;y^, -28a;yV.
^7. axy, Sbxy. - 19. acc^y^, bx*y\ cxy, 2ic^y^
j8. 49a^>Vy, 21a'¥cx\ ^20. 2(a + ^>)^ 4(rt+^»)\
^9. ^^xy, -bQxi/z\ 21. 14(a;-y)«, 21(i«-y)*.
i>^0, QSab'c^d\ -2>d'bd\ 22. 16(a + a.')VS6(a + a3)'(^> + .y)\
11. ^xYz,_^xYz\ 2xy^z\ 23. (cc-1)^ {x-l){x-\-1).
v^l2. 2Sa''b'c'd, Qa'b\ lQa'b'c\ 24. 10(a; + l)^ 5(a!-l)2(ic + l).
^5. 39(a;-l)-Xa3+l)^ 26(a;+l)(a^-l)*.
j 26. bQ{x-l)(x + 2)(x + S), (x + 2y{x-^).
"?^27. ic^-1, a;^-l.
■ 28. 8iB=' + l, (2x + l)(x-S).
29. 4(a; + l)', 20(a;-l)(a;+l), 36(ic-l)^
yio, x'-l,x'-dx + 2. Z4:. 2x\x'-Sx,x' + 4x.
31. x'-l, x' + bx-Q. \}^5. x'-Sx, 9-x\ x'-^x + Q,
32. a'-b\ a'-b\ 36. a^^ + l, a;=^ + l, tc + l.
^3. a'-2ab + b\ (a-b)(a + 2b). 37. 2a;'^ + a;-6, 6if^-7ic-3.
38. 3aa;' - Sa\ 4a V + 2a V - Qa'x. •
39. a^ - ab\ a' + a^^* + a^> + b\
^^40. a■^ + 3a2a; + 2aa;^ a* + 6a^ic + 8a V.
41. a^-9a=' + 26a-24, a''-12a2 + 27a-60.
^ 42. 7m^-2m2-5, 7m=' + 12mM lOm + 5.
COMMON FACTORS AND MULTIPLES 135
43. a' + ^ab + 2b\a' + ^ab + 4:b\ a}-^ab~U\
/ 44. 1— £c^, 1— ic^, x—x".
46. Vla? — l^ab^Zb\ ^w'-^a^b^-'lab^—W,
46. m'^ + m— 6, m='— 2m'— m + 2, m=' + 3m'— 6m— 8.
47. ic='-3a;y'-2y\ o?-x'y—^y\
JC^S. a* + b\ a^-b\ a' + aby^+b\
49. a=^-a'-5a— 3, a^-4a'^-lla-6.
60. l-x% l-2£c=* + a;«, l + a; + £c^
61. a'—bax + ^x^ d^—a^x + Zax'^—Sx^.
62. ic'-7a; + 10, 4£c='-25a;' + 20a; + 25.
63. 6x'(/ + ^xf-2i/\ Sx' + AxSj-4x}/\
i
100. A common multiple of two or more expressions is an
expression which is exactly divisible by each of them.
Two or more expressions may have an indefinitely great
number of common multiples.
Thus, a few common multiples of 2o(^ and 3xy^ are 607^1/^, 6x^1/^^
ex*y\ 6x^y\ Qx'^if, 12x^y\ 18x^y\ 24x'''y\ etc. Each of these
expressions is divisible by both 2x^ and 3xy^.
The lowest common multiple of two or more expressions is
the expression with the least numerical coefficient and of lowest
degree that can be exactly divided by each of them.
It follows that the numerical coefficient of the lowest com-
mon multiple is the least number that can be exactly divided
by the numerical coefficient of each expression ; i. e., the least
common multiple of the numerical coefficients. And the lowest
power of any literal factor that can be divided by each expres-
sion is the highest power of that factor found in any one of the
expressions. A factor appearing in any one of the expressions
136 ALGEBRA
must appear in the lowest common multiple. Hence the
rule : *
To form the longest common multiple oftico or more monom,ials^
take the product of the least common multiple of the numerical
coefficients^ and each letter raised to the highest power to v^hich it
appears in any one of the expressions.
In polynomials^ factor each expression and proceed as vnth
monomials^ treating each factor as you icould treat a single letter.
Note. — To find the least common multiple of two or more numbers,
as in arithmetic, separate them into their prime factors, and take each
prime factor the greatest number of times that it occurs in any one
of the numbers. Thus for 72 and 96, we have 72=2-2-2-3-3, and
96=2-2-2-2-2-3. Hence, the least common multiple of 72 and 96 is
2-2-2-2-2-3-3^288.
Example 1. Find the lowest common multiple of 2^aWc^ and
SOa'b'c.
The least number divisible by 24 and 30 is 120. And the highest
powers of a, 6, and c, in the two expressions, are (i\ b^, and c*,
respectively. Hence, the lowest common multiple is 120a*6V.
Example 2. Find the lowest common multiple of 15aV, 21y^z,
3&a''z\
Here the least common multiple of 15, 21, and 36 is 1260. The
highest powers of a, y, and z are a\ 2/^, and s^, respectively.
Hence, the least common multiple is 1260a*2/"^^
^Example 8. Find the lowest common multiple of 2x'^—4x + 2,
\^' + x-2, and 5x'-^10x.
* It is easily shown that the product of the highest common factor
and the lowest common multiple of two expressions equals the product
of the expressions. Hence the. lowest common multiple of two expres-
sions may be obtained by dividing the product of the expressions by
their highest common factor. Since this method could be of value only
in case the factors of the expressions could not be found, it is not used
in this book. '
COMMON FACTORS AND MULTIPLES 137
Here 2x'-4x + 2=2{x~iy;
x' + x-2={x + 2){x—l);
5x'' + 10x=5x{x-\-2).
The least common multiple of 2 and 5 is 10. The highest power
of the factor xia x; of x—1 is(a?— 1)^; and of a? + 2 is a? + 2. Hence,
the lowest common multiple is 10a7(a?— l)^(a? + 2).
/
EXERCISE 45.
Find the lowest common multiple of :
1. ^a'b, Sab\ 7. 2bm'n\ 4:cmi7i\ 10a'm7i\
2. Qa'b\ 10ab\ 8. 15^//, 6pY, Aj^Y-
' 3. M'bc\ ^ab\ y 9. SQa'% 9a'b% lQd'b\
y4. Sla!^', l^x^f. 10. 27x''i/, QxY, 4xi/\ ^xHj\
6. lxy\ ^x% GccV. 11. {x-\-y)\ 2{x + y)\
b^. 2la'bc\ 4a'bc\ Qab\ 12. (a-b)\ S{a-b)(a + by.
> 13. x{x—l)(x + l), 2x'(a; + l).
14. x'(l-{-xy(l-x), (l + a;)(l-a;).
^\6. 10(a + ^»)'(c-J), 15(c-c7)^(a + J).
^%6. 25rt^(a-^>)(2a + ^)^ 45aa!(2a + ^)l
17. aj'^-l, x'-2x + l. : 23. 2a;^ + 3a;-2, 3£c^ + 7a! + 2.
18. x'-l, x'-x-2. 24. 2iK2-aj-10, 2x' + x-S.
y 19. l-a;^a;+a;^-2ar^ 2,5. 2a;=^ + 2a;, aj=' + 5a; + 6.
]M0. a'-b\ 2a' + ^ab + b\ 26. a;^-l, x'-x, 2x\
21. a2 + 5a + 6, «^ + 7« + 12. 27. x'-l, a;^ + 3a; + 2, a;^ + a;-2
c/22. x' + x-SO, a;' + 5£c-6. 28. l-x\ l-x\ 1-x.
29. x'-fii^, 2x'-4x, x'+ 1.
' 30. a;^ + 5a;-14, 4a;^-16a;^ + 16a;, ^x\
131. a'-b\ a^-b\ a'-b\
^^32. l-x\ l-x\ l-a;«.
138 ALGEBRA
33. x + l,x^ + l,x'-l.
34. x' + 2x'-Sx, 2x' + Qx'' + 2x + 6, Sx^—^x\
35. a*-b\ a* + 2aW + b\ a'-2aW + b\
36. 27a=^-8, 9a'-4, 9a'-12a + 4.
37. x-x\ lOx + dx'-x^ x-x'-{-x^-x*,
38. 05^-1, x' + x-2, x' + bx + Q.
39. x'-l, x' + x' + x, Sx\
^40. a'-l>'\ a' + a'b-ab'-b\ a'-a'b-ab' + b\
lAi, ic2-15a; + 36, aj='-3aj2-2a3 + 6.
42. a^-a^ + a + S, a^ + a'-Sa^-a + S.
43. 6a^ + a'^-5a-2, 6a^ + 5a^-3a-2.
44. ^x^ — lx'y—2xij\ ^x^ + xy—4y\
45. a?*-13a;' + 36, aj*-a3='-7a;' + a; + 6.
^46. 633^-03-1, 2a;M-3a^-2.
47. 4a;^-4£c4-l, 4a;2-l, 4a;2 + 4a; + l.
I 48. ax—ay — bx + by, x'^—2x7/ + y^.
I 49. 6a;^ + 13aj-28, 12a^2-3l£c + 20.
I 60. 8a;^ + 30a; + 7, 12a;^-29a;-8.
CHAPTER XI.
FRACTIONS.
101. In § 52, § 53, § 54, § 55, some principles of fractions
were established which the student should now review. Other
principles of fractions in common use in the treatment of
algebraic expressions and equations are here given.
102. The algebraic signs of a fraction. In every fraction there
are three signs to consider ; the sign before the fraction, the
sign of the numerator, and the sign of the denominator.
Thus, +-rk'-' ~+7' ~^' ^^^' ^hen a fraction is standing
alone, the sign + may be omitted ; as — ^, ^, etc.
— 7 o
Since the names, numerator, denominator and fraction, are
merely other names for dividend, divisor and quotient, respect-
ively, the laws of signs for division must hold for fractions.
From division we have the following principles :
I. If the signs of the numerator and denominator of a frac-
tion are both changed, the sign of the fraction is unchanged.
II. If the sign of either the numerator or denominator is
changed, the sign of the fraction is changed.
+ 2 —2 —2 -1-2
Thus, ■— o and -^ are both positive ; 370 and 30 are both
negative.
Since the sign before a fraction may always be considered as
indicating whether it is to be added or subtracted, and since
139
140 ALGEBRA
the subtraction of a term is the same as the addition of the
term with its sign changed ; the sign before a fraction may be
changed if the sign of either its numerator or denominator is
changed.
Therefore^ any two of the three signs of a fractioii may he
changed without affecting the expressio?i as a whole.
Thus, j.=zrh~ h~~~ '—b' ^^^^ being positive ;
and
-a a a —a , , •
■jr =^= — j = — -ziy ^^^" bemg negative.
If the numerator or denominator be a polynomial, by § 40
its sign will be changed by changing the sign of each of its
terms.
Thus,
■x^-\-3x—l 'of-Sx+l x^—3x+l
103. Multiplying or dividing both terms of a fraction by the
same expression does not change the value of the fraction.
For, since -=1, by § 54 we get
a _a X _ax , . a _ax ax a
l)-\x~b^' ^^'^^ '^' l}~~b^' ^"^ Tx~b'
Example 1. Multiplying both terms of —77- by a + &, we get
a—b _ a^—b^
a+b~{a + bf
'L2a^b^c
Example 2. Dividing both terms of 10^41^2^2 by 6aWc, we get
12aWc _ 26^
18a*b'c' ~ 3ac
The processes of multiplying and of dividing both terms of
a fraction by the same expression are called reducing the frac-
tion to higher tsrms and reducing the fraction to lower terms,
respectively.
Example 2. Reduce — — 13^2 to its lowest terms.
FRACTIONS 141
104. Reducing a fraction to its lowest terms. A fraction is
said to be in its lowest terms when its numerator and denom-
inator have no common factor.
To reduce a fraction to its longest terms, divide both its
numerator and denominator by all of their common factors^ or
by their highest common factor.
This follows directly from the definition of highest common
factor, § 99, and from § 103.
Example 1. Reduce ^a^^Lyi to its lowest terms.
Dividing the numerator and denominator by their highest
common factor, 4,ix?yz,
S6x*yz^~9xz'
ocfy^z^
■x*y*z'^
Changing both negative signs to positive, and dividing both
terms of the fraction by their highest common factor, x*^V,
Qc^y^z^ _oc^y^z^ _x^ ' -
~ —x*y*z^ ~x*y*z^ ~ y^'
x^ -\- 2x 3
Example 3. Reduce ' ., ^ r. to its lowest terms.
Factoring, and dividing both terms by their highest common
laetor, a; + 3,
x^-{-2x-S _ ix-l)(x-\-S) _ x-l
x'' + 5x+6~{x + 2){x + S)~x+2'
1 -i^^
Example 4. Reduce , „ = to its lowst terms.
Changing signs, and writing both terms of the fraction in
descending powers of a?,
1-^ ^_ af-1 ^_ ix+l)(x-l) __x-l_l-x
x'' + 8x + 7~ x'^ + Scc + T" {x+l)lx + 7) x + 7 x-\-7'
Note. — The process of dividing the numerator and denominator by a
common factor is sometimes called cancellation.
142
ALGEBRA
EXERCISE 46.
Reduce to lowest terms :
^
-SaW
6.
8.
-256a^^V^
— 4:9x^1/^ w*'
3a.y-2/
28
a;^-9a; + 20
i
10 + 3£c-£c^'
29. x^-h2x'y-2xf-y''
x'—'6x^y—2xy'^^\.]f'
30
}f-2hc-\-&-a}
31.
32.
33.
?yr
10m 4 16
m'^ + m— /2
ac—bc—ad^-bd
ac^ad—bc — bd'
x'^x'.-^x-^
x'-4x' + 2x-V^'
105. Reduction of a fraction to an integral or mixed expression.
A fraction whose numerator is of a degree equal to, or
higher than, the degree of the denominator may be reduced to
an integral or mixed expression by division.
FRACTIONS 143
Example 1. Reduce ^ — -— to a mixed expression.
B S 56 Sx'-Qx + 2 _3x^_6x 2
oX oX oX oX
=x-2 + §~.
Sx
Example 2. Reduce to an integral expression.
Performing the indicated division,
X
-d^^x-\-l.
Example 3. Reduce ^ to a mixed expression.
Since the denominator will divide a^ + l, add and subtract 1 in
the' numerator.
Then, ^=^^!±i=2=a!±l__2^=„,_„ + l. S
a +1 a+1 a +1 a+1 a+1
Example 4. Reduce — ^r^^^ to a mixed expression.
x—2 ^
Dividing, we get the quotient Sor+l, and the remainder 8.
Hence, 3^JzS|+6=3x+l+ «
' X—2 x—2
The division has the effect of breaking the numerator into two
parts, such as in the preceding examples, one part of which is
divisible by the denominator, and the other part not.
EXERCISE 47.
Reduce to an integral or a mixed expression :
x^ ' / ' a; +1* ' X —y'
2a;' + 4a;4-l ^ - a;^ + 3a;^ 4- 3a; + 1 a;^-3a;'> + 2a;
ALGEBRA
4x' + 2x'-Sx + l Q x' + lQ
2x'-l ' ' x + 'I'
11.
12.
x* + U
X -2 "
x' + y' 9a' + Sab + 4:b'
x + y' ■^"* Sa-2b '
x'-y'
x-^y '
144
8.
106. Fractions reduced to their lowest common denominator.
Fractions may be reduced to equivalent fractions having a
common, or the same, denominator by § 103. Since the lowest
common denominator must be obtained by multiplying each
denominator by some expression, therefore the lowest common
denominator must be the lowest common multiple of the
denominators.
The expression by Avhich any one denominator must be
multiplied to obtain the common denominator must be the
quotient obtained by dividing the common denominator by the
given denominator of the fraction. Hence the rule :
To reduce two or more fractions to equivalent fractions having
the lowest common denominator, first find the lowest common
multiple of all of the denominators ; divide this hy the denomina-
tor of each fraction in turn, and multiijly both terms of the
corresponding fractions by the quotients thus obtained.
Example 1. Eeduce -t' #, and — to equivalent fractions
having the lowest common denominator.
The lowest common multiple of ab, be, ac, is abc. Dividing
OC CSC
this by ab gives c. Multiplying the terms of -v by c gives —r- .
Dividing abc by be gives a. Multiplying the terms of ^ by a
au
gives -#• Dividing abc by ac gives b. Multiplying both terms
of — by & gives —^. Hence the required fractions are
ex ay bz
abc abc abc
FRACTIONS 145
Example 2. Reduce ^2_|.4^^3 ^ s^^' ^^^ ^Hs *o equivalent
fractions having the lowest common denominator.
x+2 x + 2 a?— 1 x—1 X X
ie + 4a;+3"'(it'+l)(a?+3)' a?''^-9~(£i? + 3)(x— 3/ x—?r'x^Z
Hence the lowest common denominator is (a? + 3)(u7— 3)(a7 + l).
Dividing by (ic+l)(x+3) gives ic— 3. Multiplying the terms of
.T + 2 , _ . (a; + 2)(a;-3) ix^-x-^
(^•+l)(x + 3) '^y ^ '^ gi^'^^ (ic+l)(a!+3)(£C-3)' ^^af* + ar^-9x-9*
Similarly, x-1 _ (..-l)(x-fl) _ 0.^-1 .
and
(;:c + 3)(x-3) (x + l)(ic + 3)(if-3) ar^ + x^-Ox-Q'
x _ a;(a? + 3)(.r+l) __^+JteM^^^
a?-3~(^+T)(x+3)(a?— 3)"~xM^'-9a?-9*
EXERCISE 48.
Reduce to equivalent fractions having the lowest common
denominator :
2 15 « 2a^ W
^' Zx' 'Ix'' Qx ^' Wc' Ic^c' Act'b''
^' W 6^^' ia^' ^* xy' yz' xz
3 5 1 y j^ z X y
\x^
5.
6,
£C+1' X-V X^-\
3 5
a;^ + 3a; + 2' 2a;^ + 5a; + 2' 2a;=' + 3a; + l
2cc + l \—x X
10
(^+iy' (^+iy' («;+ir ^/ X '^
2 a; 2a;^ ^^^^_
146 ALGEBRA
iQ ^ •'^ ^y
J
y' '-^y—y^' x'—y^'
ii _^ y__ ^' y^
x-y' 2y-'lx' ^{x'-yy ^y' - x')'
12. _^, y -1-.
a— a y—o z—G
2a 4:c
6«-6'
1
13.
^14.
^^' (c«-5)(a=^' {b-c)(b^^y (c-a)(c^'
•|/-g y -\-z z + x x + y
^ ' (y-^)(^-»^)' (y-^)(y-^)' (^-^K^-y)'
107. Addition and subtraction of fractions. By § 56,
a+b+c_a b c
X XXX
Hence, °+* + f =^±*±£..
jr * jr JT >r
Therefore^ fractions hamng a common denominator may be
added by adding the numerators for the numerator of the sum^
and using the common denominator for the denominator of the
sum. A fraction may be subtracted by chayiging its sign and
then proceeding as in addition.
Fractions which are to be added or subtracted must first be
reduced to equivalent fractions having a common denominator.
As was pointed out in § 5, if either the numerator or denom-
inator of a fraction is a polynomial, the dividing line also
serves as a sign of grouping. Consequentlj^, in such cases, the
signs of the terms in the numerator of a fraction which is
FRACTIONS 147
preceded by a negative sign must all be changed when the
numerators are added.
Example 1. Simplify ^J+|^+^.
Reducing to a common denominator,
x+1 3 iX^-l __6 x' + Qx 9x' 2x'-2
of 2x SX" Qx' iixr' ijx'
_ ex'' + Qx + dx^ + 2x^-2
ijx'
_ 17x' + ex-2
Example 2. SimpHfy a^Ja+^ - a^-L+3 -a^-L^2
Here ^ 3 1 _
' a^— 5a + 6 a^— 4a + 3 a^—i^a + 2
2 3 1
(a-3)(a-2) (a^3)(a-l) {a-2){a-'i)
2a-2 3a-6 a-3
(a-3)(a-2)(a-l) (a-3)(a-2)(a-l) (a-3)(a-2)(a-l)
2a— g— 3a + 6-2 + 3
(a-3)(a-2)(a-l)
7-2a
(a-3)(a-2)(a-l)"
Examples. By addition reduce x + y + ~r-^. to a fractional
form.
Here the integral part may be considered as a fraction whose
denominator is 1.
Hence, i»+2/ + -^=^^ +
x—y 1 x—y
_ x^-y^ ^ y^
x-y x-y
x—y
^
148 ALGEBRA
Note. — It is best first to arrange the denominators of all fractions
according to the powers of some letter, making use of § 102 if necessary.
Example 4. Simplify >-l_ + _^-^— -J-
^ "^ x—1 1 — X^ 1 + '.
+ x
Arranging all denominators in descending powers of x, and
changing signs in the second fraction, we have,
1 a- 1 1 cc 1
1 1—x^ 1 + x x—1 x'^—l x + 1
_ x 4-1 X x — 1
~xi^—l~x^—l~~x^ — l
_ x + \—x—x-\-l
~ x'—l
2 —X
V
EXERCISE 49.
Simplify :
1. h^-^'+' ■ 9. 1 + ^.
X X^ ^X X
2. y-H 5. 10. X-\-zr-7—'
DC ac ab 1 + a;
_ a;4-l , 1 ,2^-1 ^^ , ^ x^
3. —-2- +-r-:+ -.A ■> • 11. a? + l-
x^ hx lOa?'^ ' x — \
4. 1 + i + l. 'l%l-.' ^^^
£«?/ yz xz \Zy \^rx \—x
u.
x-\
a; + l'
^ 6.
3
2 4
a^ — £C
ic^+aj iK'
7.
2a;+l
a; + 2 1
a^-2
2£C-1 X
8.
3a
^ 1 ^<
a ^ax
14.
15.
a-\-b a — b
a—b a + b'
1 1
16. o:-2
2aj^ + 7£c-4 3ic^ + 13a;+4
FRACTIONS 149
17 _« L L. 21 -^ ^-1
18 ^.+JiL_J^. V 22.^^ 1 I 2y
\-\-x x—\ 1—x^ x^-\-xi/ ^—xy x^—if-'
x^ 'A—y 2/ —4 ^ + 2/ x-\-A x—i
25 1 1
26.
ab—ac — b'^ + be bc — ab—&-\^aG
3a + l 2^>-l 46—1 6(^ + 1
12a m ^ 16c 24c? *
27. . ,A, . + 1 1
{a—b){b—c) {b — a)(a — c) {c—a){c—b)'
'*0' ;;;3:;,— ;:r3:7,+;:;23:::72- V«>"' 7777-7;;.+ or^-::.+i;F
£c— 2/ aJ + 2/ x^-Vy'^ ^ ' 2b— x 2b-]-x x^—A^b
OQ _^^ 2a; 1 2a; So;
'*^- a;=^-l a;^ + a; + l"^a;-l' *^^- ^ ^^a;"^-l x'-^V
32.-^-^+. 1 1
a;— y x + y x—'ly a; + 2y
Note. — In certain cases, like this last exercise, it is best to add only
certain ones of the fractions at a time. It saves long multiplications.
Here, add the first and second fractions ; next add the third j)nd
fourth ; then add the sums thus obtained.
33.4-,+4-o-A-^o. 34. A— ^.4 ^ ^
cc + 1 £C + 3 x—\ a?— 3* * a — b a^b c—d c-\-o
2,3 2 3
35.
a;— 5 a; + l a; + 5 a;— 1*
i - 1
36. -
a;'-^ — 1 a;''^ + l x^—a^ x^ + a^
a;* — 1 x*-\-x^ x*—x^ d'—b^
150 ALGEBRA
39. ^^~-x\ 41. a+ ^,_^, +k
■ ^ x'-hy' , x'-7/ ^^- S + 2x 2^Sx , 16.x— jc^
40. -+x ~. 42. -o —-on 2 — A~'
x—y x-\-y 2—x 2 + £c £c^— 4
108. Multiplication of fractions. Fractions may be multiplied
by the rule established in § 54. In case the given fractions are
such that their product may be reduced to lower terms,
the process of multiplication and reduction may be shortened
by first cancelling any factor of any numerator hy an equal
factor of any deno7ninator.
This is evident, since sucli factors may be cancelled after the
multiplication. See § 104, note.
Example 1. Find the product of i|^, i^, ^
12a^ Uab 5bc 'l2a^Uab-5bc
W 9a-c-* 2a~" 76'^-9aV-2a
840a362c
\2^a?b''&
Reducing, =g^
Example 2. Simplify ( ^~^ )
x^ + 2a?,
/ ^^-4 \ / .T='-9 \ __ {x-2){x\2) (x + 3)(a;-3)
\x^-^x)\x'^2xj~ x{x-Z) ^ x{x-\-2)
... „ ^ ix -2){x + 3) ^^, x' + x-e
Cancellmg common factors, = -^ -, oi — ^^
Note. — The cancellation may be indicated by drawing lines through
the common factors.
{x-2)Cs»^ (x + ^){s»^^ (x-2)ix + :^) x'-^x-6
-^^^^^' xi,^^ a-iae-t^- ~ x-x ~ x'
o a- ^^f a'- lSa + SO ^ a'-6a-7 ^a + 5
Example 3. Simplify ^..g^.go x a^-i5a + 56 ^ ^=:i*
FRACTIONS 151
a'^-Sa-SO 'a='-15a + 56'a-l~ (ii^(a,;s*4tJ) (^^^(^.*^ (a-1)
g + l
-a-l'
109. In multiplication, any mixed expression should first
be reduced to a fraction. Expressions free of fractions may
l)e treated as fractions with the denominator 1.
EXAMPLE 1. Simplify (. ^£^) (y- ^
\ i^-yj\ ^+yj \3c-yJ\3c+yj
x^—y"^'
Example 2. Multiply ^.-^^^^ by a^^ 4- 4x- 21.
x^2 x+2 (a? + 7)(^-3)
■x^ + 12x-v^h^'^ '^ ^^^-{x+^){x + 7)'^ 1
(x+2)(x—2>) x^—x—6
x + 5 ' ^' x + 5
Note. — It is clear that the product of a fraction and an integral ex-
pression, as in tlie preceding example, may be obtained by merely
multiplying the numerator of the fraction by the integral expressioiij
and placing the product above the denominator. Thus, in general,
a an
EXERCISE 50.
Simplify :
4,w2
152 ALGEBRA
x—y x^ Ay
20.-1 2y + 3y + 3
^' 7+3 ^"^^^^2a;-l*
J
- (-S)(.-^)-
J
te^-l a;'' + 3a;4 2
X 4-2^a;' + 2a3 + l'
x^-1
-^X
,2 4y2
9. ^Z^ x^ -
' £C + 2y x^—y^
a + 3 ^ a + 4 *
11 ^'-y' v ^'~y'
19. (a;
i
14
15
16
■(
ab
^ b'
y
a'-^b
d^_ _
y' ^)\ 3£c
_27a^ / 1
L+iV
J
20
xy—y
'x + y
x^—i/
b-Sa
18. ^a^-b^^(^^^^y
({x±yY\
r^X
21
23
X a?^+y^
i
/a^-hab\ / g ^\
22 /^ yy. a.^ + y^ \
Vy a;A (^+y)7'
/x^—y'^—z^ — 2yz
\ x^—xy—x:
2z
x + y+z
)
24. (^^-a^y + 2/0x5
25.
26
27
28
2a;+l
X,
^^=T6^2^M^5^+2
.2V / 4^2
5c
1. (3a+jy3.
/ x'-Sx+ 2 \ / a;^-7a; + 12 \ / a;-^-5a;^
V£c'^-8x + 15y \x'-bx+ 4/ V a;-^-4 /
'• (a?
x^—4x
bx + QjW + 2x
Sx
m-
FRACTIONS 153
Multiply the following by such expressions as will make the
products integral expressions :
Suggestion : Multiply by the lowest common denominator.
a 1
30.
31.
ci'-b' a + b'
2ic + l , 3
x"^ — 1 x^ — 1 x^ + x+1
32. ^— ^ +
110. Division of Fractions.
Since the divide7id = quotient X divisor,
.- a; a
y ^
then -= -2^ + 1
1 x—yx-\-y 9x^ — 4y^
o. 5— 7. ; 11. "o Ti — •
1 « ^~y ^ + 2/ 6x—ly
X ^"t-y ^"~y x^—y
'12. 7-
l+a3^^ a'^-a'h'^b' ^ x
Suggestion : Begin with the last complex fraction and simplify
step by step. A fraction of this kind is called a continued fraction.
13.^ 14.
a ^ x-V
a + A- 1 + ''
a-l " ' 3-a;
FRACTIONS 157
EXERCISES FOR REVIEW (III).
1. What is meant by the factors of an expression? Illus-
trate.
2. Name the typefomis that have been factored in this book.
3. Of what type form are the following? Factor them.
/ {a) Qx'-^x'y^'ixxf. (d) a'b-a'b' + aW-ab'.
,{b) ^a'b-'iaW + d'b. ^''a3^ - %^
(^0 25£c' + 70a;?/2 + 49yV. Ao) 2a;' + 5£c-12.
(o) £cy + 12a7y=' + 27. (a;) ac-bd-ad+bc.
10. , Factor:
"^{a) 9««
6a^ + l. /O*) Qx'-lSxyi-Qy\
(b) 12a' + 7ab-10b\ ^ (k) 7x'-h2bxy-12f,
Uc) a'-^a'b' + 4:b\ {I) 16«'^-48a + 35.
* {d) 4a;* - 4a;V' + 9y*. (/?i) a"'^ - ^'".
{e) IQa' + lQab-W. '' {n) a' + 216.
(^) a;^« + 2a;« + l. (/>) a' + a' + l.
(h) a'-b'-a-b. (q) Sx'y + Sxy-}-SxY.
(^) x* + xy+y\ (r) a^—a—a'^b—ab.
f^RACTIONS 150
(s) l + a—b—ab. (v) ahx^-\^{a'^^-h-)xy-\-aby'^.
(t) 4:a'b'-(a' + b'-cy. (w) a'b' + a'x'-b'x'-x*.
(u) Sbc-4ad+Qac-2bd. (x) Qa' + 9ab-3b-2a.
11. Find the 11. C. F. of :
(a) x'-S, x'-2x'-{-x-2.
(b) x'-V2x + S^, x'-2x'-19x + 20.
(d) x' + Sx' + 4x + 12, x' + 4x' + 4x + ^.
12. Find the L. C. M. of :
(a) Qx\ x'-2x, 3£c='-12a; + 12.
(b) 2a%ba'-^ab,2ab + 2b.
(c) x^—x^ — Qx, x^ + 2x\ x^ — Qx^ + dx.
(d) Gx' + 7x-^,2x'-x' + Sx—4.
13. Explain the process by which algebraic fractions are
added.
14. What law of signs must be observed when a fraction
is subtracted?
15. Simplify :
/ X 3 . 2 1
(a)
x'-l ' x'-Sx-4: x'-x-2'
(1\ 9^ + 17 , 2x-l 2x-\-l
x'-2x-4S ' 2a;+12 2a;-16
17. By what must a fractional expression be .multiplied in
order to obtain an integral expression ?
18. Multiply by such an expression as will make integral:
1 QO ALGEBRA
J. 1 3 2_
3(^-1) 3a; 9^
^ ^ i«-2 a; + 2 a; + l*
19. Sin,plify (-^-l)^(3.,-J^)
20. Simplify ^V'^'^V^ ^ x'y^—x '^
x^-\-xy-\-y^ ' y^—x^
21. Simplify 1 1 X^3ipj3-
b a
22. Divide ^ +lz:?by y^ -III? and express the quo-
J. ~T~ XX X "I X X
tient in its simplest form.
23. Simplify:
W \^, «; • \a b)^ a + b ^b\a-by
^^ \bTc~^^rFcJ\^Tb'^^^^b~' a'-b' J'
/2(^)_«+.wi ly
^ ^ \ a-^x a—xj \a xj
^ ^ Va;'-y=^ x+yj \ 2zy J x-y
ab—b '^ a^-b''
a^—ab ab 1
CHAPTER XII.
LINEAR EQUATIONS— ONE UNKNOWN NUMBER.
112. We showed in Chapter II the meaning of an equation
and how, by the use of axioms, to solve the simplest kind of
equations containing one unknown number. Now that positive
and negative numbers and fractions have been discussed, we
return to the discussion of equations.
113. An equation is integral with respect to its unknown
numbers when both of its members are integral with respect
to those numbers. Otherwise it is a fractional equation.
Thus, x-\-§=Qx-\-5 and 2^^ + 3.x^=8 are integral equations in x.
Sx^—xy + Si/=7 is integral with respect to both a? and y. The
equation - + -=10 is fractional with respect to both x and y.
X y
A rational equation is one in which both members are rational
expressions Avith respect to the unknown numbers. Otherwise
it is called irrational. The equation j/a;— i/a;— 1 = 2 is irra-
tional. All equations here to be considered are rational equa-
tions.
114. The degree of a rational integral equation is the degree
of its term of highest degree with respect to the unknown
numbers.
Thus, x + 3=4x and 3ic— 2i/=10 are of the first degree.
x^—5x+Q=0 and x^—2xy + y=9 are of the second degree.
a^ + a?"— x=0 is ot the third degree.
a?* — 1/^ = 6 is of the fourth degree.
161
162 ALGEBRA
An equation of the first degree is also called a linear equation.*
One of the second degree is called a quadratic equation.
One of the third degree is called a cubic equation.
One of the fourth degree is called a biquadratic equation.
115. Equivalent equations. Equations which have the same
solutions are called equivalent equations.
Thus, 6a?— l==4a? + 7 and 2a?=7+l are equivalent, for each is
satisfied by a?=4, and by no other solution.
In Chapter II linear equations were solved by the use of
axioms.
Thus to solve 6x— 5=4a? + l. (1)
Adding 5 to each member, 6x=4:X + 6. Axiom 1 (2)
Subtracting 4x from each member, 2x=6. Axiom 2 (3)
Dividing each member by 2, x=3. Axiom 4 (4)
It will be seen that this work consists of deducing in the
successive steps, successive equations, each one equwalent to
the preceding one. Thus, equations (1), (2), (3), (4), in the
above example are all satisfied by a3 = 3, and by no other value
of X ; hence they are all equivalent. Therefore, the solution of
the last equation is the required solution of the given equation.
116. Observe that to change the given equation of § 115 to
an equivalent one whose first term consisted of a multiple of
the unknown number, and whose second term was a known
number, i. e. to reduce the given equation to the form ax = b,
we proceeded as follows :
I. The known member in the first member^ with its sign
changed^ was added to both members to free the first member
from known numbers.
* The name linear equation is derived from the fact that the equation
of the first degree with two unknown numbers has a pecuhar relation
to a straight line. See § 134.
LINEAR EQUATIONS— ONE UNKNOWN NUMBER 163
II. The term iiwolmng an unknown number in the second
member urns added vnth its sign changed to both members to free
the second member of unknown numbers.
This process gave a new equation in which certain terms
that appeared in a member of the old equation appeared in the
opposite member of the new equation with their signs changed.
Tills result is equivalent to transferring terms from one mernber
to another and changing the signs of the transferred terms.
This process is called transposition. The term is said to be
transposed.
Note. The pupils should use the correct phraseology of -'adding
equals to both sides of tlie equation," until the thing actually done is
firmly fixed in mind. When this is thoroughly understood the briefer
form, "transpose", may be used if desired. Care should be taken^
however, that the pupil does not say "transpose", and mechanically
perform tlie process, forgetting what he has really done.
117. To solve a linear equation for one unknown number.
The following general method may be used in solving any
linear equation for one unknown number :
(1) Remove all signs of grouping., if any exist.
{2) Transform, the given equation into an equivalent one
having all the unknown numbers in the first member, and all
terms free of the tinknoion number in the second member.
(3) Unite like terms.
{J/) Divide both members by the coefficient of the unknown
number.
To check the transformed equation, see if the terms that
were cancelled in any member of the given equation reappear in
the other member of the new equation with signs changed.
Example 1. Solve 7ic+15=4a? + 3.
164 ALGEBRA
1. Adding -4a?-15, 7^-4^=3-15. (Ax. 1)
2. Uniting like terms, 3x=— 12.
3. Dividing by 3, x=—4, the solution. (Ax. 4)
Check. Substituting for a? its value —4, the given equation be-
comes —28 + 15= — 16 + 3, an identity.
Example 2. Solve 3(a + l) = 12 + 4(a-l).
1 . Removing signs of grouping, 3a + 3=12 + 4a— 4.
2. Adding -4a-3, 3a-4a=12-4-3. (Ax. 1)
3. Uniting like terms, — a=5.
4. Dividing by —1, a= — 5, the solution. (Ax. 4)
Check. Substituting for a its value —5, the given equation
becomes 3(-5 + l) = 12 + 4(-5-l),
or —12=— 12, an identity.
Example 3. Solve (5— a-)(l + a?) = (2— .T)(4 + a?).
1. Removing signs of grouping, ^ + 4:X—x'^=S—2x—x'^.
2. Adding x^ + 2x- 5, 4x-x' + 2x-\-x'=8-5. (Ax. 1)
3. Uniting like terms, 6x=3,
4. Dividing by 6, ^=h ^^^^ solution.
(Ax. 4)
Check. Substituting for x its value |, the given equation be-
comes (5-^)(l + J) = (2-i)(4 + |),
or 4^1|=l|-4^, an identity.
EXERCISE 53.
Solve the following equations for x :
1. 2ic + 7 = 14-5i«. 4. 12-Ux=Q—9x.
2. 30a; + 24 = 60 + 48a5. 5. 8a;-7 = 3a; + 3.
3. 19i«-22 = 8ic-17. 6. 14aj + 20-12= -20a; + 35i«.
LINEAR EQUATIONS— ONE UNKNOWN NUMBER 165
7. 2Sx-^b = 21x-10x-U7. 9. ^{2-4x) = 4{l-dx).
8. S{x-2)=2(x-S). 10. x-(Q-2x) = 9(x-l).
11. 2(a;-l) + 3(a;-2) + 4(ic-3) = 0.
12. 2x-b(20-x)-i) = 0.
13. 5(2a; + l)-7 = 3(2aj-7) + 51.
14. bx-Q(S-4x)=^x-7(4 + x).
15. 2(a;+12)-(a;-3)=0.
16. (a;-2)(a3-3) = (a;-4)(a;-5).
17. S(x + 4)(x-2)-^ = S(x + b)(x-S)+x.
18. (a; + 2)^-a;='=a;-5.
Solve the following equations for a :
19. l + (a-^y = (a + iy-4.
20. («-2)(a-5) + (a-3)(a-4)-2(a-4)(a-5).
21. 2.5a-6.75-1.25a-3. 22. 0.75a + 2(l-1.2«)=0.
23. 2-6.9a(l-2a)-2(6.9a'^-3).
24. \a-\a = 2. 26. Ja = 4-lia.
118. Some fractional equations may be changed to equivalent
linear equations by multiplying both members by such an ex-
pression as will destroy all fractions (Axiom 3). The neces-
sary multiplier will evidently be the lowest common denominator
of all the fractions. This process is called clearing of fractions.*
The pupil should see that the real process is multiplying both
members by the same number.
*Like the term "transpose," the term "clearing of fractions" is
often used by pupils without their knowing the real process involved
and the authority for it. A pupil sometimes thinks that clearing an
equation as ^^-—j=- of fractions consists in multiplying the first
member by 1 + a7 and the second by 2x-\-4 rather than each member by
(l+a;)(2a;-h4).
56
ALGEBRA
Thus, solve
1_
X
2 .
~3x'
5
6'
Multiplying both members by
6x, 6-
-4 =
:5X.
Adding —5x + 4-
-6,
5x=
=4-6.
Uniting terms,
5x=
r-2.
Dividing by —5,
X-
=|.
(Ax. 3)
(Ax. 1)
(Ax. 4)
If a multiplier be used whose degree in the unknov/n number
Is higher than that of the lowest common denominator, the
resulting equation will usually not be equivalent to the given
equation. (See § 173). Hence the rule :
To clear an equation of fractions 7nulti2^ly all terms in both
members by the lowest common denominator of all the fractions
in the equation.
It should be noticed in the following examples that the easiest
way to multiply a fraction by the lowest common denominator
is first to divide the lowest common denominator by the denomi-
nator of the fraction, then multiply the numerator by this quo-
2a
tient. For example, to multiply ^— by eer*, we divide ^x" by 3a?
oX
obtaining 2a?\ then multiply 2a by 2x^ obtaining 4aa?^, the product.
If a fraction is preceded by the negative sign, the sign of every
term of the numerator must be changed when the multiplication
is performed. See § 5.
Example 1. Solve -^,_-^=-A_
x—2 x+2 x^—4:
The lowest common denominator is a?^— 4. Multiplying by
x{x + 2)—x(x—2)=4. (Ax. 3)
Removing signs of grouping, x^ + 2x—x'^ + 2a?=4.
Adding, 4a7=4.
Dividing by 4, x=l, the solution.
(Ax. 4)
a?— 3 a?— 4 a?— 6 x—7
Example 2. Solve
.T— 4 x—5 x—7 x—8'
LINEAR EQUATIONS— ONE UNKNOWN NUMBER 167
Here there will be an advantage in adding the fractions in each
of the members before clearing of fractions.
1- Adding, ^__zl_^=___l_^.
2. Multiplying by (^—4)(a?—5)(a?—7)(x— 8),
-l{x-7)ix-8) = -l(x-4){x-5). (Ax. 3)
3. Dividing by -1, {x-7){x-8) = (x-4){x-5). (Ax. 4)
4. Removing signs of grouping,
x''-15x+5Q=x''-9x + 20.
5. Adding, — x'^ + 9x— 56,
x^-15x—x'' + 9x=20-5Q. (Ax.l)
6. Uniting terms, — 6x=— 36.
7. Dividing by —6, a?=6, solution. (Ax. 4)
EXERCISE 54.
Solve the following for x : .
Q 3 CC + 1 , £C^
^' ;T-ri-:;r—i+:;j-
1.
3 1
2a5 4*
2.
j« + 3 o
«.-3-^-
3.
»^ + l 2 1
a;' '^Sx X
4.
2a^+l 1 a;
4 05-1 2'
5.
1 2 3
2x + l x + l 'Ix
6.
^-2^03 + 1 ^•
7.
cc + 1 a^-l l-f2a;
3a;^l 3.r+l W-l'
R
1,1-1
£c + l a; — 1 x^ — 1
10. ?Lz| + ?i=f=,.
a;— 7 a;— o
£C — 2 05 + 2
^2 x — '^x — t)
/13.-A^l^-:i^^ + 7.
£C + 1 aJ— 1
14. 2 + -?^=^^.
05 + 3 a;+7
15 3a; + 7 _6cc-2
4aj + 8 8a.— 5*
1R 8 a;-3_a3 + l
2a; + 3 ' 2£c-3 4a;^-9' a5"^a; + 3 cc-1"
168 ALGEBRA
17. 2a;-3 ^_ a; + 5 11 ' ' ^ jjO. ^^
2£c-4 "^ dx-Q 2* ' aj' + 5a; + 6 a^^2 a3 + 3
j^g a? a;^ — 5a;_2 21 ^ ^ ^ ^
3 3a;— 7 3* x—^ x—1 x—4: x—2
X
^g ^ 3_ 4 x-\-l Qo 3a; 2a; _2a;^— 5
«„ 3a; + 5 3a;'' + 5a;— 4
24.
4a;-3 4a;^-3a; + 2*
a;— 7 a;— 9 _a?— 13 a;— 15
a;— 9 a; — ll~~a;— r5 a;— 17*
OK a;— 5 , a;— 7 a;— 4 , x
»o. — =4-
26.
27.
28.
a;— 7 a;— 9 a;— 6 a;— 10
a; a; + l a;— 8 a;— 9
a;— 2 a;— 1 a;— 6 a;— 7
13 2
a; + 3 9— a;' 3— a;
3(a;-l) 3^ _ 9
x—2 a; + 2 a; + l*
og 5(a; + 6) 2(5a; + 2) _^
a; + 10 2a;-l
a;— 5 7— a; 2a; — 15
30.
„i 9a; + 17 , 2a;- 1 2a; + l
^A. -5 — rr— r-ro +
32.
a;'-2a; + 48 ' 2a; + 12 2a;-16
5 3a; + 5 _ 8 + 3a;
l-9a;^ + 3a;-l~l + 3a;*
119. The formula. An equation often contains more than
one general number. In that case it may be solved for the
value of any one of these general numbers. It is clear that in
such cases the value found for one of the general numbers may
be an expression containing the others, and hence, the solu-
LINEAR EQUATIONS— ONE UNKNOWN NUMBER 109
tion may not be a single- valued or definite number. Such an
equation is sometimes called a formula.*
Thus, ax + b=ac may be solved for a, x, 6, or c ; but the value
thus found for either will consist of an expression containing the
others.
If the formula be linear with respect to a certain general
number, it may be solved for that number by the method of
the preceding section.
Example 1. Solve ax-\-b=ac tor x.
1. Adding — &, ax=ac—h. (Ax. 1)
2. Dividing by a, x=^^:^^' (Ax. 4)
Example 2. Solve ax-]rh=aG for &.
1. Adding — aa?, b=ac—ax. (Ax. 1)
Examples. Solve ax + h=ac fore.
1. Adding —ac—ax—b, ~ac=—b—ax. (Ax. 1)
2. Dividing by —a, c= ~ ~ (Ax. 4)
Or, changing signs, c= (§102)
Example 4. Solve ax-\-b=ac for a.
1. Adding —ac—b, ax—ac~—b. (Ax. 1)
2. Uniting terms, a{x—c) = —b.
3. Dividing by x—c^ a= (Ax. 4)
X — c
Or, changing signs, a=— — -• (§ 102)
Note. — The student will find it of great value to be able to solve a
formula in taking up the study of Geometry and Physics.
* A formula expresses a law in mathematical symbols. The type
forms of multiplication or division are really formulas. When a
formula is expressed in words it is a principle. When expressed as a
direction how to do a thing it is a rule. Thus, the formula of § 110 was
expressed as a rule.
170 ALGEBRA
EXERCISE 55.
1. ^ax—h=cy. Solve for x.
2. xy-Vx=y^~y—^. Solve for x.
3. a{x — l)-{^a=x. Solve for a.
4. «(.T + 3) + 5(£c-3) = c(£c-l). Solve for h.
^6. {a— «;)(« + 2£c) =a^-\- x^. Solve for a.
6. (« + 5a^) {h + a^c) - «5(a^^ + 1) = «' + ^'. Solve for x.
7. 2(2^-l) + 3 = «(^ + 2). Solve for t.
8. 3(^ + « + a;) + 2(^ + a— £K)=-a7. Solve for if.
9. Solve Ex. 1 for 2/. 12. Solve Ex. 4 for c.
10. Solve Ex. 3 for x. 13. Solve Ex. 7 for a.
11. Solve :Ex. 4 for a;. 14. Solve Ex. 8 for a.
The following formulae express important laws in Physics.
lb, s=vt. Solve for?;. l^. E=\Mv\ Solve for J/:
16. v = at. Solve for t. J/y2
, ,, , 20. i^=— -. Solve for r.
17. s=\at\ Solve for a. ^'
18. TT^i^s. Solve for i^. 21. 6^=i«(2^-l). Solve for ?;.
22. PD= Tr-7>'. Solve for P.
(7 i?
ic
23.---. Solve for 7?. 26. 7>=-^i-,. Solve for ^^j.
c r V!) — v")
24. ^,=^' Solve for P. 27. J^-|-(7+32. Solve for C,
25. (7=:f. Solve for P. 28.4=-+^- Solve tor 7?.
11 Br r'
120. Problems solved by use of linear equations with one un-
known.
' In § 26 we showed how problems could be solved by the use
of the equation. The important steps in the process of solving
LINEAR EQUATIONS— ONE UNKNOWN NUMBER 171
such problems by use of the linear equation with one unknown
number are as follows :
(1) Fvrst represent by some, letter one of the iinknovm numbers
mentioned in the prohle'm.
{2) Then, from conditions stated in the problem, form ex-
pressions containing this assumed letter xnhich icill represent the
values of any other luiknown numbers mentioned in the problem.
{3) By means of some other statement in the problem it should
then be possible to form an equation betioeen these expressions.
{4) Solue this equation and interpret the result.
Example 1. The difference between the squares of two con-
secutive whole numbers is 121. Find the numbers.
Let x= the less number.
Then, it' + l= the greater number, for the numbers are con-
secutive.
Therefore, x^ and {x-\- Vf will represent the squares of the two
numbers.
Hence, (^ + 1)^— x^=121, for the difference between their squares
is 121.
Removing the signs of grouping, x'^-\-2x-\-l—x^=121.
Whence, 2.r=120.
Therefore, a?=60, the less number,
and 07 4-1=61, the greater number.
Check. (61)^-(60)2=3721-3600=121.
Example 2. The length of a room exceeds its breadth by 4
feet; and if each had been increased by 4 feet, the area would
have been increased by 128 square feet. Find the dimensions of
the room.
Let x= number of feet in the breadth.
Then,. a? + 4= number of feet in the length.
Hence, x{x+^)= number of square feet in the area. Rule for
area of a rectangle.
172 ALGEBRA
If the breadth and length were each increased by 4 feet, they
would be a:+4 and x+8, respectively. And the area would be
(x + 4)(a? + 8).
From the condition of the problem,
(x -{-4){x + 8) —x{x + 4) = 128. (State the condition
that gives this equation.)
Removing ( ), x' + 12x + d2-x'-4.x=128.
Whence, 8a;=96.
a?=12, the breadth.
x+4=16, the length.
Example 3. The sum of two numbers is 21, and the quotient
of the less divided by the greater is f . What are the numbers ?
Let x= the less number.
Then 21— £c= the greater number. (Why?)
Hence, since their quotient is |, we have
x 2
21-i»~5'
Clearing of fractions, 5x=^2—2x. (What multiplier is used ?)
Whence, 7x=42,
and, • x=Q, the less number.
21—07=15, the greater number.
Example 4. A tank can be filled by one pipe in 24 minutes, by
a second pipe in 32 minutes, and by a third pipe in 40 minutes.
If all three pipes run at once, how long will it take to fill the
tank ?
Since the first pipe alone could fill the tank in 24 minutes, in
one minute it could fill ^^ of it.
Likewise, in one minute the second pipe alone could fill ^\ of it;
and the third pipe alone could fill jV of it.
Hence, in one minute the three together could fiH^V+sV + ^V
of it.
Let X = number of minutes required for the three pipes to-
gether to fill the tank. Then in one minute they could fill - of it.
X
LINEAR EQUATIONS— ONE UNKNOWN NUMBER 173
Clearing of fractions, 20x+ 15a? + 12ic=480. (What multiplier
was used ?)
Uniting terms, 47a? =480.
a[;=10{f, number of minutes.
EXERCISE 56.
1. A has 170, and B has 110. How much must A give to B
in order that he may then have just three times as much as B?
Suggestion. — If x representF- the required amount given by A, what
will each then have? What condition of the problem, then, gives an
equation ?
2. Divide 50 into two parts whose difference is 26.
3. Find two numbers whose sum is 1 and whose difference
is 15.
4. Find two numbers whose difference is 15, and whose sum
is f of their difference.
Suggestion. — If x=: the smaller, what must equal the larger? What
equation follows ?
5. Find the number the sum of whose half, third part, and
fourth part is 26.
6. Find two numbers whose* sum is 36, one of which is | of
the other.
7. Find the number such that \ of it shall exceed i of it by 2.
8. Find two numbers whose sum is 28, and such that one
exceeds 7 times the other by 4.
Suggestion. — What two conditions in the problem ? If one number
is X and their sum 28, what must the other number be? What equa-
tion follows from tlie second condition ?
174 ALGEBRA
9. Find two numbers whose sum is Gl and difference 11.
Suggestion, — When x= one number, either of two conditions will
give the other. What are the conditions? Show that you may use
either condition to get the expression for the second number and the
other condition to get an equation.
10. What number increased by i of itself and 20 is 2 more
than double itself?
11. Eight times the difference between one-fourth and one-
third of a number is 32 less than the number. What is the
number ?
12. Find the number that exceeds 20 by as much as i of the
number exceeds 7.
13. Find two consecutive Avhole numbers whose sum ex-
ceeds 25 by as much as 25 exceeds 15.
14. There is a certain fraction whose value is J ; and if its
numerator Avere greater by 2, and its denominator less by 2,
its value Avould be i. What is the fraction ?
15. What number added to the numerator and to the
denominator of j\ will give a fraction equal to | ?
16. John is six years older than James ; and in five years
John will be 3 times as old as James was 3 years ago. What
are their ages ?
17. A father's age now is 4 times as great as that of his son ;
and 4 years ago it was 6 times as great. What are their ages ?
» 18. A horse sold for $132.50, which was 6 % of the cost
more than the cost to the original owner. What did it cost ?
Suggestion. — 6^ means i^--^. If ic = the cost, then -^^ a? = the gain
and |^^a?= the selling price.
19. A man invests i of his capital at 5% and the remainder
at 6%. Ilis total income is $4080, What is his capital ?
LINEAR EQUATIONS— ONE UNKNOWN NUMBER 175
' 20. 16 is changed into 51 coins. If each coin is either a
quarter or a dime, how many of each are there ?
• 21. A train leaves a station and travels at the rate of 40
miles an hour. Two hours later a second train leaves the
station, and travels in the same direction at the rate of 55
miles an hour. Where will the second train pass the first ?
• 22. A tank is fitted with two pipes. One can empty the
tank in 30 minutes; the other, in 25 minutes. If the tank
is two- thirds full, and both pipes are opened, in what time
will it be emptied ?
< 23. A laborer was hired for 60 days. Each day that he
worked he was to receive 12.25 and board ; and each day that
he was idle he was to receive no pay, but was to be charged
60 cents for his board. At the end of 60 days he received
1106.50. How many days did he work?
, 24. A rectangular field is 6 rods longer than it is wide ; and
if the length and breadth were each 4 rods more, the area
would be 120 square rods more than it is. Find the dimen-
sions of the field.
25. What number mvist be subtracted from each of the four
numbers, 12, 14, 18 and 10, so that the product of the first two
remainders shall equal the product of the last two ?
^^ 26. A man rows down a stream at the rate of 5 miles an
hour, and returns at the rate of 2 miles an hour. He returns
to his starting point in 7 hours. At Avhat rate does the stream
flow? How far down the stream does he go? How fast can
he row in still water ?
27. Find a number such that i of it shall exceed f of it by 9.
28. B has $40 more than A, C has i as much as B, and be-
tween them they have $360. How much has each ?
29. The difterence between the squares of two consecutive
whole numbers is 23. What are the numbers ?
176 ALGEBRA
30. If a certain number be added to 8 and to 11, and the
first sum be divided by the second sum, the quotient will be |.
What is the number ?
31. What sum at 6% simple interest will amount to 1413 in
3 years ?
32. At what rate simple interest will $265 amount to $807.40
in two years ?
33. If linen costs i as much as silk, and I spend $19.25 in
buying 10 yards of silk and 15 yards of linen, find the cost of
each per yard.
34. A room is 2 feet longer than it is wide, and if its length
were increased by 4 feet and width diminished by 3 feet, its
area would not be changed. What are its dimensions ?
35. Two pedestrians started at the same time from points
44| miles apart, one traveling at the rate of 2i miles an hour
and the other at the rate of 2| miles an hour. When and
where did they meet ? /
36. Find the time between 4 and 5 o'clock when the hands
of a clock are together.
Suggestion. — Let x represent the number of minute spaces which the
minute hand lias traveled from 4 o'clock on until it overtook the hour
hand. Then j^ will be the number of spaces which tlie hour hand
has traveled meanwhile. The difference is 20. Why?
37. Find the time between 4 and 5 o'clock when the hands
of a clock are directly opposite each other.
38. John could remove the snow from a walk in 30 minutes.
James could do it in 20 minutes. John began the work, but
later James took his place, and the snow was all removed in
25 minutes from the beginning. How long did John work ?
39. A could dig a trench in 15 days, and B could dig it in 2(
i'
LINEAR EQUATIONS— ONE UNKNOWN NUMBER 177
days. If they worked together, how long would be required to
dig it? ^
« 40. A can do a piece of woi^t in ten days ; but after he has
worked two days, B comes to help him, and together they
finish it in three days. In how many days could B alone
have done the work ?
41. A solved 90% of the problems in an exercise, and B
solved I as many as A. If B had solved 6 more, he would
have solved 70% of all the problems in the exercise. How
many problems in the exercise ?
42. A man made two investments amounting together to
$6250. On the first he gained 6%, and on the last he lost
3%. His net gain was $150. What was the amount of each
investment ?
43. If 12 lbs. of iron weigh 11 lbs. in water, and 20 lbs. of lead
weigh 19 lbs. in water, find the amounts of iron and lead in
a mass which weighs 72 lbs. in air and 68 lbs. in water.
44. In an alloy of gold and silver Aveighing 60 oz., there is
5 oz. of gold. How much silver must be added in order that
10 oz. of the new alloy shall contain but \^ oz. of gold ?
• 45. There is a number of three digits, each less by two than
the one to its right. If the order of the digits is reversed, a
new number is obtained whose value exceeds that of the given
number by 83 times the sum of its digits. Find the number.
Note. — In algebra two general numbers written side by side, as ah, in-
dicate the product of the factors a and b. In the decimal system of
writing definite numbers two numbets so written do not represent a
product. By the place value principle of the decimal notation the value
represented by a figure not only depends upon its shape but upon the
place it occupies. Thus, 37 means 3tens + 7ones, or 3x10+7.
If tens' digit is represented by x and ones' digit by ?/? th^ value of
the number which they represent is lOx+y,
CHAPTER XIII.
LINEAR EQUATIONS-MORE THAN ONE UNKNOWN
NUMBER. SYSTEMS.
121. Indeterminate equations. An equation which contains
two or more general numbers, or unknowns, will be satisfied
by an indefinitely great number of sets of values of these
general numbers. Such an equation is called an indeterminate
equation.
Consider the equation x + y=10. Solving it for a?, we have
x=10—y. Now if different values are assigned to y, as many
corresponding values are obtained for x. Thus, when 1/= 0, a?= 10 ;
when y=l^ x=Q\ when y=2, x=S; when i/= — 4, x=14; when
2/=12, x=—2; etc. Since the number of values we may thus
assign to y is indefinitely great, the number of sets of values of x
and y which satisfy the equation will be indefinitely great.
122. Solutions. In an equation containing two or more un-
known numbers, i. e., an indeterminate equation, the sets of
values of the unknown numbers which satisfy the equation
are called solutions of the equation.
Thus, one solution of y—^x=2 isa?=l, y=5; because for these
values of x and y the equation becomes 5—3=2, an identity.
123. Common solutions of two linear equations with two un-
knowns. Two linear equations which involve the same two
unknown numbers will, in general, have one, and only one,
solution common — i. e., one set of values of the unknown num-
bers which will satisfy both equations.
178
MORE THAN ONE UNKNOWN NUMBER. SYSTEMS 179
This may be illustrated by the following example. Consider
the equations x + y=G and x—y=2. Some of the solutions are :
for 0^ + 2/^6, ^^g^, ^^,^, ^^4^, ^^3^, ^^g[, ^^^
x=6 ) ic= — 7 ) £c= 8 I
fnr^ «-2 ^= ) a?- 1 ) x=2 \ x=3 I jr = 4 | ic=5
it-6 ) ir=7 1 x=8 I
It is seen that of all of the solutions here calculated there is only
one common to both equations, a?=4, y=2.
That this principle is true in general will be shown in
Example 3, § 127.
Two or more linear equations with two unknowns may have
(1) all of their solutions co7nmon ; (2) just one soliftton com-
mon ; or (3) no solution common.
Two such equations having all solutions common are called
equivalent equations. One may be derived from the other by
the use of axioms.
Thus, 2x—y=3 and 4:X=Q + 2y have all solutions common, and
hence are equivalent. The second may be derived from the first
by multiplying both members of the first by 2, then addmg 2y to
both members of this equation.
Two linear equations having no solution common are called
inconsistent equations.
Thus, a? + 22/ =8 and3a?+6«/=5 are inconsistent equations. No
solution of either equation will satisfy the other.
Two linear equations having just one solution common are
called independent equations.^
* Two or more equations which liave common solutions are some-
times called simultaneous equations.
i
180 ALGEBRA
124. Systems of linear equations with two unknowns. A sys-
tem of linear equations with two unknowns is a group of two or
more equations which contain these unknowns.
Thus, Qx + y=5 and x—5y=4: constitute a system.
A solution of a system, such as defined above, consists of a
sohition common to all of the equations in the system.
Thus, a solution of the system < , ^ . is x—4:, ?/=0.
Since a system of two independent linear equations with
two unknown numbers has one solution, such a system is
called a consistent or determinate system.
Three or more linear equations which contain the same two
unknown numbers have, in general, no common solution. A
system of such equations is called an impossible system.
i2x+y=l{), (1)
Thus, in the system < 3a? — y= 5, (2) the only solution common
( x + y= 2, (3)
to (1) and (2) is x=^^ y—^\ and this solution will not satisfj^ (3).
125. Equivalent systems. Two systems of equations which
have the same solutions are called equivalent systems. Since
two equivalent equations have all of their solutions common,
it follows that two systems are equivalent if the equations of
one system are equivalent to the equations of the other system.
In general, two systems of linear equations will be equivalent if
the equations of one are derived from the equations of the other.
For example, the only solution of the system ■! U J^g^^s'
is a?=2, 2/ =3. Adding the corresponding members of (1) and (2),
we get a new equation (3) 3a? +32/= 15. And subtracting the
members of (2) from the corresponding members of (1), we get
another new equation (4) x—y— — l.
Equations (3) and (4) form a new system \ ^Z^,f^},\' which is
( X y — 1,
MORE THAN ONE UNKNOWN NUMBER. SYSTEMS 18l
equivalent to the old system, because its only solution is also
It follows that a system of equations may be solved by
solving an equivalent system.
126. Elimination. A system of two linear equations with
two unknowns is solved by a process called elimination. This
process consists of combining the two equations of the system
so as to obtain a new equation which contains but one unknown
number.
Tliere are three principal methods of elimination : (1) by
addi'ion or subtraction, ('2) by comparison, (3) by substitution.
We now proceed to discuss these three methods of elimina-
tion in solving systems of linear equations with two unknowns.
127. Elimination by addition or subtraction.
Example 1. Solve the system j l^^^^^^l' ^11
Let us first eliminate y.
Multiplying (2) by 4, 12x-{-4y = 36. (3)
Subtracting (1) from (3), 7x=U.
Hence, x=2.
The value of y may now be found in like manner by eliminating
X between equations (1) and (2).
Or, replacing x by its value 2 in equation (1), we have
10 + 42/=22.
Hence, 2/=3.
Let the student see if the solution x=2, y=3 satisfies both
equations of the given system.
Notice that the system (1) and (2) was solved by solving the
system (1) and (3). See §125.
This example illustrates elimination by subtraction.
Example 2. Solve the system ] 4x + 3w= — 3 (2)
We may first eliminate either x or y. Let us eliminate y.
182 ALGEBRA
Multiplying (1) by 3, 21x-6?/=93. (3)
Multiplying (2) by 2, 8x + 6y=-Q. (4)
Adding (3) and (4), 29^=87.
Hence, • x=^.
The value of y may now be found by eliminating x between
equations (1) and (2).
Or, replacing x by its value 3 in equation (2), we have
12 + 3y=-S.
Hence, y= — 5.
The system (1) and (2) was solved by solving the equivalent
system (3) and (4).
This example illustrates elimination by addition.
The method used in the above examples would apply to any
system of linear equations. Hence we have the following rule
for elimination by addition or subtraction :
3fulti2:)ly the memhers of each equation, if necessary, by such
a iiumher as loill make the absolute value of the numerical co-
efficients of one of the unknown numbers the same in both of the
resulting equations. Add or subtract the corresponding members
of the resulting equations.
Note. — In any method of elimination we are concerned witli only the
common solutions of tlie equations. Hence, in the addition or sub-
traction, which is involved in tliis first method of elimination, it is as-
sumed that the unknown numbers have the same values in each
equation of the S3^stem. Eor example, x stands for the same value in
each equation. Otherwise the addition or subtraction would not be
allowable. The same facts apply to every method of elimination.
Example 3. Show that the system of general equations in
X and 7/, ] ^^i^^ZS' has, in general, one and but one solution.
Given \ ax + hy=c, (1)
•. ^'""^^ \dx + ey=f. .(2)
Multiplying (1) by d^adx + bdy=cd. (3)
Multiplying (2) by a, adx + aey=af. (4)
Subtracting (4) from (3), bdy—aey=cd—af. (5)
MORE THAN ONE UNKNOWN NUMBER. SYSTEMS 183
Now (5) is a linear equation in the one unknown number y.
We learned in Chapter XII that, in general, such an equation
has one, and only one, solution.
Solving (5) we get y=M=^e
Similarly, by eliminating ?/, we get
aex—bdx=ce—bf. (6)
/^g Tf-f
Solving (6) we get ^=^^3^^-
Hence, in general, the system has one and but one solution :
^_ ce-bf cd-af
ae—bd' bd—ae ^
Note. — It will be found that if certain relations exist between the
general co efficients, a, b, c, d, e, and/, this conclusion fails ; i. e., the
equations are equivalent or inconsistent.
EXERCISE 57.
Eliminate by addition or subtraction, and solve the following
systems of equations :
l-2a3-f2/=6. ''• \'lxVy=\. "■ l8«=5^-ll.
(2£c + 5?/=15, ^ (4a-36 = l, « ( 8a=a; + 34,
(3a;-4y-ll. ^' (3a-4^ = 6. ^' (6« + 8a;=53.
10.
j7y^^ = 42,
(3y-^-8 = 0.
.. (5a-2^-35 = 0,
/^^' t^> + 4a-25 = 0.
12 n«^+iy = 30, -, (1.5a^-3.7.y = 5.4,
184
ALGEBRA
J. (3^-2^ + 4-0,
.Q (3^ + 2y=4,
^^- t4y-3a; + l=0.
16 f2a3 + y==35,
^^' |5a;-3y = 27.
on f -4a; + 3y = 45,
' ^^- \ 2y + 6^- 4.
.^ (5y-5a^ = 15,
^'- |3£c4-5//-71.
21.^
1 + ^-35,
^i^4 ^'
18. ^ ^ ^
4 + 6-^3-
[| + .==45.
22 n0a^ = 2 + 2y,
Solve for x and i/ :
24 ( aa; + ^>y = a^ + ^^
* Xbx + ai/ = 2ab.
25.
I ax + ai/ = a^ -{- b.
128. Elimination by
comparison.
EXAMPLE]. Solve |t^:^=?^6:
Let us first eliminate x.
(1)
(3)
Solving (1) for x,
^-34-2/
^ 4 •
(3)
Solving (2) for x,
a?=16-42/.
(4)-
Hence, comparing the two values of x given in (3)
and (4),
by Axiom 7,
3^;?/-16 42/.
(5)
Solving (5) for y,
• 2/-3.
Replacing y in (2) by 2, 8 + a?=16,
whence, x=8.
(6)
The above example reveals the following rule for elimina-
tion by comparison :
jSolve each of the two equations for the value of one of the
unknown niimbers^in terms of tJte other ^ and equate the resulting
valim.
MORE THAN ONE UNKNOWN NUMBER. SYSTEMS. 185
Solve for the unknown that gives the simplest expression
from each of the equations.
EXERCISE 58.
Eliminate l)y comparison, and solve the following systems of
equations.
^' \x-y = ^. L 3 4
3£c-y = l, r3m-4/2=10,
2^ + 5y^41. 13. Y'\^'\^n = \l,
1
(^ + 2y-4,
(22/-£c + 12
\^x + y = ll. ^ (3.75.^ + 2.5y = 10.25,
22/-£c + 12 = 0. .- (£c + 2//-3 = 0,
2a-^> = 5,
{
15.
1
I 7a + ^> = 265.
16.
2^3 ''
+ 26-25. l^,y
3"^2
jj j 10.T + 3« = 174,
°- 1 3a3+10«-125. 1^ (14a+66=0,
, . Lo 1 1 6^-46 = 46.
g (5a + 2y-l, ^
( 13« + 8y = ll. Solve for x and y :
10. J \2x—my=n.
L7^2 + ^^^- in C«a^+y==^,
20.
C ax + y
{bx + y-
= 30,
^' ] 3a-2^=25.
5 (3.T + 22/ = 26,
„ (3a-7a;-40 = 0,
'*• |4a-3.T=9.
g (7a;-9y = 13,
^' |5a; + 2y = 10.
3 (4a.— 5y = 26,
^' |3.T-6y-15.
7 U^+iy=i3,
- (3m + 7;?. = 16,
*; |2m + 5w = 13.
g ( 7a^ + 4y = l,
^' t9^ + 4y = 3.
9.
(3ic-
(19a;
11^ = 0,
-19y = 8.
10.
11.
MORE THA.N ONE UNKNOWN NUMBER. SYSTEMS 187
^^- t-4«c + 6y=10.
.« (55a;-15y = 270,
^^' (3l£c+19y = 262.
3 2
.^ j22a;-5y = 213,
2a-7_ 13-.y ^'* |6£c-22/ = 51.
j3 (4.. + 3y=22, 19. i^ + y^ff.
.- (5aj+ll2/=102, „ (3^ + 2.y-42,
^*- \x-Si/ + lQ = 0. '*"• |13^ + 23y = 225.
EXERCISE 60.
In solving the following systems choose the method of elimi-
nation that seems to you best adapted to each particular
system.
(x+i/ = 10, ^ (8a;+6y=10,
(ar-y = 4. **• |5£c + 2y = l.
(8a; + i/ = 60, n (5a3-2y/-63,
^- |7a3-10y/:-9. ^* t2aj + y:-18.
3 (7^+y = 42, 10. -[r^^^^n'
., (i8.-20y^i, 11. {1:^=11^
(1^/— 1^ = 2
l5«-2y = 14. f7^-3y = 26,
„ (21ffl + 86=-66, '■''■ l2x-2y = ^.
"• t.49a-15S=-53.
{
7x-4.i/ = 12, 14. -^ 4 2~^'
Sx-bi/=0. ( 2x-2i/ = W.
188
ALGEBRA.
16. J
-g-+y=18.
Solve for x and y :
^g f«a5 + 6y = l,
|&a?4-ay=l.
jQ S hx-ay = h\
\ax—hy = a^.
20. i«f-%=«'-*',
\^cry—ox=\j.
16 j2y + 79=5.x,
17.
22.
2^ {2px + ^y = 4:p' + q\
\x~'Zy = 1p'-q.
[^3"^4'
6.
130. Systems of fractional equations. Certain systems of
fractional equations may be solved by the methods of this
chapter. Such equations should not be cleared of fractions, for
the resulting equations would not be linear. Also clearing of
fractions would hitroduce new solutions, ?*.e., would give solu-
tions which would not satisfy the original system.
ExAiMPLE 1. Solve
Multiplying (1) by 2,
18_8
a h
'2fi
Subtracting (3) from (2), -^=6.
Multiplying by 5, 26 = 66,
whence.
=4.
(1)
(2)
(3)
81 36
18.
Multiplying (1) by 9,
Multiplying (2) by 2, f + f =30.
(4)
(5)
Adding (4) and (5),
117
=38,
MORE THAN ONE UNKNOWN NUMBER. SYSTEMS 189
Ml]
iltiplying by a,
117= 38a,
whence,
117
^-38-
Example 2. Solve
3x + 22/~^'
^-?-3
(1)
(2)'
9
Multiplying (1) by ^
3 27 45
2x^Sy~ 4'
(3)
Subtracting (2) from
27 2 45
Multiplying by 242/,
whence.
Similarly,
81 + 16 = 270?/-
722/,
EXERCISE 61.
Solve :
'• i i J-i
-« y
4.
^
« + 6-B' ^
15 4_
7. ^
r 1 , 7 _5
4^ + %"8'
1 3 5
^'Ix y'^28^
= 0.
1 3^6 1
6.
J •'■ y
Ix y
8. .
57. + 2Z."^'
*
3. -
^x y
6.
-
X y '
Lk y
9. .
fl-^ =4
10. <
f-l 4- ^--1
l^x'V+y '
Ll + ^ l+y .2-
11. .
^ 4
X-
1
3 9
-1-8-
131. Systems involving three or more unknowns. It is easily
shown by elimination, as in Example 3, § 127, that in general
190 ALGEBRA
three linear equations containing the same three unknowns, or
four linear equations containing the same four unknowns, etc.,
have one solution common, i. 6., one set of values for all of the
unknowns which will satisfy all the equations at once. Hence,
in general, a system of linear equations, in which there are just
as many unknowns as there are equations, will be consistent or
determinate.
If there are more unknowns than equations in the system,
in general the number of solutions of the system is indefinitely
great. Hence, such a system is called an indeterminate system.
C rjf _i_ fj t 2! -—- R
Thus, ) 2x—y + z=^ ^^ indeterminate. Some of its solutions
are ^=1, y=2^ z=3; x=3, y=S^ z=0; a?= — 1, y=l, z=6, etc.
And if there are more equations than unknowns in the
system, in general no solution common to all of the equations
will exist. Hence, the system is an impossible system.
To solve a system of three equations containing three un-
known numbers, such as x, y, and z, we may (1) eliminate any
one of the numbers from any two of the equations ; then (2)
eliminate the same number from one of these equations and
the equation not used. This will give rise to two new equa-
tions which contain only two unknown numbers. These two
equations may then be solved as a new system by the methods
of the preceding sections.
From a system of four equations with four unknown numbers
we can, in like manner, derive a new system of three equations
with three unknown numbers ; and so on.
1 x + 2y + 2z=ll, (1)
Example 1. Solve ■<2x + y + z=7, (2)
{ 3x + 4y + z=U. (3)
By subtraction eliminate x between (1) and (2).
We get Sy + Sz=15. , (4)
By subtraction eliminate x between (1) and (3).
MORE THAN ONE UNKNOWN NUMBER. SYSTEMS 191
We get 2y + 5z=19. (5)
Now by subtraction eliminate y between (4) and (5).
We get 9^=27,
whence, z=3.
Substituting this value of z in equation (5),
we get 2i/ + 15=19,
whence, 2/=^.
Now substituting the values of both y and z in equation (1),
we get x + 4 + 6 = 11,
whence, x=l.
EXERCISE 62.
^7
Solve the following systems :
(x+y + z=9,
1. ■< 2ic+y— 2=0,
( 3ic— y + s=5;
( x—2y + z=Q,
2. ■}x + Sy + 2z = U,
(2x-y + z=lS.
3. } Sx + 4y + 6z^7,
(ic + 2y + 62=4.
(bx + 6y + 7z = S,
4. •} 10a;-i2y + 2l2;=3,
( 15.93— 6y + 14^ = 4.
( Sx + y-z=2,
6. -\x-2y-Sz = Q,
(y + z+l = 0.
(x + y = l,
6. ]y+z=9,
(a; + s=— 6.
C2x + y-z = 7,
7. }y-x=l,
(z-y = l.
^ + ^ + ^=19
8. ^ ^_l-+^ = 6
10 10^6 ^'
L4 + 5 r5""^-
9. ^
10;« + -| + s = 7,
a;.+y-2s = 16,
X y
(2p-2^ + 3r = 10,
10. \ ^} + q-r=b,
' (p-q + 2r=7.
11.
12.
x+y+z + w = 10,
X — y — z-\-w = 0,
2ic + 2/4-3^—10 = 9,
Sx — y-\-z—2w= —4.
^2/9— — r+s = 13,
j(> + 5'— r— s=— 1,
jL>— (Z + r— s=— 5,
^Sp + 2q~r + 2s=-17.
192 ALGEBRA
132. Geometric picture or graph of the equation. We have
seen that an indeterminate equation has an indefinitely large
number of solutions. The relation between these solutions,
which is expressed by the equation, may be more vividly re-
presented by means of the graph of the equation, discussed in
the following sections.
133. Coordinates. The position of a city on the earth's
surface is determined by its longitude and latitude ; i. e., one
can locate the position of the
Y city, if its longitude and lati-
tude are both known. Now by
the method of § 33, longitude
east may be called negative
, longitude, and longitude west
may be called positive longi-
tude. Also' latitude south may
be called negative latitude, and
^ latitude north called positive
^^' • latitude. Thus, we speak of a
certain city as being —120° longitude and +30° latitude.
In like manner, the exact position of a point P (see Fig.
1), in the plane of this j^age may be determined, if we
know its distances and directions from two straight lines JCX
and YY\ drawn at right angles to each other and meeting at
O. We shall represent the distance from P to the line YV
i. e., iVP, by x, called the abscissa of the point P/ and the
distance from P to the line A'X^, i. e., MP, we shall represent
by y, called the ordinate of the point P.
The abscissa x and ordinate y are called the coordinates of
the point, and the lines XJT and YY are called the axes of
coordinates. A'X' is the jr-axis and YY is the /-axis. The
point of intersection is called the origin of coordinates.
M
MORE THAN ONE UNKNOWN NUMBER. SYSTEMS I93
The student will see that the axes XX' and YY' correspond to
the equator and the prime meridian on the earth's surface, and
that the coordinates x aiid y correspond to the longitude and
latitude, respectively, of a place on the earth's surface.
If an abscissa drawn to the rig/U of the y-axis ( YY') is called
positive, and one drawn to the left is called negative ; and if an
ordinate drawn above the £c-axis {XX') is called positive, and
one drawn helow it is called negative ; then the exact position
of the point is known when its coordinates are known.
Thus, to locate a point P (See Fig. 2) whose abscissa is + 1 and
ordinate +2, measure 1 unit to the right of O, then 2 units up
from XX' . This point is usually re-
presented by the symbol (1, 2), or
P(l, 2). Similarly to locate the point
Q{—\, 1), measure \ unit to the left of
O and 1 unit up from XX'. To locate
P(— 1, —2), measure 1 unit to the left
of O, then 2 units down from XX' .
To locate >S(1, —1), measure 1 unit to
the right of O, then 1 unit down
from XX'.
Let the student draw a figure and
locate the following points : (—1, 2);
(3, -2); (1,3); (-2, -3).
Q(-M
H
Rf-A-^J
y'
a^j
X'
s(''-o
Y
Fig. 2.
134. The graph. The graph, or
geometric picture of an equation, is
the line upon which are situated all of those points whose co-
ordinates, represented by x and y, satisfy the equation.
Consider the equation x—y=2. By assigning values to y and
computing the corresponding values of x, we find a few solutions
as follows :
x=2 I x=S I x=4t I x=^ \ x=Q }
y=0\ y=l\ y=2\ y = 3\ y=4.\
x= 1) x= 0) x*=-l ) x=—2)
y=.-l\ 2/=-2i y=^-d\ V = -^V
13
194:
ALGEBRA
Locating the points whose coordinates are these sohitions, we
get a series of points as in Fig. 3. It is seen that these points are
not scattered at random over the page, but all appear to lie upon
one straight line, MN. Hence, the straight line MNis the geo-
metric picture or graph of the equation x—y=2.
r
N
/
A
^v^
J
^.»/
^
\,
(-X2)
f^.^f
\
/jj)
X
^
ii.
/C"/
X'
]^
^
/-n
^
^
(^.-J/
M
-0
I
/
M
/
A
V
It can be shown, by the aid of geometry, that the graph of
every linear equation with two unknown numbers is a straight
line. We shall assume this principle in our work.
Since a straight line can be drawn when two points on it are
known, to draw the graph of a linear equation, we need to find
only two solutions, and locate the two points whose coordinates
are these solutions. Then draw a straight line through these two
points. Thus, to draw the graph of x + 2y=l, we find two solu-
MORE THAN ONE UNKNOWN DUMBER. SYSTEMS 195
tions to be a?=4, 2/=— |, and x= — S, y=2. Locating the cor-
responding points (4, — I) and (—3, 2), and joining them by a
straight line, we have the graph PQ^ Fig. 3. Any other point
whose coordinates consist of a solution of this equation will be
found to lie upon the line PQ.
If, upon the graph P §, we locate the point A, and measure
its coordinates, the coordinates Avill be found to be x= — b,
y='^. These coordinates satisfy the equation a? + 2y = l. Like-
wise, the coordinates of a second point B of the graph P Q are
seen to be 93 = 6, 2/ =— 21. These coordinates also satisfy the
equation x + '2i/ = l. Now in like manner it will be found that
the coordinates of any point whatever of the line P Q will
satisfy the equation x^-2y=l of which P§ is the graph.
It thus appears that not only do all of the points whose co-
ordhiates satisfy a linear equation lie upon a straight line^ called
the graph of the equation ; hut also the coordinates of all points
y^hich lie upon the graph of the equation satisfy the equation.
Note. This principle will be found true of the graph of an equation
of any degree. Accordingly, the graph of an equation is sometimes
called the locus of all points whose coordinates satisfy the equation. In
analytical geometry this principle is a fundamental notion.
135. The graph of a consistent system of equations. The
solutions of each of the two equations of a system in two
unknown numbers are represented by the coordinates of the
points which are situated upon their graphs. Hence their
common solution, the solution of the system, is represented by
the coordinates of a point which is on both graphs ; i. c, the
point where they intersect.
Consider the system \ J^T^—a
( 2^ 3^ s /^y^
10. 1*16327 20. i4. 2//]/ 1?~-
SURDS AND IMAGINARY NUMBERS 215
28-Vi7^.
2y. |/(«_ic)^
30.
31. iA~-'/.
|/ 4«£c^^ 4 Vlax'y + 12cea;y=^ + 4ay\
32. 2i/ 1 ,.
'«• (/j+i-
35
■Vl-l^^.
142. Addition and subtraction of surds. Only surds which,
when reduced to the simplest form, have the same surd factor
can be added or subtracted. Otherwise the addition or sub-
traction can only be indicated. Just as 3<:< + 2a==5«, so
Z\/x-\-^^/x-~b\/x.
Hence, surds vnth a common surd factor are added or sub-
tracted by adding or subtracting the coefficients of the surd fac-
tors and affixing to the residt the common surd factor.
Example 1. Find the value of i/32 + 1/50.
1/32 + 1/50=41/2 + 5.,l/2=(4 + 5) l/2=9l/2.
Example 2. Simplify Va'bc-^Vab''c-\-Vahc'.
Va?bc — vab^c + V abc'^ = a^ V ahc — V V ahc + c^ yabc
= (a''-b'-}-c')yabG.
EXERCISE 66.
Simplify the following :
1. 1/T8-V8 + 1/32: 5. 2v 3 + 1/81-V3.
2. 1/75-1/3-1/12: 6. i/40-i>320-2i/5.
3. i/20 + 1/45-i/M: 7. 4i>}-y}.
4. 1/6 -v"294 + 1/486 + 1/24: 8. 2i/^ + 3i/'^-l/^
7;s
216 ALGEBRA
9. yT^^-2i/2^' + i/W.
10. i/81a'-4i/192a^ + 3i;648al
12. 2|/V^+3v/aFc— i/VFc.
13. i/^-i/^* + V^. 17. l/F+^l/7S-2l/}-
14. 1 T + 3^|_^3. 18. I'S-V i+vV?.
15. i/27i-i/8S+l/64^^ 19. i /20 + 3i/4-2r4.
16. 3|/48-y^r_2|/|. 20. vM + i/f+i/S".
143. Change of order. A surd of a given order may be
changed to an equivalent surd of a different order by the
principle,
]' 0""= \ a".
That is, the value of a surd is not changed by multiphjing or
dimdmg the index of the root and tlie exponent of the poicer by
the same number.
This principle may be established as follows :
Let
Then
Hence,
or
Hence,
Therefore,
Example 1. Express Va^ as a sixth root.
Multiplying index and exponent by 2, we get
\/a^=y'a^\
mx
y a"' = r.
a"^=r"^^
§64.
y a''^=yr'"-%
Axiom 6.
a"=r'"-
l/a«=l/r"',
Axiom 6.
= r.
7/6^=1/ «^.
Axiom 7.
Example 2. Reduce yaWc to a tenth root.
SURDS AND IMAGINARY NUMBERS 217
Multiplying index and exponent by 5, we have
ya^b*c= y {d'h'cf=Va'Wc\
ElXAMPLE 3. Reduce yx^y^z^ to the lowest order.
Dividing index and exponent by 2, their greatest common divi-
sor, gives
y^x'y^^'=y^3(^y^z.
144. Reduction of surds to the same order. Surds of different
orders may be reduced to equivalent surds of the same order
by the principle of § 143. Manifestly the simplest common
order to which surds may be reduced is the least common
multiple of their orders.
Example 1. Reduce Va^b, v'a■'&^ Va^U" to the same order.
The least common multiple of the indices is 12. Reducing
each surd to an equivalent surd of the twelfth order,
ya'b=Va^¥; Vd'b^=ya''b^- ya'b'=Va^b'\
The values of surds may be compared by first reducing
them to equivalent surds of the same order, and comparing the
resulting numbers under the radical signs.
Example 2. Compare the values of i/l5 and y Qo.
1/15 = 1/ W=y 3375 ; V 60= ;/ 60^= 1/ 3600.
Since 3600 is greater than 3375, y 60 is greater thani/15.
145. Multiplication of surds. The product of surds which
are of the same order may be found by the principle established
in § 67; that is,
y^aiyj^yab'.
Surds of different orders whose product is to be found must
first be reduced to surds of the same order.
218 ALGEBRA
Example 1. Find the product of y'2a^, i 3a^ and F 4a.
V2a^=l 8a«; ySa'=V9a*; V 4a=V 16a^
Hence, V2d'V 'Sa^ ■V4a=\/8a'' y 9a* Vl^a^
l/Sa:>-9a*lQa'
= V1152a'^
=2aU/lSa\
Example 2. Multiply 21^5 + 5]/2 by 3V5-i\/2.
2VE+5V2
3|/5-4]/2
6-5 + 15l/l0
- 8t/10-20-2
30+ 71/10-40, or 7l/l0-10.
3 -
Example 3. Express as an entire surd 3y 4.
The coefficient may be expressed in the form of a surd of the
same order as the surd factor.
We have 3 = V 3-^ = 1/27.
Hence, 31^4= \^21^/~4= 1/^27x4= i/^I08.
EXERCISE 67.
Express as surds of the same order :
1- l/a, \^a\ 2. yx\ yx, V'x\ 3. i/5, 1^2, 1^3.
4. ]>7, //I2; 1/2. 5. l/^\ 1^^, l/V^
Which is the greater :
6. 1/5 or v^lT? 7. V 2210, l'/320, or 1^48?
SURDS AND IMAGINARY NUMBERS 219
Express as entire surds :
8. 21^5. 10- V7. 12. i^^.
9. 8i/8. 11. ||/3. 13. «T/X
3^ "• **'. ^y YT-
27a;'
^^- Fl/«- 1^' 2^V 27^
Simplify :
16. i/2v^3v 4. 18. i/61/'4l>'2. 20. i/^T^by/^Tl.
17. 2l^2-3l/5. 19. 1/^51/101/5. 21. i a^i/^^^l
22. i?^«^ + 2a^> + ^2v'^Hl>. 23. (^/2+i/3)(^2-i/3).
24. (l + i/2-i/3)(l + i/2+i/3).
25. (|/e + i/2)(i/6-i/2).
26. (^3 + i^2)(i>'9-i^6 + f'4).
27. (1 + 1/2 + 1/3)^
Simplify by reducing to lower order :
28. y'lM'. 31. jyg^^v". 33 VWy
29. ^2W. ./-. 'V^^^
_____ 32 i/^ 34 ,«/l^^«"^''
30. i/Sia^'^^v^ * K F ' V ^uy^'
35. i/(^. 36. ^s/z?:.
146. Two binomials which contain surds of the second
order and which differ only in the signs of the surd terms are
called conjugate surds.
Thus 1/2 + 1/5 and i/2— 1/5 are conjugate expressions.
Since conjugate surd expressions are of the form a + 6 and
a— 6, the product of a pair of conjugate expressions is a rational
expression. Thus, (a + y'h) x (a— i /&} =«^— 6,
220 ALGEBRA
By grouping terms, polynomials involving surds of the second
order may often be expressed as conjugate expressions.
Thus, l/2 + V^S+V5 and i/2-i/3+i/5 may be written
(l/2+ 1/5) + VS and (V2 + |/5)-i/3.
147. Division of surds. The quotient of one surd by an-
other surd of the same order may be found by using the prin-
ciple established in § 68 ; that i ;,
1/6 ^ b
But the quotient, where the divisor is a surd expression, is
reduced to. its simplest form by first multiplying the dividend
and divisor hy such an exjyression as will make the divisor
rational.
This process is called rationalizing the divisor. The mul-
tiplier used is called the rationalizing factor.
Example 1. Divide 61 2 by v 3.
Multiplying both dividend and divisor by \ 3, we have
6v /2_ 6i/2i/3 ^6| /6_3 ^g
1/3 1 3F3 3
Whe7i the divisor is a binomial involving surds of the second
order ^ the simplest rationalizing factor is its conjugate. Why?
(See § 146.)
Example 2. Simplify 1^g+l^^_ .
2i/3-|-3i/2
The conjugate of 2i/3 + 3l/2 is 2|/3-3l/2.
Multiplying by this, we have
l/2 + i/3^ _ (i/2-n/3)(2i,/3-3i/2)_ __:zi^__ 1 ^/e
2i/3-H3l/2 (2l/3-J-3l/2)(2i/3-3l/2) ~ -6 "^
SURDS AND IMAGINARY NUMBERS 221
When the divisor is a polynomial which involves surds of the
second order, group terms so as to form a binomial, and then
multiply by its conjugate. This result may, likewise, be ex-
pressed as a binomial and multipliedby its conjugate; and so on
until the divisor is rationalized.
Example 3. Divide 2 by 1/5 + t/2— 1/3.
2 _2
V5 + V2- V 3 ~ (1/ 5 + V 2) - 1/3
2(1/5 + 1/2 + 1/3")
(1/5 + 1/2-1/ 3)(l/5 + 1/2 + 1/3)
_ 2(l/5 + i/2j-y3) ^j, l/5 + l/2V|/3
4 + 21/10 ' Z+VlO
_ (l/5 + 1/2 + l/3)(2- l/lO)
(2 + l/l0)(2-l/10)
_ 2 1/3 -3 1/2 -1/ 30
-6
EXERCISE 68.
Simplify :
1 i 6 ^ 11 l/"T+^^ +^
' 1/2* • 1/-2 + 1/3' y(l+x')-x
2 V^ 1/3-1/5 1/^+6 + i/^F^
* 1/3* ' 1/3-1/2* i/^T7>-v/a=^
_ l/'2-i/3 „ 2|/7 + 3i/2 ,„ i/^Ta-i/^
«• ■ ::: • O. =: —- !«•
1/2 + 1/3
1/3-1/5
1/3-1/2
2|/7 + 3i/2
3v/2 + 2i/3
3i/5-5i/3
3|/5 + 5i/3
a''
1/7 _ ■ • 3v/2 + 2i/3 |/ic + 3 + i/a;
4 A. o 3v/5-5i/3 14. ^t^^^ .
T/2 ^' 3|/5 + 5v/r 1 + 1/2 + 1/3
^ 12 a' ,. 1/5-1/3
5. — ^' 10 . 10. — = =: — •
v3 *+l/(6^-a^) v/5-1/3-1
222 ALGEBRA
16.
17.
y'2 + yS + y7
T/2
18.
19.
yx+y
Va
y-
-yz
h
l/a+|/
b-
-yc
y2 + y3 + yb
148. Powers of surds. Powers of surds are obtained by the
use of the following principle :
(,'/iy=l'/V.
To establish this principle, we have
{l^ciy=\ya-{/a-i/a to r factors
= Paaa to r factors § 67.
3>
Example 1. Raise 2ay2a^b^ to the fourth power.
(2aV2a'b'Y=16a*y {2a'by=16a'y IQa'b"^,
149. Roots of surds. Roots of surds maybe found by the
principle,
y{\/a)=ya.
This law follows from the fact that to take the mth root of
the nth root of a is to take one of the m equal factors of one of
the 71 equal factors of a, which is one of the mn equal factors
of a.
Example 1. Find the fifth root of y^a^if.
Example 2. Simplify v^{2a''bySa¥),
First express 2a^b^/3ab^ as an entire surd.
y'{2a'by'3a^== V^{l/T2a^') = y'na'b^
Simplify :
1. (1/2)1
SURDS AND IMAGINARY NUMBERS
EXERCISE 69.
^23
2. iY'lx'yy.
3. (5a;'^i/2^')l
4. {-'lah^/^hy.
6. (^Wby.
7. {Aayya^y.
8. (3«yv^^*)l
12. i7(^/T6).
13. i/'(|^625<).
9. {4:\/d'-by.
10. i^(i/5).
11. iV{i>(^2)}.
14. y\i/21xY)'
15. \/{VM).
16. 1/^(8^4^).
17. 1/ /(a^-^>'^)^ 18. ^ ;>8^.
19. i/ 49icV«'|?'8a^
20. {^{^(f «"')}•
150. Imaginary numbers.* In § 65, even roots of negative
numbers were called imaginary nmnhers^ and it was shown
that these numbers did not belong to the series of numbers
with iMch ice were then acquainted. They constitute a new
series of numbers with some pecidiar properties.
By means of more advanced mathematics any imaginary num-
ber may be expressed in terms of quadratic imaginary numbers,
i. e., square roots of negative numbers. Hence, in this chapter all
of the language shall have reference only to quadratic imaginary
numbers.
For the sake of distinction, all numbers heretofore considered
are called real numbers. Numbers of the form i — ^, where b
is positive, are sometimes called pure imaginary numbers.
Expressions of the form a+|/^^ when a is real and y^^b
a pure imaginary number, are called complex expressions.
* We have seen that the first numbers known were whole
numbers ; that by extending the process of division fractions were
conceived ; that by extending the process of subtraction negative
numbers were conceived ; and tliat by extending the process of evo-
lution surds and finally imaginary numbers were conceived. In the
investigations of higher mathematics the imaginary number plays an
important part.
224 ALGEBRA
151. If a=0, the complex expression a+j/^—b is a pure
imaginary number i^ — b; but if b = instead, the expression
becomes a real number, a. It thus appears that real numbers
and imaginary numbers are both particular kinds of complex
expressions.
The squares of all real numbers are real positive numbers^
and the squares of all pure imaginary n^imbers are real negative
numbers.
Thus, ( + 6)2= + 3G; (-4)2= + 16; ( + l/2)2=+2; (-l/2)2= + 1^4.
And (l/"^2^-3; (-|/^)2=-5; (-4l/^)2=-32.
152. Typical form of imaginary numbers. By the principle
]/«3=]«]/ 5 any imaginary number can be expressed in the
form c]/^^, called the typical form.
Thus, i/-4=i/4(-l) = i/4i/-l=2|/-l.
And -|/-10=-i/10-(-1) = -i/10t/'^1.
Hence, the properties of any imaginary number may be
studied by studying the properties of v^^^, usually denoted
by *, which is taken as the imaginary unit.
Note. — The principle \/ah—\/a\/b does not hold when \/ a and \/b
are both imaginary, i. e., when a and h are both negative. Thus,
l/^}/— 9 does not equal i/36, or 6, but =i/a6(i/=4)2, or -6.
153.
Powers of ]
-1
or /. I
^or
the s
we have
i-
= y -
"1;
By §
64, r' =
-{V-
— 1 \2 =
= -1;
i^--
= {V~
— ~1V =
=i-i'=-
-*.
i'-
= {V~
— 1V =
=i-e=-
~f-
-+1.
i' =
= {V~
~1.\5-
=i-i'=i.
t^--
= (V~
ITi \fi =
=i-i'=i-
^ =
■■i'=-
%'--
=(y~
^1)'^
=i-f>=-
-i.
i^ =
-(!'-
3~1 \H -
^i-i'=-
-i'-
= 4-1.
i^--
-iv~
-1V =
= i-i^ = i.
SURDS AND IMAGINARY NUMBERS 225
Hence, the successim powers of i have only the four different
values: i — i, — i, — \ — i, + 1^ rej^eating in regular order.
Since *"^=— 1, and since ( — 1)^ or to any even power, is 1,
then
e-+^ = i*n.f =-1;
Thus, *25^t*«+' = i; ^l2^^='•*=l; iP=i}-'+^=-i.
154. Addition and subtraction of imaginary numbers.
Imaginary numbers to be added or subtracted must first be
reduced to the tupical fonn^ and then combined as in the addi-
tion and subtraction of surds.
Example 1 . Simphf y 1/ - 4 + V -9 + i/-25-i/ — 16.
= (2 + 3 + 5-4)j
= 6^.
EXERCISE 70.
Reduce to the typical form :
1. V~=m. 5. y-^^. 9. y~=^K
2. V^=ms. 6. i/:=r|f. 10. i/_i6a;y.
3. |/^=^25. 7. 1/^=^7.
_ 11 ,/ 9^"^^
4. 1/-^. 8. |/-98. ' K 49a;y^*
12. |/-81(£c-//)^ 13. i/-a^-2a^-6^
14. |/24a;y-9aj-'-16yl
Simplify :
15. yZIg-^ 114+^/^307 16. 1/-^+]/ =^100"- 1/^=49:
15
226 ALGEBRA
17. i/^:^-|/~4-i/^^6. 18. i/— 8 + ^rrio + |/~7.
19. a|/^=^2+-i/^^ + 3i/^:^.
20. 2xi/—xY-i'Si/y-4x'i/'.
21. Add x + 2i/^/^=ri and 2x~i/i/^^l.
22. Add 10-|/^:^9 and 6 + i/^=^.
23. Add a + b+ ^/-^h and a^-h- V~h.
24. From a + \/^-b take a — \/^^b.
26. From 3aj — 2y\/ — 1 take a; + \/ — 9yl
155. Multiplication of imaginary numbers. Imaginary num-
bers to be multiplied should first be reduced to the typical
form. Then proceed as in tlie multiplication of surds, ob-
taining the product of the factors V^-^\ by use of § 153.
Example 1. Multiply v— 4 by ]/ — 9.
Example 2. Simplify V -16V —25V —i'^.
l/^;^l/^^l/-49=4l/^-5l/^.7l/^ = 140i-^=-140j. ^
Example 3. Multiply i/ — 3 + i/ — 5 by i/— 2— i/— 4.
l/~i + V^=iV^+iV5 ; 1/^-1/^=^/2-2*.
^V34-^l/5
^/2-2i
i' 1/6 + 1' j/ 10 -2P }/S-2i' 1/ 5 =
- 1/6 - 1/10 + 2i/3 + 2i/5.
156. Two complex expressions which differ only in the
signs of the imaginary terms are said to be conjugate.
Thus, a? + 7/i/— 1 and x—yi/—l are conjugate expressions.
The product of tioo conjugate complex expressions is real.
For {x-\-y\/^-V){x-y\/^=\)=x^—y\-V)==x^-Vy\
SURDS AND IMAGINARY NUMBERS 227
15 7. Division of imaginary numbers. Imaginary numbers in
division should first be reduced to the typical form. Then, in
general, multiply both dividend and divisor by such an expres-
sion as will make the divisor real and rational.
Example 1. Divide 6 by l/^^.
Example 2. Divide ]/ —9 by y —25.
9 31/ -1
1-25 51-1
Dividing both terms by 1/ —1, =^
Example 3. Divide 2 + 1^^ by 3 - V-Tx.
2 + l/^ _ 2 + /1/5 ^ ( 2 + n/5)(3 + i]/5 )
3-l/^~3-^l/5~(3-^l/5)(3^-^V5)
_6 + 5/l/5 + 5/^ l + 5zi/5 1 5iVy
- 9_5i-^ - 14 -14"^ 14 •
EXERCISE 71
Simplify :
1. i/=25i/^^. 3. |/^=T00i/-49.
2. i/^=:9]/^=l6. 4. y^^^i/^^i-Oe.
5. i/^=25i/^4i/^^36i/^=49.
6. i/^=^l6i/"=^|/-25i/-81i/-4.
7. -i/=7^V^=^^^ ^' (l + l/^=4)(l-l/^.
8. -i/ir^^(_v/^. 10. (3 + i/"=2)(2-i/^=^).
11. (1/^=2 + 1 /=3)(l/'=^-l/^).
12. (2i/^=^ + 3|/^(3i/^=^-2i/^^).
228 ALGEBRA
13. (^y^^ + S]/^=l.)(2i/^^b-i'^).
14. (1/^=^ + 1/^(1/ -^-l/^.
15. {xy'^-yy^)(x}/^=^+(/i/^^).
— b}/—x^
2i/ —a;
2
^^- 1 + 1/^2'
l + i/^^F
30. ^, -==y
1-V -2
3-|/==~5
^^•sTi;^-
16.
5
1/-1
3
17.
2i/-l
18.
10
1/-4
19.
21
l/=9*
20.
a;
1/-X
91
4,/-l
22.
23.
24.
1/-16
1/-9
1/-100
i/'^2r
— X
y'-9x'
25.
26.
2i/-25x*
3i/ — 16a;^
1/-16
1/-25
07
l/-25a;«
32.
6 + 1/-
-3
3-1/=
1
=^
33.
x+^
5 + 2i/
/ -y
-5
■ |/-4 '^" i/ — 49a;'
34. ' ^ ' -^ ' 35.
1/-5-1/-7 .T-i -2/
158. Geometric representation of imaginary numbers. Ac-
cording to the method of Chapter XIII all real Clumbers may
be represented by distances measured along the line XX'
from the point 0, positive numbers by distances to the right,
and negative numbers by distances to the left.
Let us consider any positive number, say 2, represented by
OA. By revolving OA counter clockwise about the point
it may be made to take the position of OB^ C, OD^ or any
other line drawn through O.
Now, multiplying 2 by y^A gives 2y ^^1. Multiplying
again by v— 1 gives —2, which is represented by OC in
SURDS AND IMAGINARY NUMBERS
229
the figure. Hence, multiplying 2 twice by |/^^1 gives the
same result as revolving OA into the position OC ; i. e., reverses
the direction of the line
' which represents the mul-
tiplicand 2. Since mul-
tiplying twice by i/^
revolves the line OA into
the position C\ multiply-
j^' ing once by |/ — 1 may be
interpreted as revolving
OA through only half of
the distance, i. 6., into,
the position OB. Mul-
tiplying three times would
revolve it into the posi-
tion OD. And multiply-
ing four times would re-
volve it into the position OA again. But multiplying once
gives 2|/ — 1 ; multiplying twice gives —2 ; multiplying three
times gives — 2]/ — 1; multiplying four times gives 2.
Hence, 2|/ — 1 is represented by OB^ and — 2]/^^l is repre-
sented by OD. The same interpretation would hold in case
of any other real number as well as in the case of 2. Hence
the following principle :
If all real numbers are represented hy distances measured
along the line XX' from, 0, positive numbers to the right and
negative numbers to the left,, then all imaginary numbers will
be represented by distances 7neasured from along the line YY'
perpendicular to XX\ those with positive coefficients above 0, and
those with negative coefficients below.
CHAPTER XV.
QUADRATIC EQUATIONS— ONE UNKNOWN NUMBER.
159. Quadratic equations. Any quadratic equation^ or
equation of the second degree^ in an unknown number x must
have terms in x'^, and may liave terms in x, and terms free ofx,
but 710 other kinds of terms.
A quadratic equation that contains a term of t\\Q first degree
in the unknown number is called a CDinplete quadratic equation.
A quadratic equation that does not contain a term of the first
degree in the unknown number is called a pure quadratic equa-
tion.
Thus, 3x^ — 5x=6 is a complete quadratic equation. And
5x^—4=0 is apui^e quadratic equation.
By grouping terms, any complete quadratic equation can be
written in the type form
in which x is the tmknown 7ium,her, and A, 7?, and C are hnovin
numbers having any finite values.
Any pure quadratic equation can be written in the type
form
Ax'^B = 0,
in ichich x is the iinhnovin 7ium,her.
160. To solve a pure quadratic equation. Any pure quadratic
equation can be solved like the following :
Example 1. Solve 23(^ — 12=2>Q-x\
Adding a?^ + 12, 2x^-\- x' = 36 + 12. (Authority ?)
Uniting terms, 3a!^=48.
230
QUADRATIC EQUATIONS— ONE UNKNOWN NUMBER 231
Dividing by 3, x^=16. (Authority ?)
Hence, x must be a number whose square equals 16 ; i. e.,
x=i/T6
=4 or —4.
This may be written x= ± 4.
2
EXx\MPLE 2. Solve <^ + l=~zrTv
Multiplying both members by 'a — I , a'^ — l=2.
Adding 1, 0^=2 + 1.
Uniting terms, a^=3.
Hence a must be a number whose square equals 3 ; i. e.,
a= ± I 3
The solutions, or roots, are yS and — ]/3.
These examples illustrate the following rule :
(1) Clear the equation of fractions^ if necessary, and rentoce
all signs of grouping.
(2) Transform the equation into an equivalent equation
having all the unknown numbers in the first member , and all
terms free of the unknoimi yiumber in the second 'member.
(3) Unite like tertns.
(4) Divide both members by the coefficient of the unJcnown
number .^ thus giving an equation of the form x'^=A, where x is
the unknoimi number.
{5) Extract the square root of both ynembers of tJie resulting
equation.^ attaching the double sign, ±, to the second member.
EXERCISE 73.
Solve for the general number :
1. 7a.'^-28 = 0. ■ 3. r^ = 845-4rl
2, 10.r^-150-4a;l 4. «(rt + 4) = 4a + 49.
232 ALGEBRA
e. Sx' = 2x{x-b) + 10{x + 2).^ ' ^ ^
^ _^ 16. ^-^ + ^J^1.
7.
125 X
3^4 17 3a;^-7 a;-l_a; + l
8. I g^- ' CC^— 1 £C+1 £C— 1
9 ^ -''/-^ 18. ^^-4l=l.
^- ^rn — 6^" 6 "+1
10 ?^=i. 19. r^.-^^r
• 8 6s+l 2 +
11 »^' + 8 o on 4a3 + l_8^-19
11- -2""^^- ^"' 3^=4"7^=24-
.. 9x^-l_3 21 % + ^- 1^
t/l-i. — 3 ^. -61. 2 + y 2y + 3'
/l3. -^^4. 22. 4^+i = l^.
^)^ — 1 5 3a;— 2
J^_l_9 3s + 6 2s + 5
/^
2/ y^ 4* ""• 22-6 32-6'
^ ' a; + 2
25. 4£z!=^. 27. (y + i)(y-4) = 2.
5£e-
-4
4x-
-5'
2a;
1 +
x'
3
5"~
2x-
f3"
-5"
X^ + X-^1 £C- — £C+1
x—l x+l
161. Ill the following sections three methods of solving a
complete quadratic equation are discussed : (1) by factoring ;
(2) by completing the square ; (3) by the use of a formula.
Note. — The factoring method, it will be found, may be used to solve
an equation of any degree higher than the first in whicli rational fac-
tors can be found. But, since comparatively few expressions have
rational factors, the method of solving equations by f^,ctonng may
be used in only a limited number of cases.
QUADRATIC EQUATIONS— ONE UNKNOWN NUMBER 233
162. To solve complete quadratic equations by factoring.
This method is based upon the principle, that, in general^ a
product is zero lohen one^ or tnore^ of its factors is zero^ and
not otherwise.
Thus, in general, ah is if a is 0, or if h is 0. Also if ah is 0,
then either a or 6 must be 0.
Note. — This principle fails if one of the numbers is indefinitely great
at the same time that another factor is 0. See (4) § 219.
For example, the product
8
(^-^>{^)
Now when x=2 the product is equal to |, or 2, and not to 0, although
the factor x — 2 becames 0.
Example 1. Solve x^ + Q = 5x.
Subtracting 5x, x^—5x + Q=0.
Factoring, {x—S){x—2)=0.
This equation is satisfied by a value of x that Avill make either
factor of the first member 0. Such values of x are obtained by
equating each factor to 0, and solving the resulting equations.
Equating each factor to 0, we have
a*— 3=0 ; whence, x=S.
0?— 2=0 ; whence, x=2.
Check. 3' + 6 = 5 • 3 and 2' + 6 = 5 • 2, identities.
Example 2. Solve Qx^=x+2.
Adding -x-2, Qx^-x-2=0.
Factoring, (3j;-2)(2ir 4- 1)=0.
Equating each factor to 0,
3x—2=0 ; whence, a?=|;
2^^+1=0 ; whence, x=—}.
Let the student check the two solutions.
231 ALGEBRA
To solve an equation by factoring we evidently have the
following rule :
(1) Transform the equation into an equivalent equation
having all of its terms in the first member.
(2) Then find the rational factors of the first member.
(3) Equate each of these factors to 0^ and solve the residting
equations.
EXERCISE 73.
Solve by the factoring method :
1. x'-1x^Vl = ^. ^- xll -- r:_4 , 33
2. i«M-3£c-10 = 0. A 1 X ■ XX'
/3. a3^ + 8.T + 7 = 0. j3 14-??=^:. ^^' «' + i«=¥-
x" x' 22
22. 1+--Gy = 0.
6. 2a;^-5£c + 2 = 0. 14. l5a; + 4=-. ^
X 23 6f^2 = 11^ + 7
7. W + 12.^24. 15- -+1^1-22.. 24.17.^^70.-8.
^ 8. hx^=l^Ax. ^^' 6^^ + 7^=-2- ;J5. 12/-7y + l = 0.
* 9. 2Lt^ = 10 + 29«!. ^'''- 4^-^l'=-77. 26. 96 + 22/=.3yl
10. 6.«'''-ll^=2. 18. P^-llP--30.27. 10^-3 = 3^^
11. 3.^^ + 5i« = 2. 19. i«='-3£c-10 = 0. 28. lO-s^-Ss.
29. ;«^- 23a; + 132 = 0. ^. 32^-4 ^ ,,
^ 34. ^ , ,) -=2g4l.
30.6-^^4^ = 0. t'
21 y+3 '^
31. 5r— 8= — . o 1A , ^oA
r „^ 3ic— 10 £c + 120
OD. =-i = .
32. 2/^ = 13y-36. 14 x
«o , , 10 , 1 ^ OP, 1 ^-7 , a; + 24 _
33. .^+-. + -=0. 37. 2-:z2-,rxi + 5^^^=^-
38. -^,-^0=2.
QUADRATIC EQUATIONS-ONE UNKNOWN NUMBER 2S5
163. Completing the sq[iiare. From the identity
a?^ -\-^ax-V a? = {x-{- a) ^,
we have x^^-^ax-^-a^ as the general form of the perfect
square of a binomial. Hence, if we have given the binomial
x^ + 2a£c, to make it a perfect square, we must add a^^ ^^e., the
square of half the coefficient of x. The process of adding a
term to a binomial such as a^^ + 2a.T, in order to form a perfect
square^ is called completing the square.
Example 1. Form the perfect square whose first two terms
are x^ — V)x.
/10\ ^
Here we addf -^\ , ov 25. This gives a?^— 10^ + 25.
Example 2. Complete the square of which two terms are
x} 4- 3a?.
Here half the coefficient of x is |. The square of | is \. Add-
ing I, we have x'^ + 3a? + 1 . a?' + 3a? + 1 = (,r + 1)^
Note. — It would be well here for the student to review § 61.
164. Solving the complete quadratic by completing the square.
Any complete quadratic equation whatever can be solved
by use of the principle of § 163.
Example 1. Solve a?^ + 3x— 10=0.
Adding 10, x"- + 3x- 10.
Adding Q)^ to both members, £C^ + 3a? + |=— .
Extracting. the square root, a? + |=±2-
Solving for a?, '■^=—1 ± I-
Using the + sign, a?= — 1 + |, or 2.
Using the — sign, a?=— |— |, or —5,
Hence the two roots are 2 and —5.
Let the student check the results.
236 ALGEBRA
Example 2. Solve 2x^ — 5x=7.
Dividing by 2, x^—§x=l.
Adding the square of half of |, a?^ — |x + f |=| + ||,=|^.
Extracting the square root, x—^= ± |.
When £c— 1=1, x=l. When ic— f^— |, x= — l.
Check the solution.
Examples. Solve x^—4:X + 5=0.
Adding —5, x^ — 4a?= — 5.
Adding square of half 4, x^ — 4x-\-4^= — l.
Extracting the square root, .r — 2 = ± y/ — 1 .
When ir— 2=1 ^, ir=24- 1/^.
Whenir-2=-i ^, x=2-l ^.
These may be written x=2± ]/ — 1.
Here the two roots^ or solutions, are complex numbers.
Check the solution.
Evidently to solve any quadratic equation we have the fol-
lowing rule :
(/) Reduce the equation to the form Qt?-\-px = q.
(2) Add to each member the square of half the coefficient of x.
(3) Extract the square root of both members of the restdting
equation., attaching the double sign ± to the resulting second
member.
(4) Solve the tico restdting linear equations.
Observe that the double sign might have been placed before the
resulting first member as well. The reason for not doing so may
be seen from the following example.
Let x^=a^.
Extracting the square root of both members, ±_x=±_a, that is
x=-\-a or —a and — a?= +a or —a.
Now if in this second equation we divide by — 1, we have x= —a,
or + a, the values in the first. That is, .t= ± a is the same as —x=
±a; hence we use the double sign before the second member only.
QUADRATIC EQUATIONS— ONE UNKNOWN NUMBER 237
EXERCISE 74.
3. x'
4. x'
Solve by completing the square :
1. x}-%x + ^ = ().
2. x'- + 4:X-=V2,.
7. x'-^x^^.
8. a;^-9£c=22. ^ 4 + a;
9. x'-lbx^U. 31. -Sit-^
10
. X'
12a; + 20 = 0.
11. 2a;2 + i«=l.
12. "Ix'-bx^^.
13. 6a;^ + 6 = 13a;.
14. 4a;2-lla;=3.
16. 10a;2-29£c + 10 = 0.
16. 3a;^ = 17a; + 28.
17. 2a;^ + 3a;=5.
18. f>x'-x=^l.
19. 9ic^ + 3£c + 18 = 0.
20. 12£c^-14a; + 3 = 0.
21. 0^=^ + 80; + 21 = 0.
22. aj^ + 6aj+ll = 0.
23. ^x^-'^Q = Qx,
24. i«^ + 6a; + 25 = 0.
25. 16a;^-96«=1792.
26. ic^-8a;=-15.
27. x' + Ux=-^h.
28. 39icH96 = 51a;='-96.
1 2 18
29.
ic4 1 1— JC 4ic— r
14a; + 40=-0. 5. a;^-lla; + 30 = 0.
6. £c- f 5ic = 14.
5 _8
4:-x 3"
ic=6.
30
36.r-105 = 0.
32.
20,
42|.
33. 6.3+?^:::^=44.
34.
7a.-=7(fc + 3) + 4.
X—^ X+4:
36. ^5-,^7-2i.
36.
37. ,
1
£C+1 X—1
2
= 1.
38. 3
2a!-3 ' a;
1
+ 2 = 0.
39.
40.
icH 2 a;-2'
£C — 2 37 — 3
JK-l
a; + 2 4-£c 7
0.
ic-1 2aj 3
41. (a;-5)^ + (ic-10)^ = 37.
42. ^^ + ^ =0.
43.
if + 9 ' a;-l
• +
x'-\ ' 2a;-2 4'
or THi ^
238 ALGEBRA
44 __i L = _l 45 _A . _1=11
**• £c + 2 x-'l l-x ^''' x-l^x-3 2*
46. {4:i^xy + (x-4y = Q(x-4:)(x + 4).
2.;^-l a;— i Sa;— 10 a;— 4 _ a— 4 11
*'• "^^^£^-2"" a;- 3' *^' 2a3-12 3.t-16~^'
165. Solving a quadratic equation by the formula. Since any
complete quadratic equation may be thrown into the type form
the solutions of this equation will give 7i formula by which the
solutions of any particular quadratic equation may be written
down at once.
Solve Ax^ + Bx + C= by completing the square.
Dividing by A^
-'+a-+3=«-
Adding -^,
Completing the
square,
^., , ^ ^ B'- B^ a
^+yl^ + 4.P 4.4^ A'
,^B , B' B'-4A0
Extractmg square root, ^ + w~i= ±|/ - — j-p
Solvnig for x, ^^~~2A^ 2A — '
/ -B±yB'-4A(y
or ^J x= 271 -•
.p, ,. w- • -i?+i/i5^-4^6^
1 hat IS, one solution is 2~4~~ — ~ '
ifi.fl- -B-rB'-4A0
and the other is 0-3 •
QUADRATIC EQUATIONS— ONE UNKNOWN NUMBER 239
Now, by replacing A, B, and C in this formula by the par-
ticular values which they have in any given equation, the
solutions of any quadratic may be written out. Thus, the long
process of completing the square, etc., in every equation may
be avoided. Hence, the student should master this formula and
tise it in all future VDorh where the quadratic eannot he factored
at sight. He should be able, hoAvever, to solve any quadratic by
completing the square, and in this way to derive the formida.
Example 1. Solve Sa?^— 4ic=15.
Written in the form Ax'^ + Bx-\-C=0, this becomes
Here A=3, 5= -4, C=-15.
„ 4 ±1/16 + 180
Hence, x= = — ^
^4±14
6 •
' 18
Using the + sign, ^~"~6' ^^ ^'
Usmg the — sign, x= g, or — o-
Let the student check the result.
Example 2. Solve 4ic2 + 2=— Sa:*.
Adding 3x, we have 4^?^ + 3a? +2=0.
Here, A=4, 5=3, C=2.
-3 ±1 /9-32
Hence, x= ^-g
_-3±j/^
~ 8
-3 + 1-23 -3-1/-23
= '-^ or '
Let the student check the result.
8 ""^ 8
240 ALGEBRA
EXERCISE 75.
Solve by use of the formula :
1. a;''-3a;-10. 8. Qx' + Q = Ux. J,5, 2b'-7b + Q = 0.
2. aj'' + 10ic + 21 = 0. 9. 10x' + U = ^9x. 16. W = Qk-b.
3. 2x'-i^x=Q. 10. 8£c=^-65£c+8 = 0. ^^ 21b'-b = 4:
4. ^x' + bx=2} 11. 2«^ + 3a + 4 = 0.
1ft q^ o'2z=fi
5. 6i«^ + a;-5 = 0. 12. ba'-2a + 7 = 0. / ' '
6. 7a;^=50a!-7. 13. 15y^-y + 3 = 0. l^. 12 + a; = lla;l
7. 4x'-}:x=b. 14. y'-8y + 4 = 0. 20. l-a^-Sa^O.
21 ?^^±I
y 2
22. — ^~^ =0.
1-t 3
23. i— i =2.
24 ^-l_.9-+l_i
26. l+v = li
26. r + 3 = -^.
7*— 1
27. #-i=3-i.
28 2^ + 3 _ «^ + 3
• x-2 2x-r
30. (3a;-l)2-7i«.
^, ^„ 3a;^-4
3^- «^+2=2^-q^r
32.
(5.T + 2)2 -(2cc- 3)^ = 0,
33.
l-(2x-iy = 0.
34.
10v' = bv.
35.
Qx-m=^x\
36.
1-1-221
37.
2a! + | = 2iK^
38.
^x' = 42-6x.
39.
■+"m-
>40.
„=j-..
41.
a-a' = 0.
42.
3 _ 6 8
x—4 x — b x — S'
43.
x-2 x + 2 10a;-8
x + 2 ' x-2 4-x'
44.
, l + 2a;2
1— CC = -v— .
5 + a;
0.
QUADRATIC EQUATIONS— ONE UNKNOWN NUMBER 241
45 2 l-xx+l -g 8 + a; 4-2a; _8
5(a;-2) a; + 2 a;^-4' * 8-ic + 4 + 2a;~3'
-c 7 — 3£c , 7-£c 2x + S ^^ ^ Sx-S ^ , dx-6
40. -j — 0-4-0 — TTJ^^'^i rt- 50. ox— ii = 2x-{- — o —
4— 3a; 2a; + 2 Ax— 2 x—S 2
47 1 , y _ y + 3 51 6 2
^^- g + 4(^) = 2^- ^^- ^^-^=14-
2^4
Hmf. — Consider (x — 1) the unknown and this is a quadratic in
166. The discriminant. In the use of the formula of § 165,
x= k-i 5 it was observed that the character of tUe
2A
roots depended upon the part under the radical sign. This
quantity, B'^—AAC, for that reason, is called the discriminant of
the quadratic*
Observing the formula, we see that
(i) When B'—JfAG is a perfect square, the roots are real-,
rational, and unequal.
(2) When B'—Jf^AC is equal to zero, the roots are equal.
{3) When B-—j^AC is positive but not a perfect square the
roots are real and conjugate surds.
(4) When B^ — J^AG is negative the roots are conjugate com-
plex numbers^
* It is assumed in principles (1), (2), (3), and (4) that the co-
efficients ^, J5, and Cof the general equation Ax'^-{-Bx^rC=^ are all
real numbers. The principles may or may not be true when one or
more of the coefficients are imaginary.
16
242 ALGEBRA
From these observations we can determine the nature of the
roots of any quadratic equation without solving it.
Example 1. Determine the nature of the roots of a?^—5a^ + 6=0.
Here-B= — 5, A=l and C=Q ; hence, B'— 4AC=1, a perfect
square; hence, the roots are real, rational, and unequal.
Example 2. Determine the nature of the roots of x^-\-x-i-3=0.
Here JB^— 4AC= — 11, hence the roots are conjugate complex
numbers.
Without solving tell the nature of the roots of Examples 1 to
20 in Exercise 76.
167. The relation of the roots to the coefficients. By dividing
both members of the general quadratic, Ax^-}'JSx+ 0=0, by A,
B C
we have x^ + -ix-\--j=0, which is of the form x}+px + q = 0,
where ^ and q may have any finite values, integral or fractional.
Solving 9?H7>a3 + 5' = by the formula Ave have
_ —p ± y'jf — 4:q
X 2 •
Calling one root r^ and the other r^, we have
_—p + l/2f—4q
and r.=^I^PJZ^,
Adding these we have
r^ + r^=—p.
Multiplying, we have
.^^^^ Tzi^+vVzlil X r -i^-ri^'-4( y1
p'-(p'-4:q)
4
QUADRATIC EQUATIONS— ONE UNKNOWN NUMBER ^43
From these relations we can state the following principles :
When an equation is throicn into the form x^-\- px-\-q = ;
(1) The sum of the roots is the coefficient of x with its sign
changed ;
{2) The product of the roots is tlie term free from x*
From these relations we may form equations with given
roots.
For example, form an equation whose roots are 5 and 3.
Since the coefficient of a? is — (5 + 3) and the term free from x is
3 • 5 , the required equation is x^ — 8.r + 1 5 = .
EXERCISE 76.
Write the quadratic equations whose roots are :
1. 6, -5. ^ _3^ 6. 2, -f.
2. -2, 7. * -^' 7. -4, -6.
3. \. 5. i, -|. 8. -f, -4.
9. i, 1. 10. -i -i
11. By the use of principles 1 and 2, § 167, show what the
signs of the roots of ic^ — 11.^ + 24 = must be.
12. What are the signs of the roots of £c' + 5a;— 24==0? Of
a;'^-7i^-18 = 0?
13. For what value of a will the equation aic^ + 3.7;— 5 =
have equal roots ?
14. For what values of m will the equation 2a;^ + mic + 32 =
have equal roots ?
15. For what values of c will 3a;'^— 2£c + c=0 have real roots ?
For what value of c will the roots be equal ?
* The term of an equation free from the unknown is called the
absolute term.
-2h±y4b' + 16x
-8
-2b±2]/b'+4x
-8
24:4 ALGEBRA
168. Solving a quadrate formula. In § 119 we solved some
formulae which were linear with respect to certain general nmn-
bers involved. By the methods of the preceding sections,
formulae which are of the second degree with respect to a
certain general number may now be solved for that number.
Example 1. Solve x=2a{2a—h) for a.
Removing the sign of grouping, x=4a^—2ah.
Arranging in type form, — 4a^ + 2a6-f a?=0.
Here A=-4, B=2b, C=x.
Hence, by § 165, a=
Example 2. Solve 1=-^ for f .
Multiplying by c^ Ic^—af^.
Hence, —at^=—lc^.
Dividing by —a, ^^=77'
Hence, f=±l/— , or ±c|/Z, or ±^y^W,
^ a f^ a a^
EXEBCISE 77.
1. ax'^—c=Q. Solve for x.
. 2. aW-a'=^. Solve for h.
3. ah = a'^—A. Solve for a.
4. 2ic^ — 3a' = bax. Solve for a;, and for a.
5. x'^^-2a^ = Mx. Solve for aj, and for a.
6. a3^ + 2«£c=^- + 2«A. Solve for 5, and for x.
7. £c' + ra; + s = 0. Solve for a;.
8. mx^-\-nx^t=^. Solve for aj.
QUADRATIC EQUATIONS— ONE UNKNOWN NUMBER 245
9. a?+-=aH— Solve for a.
' a X
The following formulse express important laws of physics.
10. 5=^/7^. Solve for ^. 14. i^=-^. Solve for f?.
11. /='^1 Solve for t. 16. E=\mv'. Solve for v.
12. F=^ — . Solve for v. 16. r=h-^. Solve for d.
r d-
13. F=^. Solve for v. 17. s^vt +\gt\ Solve for t.
18. ?i^=-T2--T. Solve for ,9, /, and n.
169. Problems which lead to quadratic equations.
Some problems can be solved by the use of quadratic equa-
tions. It has been seen in § 166 that the solutions, or roots, of
a quadratic equation may be whole numbers or fractions,
positive or negative, rational numbers or surds, real numbers
or imaginary numbers, depending upon the relation among the
coefficients. A problem may by its nature require for its
solution a certain kind of number. For example, a certain
problem might require for its solution real numbers. Now, if,
by solving the equation, imaginary solutions are obtained, they
must be discarded. Plence, so7ne of the solutions of ayi equation
may satisfy the equation and yet not satisfy all of the requirements
of the problem. It is necessary, therefore, in solving problems
by the use of equations, to examine the solutions to see if all of
them satisfy the requirements of the problems.
Example 1. Sixty-four times the number of students in a
class exceeds 3 times the square of the number by 21. Find the
number of students.
246 ALGEBRA
Let x= number of students.
Then Ux=3x^ + 21.
Solving, we obtain x=21, or x=^.
Evidently the problem requires for answer a ivhole number.
Hence, the solution x=:^ must be discarded ; and the required
number of students is 21.
Example 2. A man walks 25 miles at a uniform rate. If he
had walked f of a mile per hour faster, the journey would have
taken 2 hours less. Find the rate of his walking.
Let x= number of miles traveled per hour.
Then — = number of hours required for the journey.
25
^nd ——5— number of hours the journey would have re-
x-j- ^
quired had he walked | mile per hour faster.
25 25
Hence, ^=;F4:^ + 2.
. Solving, we get x=2l, or — 3^.
The rate must be an arithmetical number. Hence, the solution
a?=— 3| must be discarded ; and the required rate is 2^ miles per
hour.
Example 3. Find the real number whose square increased by
32 equals 8 times the number.
Let x= the number.
Then, £cH32=8iC.
Solving, we get
a?=4 + 4l/^ and 4— 4|/^.
These expressions are not real.
Hence the problem is impossible.
QUADRATIC EQUATIONS— ONE UNKNOWN NUMBER 247
EXERCISE 78.
By the use of quadratic equations solve the following prob-
lems.
1. Find two arithmetical numbers whose difference is 7 and
product 330.
2. Divide 31 into two parts the sum of the squares of which
is 541.
3. Find two arithmetical numbers whose sum is 50 and
product 336.
4. One of two numbers exceeds 25 by as much as 25 exceeds
the other, and their product is 561. What are the numbers ?
5. Find two consecutive integers whose product is 5852.
6. Find two consecutive whole numbers the sum of whose
squares is 313.
7. The difference between two numbers is 10, and the sum of
their squares is 212. Find the numbers.
8. The denominator of a certain fraction is 5 more than the
numerator. If the fraction be added to the fraction inverted,
the sum will be 2J|. Find the fraction.
9. If 64 be divided by a certain number, and the same num-
ber be divided by 2, the second quotient will exceed the first
quotient by 4. What is the number ?
10. A pupil was to divide 12 by a certain number, but by
mistake he subtracted the number from 12. His result was 5
too great. Find the number.
11. One-half a number plus the square of the number is 68.
What is the number ?
12. The side of one square exceeds that of another by 3
inches, and its area exceeds twice the area of the other by 17
square inches. Find the lengths of their sides,
248 ALGEBRA
13. A rectangular field is 96 feet longer than it is wide, and
it contains 298,000 square feet. What are its dimensions ?
14. One side of a rectangle is 4 inches longer than the other,
and its diagonal is 20 inches. How long are the sides ?
15. A lawn 25 feet wide and 40 feet long has a brick walk
of uniform width around it. The area of the walk is 750
square feet. Find its width.
16. There are two square lots ; the side of one is 42 feet
longer than the side of the other. The two together contain
2146 square yards. What are their dimensions ?
17. A floor can be paved with 200 square tiles of a certain
size ; if each tile were one inch shorter each way, it would re-
quire 288 tiles. Find the size of each tile.
18. The printed portion of the page of a book is 4 inches
wide and 6 inches long. How wide must the margin be in
order that the whole page shall contain 48 square inches ?
19. In the center of a rectangular room is a rug 9 feet by 12
feet ; around this is a border of uniform width. The area of
the floor is 208 square feet. What is the width of the border ?
20. The length of a rectangle is 6 inches greater than its
width ; and if its width be doubled and its length diminished by
3 inches, the area will be increased by 36 square inches. What
are the dimensions ?
21. The perimeter of a rectangular field is 184 rods, and the
field contains 12 A. What are its dimensions ?
22. Two men start at the same time from the intersection of
two roads, one driving south at the rate of 3 miles an hour,
and the other west at the rate of 4 miles an hour. In how
many hours will they be 25 miles apart ?
23. Two trains are 100 miles apart on perpendicular roads,
and are running toward the same crossing. One train runs 10
QUADRATIC EQUATIONS— ONE UNKNOWN NUMBER 249
miles an hour faster than the other. At what rates must they
run if they both reach the crossing in 2 hours ?
24. There are two numbers whose difference is 4. If 240
be divided by each of them, the difference between tlie quo-
tients will be 10. What are the numbers ?
25. There are two arithmetical numbers whose sum is 33.
If 36 be divided by the smaller number, and the larger number
be divided by 4, the sum of the quotients will be 10. Find the
numbers.
26. A man bought a certain number of books for $7.50. If
he had paid 25 cents apiece more for them, he would have ob-
tained 1 fewer for the same money. How many did he buy ?
27. A man finds that by increasing his speed 1 mile an hour
it takes 6 hours less to walk 36 miles. How fast does he
walk ?
28. A grocer paid $2.70 for eggs. Pie found that if he had
paid 3 cents less for a dozen, he would have received 3 dozen
more for the same sum. Find the price per dozen and the
number of dozen.
29. A merchant paid 12160 for some carriages, all of the
same price. By selling all but 2 of them at a profit of $36
each, he received the amount he paid for all of them. How
many did he buy ?
30. A man sold a lot for $1125, thereby gaining i as many
per cent as the lot cost him dollars. What did the lot cost ?
31. In a certain number of two digits the tens' digit exceeds
the ones' digit by 2, and when the number is divided by the
sum of its digits, the quotient exceeds twice the ones' digit by
3. Find the number.
32. A tree was broken over by a storm so tliat the top
touched the ground 50 feet from the foot of the stump. The
250 ALGEBRA
stump was | of the height of the tree. What was the height
of the tree ?
33. A tree which stood at the edge of the bank of a stream
fell with its top in the water. The tree was 60 feet high, the
\^ bank on which it stood was 15 feet above the water level, and
the body of the tree passed under the surface of the water at a
point 20 feet from the bank. What portion of the tree was
under water?
34. The hypotenuse of a right triangle is 4 inches longer
than one leg and two inches longer than the other. Find the
sides of the triangle.
-y 36. One pipe can fill a cistern in 6 minutes less time than is
'required for another pipe to fill it. The two together can fill
it in 10|^ minutes. Find the time required for each pipe alone
to fill the cistern. ^ '^"^ . \ ' ^- "^
36. One of two pipes can fill a tank in 28 minutes, and the
time required for the other pipe to fill it is 19| minutes longer
than is required for the two pipes together to fill it. Find the
time required for the two pipes together to fill the tank. -^ >
37. It would take B four days longer to do a piece of work
than it would take A to do it. The two together could do it
in 6|^ days. In how many days could A alone do the work ?
38. With John's help Henry could remove the snow from a
piece of side- walk in 6^ minutes less time than would be re-
quired for Henry to do it alone. John could do it alone in 21
minutes. In how many minutes could Henry do it alone.
39. A could do a piece of work in 2 days less time than
would be required for B to do it. A works 4 days, then
leaves; and B takes his place and finishes the work in 5
days more. In what time could each one do it alone.
40. The sum of two numbers is 24, and the quotient of the
QUADRATIC EQUATIONS— ONE UNKNOWN NUMBER 251
less divided by the greater is f| of the quotient of the greater
divided by the less. Find the numbers.
41. Two trains on the same road start at the same time from
stations 225 miles apart. One takes | minutes longer than the
other to run a mile, and they meet in 3 hours. Find the
speed of each train ?
42. A teamster having 12 miles to drive increased his speed
one mile an hour after he had traveled two miles. He thus
finished the distance in half an hour less time than it would
have taken had he not increased his speed. How long did it
take to drive the 12 miles ? 7^0
Suggestion. — Let his first rate be x miles an hour.
43. A merchant bought a certain number of mirrors for 1 36,
and, after breaking one, sold the rest for 50 cents apiece
more than they cost him, thus making $2.50 by the transaction.
How many did he buy ?
. 44. The rate at which a man can row in still water is twice
the rate of the current of a river. He rows 6 miles down the
river and back in 4 times as many hours as he could row miles
per hour in still water. Find the rate of the current and the
rate of his rowing in still water.
45. The number of square inches in the area of a square ex- ry
ceeds the number of inches in its perimeter by 32. What is
its area ?
46. The side of a square is the same as the diameter of a
circle. The area of the square exceeds that of the circle by ap-
proximately 3.4336 square inches. Find the diameter of the
circle.
47. The base of a triangle exceeds its altitude bj^ 2 inches,
and its area is 112 square inches. Find its altitude.
48. An engraving whose length was twice its width was so
i-
U
252 ALGEBRA
mounted on Bristol board as to have a margin 3 inches wide,
and equal in area to the engraving, lacking 36 square inches.
Find the width of the engraving.
49. A certain number consists of two digits whose sum is
12 ; and the product of the two digits plus 16 is equal to the
number expressed by the digits in reverse order. What is the
number ?
50. The glass of a mirror in the shape of a rectangle whose
length Avas twice its breadth cost $1.25 a square foot. The
frame, measured on the inside, cost 10.75 a linear foot. If the
glass cost $22 more than the frame, what were the dimensions
of the glass ?
61. A certain farm is a rectangle, whose length is twice its
breadth; If it were 20 rods longer and 24 rods wider, its
area would be doubled. How many acres does the farm con-
tain ?
52. A square lot has a gravel- walk around it. The side
of the lot lacks one yard of being six times the breadth
of the gravel-walk, and the number of square yards in the walk
exceeds the number of yards in the perimeter of the lot by
340. Find the area of the lot and width of the walk.
CHAPTER XVI.
HIGHER EQUATIONS. EQUATIONS INVOLVING
SURDS.— ONE UNKNOWN NUMBER.
1 70. It is not within the limits of this work to consider
general equations of degree higher than the second^ but some
special equations of higher degree may he thrown into a
quadratic form^ i. e., we may consider tlie unknown as being a
power of some single unknown, or as an expression involving
an unknown.
Thus, x^ + 3a?^ + 8=0 is a quadratic in iir^, for if we let J(^=y, an
unknown, we have y'^ + Sy + 8=0, a quadratic in y.
{x' + 3x—2y + 2{x^ + 3x—2) — 3=0 is a quadratic in x' + 3x—2.
Letting y=x'^ + 2x—2, we have y^ + 2y—o=0, a quadratic in y.
Such equations may be solved as quadratics.
Example 1. Solve x^-13x'2 + 36=0.
This equation has the form of a quadratic in x^. Solving
for x^
Now solving
Solving
Hence,
Example 2. Solve
_13±T 169-
Ml
x- 2
13±5
- 2
= 9 or 4.
x'^d, *
x=±3.
x^=4,
x=±2.
x=S, -3, 2, or
-2.
x^ x + 1 ,
X+1+" x^ -^
253
254 ALGEBRA
Here the second fraction is the first fraction inverted.
Putting y for -— -^, the equation becomes
Multiplying hj y, y''—3y + 2=0.
Factoring, (y—2){y-l)=0.
Hence, y=2 or 1.
Therefore the given equation is equivalent to the equations
-— -7=2 and — r^=l.
x+1 x+1
x^
Solving ^+1""^' ^=^ ± y^-
Solving ^^=1, .=1^.
Example 3. Solve 2x^ + 2x + l ^^
x^ + x'
This may be written 2{x'^ + x) + l= .^ .
X -\- X
Multiplying by x"^ + x, 2{x^ -{-xY-\-{x^ + x) — \Q=Q.
This is a quadratic in x^ + x. Solving for x^-\-x,
x^ + ^.= -l±l/l + 80
4
=2 or -f.
Hence, the given equation is equivalent to the two equations
x^ + x=2 and x^-{-i
Solving x^ + x=2, x=l or —2.
Solving x'^ + x=—^,x= 2~^ *
Note. — In every equation which we have solved in this book, the
number of roots, or solutions, is equal to the degree of the equation. In
the Theory of Equations it is proved that this is true in general. A
linear equation has one root, a quadratic tico roots, a cubic three roots,
etc. The student should see that in every equation he gets a number
of roots equal to the degree of the equation.
HIGHER EQUATIONS. EQUATIONS INVOLVING SURDS 255
171. Equations of higher degree than the second can some-
times be solved by factoring.
Example 1. Solve xHl=0. (1)
Factoring, {x+l){x^—x+l)=0.
Hence (1) is equivalent to a:? + 1=0 and x^—x-\-l=0.
Solving a? + 1=0, x= — l.
Solving x^-x + l=0, x=^±^]/'^.
Example 2. Solve x^-U=0. (1)
Factoring, (x + 2){x- 2) {x^ + 2x+4)(x'-2x-{-4)=0.
Hence, (1) is equivalent to
i»+2=0, a?— 2=0, x^ + 2x + 4:=0, and x^—2x + 4=0.
Solving each of these, x=—2, 2, — 1 ± j/— 3, or 1 ± ]/— 3.
Examples. Solve ir' + 5a?=6a?^— 12. (1)
Adding -6.^2 + 12, ar'-6icH5if + 12=0.
Factoring by the remainder theorem,
{x-3){x + l)(x-4)=0.
Hence, (1) is equivalent to a?— 3=0, a7+l=0 and a?— 4=0.
The solutions of these equations are 3, —1, and 4, respectively.
Solve the following :
1. i«*-16-0.
2. x*-2x' = l^,
3. 2a;*-5a;^-12 = 0.
4. Q^ + 2x'=x + 2,
5. ar*-a;'-4a; + 4 = 0.
6. a;^-2a;^-5a; + 6 = 0.
11. 1--X
.X
EXERCISE
! 79.
7.
1 .x'^-l
x^'^ 6 "
= 1.
8.
0.^ + 1=3.
9.
(a;^-iy + 24=
= ll(a;''-
-!)•
10.
{x^^tcy-
-2(a;^ + 2a;):
-3.
-11(
n ^
X ,
1+30 = 0.
256
ALGEBRA
1^ l4s + =^'-?'=B^
17. x' = \.
18. a;*-81 = 0.
19. i««-l = 0.
j4 2x + l «' _„
20. .T^ + 1 = 0.
^*' x' ' 2a! + l ^•
15. .^ + 1-^4,.
16..' + . 1-/^^.
21. £c^-256-0.
22. i««-16 = 0.
23. ;«'' + l = 0.
24. (Saj'^ + Ga;)^
= 1
-(.?.''^ + 3a^ + 2).
25. {x^-x + 4.)
^ +
{x'-x) = 2.
26. (. + ?)' +
('
4)=-
27. a3H-a^ + -4-
X
1
= 4.
Suggestion, x'
«-|.=("S'-
28. a;* + 3ic^-2ic^-3a;+l = 0.
Suggestion. Divide by x^ and arrange as in Ex. 27.
29. x'-%x^^^^x^-Zx-\-l = ^.
172. Equations which involve surds. To solve equations
which are *>ra^/o?ia^ with respect to the unknown number, it is
necessary to free the unknown number from the radical sign.
To do this both members of the equation must be raised to the
same power. (Ax. 5.)
The following rule may be used in general :
(1). Transform the equation into an equivalent one in v^hich one
member consists of a single surd. This is called isolating the surd.
(2). Raise both members to such a power as will free this iso-
lated surd of the radical sign.
(3). If surds still remain., repeat the operation.
HIGHER EQUATIONS.— EQUATIONS INVOLVING SURDS 257
173. Introductionof new solutions.
In general^ icheri both members of an equation are raised to
a higher power ^ new solutions are introduced.
For, let any equation be represented by
A = B.
Then, squaring both members,
A' = B\ or A'-B' = 0.
Factoring, {A-B){A + B)=0,
This is equivalent to A — B = ^ and A^^B = 0.
Of these two equations A—B=0 alone is equivalent to the
given equation. Hence, all of the solutions of /l+B = are
introduced by squaring.
It is necessary, therefore, in solving equations which involve
surds, to examiyie all solutions and discard those which do not
satisfy the equation in its original foi'm.
Example 1. Solve |/ 2^—3 = 5.
Squaring, 2x—S=25.
Whence 2x=28.
And ir=14.
The solution 14 satisfies the original equation, and therefore
was not introduced.
Example 2. Solve ^ + y'x-\-7=x.
Isolating the surd, \/x + 7 = a? — 5.
Squaring, x+7=x'^ — 10a:j + 25.
Or x'—llx-^l^^O.
Factoring, {x—^){x—2)=0.
Therefore, ic=9 or 2.
Now 9 satisfies the given equation, but 2 does not. Solution 2
was introduced by squaring.
17
258 ALGEBRA
Example 3 . Solv^e y 2ic -f 8 + 2 + 2 ]/a? + 5 = .
Isolating second surd, i/2x + 8-{-2= — 2yx-{-5.
Squaring, 2x + 8 + 4: + 4^/2x + S=4x + 20.
Therefore, 2\/2x-\-8=x + 4..
Squaring again, 8aj+32=ic'' + 8^7+16.
Or, 0^2^16.
Whence, 07=4 or —4.
Now neither 4 nor —4 will satisfy the given equation. Both
solutions were introduced by squaring. Hence the equation
has no solutions ; it is an impossible equation.
Example 4. Solve 3ic'— 4ic+]/3a;'— 4a;— 6 = 18.
Equations such as this, where the unknown enters alike in the
part free from radicals and under the radical, may be solved as a
quadratic.
By adding —6, we have 3a?^— 4a?— 6 + ]/3a?''— 4ic— 6=12, a qua-
dratic in y2,x^ — 4:X—%.
l±l/l + 48
Hence, ]/3x^— 4a7— 6 = -^ = 3 or —4.
The solutions of |/3x^— 4£C— 6=3 are 3 and — |.
And |/3ic^— 4ic— 6=— 4 will be found to be an impossible equa-
tion.
Hence the required solutions are 3 and — |.
Example 5. Show that 1 has three cube roots, and find their
values.
We are to find what numbers cubed will equal one. If we let
X stand for the cube root of 1, then
a^=l.
Adding -1, a^-l=0.
Factoring, {x—l){x^ + x-\-l)=0.
Therefore, a;— 1=0, 3(^ + x+l=0.
The solution of a?— 1=0 is 1.
The solutions of a;Ha?+l=0 are — | + ^i/— 3 and — |— ij/— 3.
HIGHER EQUATIONS.— EQUATIONS INVOLVING SURDS 259
Hence, there are three cube roots of 1, one a real number and
the other two complex expressions ; viz. :
1, — 2 + ^1/— 3, —\ ^]/ — 3.
EXERCISE 80.
Solve the following :
1. i/aj+l==2. 4. 4i/ic + 5 = 3]/3i«4-4.
2. 8-|/a;-l = 6. 5. 5i/ic + 2-3i/4ic-3 = 0.
3. 6 = 10-2i/5a;-l. 6. iri3 + «=2.
7. f/3^M=T6-if2^M=3^T20 = 0.
8. i/15 + 3a; + 2 = i/23-a-. 17. yx + 2-l = \/x-^.
9. 2i/9T^=3i/^T24-10. 18. ^=-f2x-^.
10. Vx' + ^x-r^=x + l. 1^' V^=^=J-^'
20. i/aj+i/32 + iK = 16.
11. cc— 3 = i/2i«"'— ic + lO — 5.
12. i/5iK + 4-i /12ic + 21.
21. i/£c + 4=-2 + |/a;.
13. \/x=yx + 'lb — l.
22. \/x+Q = vl2 + x,
23. ^/x-S2 = lQ-i/x.
14. i/«.-12 = i/-«-2. 24 1/^^3-1/^+12::=-^
15. 1/^-7 + 1/^+7 = 0. 25. a!+i/^T5 = 5.
16. 2j/^+2 = i/^. 26. 1/2^^ + 1/2^+9 = 8.
Ql
27. i/3a; + i/3a; + 13
28. i/aj-i/a;-3-^-
l/3a; + 13
2
29. i/a;^-3a; + 5-i/a;^-5ic-2 = l.
30. ■~T~ = \/x+i/x-l, 31. i/cc + i/ic-6 =
l/aj •^-•-- - -"• ^^•^^'/^-"-l/-^:=r6-
260 ALGEBRA
32. x'-Sx + 1 + 2yx'-^x=4. 39. y^^:f^j^^y^:f^ = Q^
40 ^—= — =2.
34. i/i« + 2-5--i/ic-3. ' yl + x-yl-x
Vx-\ _ X 41. :^— =__!—._ 8 _^
00. ■ ,— I -, — riTT' 1 — |/£c 1 + 1 /cc 1— .«
y'x + 1 iG — 1
y'x+\ — \/x—\
36. ■\/x+yl+x=^/Yjf^: ' yx + l + y'x—1
Vl + x
2 43. ^/x+Vl — i/x + x^^l.
37. i/£c— 1 i«—
l/a;— 8 44. i/l—x+i/2—x = yl—4x
38- |/^^^^a^-r*/^H^"3V. 45. i/a^^ + 3ic + l-l-2a;^-6a;.
47.
46. yx + 2 + i/x—S = y'2x+ll.
20 _ ._ 12
= -l/«.= v/15+a.. 48. -^^==.= ,/^^ + s + x.
1/15
49. i/xTlO + i/x^^=yQx.
Solve for a? :
60. i/ic + a'^— i/ic— a^ = l/26. 52. y^x+ya—x=y'a-\-b.
_ 2a3
61. i/x-\-\/x + a=—=^- 53. i/aj— a = ^>— i/^.
54. Find the four fourth roots of 16.
66. Fhid the six sixth roots of 1.
CHAPTER XVII.
SYSTEMS INVOLVING QUADRATIC AND HIGHER
EQUATIONS.
174. As in the case of linear equations, so in general, solv-
ing a system of quadratic equations requires, first, the elimina-
tion ofallhutone of the unknown numbers ; and second, so?y*?i^
the resulting equation for that unknown number.
When a system involves quadratic or higher equations, the
method of elimination depends upon the forms of the equa-
tions.
175. One equation linear and one quadratic.
When one equation is linear and one is quadratic, either of
the unknown numbers may be eliminated by substitution ;
and the elimination leads to a quadratic equation. Since a
quadratic equation in one unknown number can alicays be
solved, then a system consisting of a linear and a quadratic can
always be solved.
Example 1. Solve | ^.-^I?7. g|
Solving (1) for a?, ir=6 + 22/. (3)
Substituting this value of x in (2),
42/' + 241/ + 36 4-2/'= 17. (4)
Solving (4) , 2/ = - 1 or - »/. ^ v
Substituting these values of y in (3), \X.""*v
when 2/=— 1» ic=6— 2=4; ^
when 2/=-¥, a^=6-^8-=-f- ^^ ^
Hence, there are two solutions : v-^ "*\
a?=4, 2/=-l; ^=-1, 2/=--/.
Let the student check the result.
261
262
ALGEBRA
EXERCISE 81.
Solve:
4.
5.
7.
10.
11.
y=-l/12; {y=V7; \y=.-y7- \y\\V'^\
J x=-h _ j x=2, j x=-2,
i y=-v7; 1 2/=o; 1 2/=^; etc.
QUADRATIC AND HIGHER EQUATIONS
2Y5
Now proceeding as in Example 1, we get the curve shown in
Fig. 8.
This curve is called an ellipse. Astronomers have found that
all of the planets move in ellipses.
Example 3. Draw the graph of a?H i/^=9.
By proceeding as in Example 2, we get the circle shown 11
Fig. 9.
Fig. 9.
183. Solutions of systems. The solutions or roots of a sys-
tem are the coordinates of the points where the graphs of the
equations cross^ or intersect.
Example 1. Interpret the solutions of the system
a? + ;
4.
276 ALGEBRA
The solutions are ^=l±?i^, ^^ 8-1/29 ^^^
5 5
4-21 29 ,._ 8 + l/29
^ 5 ' y- 5^'
The graphs of these equations are the circle and the straight
line PQ in Fig. 9.
The points whose coordinates are their two solutions are the
points P and Q^ the points where the graphs of the equations
intersect.
Observe that the straight line intersects the circle twice. This
is as tnany times as the system has solutions.
Example 2. Interpret the solutions of the system
ix^ + y'=9,
x4-l/32/=6.
The solutions are, x=§, y=§i S\ and x=^, y=^i/3.
These solutions are the same, i. e., the system has a pair of
equal solutions
The graphs of these equations are the circle and the straight
line AB, Fig. 9. It is seen that the line AB does not cross the
circle at all^ but touches it at one point -K, whose coordinates
(I, |l/3), are the solutions of the system.
Example 3. Interpret the solutions of the system
( icH 2/^=9,
I x—2y=S.
The solutions are, ^^8 + 2i/^19^ ^^-16 + l/-19. ^^^^
8— 21/-19 -16-1/ -19 -D 4-1 1 4.-
x= ^ y= -• . Both solutions are imagmary.
The graphs of these equations are the circle and the straight
line MN, Fig. 9. It is seen that the circle and the line MN do
not intersect or touch at any point. In general, ivhen the solu-
tions of a system are imaginary expressions., the graphs of the
equations neither cross nor touch.
QUADRATIC AND HIGHER EQUATIONS
277
Example 4. Interpret the solutions of the system
This system has four solutions :
a?=T?^l/l3, 2/=Al/l3; x=-^\\/n, I/-TJ
y= _-^6j|/i3 ; and ic= — t«^i/13
The graphs of these
equations are the two
ellipses in Fig. 10. The
four solutions are the co-
ordinates of the four
points, A, B, C, X), where
the curves meet.
Example 5, Interpret ^
the solutions of the sys-
J 4xH9?/^=36,
jl/lo; X — y^
•Vt/13,
tern
X^ + '
This system has four
solutions :
x=i\/-15, y=l\/lO
x=iV--15, y=-ii/lO
x=-ii/-15, y=iVlO _
and a?=— f]/ — 15, y=—^i/l{). These solutions are imaginary.
The graphs of these equations are the ellipse (1) and the circle
in Fig. 10. It is seen that the circle lies entirely within the ellipse,
and does not meet it at all. This is to be expected, since the
abscissas of the points where they should meet are imaginary.
(1)
c=0. (2)
Eliminating y from this system, observe that we have the
general quadratic equation ax'^ + bx + c = 0. Since all solutions
of the system are solutions of the quadratic ax'^ + bx + c=0,
The graphic solution of the system j ^^ f^'^.,
278
ALGEBRA
it follows that the abscissas of the intersections of the graphs
of equations (1) and (2) will be the solutions of the quadratic.
Now since the equation y=x^
will be the same for all systems
formed from the quadratic, to
solve any quadratic we need
merely to find where the
straight line ay + bx + c = cuts
the fixed parabola.
Example 6. Consider the
equation x^—x—2=0. This is
equivalent to the system
I y-x-2=0.
The graphs are (a) and (&), Fig.
11. The abscissas of their points
of intersections are —1, and 2,
the solutions of the quadratic.
Example 7. To solve the qua-
dratic, ic^— 6a?+9=0, we find the
intersection of the line y—Gx +
9=0 with this same parabola.
The straight line touches the
parabola at the point x=3 ;
hence, the solutions are 3 and 3.
It is thus seen that to solve any quadratic we have simply to
find the graph of ay + bx + c=0, the parabola remaining fixed
for all quadratics.
In using this method make on coordinate paper, ruled to
centimeters and millimeters, a perfect graph of y=x'^ for your
fixed parabola. Now find two points on the graph otay + bx-\-c=0.
Connecting these by your ruler find the abscissas of the points
where the ruler crosses tlie parabola, and these abscissas will be
the solutions of the given quadratic.
-1—
Y
: 1
. 1
qi
' L
J
. I
r- i -
1
I
: T
f .
it _
i ^2
t
4 2_
Y
I._Z .
T
t^ -
^
_^K
I
2
t
71
V
t
'V Zt
2 V
2
(tjZ
z
)•
Fig. 11.
QUADRATIC AND HIGHER EQUATIONS
The solution of the system } p ^'^2 + ^^ _|_ ^^
279
Here again the elimi-
nation of y gives the
quadrati c, ax^ + ^aj + c =
0. Hence, the abscissas
of the intersections of
the graph of 2/=0,
which is the x — axis^
with the parabola^
y — ax^ -^hx-\- c, will be
the solutions of the qua-
dratic equation.
Example 8. Solve the
equation a;'-^ ^ 4x' + 3 = .
The graph of y=x^—
4a? + 3 is the parabola of
Fig. 12 which is cut by^
the x—axis, or y=0^ in
points x=l and x=3
which are solutions of
Y
*. .g the quadratic.
Note. — The method of the preceding system is preferable to this
one, since by the method of this system a new parabola must be
found for each equation.
It is evi(ient that the graphic solution of equations may be
employed with equations of degree higher than the second.
Example 9 . To sol ve x'' — 3x + 2 ^ .
To solve this, we may solve the system
( 2/=^-3^ + 2, (1)
t 2/=0. (2)
280 ALGEBRA
Some solutions of (1) are as follows I:* >
y
^
ik
x= — 3,
i=-16;
x=l,
y=0\
x=—2,
y=0;
x=0,
y=2\
j x=S,
I y=2o
X
20.
From these we get the
graph in Fig. 13. The
x—axis cuts this graph at
the point— 2 and touches it
at the point 1. Hence, the
three solutions of the equa-
tion are —2, 1, and 1.
Y Note. — In this and pre-
Fig. 13. ceding chapters it has been
seen that an equation in two
unknowns may represent some kind of line, straight or curved, and
that the solution of a system of such equations may be obtained by
carefully plotting the curves and measuring the distances of theii-
points of intersection from the two axes. Since the accuracy of tlie
solutions depends upon the accuracy with which the curves are con-
structed and the accuracy with which the coordinates of the points of
intersection are measured, the graphic method is useful chiefly in dis-
cussing the nature of the solutions rather than in determining the
actual solutions.
That an algebraic equation may represent a curve was first discov-
ered by Descartes in 1637. The subject of Analytical Geometry is
founded upon his discovery and is a discussion of curves by use of the
equations which they represent.
QUADRATIC AND HIGHER EQUATIONS 281
EXERCISE 87.
Draw the graphs of the equations :
1. x' + i/' = 16. 4. a;'^ = 8y + l. 7. x'-f = ^.
2. Wx'+f = 16. 5. 2/^ = 8iK + l. 8. x'-\-^x + f = S,
3. 4ic2 + 25/ = 100. 6. x' = ii/ + 4. 9. x'-^xy-2y' = 0.
Draw the graphs of the following systems, and interpret
their solutions :
10. |"+'^r^' 11. |»^ + .'/=25, ,2^ |-^+.V'=4.
^'^' |i«-22/ = 4. ■^*'* |a;^ + 36/-36.
(4a^^ + V = 36, .y (2/^-2^-1,
14,
16,
f a;'+a;y-6y2 = 0, ^g ( £c^ + 16y^ = 16,
lQx' + i/ = U.
By the use of the graph find the solutions of
Tl^ ic^ + 4« + 4 = 0. 22. a;2-2a5-4 = 0. 26.^ x'-2x-l = 0. X
"2©» £c^-2a;-8 = 0. 23. x' = lQ. 26. a^='-5i« + 3 = 0.
21. a!'^-2a; + 4 = 0. ^V + £c-l = 0. 27. a;''-2a;=0.
28. What is the geometric interpretation of an imaginary
solution of a system ?
29. What is the geometric interpretation of equal solutions
of a system ?
30. How does the graph show that in general a system
which is composed of a quadratic and a linear equation has
two solutions ?
31. How does the graph show that in general a system
composed of two quadratic equations has four solutions ?
282 ALGEBRA
184. Problems solved by systems involving quadratics. Some
problems may be solved by solving systems involving quad-
ratic equations.
Example 1. Divide 9 into two parts the sum of the squares of
which shall be 45.
Let the parts be represented by x and y.
Then from the conditions of the problem we get
x^y=^, (1)
and a^' + 2/'=45. (2)
Equations (1) and (2) form a system whose solutions are
These two solutions of the system give only one solution to the
problem.
The required parts are 3 and 6.
Example 2. In running a mile a drive wheel of a locomotive
makes 264 revolutions fewer than a wheel of the tender ; but
if the wheel of the tender were 2 feet greater in circumference,
the drive wheel would make only 176 fewer revolutions in a mile.
Find the circumferences of the wheels.
Let x= number of feet in the circumference of the drive wheel,
and 2/= number of feet in the circumference of the other wheel.
Then in running a mile the drive wheel makes revolu-
tions, and the other wheel makes — - — revolutions.
Hence, from the first condition,
^-5|«=264. (1)
If the wheel of the tender were 2 feet greater in circumference,
it would make — — ^ revolutions in a mile.
Hence, from the second condition,
2/ + 3 X '
QUADRATIC AND HIGHER EQUATIONS 283
Equations (1) and (2) form a system whose two solutions are
x=20, y=10 ; x=-7^, y=-'^2.
From the nature of the problem the solutions must be positive.
Hence, the second solution of the system must be discarded.
Therefore, the circumference of the drive wheel is 20 feet, and
the circumference of the other wheel is 10 feet.
EXERCISE 88.
1. Find two numbers such that their sum shall be 13, and
the sum of their squares 97.
2. Find two numbers such that the sum of their squares
shall be 146, and the difference between their squares 96.
3. Divide 8 into two parts such that the sum of their cubes
shall be 152.
4. Find two numbers such that their difference shall be 4,
and the difference of their cubes 208.
5. Find two numbers such that the square of the greater
shall exceed the square of the less by 64, and such that the
square of the less shall exceed twice the greater by 16.
6. The sum of two numbers multiplied by their product is
160 ; and their difference multiplied by their product is 96.
Find the numbers.
7. The sum of two numbers is 58;. and the sum of their
square roots is 10. Find the numbers.
8. The sum of two numbers equals the difference of their
squares; and their product exceeds twice their sum by 8.
Find the numbers.
9. The difference of the fourth powers of two numbers is
255 ; and the sum of their squares 17. Find the numbers.
10. If a number of two digits be multiplied by its ones'
digit, the product will be 124. If the digits be interchanged
284 ALGEBRA
and the resulting number multiplied by its ones' digit, the
product will be 156. Find the number.
11. Find the number of two digits which equals the square
of the ones' digit, and which also equals 4 times the sum of its
digits.
12. If the suni of the squares of two numbers be divided by
the first number, the quotient Avill be 11 and the remainder 6.
The first number exceeds the second by 6. Find the numbers.
13. The area of a rectangle is 36 square inches. If its length
be increased by 3 inches and its width by 2 inches, its area will
be doubled. Find its dimensions.
14. A rectangle is 20 inches long and 16 inches wide. How
much must be added to its width and how much must be
taken from its length, in order that its area may be increased
by 22 square inches and its perimeter by 2 inches?
15. The hypotenuse of a right triangle is 10. If one leg
be increased by 3 and the other leg by 4, the hypotenuse will
become 15. Find the sides of the triangle.
16. A rectangular lot which contains 880 square yards is
2.2 times as long as it is wide. How mucl\ will it cost to build
a fence around it at 25 cents a linear foot ?
17. A guy-rope to a derrick is attached to a stake 30 feet
from the foot of the derrick. If the rope were 16| feet longer,
it would reach to a stake 53 i feet from the foot of the derrick.
Find the height of the derrick and the length of the rope.
18. A flower garden contains 1000 square feet, and is sur-
rounded by a path 5 feet wide. The area of the path is 750
square feet. What are the dimensions of the garden?
19. If the numerator of a fraction be increased by 1 and the
deuominatov be diminished by 1, the resulting fraction will be
QUADRATIC AND HIGHER EQUATIONS 285
equal to the given fraction inverted ; and 3 times the numer-
ator of the given fraction exceeds 2 times the denominator by
3. Find the fraction.
20. A laborer received $32.50 wages. If he had worked 5
days longer, and had received 50 cents a day less, he would
have received $41.25. How long did he work, and what were
his wages a day ?
21. A certain number of men do a piece of work in a certain
number of days. If there were 2 fewer men, it would take 4
days longer to do the work ; and if there were twice as many
men, it would take 4 days less to do the work. Find the num-
ber of men and the number of days required for them to do the
work.
22. A grocer bought apples and potatoes for 154. He sold
the apples for $36.80 and the potatoes for $18.70. He gained as
many per cent on the apples as he lost on the potatoes. How
much did he pay for each ?
23. A certain sum of money placed at simple interest for
one year amounted to $265. If the principal had been $50
more and the i*ate 1% less, the interest would have been the
same. Find the principal and the rate.
24. Two men can do a piece of work in 4| days. It would
take one four days longer to do the work than it would take
the other. How long would it take each of them to do the
work ?
EXERCISES FOR REVIEW (V).
1. What is the difference between a surd and a rational ex-
ession f
%, What determines the order of a surd ?
pression f
286 ALGEBRA
3. When is a surd in the simplest form? By what prin-
ciple may a surd be reduced to the simplest form ?
4. Simplify y'x'ip ; 2 1 1 65* {a —lif\ v^x^ — ^x^y + ^xy"^ ;
y_l_ /a + 6
(x—yy y a—Q
5. What kind of surds may be added or subtracted ?
6. Simplify i/Sa-2VSla' + Syl92ab\
7. Add i/a'x\y^z)\ i/9aV(y + 2;), and 3i/16(y + s)^
8. Add tVv "72, -ii i, and 6i/21i.
9. By what principle do you change the order of a surd ?
10. Change i/5, y 11, i/13, to surds of the same order.
2d
3/27^
11. Write as an e7itire surd o
Sx
12. Which is the greater, i 5 or v TO"?
13. Write as one surd|/^ y 257
14. By what principle are surds multiplied ? Illustrate.
15. Find the product of i/Sx, y 27x\ and y9x\
16. Find the product of Si/i + S]/ 5 and Qy2 + 7yW.
17. What is the corijugate of 2 + |/5 ?
18. Give a method for dividing by a surd. Illustrate.
15 + 3i 3
19. Simplify
15-2|/3
20. Find the value of (2i/a=^6)^; {bx\ 96a;«y; (v 48icy)l
21. Find the value of y^/^lxY^i i/vV + 6a;+9;
l/^lx'yxXx-yY.
QUADRATIC AND HIGHER EQUATIONS 287
22. Simplify:
(''> 7I/3-5V2- ^^ Vf^x-
31/5 + 51 g (/) (I'/^^M?)/
^ '' V/5-V3
^) 2)/n5-3i/63 + 5v"28.
W
^/s-!v^^)
^(f7)3i/f+i/TV + 4i/^V W l/2Xi/3xi/iXl/i. ^
23. What is an imaginary number ? Illustrate.
24. What is the typical form of an imaginary number ?
25. Reduce ]/— 4ic* to the typical form.
26. What are the values of the successive powers of i/— 1?
How is ^/^-i usually represented ?
27. Simplify:
(a) V-
-36a^-
-l/-49a^ + l/-
81«^
ip) v~-
^•1/-
-4 i/-l(>.
(^) (l/'
-2 + 1
/-5)(i/-2-i/
-5).
(^0 (1/
-12-
-l/-15)-i/-3.
28.
, Write the conjugate of Z-\-y ■
-5.
29. What is a complex number f Illustrate.
30. Square i/^^S-j/'^^.
31. Divide 3 + |/^=5 by 4-i/^=^.
1/^^+3
32. Simplify ^-^_^^-=^'
33. How many roots has a quadratic equation in one
unknown ? What kind of numbers may they be ?
34. Give an example of a 2yf(re quadratic equation in a.
288 ALGEBRA
35. How do you solve any pure quadratic ? How is the
solution of any equation checked?
36. Solve and check :
{a) 4£c^-3 = 0. {h) 3«2 = 9a2-54. (c) ^{t^\)-t{t-\)=^t.
37. What is a complete quadratic equation? Give two
general methods of solving such an equation.
38. Solve by factoring :
{a) 2a;^ + 9a;-5. {h) 6y^ + 5y-6 = 0. (c) 1-\%x = ^x\
39. Solve by completing the square. (Always check your
solutions.)
(a) i«^-4£c = 32. {h) 2ic^ + 9;K-5 = 0. (c) SGaj'^-aea. + S^O.
40. Solve by the quadratic formula :
{a) 8a;^ + 2i« = 3. (c) 3ic2 + 35-22a; = 0.
{h) x^-\x=\. (d) 4x' + 17x=U.
Solve the following for the general number :
41 J_ = _L+^ ' 42 ^zL^.?:z8=_?^+l \\
• a-\ a-2^a-'f ^- x+3^ x-3 x^-9^'2'
■ y-\ 2y ■
44. In changing a, /*r«c^/o/2a/ equation to an integral equation,
by what expression, in general, must the members of the
equation be multiplied ?
45. How can you tell the nature of the roots of a quadratic
without solving the equation? What is meant by the dis-
criminant of a quadratic ?
46. Solve ax^-^hx^c = ^ and give the relations that must
exist among the coefficients for the different kinds of roots.
47. What must be the value of m if the roots of ^tx^—lx-^m
= are equal ?
QUADRATIC AND HIGHER EQUATIONS 289
48. What relations exist between the roots and coefficients
of x''+px + q = ?
49. Construct an equation whose roots are 3 and — ^. One
whose roots are l + j/f and 1 — i |.
50. Explain the method of stating and solvmg a problem by
means of an equation.
51. Can you solve an equation of a?i(/ degree ? What special
kinds of equations of higher degree can you now solve ?
52. Solve the following :
(a) x*^Q = bx\
53. How many solutions has an equation of the 4th degree
in one unknown ? One of the 5th degree ? One of the nth
degree ?
54. What is an irrational equation in a; ?
55. Give the general method of solving an irrational equation.
56. When is a solution said to be introduced in a derived
equation ?
57. Show that, in general, when both members of an equation
are squared, 7ieia solutions are introduced?
58. Solve and check :
(a) '6\/x=i/xTS + VxTQ.
{b) 1/7 -a; 4-1 = 1/205 +
l-l/l+a; i/l-a;
^''^ i+i/rT^~i/rF^*
59. How many cube roots has a number? How nmny fourth
290 ALGEBRA
roots? How many nth roots? Find the cube roots of 1.
Of 8. Of 27.
60. How many solutions, in general, has a system of one
linear and one quadratic equation ? A system of two quadratic
equations ? How is this illustrated by the graphs ?
61. What is a defective system? Illustrate.
62. What represent the solutions in the graphs of a system
of two quadratic equations ?
63. Solve the system { f +/;t^lr''
Cr
64. The area of a circle is equal to -^, where G represents
the number of units in the circumference, and r the number of
Cr
units in the radius. From ^=_-, and (7=2;:r find ^ in terms
of r and r.
Sr
65. From F=-q-j ^==4-^^, and r=\I>^ find V in terms of r
o
and TT ; also in terms of U and r.
CHAPTER XVIII.
INEaUALITIES.
185. Definitions. If a— ^ is positive, a is said to be greater
than h. If a— ^ is negative, a is said to be less than b.
The symbol for ''Hs greater than^^ is >, and the symbol for
"is less than^^ is <.
Thus, a>6 is read, "a is greater than 6." And a<&isread,
" a is less than 6."
From the above definitions it follows that when a — h is
positioe^ ay>b, and when a—b is negative, a<^b.
Thus, since 6—4=2, therefore 6>4; since 5— 8=— 3, therefore
5<8.
The statement that two expressions are not equal, i. 6., that
one of the expressions is greater or less than the otlier, is called
an inequality. The expression at the left of the inequality
sign is called the first member of the inequality, and the expres-
sion at the right of the inequality gign is called the second
member.
Two inequalities which have the same sign of inequality are
called inequalities of the same species. Two inequalities which
have opposite signs of inequality are called inequalities of
opposite species.
Thus, 15>12 and 7>— 2 are of the same species ; and 6>— 4
and 9<17 are of opposite species.
In this chapter, the letters used in the members of an
inequality represent 07ily positive real nimibers. A negative
number will be denoted by a negative sign.
291
292 ALGEBRA
186. Principles of inequalities. The following are some
useful principles of inequalities.
(1) If equal numbers be added to, or subtracted from, the
members of an inequality, the result will be an inequality of the
same species ; that is, .
if fl>6, then a-\-c^b^c, and a—d^b—d.
For, since a> b, then a — b\& positive. Hence, (a + c) — (6 + c),
which equals a — b, is also positive. Therefore,
a + c>^» + c. § 185.
In like manner, a—d^b—d.
Evidently the proof would have been similar, had the given
inequality been a7, therefore 10 + 6>7 + 6, and 10-4>7-4.
{2) If the members of an inequcdity be tnidtiplied, or divided,
by equal positive numbers, the result ivill be an inequality of the
same species ; that is,
if a>6, then ac^bc, and->-.
c c
For, since a>^, therefore a — ^ is positive. Hence, c(a — b),
or ac — bc, \s positive when c is positive. Therefore,
acybc. § 185.
In like manner, ->-, when c is positive.
Thus, if o + 2>4,then multiplying both members by 3, we have
i» + 6> 12. And if ocf -f 2xy Qx, then dividing both members by a?,
we have aj + 2>6.
(3) If the corresponding members of tivo inequalities of the
same species be added, the residt will be an inequality of the same
species ; that is,
if fl> b, and c> (/, then a + c> 6 -h (/.
INEQUALITIES 293
For, since a>^, and c>r/, therefore a — h and c—d are both
positive. Hence, their sum, a— b + c—d, is positive ; i. e.,
(a-\-c) — (b + d) is positive. Therefore,
a + c>i + (^. §185.
Thus, if 2a?— 2/>10, and x+y^4^ by adding them we have
3x>U.
(4) If the members of an inequality be subtracted from the
members of an equation^ the result loill be an inequality of oppo-
site species ; that is,
if a>6/ and c = d, then c—a9, therefore 20-15<20-9. •
(5) If the members of an inequality be subtracted from^ or
divided by the members of another inequality of the same
species., the residt is not necessarily an inequality of the same
species ; that is,
if fl>6 and c>(/,
then a—c may or may not >6— (/
and - may or may not >;^.
C Q
Thus, 8>4 and 7>2, but 8-7<4-2, and |<|.
Also, 8>4 and Q>2, but 8-6=4-2 and |<|.
{6) If the members of an inequality be multiplied^ or divided.,
by equal negative numbers., the residt will be an inequality of
opposite species ; that is,
if fl>6, then acJ, therefore a—b is jyositive. Hence, ac—bc is
negative when c is negative. Therefore,
ac<,bc. § 185.
Similarly, -<-•
Thus, since 9>6, therefore 9(-2)<6(-2), i.e., -18<-12.
(7) If the signs of all terms of an inequality be reversed., then
the symbol of inequality must also be reversed
This follows directly from (6) when c equals —1.
Thus, if 3ic-22/ + 6>a-46, then 2y-^x-Q<4.b-a.
Note. — These are but a few principles of inequalities. There are
many others which could be established by similar processes.
187. Identical and conditional inequalities.
Inequalities which are true for all values of the general
numbers involved are called identical inequalities.
Thus, a + 10>a is an identical inequality.
Inequalities which are }iot true for all values of the general
numbers involved are called conditional inequalities.
Thus, 6a? + 2>0 is true only under the condition that x is
greater than —^.
Note. — It is to be observed that identical and conditional inequalities
are analogous respectively to identical and conditional equations.
188. Proofs of some identical inequalities. Some identical
inequalities may be established by use of the pJ-inciples in § 186.
Example 1. Show that a'^ + b^^2ab, if a and b are unequal.
Since a and b are unequal,
therefore, {a—byy>0,
or a'-2ab-i-b'>0.
Adding 2a6, a' + b'>2ab. §186, (1)
INEQUALITIES 296
Example 2. If a and h are unequal, and a + 6>0, show that
We have a^ + h''>2db. Ex.1.
Subtracting a6, a^-ab + W^ah. §186, (1)
Multiplying by a + 6, d'-\-lf^a'h^ah\ §186, (2)
Example 3. The sum of any positive number n and the quo-
tient - is greater than 2.
We have a' + h'''>2db. Ex.1.
Now let d^ be n and let If be -.
n
Then, substituting in the above inequality, we have
- » + i>2-
189. Solving conditional inequalities. A conditional in-
equality is said to be solved when the possible values of the
unknown numbers which will satisfy it are found. The range
of these values is discovered by means of principles such as
those established in § 186.
Example 1, Find the possible values of x in
37-2^ ^ 3x-8 ^
3— + a'> ^-9.
Multiplying by 12, 148-8^+ 12a;>9x-21-108, § 186, (2).
or 148 + 4;r> 9^-132.
Subtracting 148 + 9aj, -5ic>— 280. §186,(1).
Dividing by -5, xQx+16.
Subtracting 6ic + 16, x''-6x-16>0. §186,(1).
Factoring, (x-8){x + 2)>0.
Hence, the factors a?— 8 and x + 2 must both be positive or both
be negative,
296 ALGEBRA
To make both positive, x must be greater than 8.
To make both negative, x must be less than —2.
Hence, a^>8 or <— 2.
Example 3. Find what values of x will satisfy the system
10a?>3x + 49, (1)
a7+5<|+55. (2)
From (1), x>7.
From (2), x<75.
Hence the system is satisfied by any value of x between 7 and 75.
Example 4. Find what values of x and y will satisfy the
system
\ 3j! + i/>10,
\x^y=Q,
Subtracting (2) from (1), 2x>2.
Whence, ic>l.
Multiplying (2) by 3, 3a: +3?/= 24.
Subtracting (3) from (1), -2i/> -14.
Dividing by —2, 2/<7.
Hence, any solution of (2) in which x>l and y2i/aJ. 6. a* + b*:^a'b\
Suggestion : Let a=x^ and h=y'\
(1)
(2)
§186,
(1).
§186,
(2).
(3)
§186,
(1).
§186,
(6).
INEQUALITIES 297
Find what values of x will satisfy the following:
7. 2aj-3>7. 11. a;2-2a;>3.
8. ^-hx<:^x-l\. 12. _l_3- 13. 6 _ 3 8
a?— ic— 4 tc — 3
10. -1- <_L 14. r^^a<^_\
Find the values of x that will satisfy the following
systems :
.. (3ii;-5<2a;+l, r3a;-16 5
^^- l3a^ + 4>x + 8. ^g I— ^<3'
17 (3(0^-2X2(0^-3), r4^ ^ a; ^
l2a3-7>5.^+5. !^l>^=r2 + 3,
.„ j^(^ + 3)>ic(i«-5) + 16, 1 2a; , p^ 4x
^°- |3a;(a; + 2)<3a;(a;-l). L.'« + 3'^'^^a; + 7'
Find the values of a? and y that will satisfy the following
systems :
21 ji«+2/=10, ric + 2
24
22. ^ 3a; + 2/>14,
3" +4y>2,
y + 11 ^+1^-,
11 2
|a; + 2y=13.
(7a; + 4y = l, , __ _
23. Il^ + f^^i' 25. |?^ + *y>l.
CHAPTER XIX,
RATIO AND PROPORTION.
190. The Ratio of one number to another number of the
same kind is the quotient obtained by dividing the first num-
ber by the second. It follows from the definition that the
ratio of two numbers is an abstract numher indicating the num-
ber of times one contains the other^ or the part one is of the
other.
Thus, the ratio of 6 to 2 is 6-^2, or 8. The ratio of $8 to $4 is 2.
The ratio of 5 ft. to 8 ft. is 5-i-8, or |. The ratio of a to Z> is a-=-6,
• a
OV y
It follows that an indicated ratio is a fraction. Hence, a
ratio may be expressed by any of the signs used to express a
quotient or a fraction.
Thus, the ratio of 3 to 4 may be written 3^4, 3/4, |, or 3:4.
In the ratio between two numbers the dividend, or numera-
tor, is called the antecedent, and the divisor, or denominator, is
called the consequent. The two together are called the terms
of the ratio.
The ratio of a to b is sometimes called the direct ratio of a
to b. And the ratio of J to « is called the inverse ratio of a
to b.
Thus, the direct ratio of 3 to 7 is f . The inverse ratio of 3 to 7
is |. The inverse ratio of one number to another is the direct
ratio inverted.
298
RATIO AND PROPORTION 299
191. It is clear from the definition of a ratio that all of
the laws established for fractions must apply to ratios.
Ratios may be reduced to higher or lower terms. They may be
added, subtracted, multiplied, divided, raised to powers, and have
roots extracted. Ratios may be compared by reducing them to a
common denominator, and comparing the resulting numerators.
Ratios are compounded by taking their product.
Thus, the ratio compounded of | and f is -^%.
192. Commensurable and incommensurable numbers.
Two numbers whose ratio can be exactly expressed by two
whole numbers are called commensurable numbers, i. e., two
numbers are commensurable when there can be found some
third number of the same kind, called their common measure^
that is contained an integral number of times in each. Two
numbers whose ratio can not be exactly expressed by two whole
numbers, i. 6., two numbers that have no common measure^ are
called incommensurable numbers.
Two fractions are commensurable if their numerators and
denominators are commensurable. For the ratio of two such
fractions can be expressed as the ratio of two whole numbers.
Thus, the ratio of | to A=l-A-|-V-=!|.
If the ratio of two numbers is a surd, the numbers are incom-
mensurable.
Thus, the ratio of v 2 to V^=y-^ = ^—^- The ratio can not
be expressed by two whole numbers, hence \ 2 and i/5 are m-
commensurdble.
The ratio between two incommensurable numbers is called
an incommensurable ratio.
193. Proportion. A proportion is an equation each of
whose members consists of a ratio. Four numbers, a^ b, c, d.
300 ALGEBRA
are said to be in proportion when the ratio oi ato b equals the
ratio of c to d.
The most used forms in which the proportion may be writ-
ten are
(1 c
T = -vand a : b=c : d, sometimes written a : b::c : d.
This proportion is read " a divided by b equals c divided by
f?," or " a is to ^ as c is to t/."
The terms of the ratios in a proportion are called the terms of
the proportion. The first and fourth terms are called the
extremes, and the second and third terms are called the means
of the proportion.
Thus, in 2:x=Qx'^:Sixf, 2 and So(^ are the extremes, and x and
6x^ are the means.
a c
In the proportion T=-p d is called the fourth proportional to
ay by and c.
194. In the following sections are given the most im-
portant principles in proportion. The student should master
these principles. • • .
195. In any true proportion the product of the extremes
equals the product of the means.
Suppose a : b=c : d.
a c
Written in fractional form, ^=^.
Multiplying by bdy ad^bc,
which proves the principle.
Thus, in 2: 4=6: 12, 2x12=4x6.
196. If the product of two iiumbers equals the product of
two other numberSy then these four numbers form a proportion
in ichich the extremes are the factors of either product and the
means are the factors of the other product.
ad= be.
a c
b-d'
c a
d b
i> d
RATIO AND PROPORTION 301
Suppose that
Dividing by bd,
or dividing by ac, -=7:^ or 7, = -j etc. ;
which proves the principle.
Thus, since 2x10=4x5, therefore 2: 5=4: 10, or 5: 10=2; 4.
It follows from this principle that to test the correctness of a
proportion it is sufficient to shoio that the product of its extremes
equals the product of its means.
Thus, 2 : x=6a?^ : Zj? is a correct or true proportion, for
2 • Zd^=x ■ Qx\
197. Remark. It should be noted that by the principle
of § 196 we get from the equation ad=bc the eight proportions :
1. a : b = c : d; 3. d : b = c : a; 5. b : a = d : c ;
2. a : c=b \ d\ ^. d : c = b : a; Q. b : d=a : c;
7. c : a = d : b; 8. c : d=a : b.
It should also be noted that sii>ceany one gi\esad=bc, from
any one of them all the others follow.
Observe that (2) was obtained from (1) by interchanging the
means, and that (3) was obtained from (1) by interchanging
the extremes. The old mathematical terms for such changes
are, respectively, mean alternation and extreme alternation.
Observe also that (5) was obtained from (1) by interchanging
the terms of each ratio. This process is called inversion.
In (4) we observe that it is obtained from (1) by inter-
changing the means and the extremes.
Illustrate each of the 8 proportions given above with numbers.
198. The products of the corresponding terms of two or
r>iore proportions form a proportion.
302
ALGEBRA
Suppose that
a c
b-7(
mp
and
x_io
y~lj'
Multiplying,
am X cpw
buy d q V
or
amx cpw
bny cJqv
Hence, amx^ buy,
cpw, and dqi^ are in proportion.
Axiom 3.
199. Like powers or like roots of the terms of a proportion
are in proportion.
Suppose
that
a : b = c : d.
or
a c
1>-71'
Then,
it)- -($)-.
Axiom 5.
or
\ / \ /
b"" d"'
Again,
n-vi
Axiom 6.
n — n —
\/a 1 c
or „ _
yb yd
Thus, since 2 : 3=4 : 6, therefore 2=^ : ^^=^'' : 6^
or 8 : 27=64 : 216.
And since 4 : 25=16 : 100, therefore |/4 : |/25=i/T6 : i 100,
©r 2 : 5=4 : 10.
200. The terms of a proportioii are in proportion by addi-
tion ; i. e., the sum of the first two terms is to the first (or second)
as the sum of the last tico terms is to the third (or fourth).
RATIO AND PROPORTION 303
§ 195.
Suppose that
a c
b-d'
Then,
ad— he.
Adding bd^
ad-^hd=bc-\^bd.
or
d(a + b)=b(c-i-d).
Hence,
a-\-h c + d
b d ■
In like manner,
a-{-h c^d
§ 196.
§ 196.
a c
Thus, since 3 : 5=12 : 20, therefore (3 + 5) : 5= (12 + 20) : 20,
or 8 : 5=32 : 20.
The old mathematical term for this process is composition.
201. The tenuis of a 2^^02)ortio)i are in pro2^ortion by svh-
tr action ; i. e., the difference of the first tico terms is to the first
{or second) as the diff'erence of the last tvm terms is to the third
{or fourth).
a c
Suppose that t— ;/•
Then, ad=bc. § 195.
Subtracting bd^ ad—bd=bc — bd,
or d{a — b)= b(c — d) .
Hence, ^^JlI^. § 196.
T vi a—b c—d
In like manner, =
a c
Thus, since 10 : 3=30 : 9, therefore (10-3) : 3=(30-9) : 9,
or 7 : 3=21 : 9.
The old mathematical term for this process is division.
202. The terms of a proportion are in proportion by addl-
tion and subtraction ; i. e., the sum of the first two terms is to
their differenae as the sum of the last tico terms is to their dif-
ference.
304
ALGEBRA
Suppose
that
a c
Then,
h d '
And
Dividine'
a—h c—d
b d '
a + b_c-{-d
a—o c — d'
§200.
§201.
Axiom 4.
This and the preceding sections will enable us to obtain an
equivalent proportion from a given proportion.
(X c
Example. Suppose T,~d' ^^^
By §199, p=^. (2)
By § 201, ^JL^. (3)
By § 199, ^ -^ (4)
203. Continued proportion. When, in a series of numbers,
the first is to the second as the second is to the thirds and so
on, the numbers are said to be in continued proportion.
Thus, a, 5, c, d, e, are in coyitinued proportion, if T— 7=^— J'
In the continued proportion ^=-, which is called a mean pro-
portion, b is called a mean proportional between a and c, and c is
called the third proportional to a and b.
Thus, 4: 10=10: 25 is a mean proportion. 10 is the mean propor-
tional between 4 and 25. And 25 is the third proportional to 4 and
10.
204. The tnean proportional between two mimhers equals the
square root of their product.
RATIO AND PROPORTION 305
Let b be the mean proportional between a and c.
mu ah
Then, -t = -
* . he
And }f = ac. §195.
Hence, h=\/ao.
a
m
_/>_
X
b
71
<1
y
a
= 7*.
Thus, in the proportion 9: 18=18: 36, we have 18 = 1/9 x 36.
205. In a series of equal ratios the sum of the antecedents is
to the sum of the consequents as any antecedent is to its con-
sequent.
Suppose that
Let
Then —=^r.~ = r.- = r. etc. Axiom 7.
n ' q ' y
Clearing of fractions, a=rb^ m=rn^ p = rq^ x — ry^ etc.
Adding, a + m-fp + a;+ • • • • =rb^rn^rq^ry^r • • • •
Dividing by b^rn-\-q-^'y^r • • • • »
a-Vm-\p^-x■^- am
r^"—. , =r=7; =— =etc.
o-\r n-^-q^y^ o n
2_4 8 16 2 + 4 + 8 + 16 30
ihus, 3~6~12~24'-3 + 6 + 12 + 24 ^^ 45'
206. The representation of a ratio by a single letter., as in
§ 205, is often a useful process. In order to test the accuracy
of a proportion, we may proceed as in § 196, or as in the
example in § 202. But often it is more convenient to rep-
resent the equal ratios by a single letter, and then show that
the proportion reduces to an identity,
20
306 ALGEBRA
Example. If -r = :^, show that / .. , ~ 3. -
Let r=r. Then ^=r.
Hence, a=br^ and c=dr.
Substituting, }AV+£^vV^+£
VdV + d' y^d'r^ + d'
Factoring and reducing,
b\/?Tl b V'r^ + l
dVr^^Ti~d^^r^
or d~d' ^^ identity.
Hence, the proportion is true.
207. Since a proportion is an equation, it may be solved
for any number in it.
Example 1. a : b=c : d. Find the value of d.
a c
Writing in fractional form, x=^'
Multiplying by 6d, ad=bc.
be
Dividing by a, ^~a'
Example 2. Find the fourth proportional to 3, 2ic^ and 21a?.
Let a represent the fourth proportional.
Then. ^3=--
Multiplying by 2ax^, Za=42o(?.
Dividing by 3, a=14a?^.
Example 3. Find the third proportional to a' and a6'.
Let X represent the third proportional.
Then, • 4=2^.
' db^ X
Multiplying by db^x^ a^x=a^b*.
Dividing by a^, x=b^.
RATIO AND PROPORTION 307
EXERCISE 90.
1. What is the inverse ratio of 6 to 11?
2. Reduce the ratio of 10 to 18 to lowest terms.
Find the simplest expression for the ratio of :
3. bx to Ihx^. 4. - to -T. 5. 7 rr, to
a h' ' \x—yy {x—yf
6. Which ratio is the greater, -J^ or || ?
7. Write in descending order of magnitude |, i|, f ^.
8. Write the ratio co^npounded of the ratios J, f^^ |.
9. Write the ratio compounded of the ratios -, 'A, tt-t^.
(a\l>V 12
10. Write the ratio compounded of the ratios ^^. — --, —ry
11. What must be the value of a, if the ratio -— -^ equals
the ratio | ?
12. What must be added to each term of I to make it equal
to I?
13. Two numbers are in the ratio of 3 to 4, and their sum
is 35. Mn4 the numbers.
14. Two numbers are in the ratio of 7 to 5, and their differ-
ence is 6. Find the numbers.
15. Two numbers are in the ratio of 2 to 9, and the difference
of their squares is to the square of their difference as 11 is to 7.
Find the numbers.
308 ALGEBRA
Find i\\Q fourth proportional to :
16. 3, 4, and 36. 18. p>, v, pv. 20. «, h, c.
17. X, ^x\ bx\ 19. 1, 7, 10. 21. x, y, z.
Find the third proportional to :
22. 1, 2. 24. p, pq. 26. a, Z>.
23. X, '6x\ 25. 10, 1. 27. ^, y.
Find the mean p)roportional between :
28. 2, 8. 30. x\ ^x\ 32. a, h.
29. 15, 135. 31. 6, 9. 33. x, y.
34. In the proportion T^p'^ 17772' solve for h.
35. Show that | and 5 are com.mensnraMe numbers.
36. Show that ^ and 2J are commensurable numbers
37. Show that 7/2 and \^\^ are commensurable.
38. Show that f^'I and f/3 are incommensurable.
39. Show that ]/15 and 7/5 are incommensurable.
40. Show that 7 : 12 = 21 : 36 is a true proportion.
\i a : b=^c : (7, show that :
41. a^ : b''=-ac : M
42. a& : bd'^ = a^c : ^V?.
43. (5a + Z>) : ^>=(5c + (^?) : ^?.
44. a : (a + c) = (a + ^>) : {a^b^c^d).
45. c : d=-f : -.
46. 2a : 5c = 2J : 5r?.
47. a : b=\a~c : 1/^.
48. {a^b) : (c + f?) = 7/aMT' : v"eTd\
49. 7/c«^-^»=^ : 7/c^-^ = 7/«^ + ^>' : yc'^d\
RATIO AND PROPORTION 309
50. {a-b) : {c-d)=^d^^' : ^&^^\
51. {a'c + ac') : {b'd+bd') = {a + cf : {b^-d)\
62. The sides of a triangular field are as 4 : 5 : 6, and the
distance around it is 1,200 yards. What are the sides ?
53. What must be added to 2, 5, and 12, in order that the
fourth proportional to the sums may be 24.
54. Express the ratio of 3^ to 1^ by the ratio of two whole
numbers.
55. The difference between two numbers is 4, and their
product is to the sum of their squares as 6 is to 13. Find the
numbers.
56. A tree casts a shadow 80 feet long, when a rod 4 feet
high casts a shadow 5 feet long. How high is the tree ?
CHAPTER XX.
VARIATION. ALGEBRAIC EXPRESSION OP LAW.
208. Variables and constants. In many problems in mathe-
matics, numbers are involved %chose values are changing.
Such numbers are called variable numbers, or variables. For
distinction, numbers whose values do not change during the
investigation of a certain problem are called constants.
For example, one's age is a variable magnitude, expressed by
a 'number which is always increasing. The weight of a stone
lying on the ground is a constant magnitude, expressed by a
r>umber of pounds which does not change.
Two or more variables may be so related that a change in
the value of one will cause corresponding changes in the
values of the others. This relation between the variables is
expressed by means of an equation.
Thus, in case of a running train, the distance, time, and speed
are varnables, and their relation is expressed by the equation
vt=d, where v, ^, and d represent the speed, time, and distance,
respectively.
If the variables, x and y, are connected* by the equation y=x^^
then when x assumes the successive values 1, 2, 3, 4, 5, 6, 7, 8,
9, etc., y assumes the corresponding values 1, 4, 9, 16, 25, 36, 49,
64, 81, etc.
If one variable depends for its values upon the values of
another, the first is called a dependent variable, and the second
is called an independent variable.
810
VARIATION. ALGEBRAIC EXPRESSION OF LAW 311
Thus, in the example above, if x is an independent variable, y
is a dependent variable, for as x changes in value y also changes,
and the value which y assumes depends upon the value which
X assumes.
209. Direct variation. If two variables are so related that
during all of their changes, their ratio remains constant^ each is
said to vary as the other, or to vary directly as the other. As
the first increases, the second also increases ; and as the first
decreases, the second also decreases.
If X varies directly as y, their relation is expressed by the
equation
~=k, or x=ky,
where ^ is a constant.
The symbol oc is called the siyn of variation, and is read
" varies as." Thus a cc b is read " a varies as 5," and is equiv-
alent to the equation t=^, or a=kb-
Example 1. xcc y^ and when y=4t, x=20. Express the rela-
tion between the variables.
Let x=hy.
This equation must be satisfied by 2/=4 and x=2Q.
Substituting, 20=4/b.
Hence, A?=5.
Therefore, x=5y is the required expression.
210. Inverse variation. One variable is said to vary in-
versely as another when, during all of their changes, their j^roduct
remains constant. As one increases, the other must decrease.
If X varies inversely as y, their relation is expressed by the
equation xy=k, where ^ is a constant.
Example 1. Given that x varies inversely as y, and when
a?=2, 2/=4. Find the relation between them.
312 ALGEBRA
Let xy—k.
This equation must be satisfied by ic=2, 2/=4.
Substituting, 2-4=A;.
Hence, fc=8.
Therefore, xy=8 is the required equation:
211. Joint variation. One variable is said to vary as two
others jointly when the first varies directly as the product of
the other tico.
If X varies as y and z jointly^ their relation is expressed by
the equation x=kyz.
Example 1. Given that a varies as h and c jointly, and when
a=36, 6=3 and c=2. Express the relation between the variables.
Let a=kbc.
This equation must be satisfied by a=36, &=3, c=2.
Substituting, 36=:fc-3-2.
Hence, k=6.
Therefore, a=66c is the required equation.
212. One variable is said to vary directly as a second and
inversely as a third, when it varies as the quotient of the second
divided by the third.
If t varies directly as d and inversely as y, the relation be-
tween them will be expressed by the equation ^=Z;-, where
^ is a constant.
Example 1. Given that t varies directly as d and inversely as
r, and when f=20, d=8, and v=2. Find d when t=\i) and v=^.
Let t=k -■
V
This equation is satisfied by f =20, cZ=8, u=2.
Substituting, 20=|A?.
Hence, k=5.
VARIATION. ALGEBRAIC EXPRESSION OF LAW 313
Therefore, the equation becomes
When ^=10 and i;=3, this equation becomes
10= sf.
Hence, d=6.
EXERCISE 91.
1. fcocy, and when 2/==5, x=lb. Write the equation be-
tween tliem.
2. tcoc 2/, and when 2/ = 2, x=VI. Find x when y = l.
3. X varies inversely as y, and when 2/ = 3, a; = 8. Write
the equation between them.
4. X varies inversely as y, and when i/^lO, £c=5. Find a;
when 2/ ==2.
5. X varies jointly as y and ^, and when 2/ = 5, and^=l,
£c=40. Write the equation between them.
6. X varies jointly as y and z^ and when y = ^ and 2 = 5,
05=150. Find z when a; = 120 and 2/ = 6.
7. X varies directly as y and inversely as ^, and when y=l^
and 2=2, 93 = 24. Find y when a:; = 14 and 2; = 2.
8. «^cc2/^ and when 2/^4, £c = 16. Find the equation be-
tween X and 2/.
9. The velocity of a body let fall toward the ground varies
as the time during which it has fallen from rest, and the
velocity at the end of 3 seconds is 96 feet per second. Write
the equation between the velocity and time.
10. The area of a triangle varies jointly as the altitude and
base. When the altitude is 4 inches and the base 9 inches,
the area is 18 sq. inches. What will be the area when the
base is 6 inches and the altitude 10 inches ?
3l4 ALGEBRA
11. The area of a circle varies as the square of its radius.
When the radius is 6 feet the area is 113.0976 square feet.
What is the area when the radius is 8 feet ?
12. The vohime of a gas varies inversely as the pressure
upon it. When the pressure is 8 lbs., the volume is 8 cubic
inches. What is the volume when the pressure is 4 lbs.?
13. The distance a body falls from rest varies as the square
of the time. In 2 seconds it falls 64 feet. How far will it fall
in 4 seconds ? How far will it fall during the fourth second ?
14. Tlie number of oscillations made by the pendulum of a
clock in a given time varies inversely as the square root of its
length. A pendulum 39.1 inches long makes one oscillation
in a second. How long must a pendulum be to make 4 oscil-
lations in one second ?
15. In playing see-saw the two persons must sit on the
board at distances from its point of support which vary
inversely as the weights of the persons. Where must the
support be placed under a board 16 feet long in order that two
girls Aveighing respectively 65 lbs. and 80 lbs. may play see-saw?
THE ALGEBRAIC EXPRESSION OP LAW.
213. Law expressed by the equation. An equation contain-
ing two or more variables expresses a law according to which
the variables change their values. Since there is a relation
between the variables, they can not each assume any values at
random. There is a fixed laio according to which they must
change. This laio is expressed hy the equation.
Thus, Newton's law of bodies falling from rest is expressed by
the equation
s=^t'\
where a is a constant, and s and t variables, s representing the
number of feet through which the body will fall in t seconds.
VARIATION. ALGEBRAIC EXPRESSION OF LAW 315
It evidently expresses the law that the distance through ivhich
a body will fall from rest varies directly as the square of the time.
Again, Boyle's law of gases is expressed by the equation
VP=k,
where Ar is a constant, and V the volume of a gas when under the
pressure P.
The equation evidently states that the volume of a gas varies
inversely as the pressure upon it.
When a law is known, the equation which expresses it can
be written.
Example 1. The distance through which an object will move
with uniform motion varies as the velocity and time jointly, and
it is found by trial that the object with a velocity of 20 feet per
second moves 60 feet in 3 seconds. Write the equation which
expresses the law of motion.
If s= distance, v= velocity, t= time, we have
s=kvt.
Since s=60 when v=20 and ^=3,
60=A;-20-3.
Hence, k=l.
Therefore, s=vt expresses the law.
EXERCISE 92.
Write the equations which express tlie following physical
laws.
1. The force with which moving bodies having the same
velocity strike a stationary body varies directly as the masses
of the moving bodies.
2. The attraction between two bodies varies directly as the
product of their masses, and inversely as the sqmire of the
distance between them.
316
ALGEBRA
3. If an object is revolved at the end of a string, the tension
exerted upon the string by the object varies directly as the
product of the mass of the object and the square of its velo-
city, and inversely as the length of the string.
4. The time of vibration of a pendulum varies directly as
the square root of its length.
5. The downward pressure upon the bottom of a vessel con-
taining a liquid of given density
varies jointly as the depth of
the liquid and the area of the
bottom of the vessel.
,
Y
,
\
/
\
/
\
/
^
/
y
/
\
/
\
/
A.
V.
y
V
Fig. 14.
6. The amount of heat re-
- ceived by a body from % radiat-
ing source varies inversely as
the square of the distance of
- the body from the source of
heat.
7. The resistance offered by
a wire to a current of electri-
- city varies directly as the
X' length of the wire, and inverse-
ly as its cross-sectional area.
214. As another illustration
- of the expression of law by an
equation, let us consider the
graph of an equation with two
unknown numbers.
The graph of the equation y=x^ is shown in Fig. 14. Here x
and y represent the coordinates of any point P on the graph.
If we consider x and y as variables, by changing their values
the point P can be made to move. As the absolute value of
VARIATION. ALGEBRAIC EXPRESSION OF LAW 317
X increases, the point moves away from the line FP, and as the
absolute value of y increases the point moves away from the line
XX'. By making x and y assume as values all possible solutions
of the equation, the point Pcan be made to take up in succession
all positions on the curve.
Hence, the graph may he considered as the path of a point
moving subject to a law. And the equation between the coordinates
expresses the law governing the m,otion of the point. The point
must so move that its coordinates continually satisfy the equation.
This principle is of great use in mathematical sciences such
as astronomy and physics.
Example 1. A body is ||^trn horizontally from a high point,
at a velocity of 16 ft. per second. Find the path of its fall to the
ground.
Let the distance which it moves horizontally in t seconds be
represented by a?, and the distance which it falls vertically in the
same time be represented by y.
Then, x=im.
By Newton's law of falling bodies, y=lW^.
Eliminating t between these two equations, x^=16^.
Measuring posi-
tive ordinates
downward, and
positive abscissas
to the right, the
graph of this
equation, which
is the path of the
moving object, is
found to be the
portion of the
parabola shown in
Fig. 15.
oJ
1
I +
^^
\
\^
N
\
+
\.
1
V
Fig. 15.
318
ALGEBRA
Example 2. A body is thrown from the ground at an angle of
45°, and with a force such that its horizontal velocity is 16 ft.
per second. Find the point at which it strikes the ground.
Let X represent the distance the object will move horizontally
in t seconds, and y the vertical distance it will move in the same
time.
Then x=lQt.
The vertical distance would be the same as the horizontal dis-
tance if the force of gravity did not make the body tend to fall
the distance 16^^ Hence,
y=16t-iet\
Eliminating f, x^=lGx—lQy.
The graph of this equa-
tion will be found to be
- the portion of a parabola,
- shown in Fig. 16. Hence,
this curve is the path of
the object's flight.
^ Now the distance from
f
^
■ —
....
>
y
N
v
/
\
^
—
—
—
—
^
^ the point of ascent to the
Fig. 16. required point is the dis-
tance AB of the figure.
At the point 5, 2/=0- Hence, the distance is the value of x ob-
tained by making 2/=0 in the equation
x'=16^-16i/.
When 2/=0, x^=z\^x.
Hence, a?=0 or 16,
Therefore, the body strikes the ground 16 feet from the point
of ascent.
215. The limit of a variable. If the value of a variable
changes according to some law, such that the difference be-
tween the variable and a constant may become and remain less
than any assigned small value, which may be taken as small
as one wishes, the constant is called the limit of the variable.
VARIATION. ALGEBRAIC EXPRESSION OF LAW 319
Suppose a point to move from X toward F, moving | the distance
X Y
XY the first second, arriving at A ; then ^ the remaining dis-
tance AY the next second, arriving at B ; then }^ the remaining
distance 5F the third second, arriving at O ; and so on indefi-
nitely. In an infinite number of seconds, the point may be
brought as near as we please to F, since we may continue the
process of halving the distance from the point to Fas long as we
choose. Evidently the distance from X to the moving point
gradually increases^ and approaches as near as we please to the
distance XY. Also the distance from the point to F gradually
decreases and can be made less than any fixed small distance,
which can be made as short as we please.
Hence, the variable distance through which the point has moved
from X approaches the whole distance XY as a limit ; and their
difference^ the distance from the point to F, approaches zero as a
limit.
The sign, = , placed between a variable and a constant, indi-
cates that the variable approaches the constant as its limit.
Thus, a? ^« is read " ic approaches a as its limit " and means
that, as X changes according to some law, the difference be-
tween X and a can become and remain less than any small
value whatever that may be assigned.
216. It follows from the definition of the limit of a vari-
able, § 215, that if the limit of x is a, the limit of x — a is zero.,
i. e., the difference between a variable and its limit is another
variable whose Vimit is zero. That is,
If x=a, then x—a = x', where x'=0.
Conversely, if x—a = x\ or jc=a + jr', vjhen x' is a variable
whose limit is zero and a is a constant^ the limit of x is a.
That is.
If x=a + x', where jr'=0, then x=a.
320 ALGEBRA
217. Finite, infinitesimal and infinite num'bers.
A variable whose limit is zero is an infinitesimal.
Thus, if the value of a variable can be made to decrease indefi-
nitely, and become and remain less than any assigned value,
which may be taken as small as we please, it is an infinitesimal.
A number, however small, which does not approach zero as a
limit is not an infinitesimal.
If the value of a variable can be made to increase indefi-
mtely according to some law, and become greater than any
assigned (or fixed) number, which may be taken as great as
we please, it is called an infinite number, or infinity. An infi-
nite number is represented by the symbol go .
An infinite number does not approach a limit, hut increases
without limit. The statement that x increases without limit is
expressed by a?= go .
Any number which is neither an infinitesimal nor an infinite
number is called a finite number.
All^ice^or definite numbers and all numbers heretofore con-
sidered ?iYe finite numbers.
218. If.) in a fraction inhose numerator is any finite num,-
ber^ except 0^ the denominator increases without Ibnit^ then the
fraction approaches as a limit / that is,
if jif=oo , then ~=0.
X
For, evidently as x increases, the fraction - decreases in
value, and, by making x sufficiently great, - can be made less
X
than any assigned value, which may be taken as small as we
please.
Thus, y\, T^o, iwo, TO loo, TTTo^offo-) ©tc. , are a series of fractions
in which each fraction is one-tenth the preceding one, and
each denominator is ten times as large as the preceding one.
VARIATION. ALGEBRAIC EXPRESSION OF LAW 321
If this series of fractions is carried out indefinitely, a fraction
will be reached whose value is less than any assigned value,
however small.
219. Properties of zero. Zero may be defined as the differ-
ence between two equal numbers ; that is,
0=n—n.
Thus defined, it leads to some interesting results.
(1) Let a be any finite number. Then
a(i = a{n—ri)
=^an—an
= 0.
Hence, the product obtained by inidtiplying any finite number
by zero^ equals zero.
(2) If a be any finite immber, then
^ _n — n
a a
n n
~^ a a
= 0.
Hence, the quotient of zero divided by any finite number^ equals
zero.
(3) The symbol ^ represents a number which multiplied by
gives 0.
But, according to (1), any finite immber multiplied by
gives 0.
Hence, ^ represents any finite number ichatever.
The fraction ^ is called an indeterminate fraction.*
* In like manner it can be shown that the symbol § is indeterminate.
322 ALGEBRA
(4) The fraction - where a is any finite number^ represents
a number which multiplied by gives a.
But, according to (1), no finite number multiplied by
gives a.
Therefore, tt has no finite value lohatever.
If a is finite and a? is a variable whose limit is zero, then by
taking £c smaller and smaller, - can be made to exceed any fixed
number which may be chosen as great as we please. Hence,
the limit of - when cc = 0, cannot be found; that is, it exceeds
all definite fixed numbers^ or is infinite. This is expressed by
a.
Mut this is not an equation in the ordinary sense.
(5) Since ^ is indeterminate, and ^y is not finite, therefore,
it clearly is not allowable to divide by 0. See foot-note to § 24.
Let the student show the fallacy in the following reasoning.
Let a=h.
Multiplying by a, a^=ab.
Subtracting 6^, a^—b'^=ab—¥.
Factoring, {a-\-h){a—b)=h{a—h).
Therefore, 2b{a—b)=b{a—b).
Divide by 6(a-6), 2=1.
220. Indeterminate fractions. For certain values of the
variable, a fraction will sometimes take the indeterminate
.
form -^
VARIATION. ALGEBRAIC EXPRESSION OF LAW 323
Thus, when a = 2, _i) = m ^^ indeterminate form. This
fraction, however, is equal to ^^ _,^ Now, as long as
a is 7iot equal to 2, we may divide both terms by a— 2,
hence -^^ ^^^ - = a + 2, when a is not equal to 2.
If a=2, by the definition of a limit, a cannot become 2,
although it differs from it ever so little ; hence we may divide
by a— 2. Now, as a approaches 2, evidently a + 2 approaches
(f 4
2 + 2 or 4. Hence we say ^=4 as a=2. Hence,
•^ a —2
a'^—4.
4 is called the value of -. when a=2.
a —2
The value of such a fraction, for any particular value of its
variable, is defined as the limit which the fraction approaches
when the variable approaches this particular value as its limit.
Example. Find the value of — r-— - when x=a.
oc — a
When£c=a we have =-, the indeterminate form.
x—a
For any value of x other than a, we have
x—a
Now, as X approaches a this expression evidently approaches
n^^ O'i
a^ + aa + a^ or 3a^, the value of -— — — when 0?= a.
X ^~a
CHAPTER XXI.
FRACTIONAL AND NEGATIVE EXPONENTS.
221. The following five fundamental laws of exponents
have been established in Chapter V, Chapter VI, and Chapter
VII:
(A) «"'a" = a'^+";
(C) («'")" = «'»";
(D) (aby = a''b''l
(-) (!)"=?
In all the work heretofore, the exponent has been a positive
integer. Where the exponent is a positive integer, by defini-
tion it indicates the number of times that the base is to be
used as a factor. Thus, d^ means a- a- a. The proofs of these
five laws were based upon the definition of an exponent. And
therefore these laws were established only when the exponents
were positive
22t2t. Fractional and negative exponents. We have seen
that by extending the operations with numbers so as to make
them general, new kinds of numbers have been conceived ;
viz., fractional, negative, and imaginary numbers. Likewise in
the attempt to extract a root of a power of a number by the law,
a fractional exponent, a neAV kind of exponent, is sometimes
824
FRACTIONAL AND NEGATIVE EXPONENTS 325
obtained if this law be extended to hold for all j^ositive integral
values of m and n.
Thus, y'd^=a^^'^ or a^-
If the law a'"^«" =«"»-" be extended to hold when n is greater
than m, then a negative exponent will result.
Thus, a^-^-a5=a=*"^=a"2.
It thus appears that the extension of one process gives rise
to fractional exponents, and the extension of another to negative
exponents.
It is clear that a fractional or negative exponent cannot he
given the same meaning as an exponent which is a j^ositive
integer.
3
For example, a^ cannot indicate that a is to be used as a factor
I of a time. Such a statement is absurd. Also, a-* cannot in-
dicate that a is to be used as a factor —4 times.
We shall define operations with negative and fractional ex-
ponents so that fundamental law A, § 221, shall be true also
Avhen m and n are fractional or negative. We may then in-
vestigate the meanings for such exponents as are required by
this law.
223. Meaning of a positive fractional exponent.
Since fractional exponents are to be so defined, that law A
shall be true for such exponents, then if m and n be any positive
integers.
m , m . m
• to n terms
a".a".a" . . . to ^i factors =«" " " (A),
m
or.
Hence, a*'~|/a'", by the definition of a root,
= (f'ar. § 147.
326 ALGEBRA
Therefore, a positive fractional exponeiit indicates a root of
a power ^ or a poioer of a root^ of its base ; the numerator indi-
cating the power ^ and the denominator indicating the root.
Thus, a"^=^i^a^, or (i^a)^ a^=i^^ 16^= (1^16)3= 2^ =8.
It follows that in any expression, any indicated roots may be
changed to fractional powers, and vice versa.
^ g .___ 3/"~" -2 11 2
Thus, y x'^ + y xy + y y^ meiy be written x^^+x^^'y^ + y^.
Note. — Keducing a fractional exponent to higher or lower
terms does not change the value of the expression.
TO
For, a" = V^
=17 «^^ § 148.
mx
Hence, fractional exponents that may occur in using laio A
tnay he reduced to a common denominator^ then added.
224. Meaning of a negative exponent. Since a negative ex-
ponent is to be so defined that law A holds, if n be any positive
integer or a fraction, then,,
a" •«"" = «""" Law A
48.
Dividing both members by w\ «""=—-.
a
Hence, a negative exponent indicates a power., or a power of a
root, of its base that is to be used as a divisor.
Thus, a-^=\ or in-a*; a^lr^ ^ — ova^-^lf; a~^=~^-^.
« '6^ at pa'
FRACTIONAL AND NEGATIVE EXPONENTS 327
225. Atiy factor may he changed from the dividend to the
divisor^ or from the divisor to the dividend, by changing the
sign of its exponent.
This follows from the interpretation of a negative exponent.
Example 1.
7y^z-.^ 7a'
Example 2. ^i^=^ip or a-^h-x-y-^,ov ^,^-,^,y, ,
_6 1 1 11
Examples. 4 ^=— = — ^~— 17^—7:7^'
4I (^4)5 25 32
EXERCISE 93.
Express as roots of integral powers ;
4 2 2
1. a'T. 4. b^. 7. x'^.
i i_o 5
2. a2. 5. ^7 . 8. 83.
3. x^, 6. yi 9. 27i
10. it\i.
Write with positive exponents :
11. a;2y-3. ^^ lar^b-^c ^^ 2x(x'^-y^)-^
12 a-i^>2c-2 ' ^xy-'^z-'^' ' Sy(x-y)-'' '
*^^ ^ (a— ^)(a4-^)^
14. 3x-5y22-3. ^_1 05-4^-1 2 _3
ly. -. — 5 — . ^, x^y '^
2 xy-^z-^ 24. ^
15 -r27/-is-2 ^ ^-1
16 5^' 21 (^^-^)"' 25 <^'"^'>"^ .
328
ALGEBRA
Write as
roots of positive integral powers.
26. a~t
2
29.
■■ 2 ■*
31.
1
i*
27. x~^.
^
2/ «
28. A.
C
30.
33. (^^-^)"f.
32.
2(a + 5)-t
Find the value of
:
34. di
37.
625"*
40.
(-64)-*.
35. si
38.
(»)-'■
41.
125~*X3^
36. 27" i
39.
©-'■
42.
8*x9-t.
43
25"^
• 81""^
" &
/1\|
226. Having defined operations \oith fractional and negative
exponents as being required to satisfy the laic a'"-a" = a'"^", toe
loill 71010 shoio that they icill satisfy the other latos of exponents^
also : viz : .^''' . '■'
Law B^ ar-T-a"- a'""".
Law (7, (a'"y = a'"".
Lawi>, (aby = a"b".
Law JS;
(» -F
I. When the exponents
are fractional.
(1) a-n-^.
-a'» X>, § 221.
=i;^^i/ F § 67.
Hence, law D is satisfied.
=K^ js;§22i.
=^r^ § 68.
y b'"
m
Hence, law J^ is satisfied.
II. When the exponents are negative.
(1) 6«-"*^a-"=a-'"a'\ § 225.
=«-'»+". § 222.
330 ALGEBRA
Hence, law B is satisfied.
(2) (^a-)-^=(L}j-^ §225.
. = (<:r)" § 225.
= cr\ or a'-"'''-''' E, § 221.
Hence, law G is satisfied.
(3) W-"-=(i'^ §225:
1
1 1
§221.
= a-''-b-\ § 225.
Hence, law D is satisfied.
(4) vv y«y' §225.
^*
= ^*
a"
Hence, law E is satisfied.
225.
227. The consideration of the laws when the exponents
are surds or imaginary numbers will not be taken up in this
book.
228. In the following exercise we may make use of the five
laws of exponents, which have now been interpreted for nega-
tive and fractional exponents, as well as for integral exponents.
FRACTIONAL AND NEGATIVE EXPONENTS 331
All radical signs should first he replaced by fractional expo-
nents.
Example 1. Simplify a-^- a*.
2 3 8 9
1 7
=a;^^.
' Example 2. Simplify a?~^-^a7"^".
-2 _ 3 _3_/'_ 3\
4,3
■x
To
Example 3. Simplify (4a"*6~^)~^.
3'
4^
8
3/-^ / I-. 2 _1
Example 4. Simplify y=I:-(^) -'^•
_i
332 ALGEBRA
3,— 3, 3,—
Example 5. Multiply yx^—Vxy + vy^ by ^/x+^/y.
3/— 3 3 — 2 1_ 1 a
Vx^—Vxy+Vy^=x^—x^y''^+y^.
5./73.r/2
3 3 -J ^
8 It 2
x^—x^y^ + y'^
x^ + y^
2 1 12
_? 1 12
+ a?^ 2/ x^y'^ + y
X +y
EXERCISE 94.
Simplify :
1 9^-2.7.-3.^7 _7 3 1 1
2. a
1 .3 16. 6a 2^.3^(3a */> i).
3. a2.^-2.«2. ^^ ^^
2 1 16. (« 2^ V^-
4. 5iK a-3ie2. _2 1 3
4 1 17. (4.^ 3y 2^)-^^.
6. a-2-^a-4. 18. {-^la-^x 4) 3.
7. x-i^x"^. 19. (32^-tyH^)"i
8. a2^-3_^f^-i.7.-i, 20. i/av'aae.
9. a-lJ-2c-3-^«3^;2e-4. 21. T^-P^-^1>^.
_4 _7. 5 , 5
10, 711 ^—771 5. 22. i/a-2a;-i---va-i 23.-6.
1^- y'-^y-'- 23. i>v^-^-2.
12. 5tV^(^-i^.-A). 24, |/^3^^^2.
13. aHic~^--(aZ»"3c"i). 26. y x^i/^x^y'x^^.
FRACTIONAL AND NEGATIVE EXPONENTS 333
o/j V^^ ya^b^ 30. Qi/a-^---2V cr^.
26. "i/—^-^ 12
Va^b Va^b^ „^ _i
^ 31. (0^2 _y2) 2.^/(^_^.y)3.
28. i^a/«V«^)^ 33 ^/^JaW
29. a"(-«-'^). • I ^-3 • \^,-i^
4 2.2. 44 22 4
34. {x^^-x^y^^y^){x^—^y^-^y^).
35. (a^— a^ + a^-a~i)(a^ + a^).
36. {x^ ^x^y^ -\-y^){^ —y^).
37. (2aj-i-3cc-7)(5 + 4i«-i).
38. (i>a4 + T/a2i>^2 4_-^^4-)(^^^^2_^V^2).
39. {Vx'^ — \^'xy^\/y''--){Vx^ yy).
/ 4 6 \ / 2 \
40. (-X^— 3^ + 9)(-x:r-+3].
\l/a;4 yx'^ JWx'^ J
41. (a + 2i^a2_3^>--)(2-4a~i-6a~4).
42. {x^+y^)^(x^ + yi),
44. (x'^—2 + x~^)~{x-^—x~'^).
45. (^>5 + 2^»-5 + 7)(5-3/>»-5+2^>5).
46. (i/^ + v'^+l)(£c~^+a;~i + l).
47. (8a-2-8a2+5a6_3^-6)^(5^2_3^^-2)^
48. (a^ + Z>~*)2. 50. (x^ + xi + l + x~~^-{'X~^)^-.
49. (a;~i-2/^)2. 51. (ai + ic~^)(a^-a;~4).
334 ALGEBRA
62. (3£c~^-4y^)(3a;~* + 4y^). 57. {a'' -\-b^)-^{a^-\-h^),
53. {a-4:h)^{ya-'li/b). 68. {x-y)^{\/~x-vy).
^ ^ ^ ^ 59. 8 3(252-81M.
55. (a-2--9)--(a-"-3). ^ ^
66. (a-4-l)--(a-i-l). 60. 27^(27^+3^).
EXERCISES FOR REVIEW (VI).
1. Define ratio ; terms of a ratio. Illustrate each.
2. What are commensurable numbers? Incommensurable
numbers ? Illustrate.
3. What is a j^?rop(9r^^o?^? Illustrate. Name the ^erws of a
proportion.
4. Show that the product of the extremes in any proportion
equals the product of the means.
6. What is a test of the correctness of a proportion ?
6. Is 2a V : 3aic^=6aic^ : Stc^" a true proportion? Give
proof.
7. Give five fundamental principles in proportion, and illus-
trate each.
8. Define mean proportional ; third p>roportional ; fourth
proportional.
9. Find the mean proportional between ^x and ^x^ ; between
72aj^y and ^x^\f\ between 4(a + ^) and 9(a + ^)^; between a
and X.
10. Find the third proportional to 1 and 8 ; tox* and l^x^a^ ;
to -l^^y ^^^ 25y^; to — t~x ^'^^ {a-\-by.
11. Find the fourth proportional to 9, 7, and 3; to 1, x^^ and
2y ; to !«, 2aJ, and bx ; to x—y^ ^^~y^ and 1.
12. What is a continued proportion ?
FRACTIONAL AND NEGATIVE EXPONENTS 335
13. Show that if -=-==-=-, then — , — =—, —
X y z w* x-\-y z + w
14. Show that if t = ^i^ then -j= \ ^
b cV d ^c'-^d'
16. Define a variable. A constant. Illustrate each.
16. What is direct variation?
17. If £c oc 2/, and £c=6 when y=4, find y Avhen £c=2.
18. If icoc y, and a;=10, when y=l, write the equation con-
taining X and y.
19. If icocy, and when i»=f, y=9, express the equation
between x and y.
20. What is inverse variation ?
21. Given that x varies inversely as a^ also that when iK = 2,
a =5. Find aj when a = 10.
22. Given that x varies directly as y and inversely as z.
W^hen jc=3, y=5, and z=l. Find 2 when aj=3 and y=15.
23. What is meant by the limit of a variable ?
24. What is an infinitesimal? An infinite number? A
finite number ?
25. What is the value of - ? Of^? Of^? Of—?
iC QO
26. Express in symbols the five fundamental laws of ex-
ponents. Prove each when m and n are positive integers.
27. Simplify ««•«"; a'^a'-, {ay-, {ct'b'f; (^^'
2, m 3
28. What is the meaning ofa^? Of «'» ? Show that a^
means ya^. ^
29. Simplify (a^)^^' ; {x'Y ; (2aM)«.
30. Find the value of 4* ; 8^ ; 25^ ; (J^)i
336 ALGEBRA
31. What is the meaning oia-'^ ? Of «-" ? Show how this
interpretation of a negative exponent is derived.
ct ^ 2 1* (/ — ^
32. Write with positive exponents -Trr2 ; q-^'^^zTg ; 2(a—b)-^.
33. Find the value of 4~^; 5-2^5-3; 4~^; 25"i;(-8)t-
34. Simplify (a-^y^ ; (a'^fi; (i/^^T^; (aJ^y-^)-^ ;
35. Write the value of (x'i-^-y-^y; (2a-^-Sb~^y;
36. What is an ineqiiality ? Distinguish between con-
ditional inequalities and identical inequalities. To which
class does a^-^b^^^ab belong? To which class does
^3>^| + 5 belong?
37. What is meant by the solution of an inequality ?
38. Solve the conditional inequality in Exercise 36.
39. Given | ^""12^=-^^' ^^"<^^ ^^^ ^"^^^''^ ^* ^ ^^^^ V-
40. A teacher, being asked the number of his pupils, replies
that twice their number diminished by 7 is greater than 29,
and three times their number diminished by 5 is less than
twice their number increased by 15. Find the number of
pupils.
41. Three times a certain number plus 16 is greater than
twice that number plus 24, and | of the number plus 5 is less
than 11. Find the* number.
ia
— 23
CHAPTER XXII.
PERMUTATIONS AND COMBINATIONS.
229. In this chapter we shall discuss an important class of
problems which the following examples will illustrate.
Example 1. In how many ways can a program be arranged
consisting of a solo, a debate and an oration ?
By putting any one of the three numbers first and each of the
remaining two numbers after it, we get the following six ar-
rangements : (1) Solo, debate, oration ; (2) solo, oration, debate ;
(3) debate, solo, oration ; (4) debate, oration, -solo ; (5) oration,
solo, debate; (6) oration, debate, solo.
Example 2. How many different numbers of two digits each
can be formed from the digits 2, 3, 4, 5, using each digit but
once in the same number ?
We can take each with one of the others ; hence, we get the
following twelve numbers:
23, 24, 25, 32, 34, 35, 42, 43, 45, 52, 53, 54.
These examples illustrate the general problem of finding the
number of arrangements of a certain number of things taken
from a given number of things.
230. Permutations. All the possible arrangements that can
be formed from the different groups of ..r things which can be
taken froni n different things, are called the permutations of
n things taken r at a time.
Thus, the permutations of the three letters a, &, c, taken one at
a time, are a, 6, c Taken tn^o at a time, they are a6, 6a, ac^
22 337
338 ALGEBRA
ca, he, cb. Taken three at a time, they are abc, acb, bac, bca,
cab, cba.
It is clear that two permutations are different unless they con-
tain the same things arranged in the same order. Thus, ab and
ba are different permutations.
The number of permutations of n distinct things taken r at
a time is represented by the symbol ^Z'^.
Thus, 3P2 represents the number of permutations of 3 things
taken two at a time.
4P2 represents the number of permutations of 4 things taken
2 at a time.
jf^Pg represents the number of permutations of 10 things taken
6 at a time.
231. The value of ^Z',., From the illustrations in § 230 we
have seen that ^P^ = S, ^P^—^^ 3^3=6. There is a law by
which the number of permutations in any case may be writ-
ten. The law may be derived by the use of the following
principle :
Jff" a thing can be done in a dijlfere7it tcai/s, and a second
thing can he done in b different ways without interfering with
the firsts there vnll he ab wags of doing the tico things.
The truth of this principle is evident.
To illustrate it, suppose that 5 boats are plying between two
cities. Find the number of ways in which a person may go and
return from one city to the other by a different boat.
Evidently, in going, he has the choice of any one of the 5
boats. In returning he has a choice of any one of the 4 not used
in going; hence, with any one of the five choices he has four
others, or 5 x 4 in all.
The number of ways that n things can be taken from n
things one at a time is evidently n. Hence,
nP.=n. (1)
PERMUTATIONS AND COMBINATIONS 339
From 9\ things one thing can be taken in n different ways,
and after tliis is done a second thing can be taken from the
remaining n—1 things in 71 — 1 different ways. Hence, from
the preceding principle, there are ?i(/i—l) ways of taking the
two things. That is,
^P,=?i(n-1). (2)
Thus, 5P2=5-4=20; i2P2=1211=132.
After the first two tilings have been taken in any one of the
n(n—l) ways, the third thing can be taken from the remain-
ing ?i—2 things in. n—2 ways. Hence, there are n{?i—l)(n—2)
ways of taking the three things. That is,
,,P,=n(n-l){n-2). (3)
Thus, 4P3=4-3-2=24. ^,P^=10-9-8=720.
In like manner,
,P,=n(n-l){n-2)(n-8)(n-4.); ^' (5)
,P,=n(n-l)(?i-2)(7i-S)(n~4:)(n-b) ; (6)
and so on. ^5 4 ^ X /
Kow from (1), (2), (3), (4), (5), and (6), Ave see that ^P, has
1 factor ; ^^P^ has 2 factors ; ,,P^ has 3 factors ; ^P^ has 4 fac-
tors ; jjPg has 5 factors ; ^Pg has 6 factors ; and so on. And in
general „P^ will have r factors, the last one being 7i—(r—l),
or n — r-\-l. That is, ^ <>- „ = /7(/7-l)(/7-2) 4 3 21. (8)
340 ALGEBRA
232. Factorial-/?. In (8) of § 231, the product .
?i(7i-l){7i-2) 4-3-21
is called factorial—/?, and is usually designated by the symbol
I )i, or n\ .
Thus, |^ = 6-5-4-3-21=720; 5! = 5-4-3-21 = 120.
Hence, from § 231, we have
Example. In how many ways can 5 books be arranged on a
shelf ?
The number =,P,, or |_5=5 4-3-21=120.
233. When the things permuted are not all different.
In many problems the things permuted are not all different.
We sliall now determine the number of permutations in such
cases.
Suppose that a of the n things are alike, and suppose that
we form the iV^ permutations of the ??- things taken ?i at a time.
Now, if in any one of these permutations the a like things be
replaced by a unlike things, different from all the rest, then by
changing the order of these a new things only, we can form
I a new permutations from the one permutation. This can be
done in the case of each of the JV permutations. Ileuce, in
all, ]}^\a new permutations can be found. Therefore,
JSr\a = „P,, (all different),
and N = '^=^.
Similarly, it can be shown that, if a of the n things are alike,
and h others alike, then
PERMUTATIONS AND COMBINATIONS 341
And if a of the n things are alike, h others alike, and c others
alike, then
\n
^ \a \h \c '
and so on.
Example 1. How many different numbers can be formed by-
using all of the figures 2, 2, 2, 3, 3, 4, 5, 5, 5, 5 ?
110
The number = -• =12600.
li It li
Example 2. Find the number of permutations of the letters
of the word Indiana.
Here there are two e's, two n'fe, two a's, one d. Hence the
number = j^ g ^^ =630.
EXERCISE 96.
Find the value of :
1- li- L^ li 12. ,p,. '
2. 16. '• ^ ■ ''■ '^^- . ,
|2 13 14 14. 1.A- V'
3. |3|5. 8.^^- 15. „P,.
4. |4 |2. [6 |8 16- i«^«-
17
9. -n^- 17. ,,A.
'• T 14 15 16 ''• 'J-
[8_ ^°- Tir' 19- #
342
ALGEBRA
Show that
21. n(n-l){n-2) \n-d
\n-l
22. 1 --\n-2.
y^
!y('^^y^^X
n-l
23. \a • \a • (a + l)~a = \ a + l ■ \ a — l.
24. How many numbers of three digits can be formed by
using the digits 1, 2, 3, 4, 5, using each digit but once in tlie
same number ?
25. How many numbers of two digits can be formed bj^ using
the digits 2, 4, 6, 8 ?
26. How -many permutations can be formed of the letters of
the alphabet taken twji-at a time ?
27. In how many different ways can 5 boys stand in a row ?
28. If 8 steamers ply between Liverpool and New York, in
how many ways can I go by steamer from New York to Liver-
pool and return by a different steamer ?
29. Six ladies and six gentlemen are to be seated about a
circular table. In how many different positions can they be
seated so that there shall be a gentleman at the right of each
lady?
30. In how many ways can a class of 15 pupils be seated in
15 seats?
31. In how many ways can 2 different prizes be awarded to
10 boys so that no one boy gets both prizes?
32. How many permutations can be made of the letters of
the word A?ina ? The word Missouri ?
234. Combinations. The different groups of r things that
can be taken from n things, when the arrangement is not jcmi-
sidered, are called the combinations of the ?^ things taken r at a
time.
PERMUTATIONS AND COMBINATIONS 343
Thus, the combinations of the letters a, 6, c, d^ taken two at a
time, are ah, ac, ad, be, bd, cd. Taken 3 at a time, they are
abc, abd, acd, bed.
It is clear that two different combinations can not contain
the same things arranged in different orders.
Thus, abc and acb are the same combination.
235. The number of combinations of n things taken r at a
time is represented by the symbol „Cr.
Thus, 4C3 represents the number of combinations of 4 things
taken 3 at a time.
236. The value of „^^, The number of combinations of n
things taken r at a time is easily found by establishing the
relation between „ C^ and „P^.
Suppose n different things combined r at a time. . Every
combination of the r different things will have |r permutations,
taken r at a time. Hence, the total number of permutations
will be „CV Ir ; that is,
Therefore,
Thus,
237. If, in the value for ^^C\ found in § 236, we replace
„P,. by its value found in § 231, we get
_ n{n-1)(n-2) ■ (/i-r+/)
nC/^ It;; • v-"-;
It is sometimes useful to express the value of „ C^ in a dif-
ferent form.
nCr
L=nP.
,Cr =
C?a =
5-4-3
~3-21
10.
344 ALGEBRA
Multiplying the numerator and denominator of the fraction
in (1) by \?i—r, we get
7i(n—l){n—2) (n—r-i-l)\?i—r
r \n—r
or ^C=r-^=— (2)
r \n—7'
n^ r
If we replace rhjn—r in (2), we get
\n In
C = ^= =_!=___. (3)
^ ""'' \7i—r \7i—n + r \n—r \r
From (2) and (3) it follows that
nCr = nCn-r' (4)
Thus, 50C48 = 50C2=^ = 1225.
EXERCISE 96.
Find the value of :
1. ^Cy 3. 80 ^n-
5. 12 Cg. 7. gOg.
9' 25620-
2. joCg. 4. 15612.
6. 8 63. 8. 206^3.
10. ,C',.
11. Find the number of combinations of 12 things taken 2
at a time ; taken 3 at a time ; taken 9 at a time.
12. How many selections of 3 books can I make from 5
books ?
13. How many combinations can be made from the letters
a, b, c, d, e, taken three at a time ?
14. How many combinations can be made from the 26 let-
ters of the alphabet, taken 2 at a time ?
15. How many . different products can be formed from the
numbers 2, 4, 6, 8, if each product contains 3 unequal factors ?
PERMUTATIONS AND COMBINATIONS 345
16. In how many ways can a committee of 3 be selected
from 8 men ?
17. How many different committees can be formed from 10
Republicans and 6 Democrats, if there are 2 Republicans and
1 Democrat on each committee ?
18. There are 8 points in a plane, no three of which are in
the same straight line. Find how many lines can be drawn,
each connecting 2 points.
19. In how many ways can 3 red balls and 2 white balls be
selected from 8 red balls and 5 white balls ?
20. In an algebra class of 25 students, 20 recite each day.
In how many ways can these 20 be selected for one day ?
21^' A farmer has 7 Borses. In how many ways can he
hitch a two-horse team ?
22. In an examination a teacher gives a pupil the choice of
any 8 questions out of 10. In how many ways can the pupil
choose his 8 questions.
9
^
C.
^ 1°
Ti
n
^/
CHAPTER XXIII
THE BINOMIAL THEOREM.
238. In § 63 it was shown that any positive integral powier,
of a binomial can be written clown by some laws which taken
collectively constitute the binomial theorem.
Thus, by these laws,
{a' + 2by={ay + 4{a'f{2b) + Q{a'y{2bY + 4{a'){2by + (26)*
=a^ + 8a^b + 24a*b' + 32aW + H)b\
In the general case, it will be found that the laws in § 63
will give
It will be recalled that no rigorous proof of this theorem
has been given. The binomial theorem can be proved to
hold true for all exponents, integral or fractional, positive or
negative. We shall prove the theorem here for the case when
the exponent is a positive integer.
239. Proof when the exponent is a positive integer.
From the rule for obtaining the product of two expressions,
which is based upon the distributive law, it follows that a term
in the product of any finite number of expressions can be ob-
tained by multiplying a term of any one of the expressions by a
term from each of the other expressions.
A repetition of this process in every possible way will give all
of the terms in the product of the expressions.
346
THE BINOMIAL THEOREM 347
For example, consider the product {a + b){p + q){x + y). In
finding the product ot a + b andp + g we multiply each term of
a + bhy each term ot p + q. Each of these resulting terms is then
multiplied by x and by y to obtain all of the terms in the product
of the three binomials. This process evidently amounts to the
use of the above rule.
. Consider the expression (a + Z>)^ This may be written in
the form
{a + b)(a-\-b)(a-\-b) to ^z factors.
If we select a tenn from each of the n factors in this product^
and multiply these terms together^ and do this in every pos-
sible way^ we shall obtain all of the terms in the product.
(1) Now the term a can be selected from all of the factors
in just one way. Hence, the product of these a's, which is
a'\ is iho, first term of the product.
(2) The term b can be selected from one factor and the term
a from each of the other n — 1 factors ; and this can be done in
as many ways as b can be selected from the n factors, which
is n or „ C^. Hence, a"~^6 can be selected in n ways, or ^ C^
ways ; that is, na'^~^b^ or „ C^a'^~'^b^ is the second ter^tn of the
product.
(3) The term b can be selected from each of two factors and
the term a from each of the other n — 2 factors ; and this can
be done in as many ways as two 5's can be selected from the n
factors, which is „ C^. Hence, a'^-'^b^ can be selected in „ C^ ways ;
that is, „ C^a'^'^b'^ is the third term of the product.
(4) The term b can be selected from each of three factors
and the term a from each of the other n—^ factors; and this
can be done in as many ways as three ^'s can be selected from
the n factors, which is „ C^. Hence, a'^'W can be selected in
„ C3 ways ; that is, „ O^a/^'W is th-e fourth term of the product.
348 ALGEBRA
(5) Let this process be continued. In general, the term h
can be selected from each of r factors and the term a from
each, of the other n—r factors ; and this can be done in as
many ways as r i's can be selected from the n factors, which
is „(7^. Hence, a'^-^'W can be selected in ^G^ ways; that is,
^C/i^'-'lf is the (r+l)^A term, of the product.
(6) Finally, the term h can be selected from all of the fac-
tors in just one way. Hence, If can be selected in just one
way ; that is, 5" is the last term of the product.
Hence, (aH-6)"=fl'* + „(?ia'^-^6-h„(?2a"-='6'+„(?3a""'6'+
+Xa"~'*6"+ +6"- (1)
The expaiision in (1) expresses in symbols the binomial
theorem. If, in this identity, „ Ci, „ C^^ „ G^ etc., are replaced by
their values, the identity becomes
^ . . ^^ n{n~1){n-2)-^ ■ ■ ■ („-.+ / )^_^^^ _^^„ ^^^
It is seen that (2) conforms to the laws of § 63.
When n is a positive integer it is easily shown that there are
always /? + 1 terms in the expansion.
When r^ 71, the coefficient of the (r+l)th term (t. 6., the
(n + l)th term) is 1 ; but when r=?i + l, the coefficient of the
(r + l)th term becomes
n{n—l){n—'^) {n—n — 1 + 1)
n + 1
which is 0, since the last factor in the numerator is 0.
Hence, a term does not exist in which r is greater than n ;
that is, there are only n-\-l terms.
Example 1. Expand (a^ + i/^)^
Here n=5. And fi^=^, 5^=10, 503=10, ^C,=^.
THE EilNOMIAL THEOREM 349
Hence, {x' + y'f={x'r + Mx^W) + ,C,{aff{yy 4- .C^ix^Yifr +
\c\{x^){yy+(yr
=ic" + 5x^y^ + lOx^y* + lOxV + 5x^y^+ 2/^^
Example 2. Expand (2— ar*/.
Here, ti=4. And ^0^=4, ^0^=6, fi^^^.
Therefore, {2-x^)'=i2y + ^{2f{-x^)+^C^{2fi-0(^Y +
,C3(2)(-^/ + (-^r
=lQ-32x^ + 2Ax^ - %x^ + x^\
Examples. Expand (a — 2& + c)^
Grouping terms, this becomes [(a— 26) + c]^
Hence, [{a-2h) + cY={a-2hf + fi^{a-2hfc-\-fi^{a-2hy + (^
=(a-26)» + 3(a-26)2c + 3(a-26)cHc^
Expanding each term, we have
[(a-26) + c]^=a3-6a2& + 12a&2-86^ + 3a='c-12a6c + 126='c + 3ac^-
66cHc^
240. The general term. The (r+l)th, or general term,
in(l)of §239is„(?^a"-'-6'-,or-^ ^- ^— — ^ ' —
By substituting the values of a, ^, ?i, r, in this expression,
for the (/•+l)th term, we may write down at once any desired
term of the expansion of any power of any binomial.
Example 1. Find the 8th term in the expansion of {2x—iy^.
Here, a=2x, 6= — 1, w=10, r=7.
Hence, the 8th term =^^C,{2x)\-iy=^^C.,{2xf{-iy
=i|^-8x»(-l) = -960a^.
EXEBCISE 97.
Expand :
1. {x-Vyy. 3. (l + 2aj2)*. 6. (l-3a=')».
2. (2ic-3y)«. 4. {^a'-Vhy. 6. {x'-ay..
350 ALGEBRA
7. (i+2xy. ^^ ^^_^i^e. . 14. (^^-wy.
8. (l-ixy, ^^ (a-2+5-2)^ 15. ra-i + dy,
9. {x-^+xy. /,j 2by ^/ 5
10. (2a;-2 + l)^ -^^V V^^"^/ * 16.' {x'-y'^)\
17. Find the third term in the expansion of (l + 2a^y.
18. Find the sixth term in the expansion of (cc^ — 2y)^*'.
_i 2
19. Find the eighth term in tlie expansion of (x '^—a^y.
20. Find the fifth term in the expansion of (-^ — *^
21. Find the fourth term in the expansion of (|«^^ + |c~2)i^
-2. 2.
22. Find the seventh term in the expansion of {x '^+x'^y^.
f V^ ^*^
23. Find tlie sixteenth term in the expansion of { 1
X
24. Find the twelftli term in the expansion of (2— la?^)'*.
By grouping terms, express as binomials and expand :
26. {\-^x-xy, 27. {2-a-\-hy. 29. {ci~h-\-c-dy.
26. (£6^ + 2/' + ^')'. 28. {X^x^-x^-xy. 30. {2a-h^Zey.
24 J. Binomial theorem,— exponent negative or fractional.
When the exponent -is a negative number or a fraction the ex-
pansion of a power of a binomial as in § 239 gives an inde-
finitely large number of terms. This follows from the fact
that, for such an exponent, the coeflBcient,
n{n-V){n-2) • • '\n-r^-\)
of the general term can never become zero.
It can be shown, however, that the use of the binomial theo-
rem in such cases is allowable, pravided that the absolute
value of a is greater than that of h. It is advisable to exclude
the proofs from this book. The student may assume that the
theorem holds in the exercises that follow.
THE BINOMIAL THEOREM 351
Example 1. Expand {2—x^)-^ to 5 terms.
We have i2-x')-^={2)-'+i-3)(2)-'{-x') +
(_3)(--4)r-5)(-6)^^^_,^_^y^
— 1 I 3 '«»2 1 3 /y.4 I 5 ^6 r 1 5 /y.8 I
— ¥ ' T'S^*^ ^TS"^ ^H'S'^ ^TS'^'^ ^
_3
Example 2.- Expand (a + 2a?)* to 5 terms. ^
We have {a + 2x)^={a)^ -\-{f){a)~^{2x)+^i^i:^{a)~^{2xy +
ci)(z
=a*+|a ^a?— fa ^a^^ + ^a ^o?-*^— jVgCt ^ a?*+
1 2- O , ^
Example 3. Expand (i—x) ^ to 4 terms.
We have (l.-a?r^==(l)"^ + (-i)(ir"2(-a^) + ti)H)(l)-|(_ic)
4-tMJ.Kd)(l)-i(-;rf
• (iK=i^t:4)(«)-l-(2^)3^(J0G_^ . . . .
' +
=l+4a?+|a?'.+ A^ +
EXERCISE 98.
Expand to four terms :
1 /I — oA-2 2 4
^ ^ * 6. (1-x'y. 10. (ic + 2r's-.
3. K-.r^ '' ^^^'^^^' . 11. (8 + .)i
4. (2x.-3y)-^. 8- (20.^-1)"^.
6. (1 + a;^)-^ 9. (i + a;)"^ 12. (2-ic2)-J.
In its expansion find :
13. Tlie 6th term of (2-a;)-3.
352 ALGEBRA
14. The 10th term of (a'^c^)^.
15. The 5th term of {ei-e'^)-^.
16. The 8th term of {l-2x^yi.
242 . Extraction of roots by use of the binomial theorem. The
binomial theorem may be used to extract roots of arithmetical
numbers. The process is best shown by an example.
Example 1. Find to 4 places of decimals the value of y'W.
(i)H)H)(25)-V-(_3).+ . .
2- 3 9 81
80 6400 1024000
=2-.0B75-.0014-.0001-
=1.9610 approximately.
A similar process may be used in any case. Hence the fol-
lowing rule :
^Separate the number into two parts^ the first of ichich is the
nearest possible perfect power of which the required root can be
found ; then expand the resulting binomial by the binomial
theorem^ and combine the values of the terms thus obtained.
Thus, v65 = v'64 + l = (43 + l)^; i>623=|/625-2=(54-2)*.
It is evident that the first few l^erms of the expansion will
give a close approximation to the value of the root, if the suc-
cessive terms decrease in value rapidly ; i.e.^ if the second
term of the binomial is much smaller than the first. By
§ 241, no correct approximation of the root can be found by
. THE BINOMIAL THEOREM 353
this method unless the given number is expressed as a bino-
mial in which the second term is less than the first.
Note. — A shorter method of finding the approximate value of any
root of a number is by the use of logarithms, discussed in Chap-
ter XXVI.
EXERCISE 99.
Find to four places of decimals the value of :
1. 1^15. 3. v2T9. 5. vMT. 7. 1/3120. 9. t>730.
2. 1/240. 4. yd. 6. y^M. 8. i/26. 10. 1/21.
23
CHAPTER XXiy.
PROGRESSIONS.
243. Series. A succession of terms, in which each term after
the first may be obtained from one or more of the preceding
terms by some fixed law ; ^.e., obtained in the same way for all
terms, is called a series.
Thus, 2 + 4 + 6 + 8 + 10 + 12+ is a series. Each term
after the first may be obtained by adding 2 to the preceding term.
Also, l + x-]-x^ + oc^ + x*+ ' ■ ' • is a series. Each term after
the first may be obtained by multiplying the preceding term by x.
A finite series is one that has a finite number of terms.
An infinite series is one that has an infinite number of terms.
A series is called convergent either when the sum of all of
the terms equals a fixed finite number ; or when the sum of
the first n terms approaches a certain fixed number as a limit,
when n is indefinitely increased.
A series is called divergent when the sum of the first n
terms can be made greater than any assigned number which
may be as great as we please, by taking n sufficiently great.
A finite series is ahoays convergent.
In the series 2 — 2 + 2 — 2 + - • • •, the sum of the first n
terms is either 2 or according as 7i is odd or even. Such a
series is called an oscillating series.
244. We shall discuss in this chapter three special forms of
the simpler series, known as arithmetical progressions, geometri-
cal progressions, and harmonical progressions.
354
PROGRESSIONS 355
ARITHMETICAL PROGRESSIONS.
245. An arithmetical progression is a series of terms in
which the difference between any term and the preceding
term is the same for all terms of the series. This difference is
called the common difference, and may be either a positive or a
negative number, integral or fractional. The name arithmeti-
cal progression is usually abbreviated to A. P.
Thus, the series 1 + 3 + 5 + 7 + 9 + 11+ • • • • • is an A. P. in
which the common difference is 2.
And the series 12 + 8 + 4 + 0-1-8-12—14— • • • is an A. P.
in which the common difference is —4.
246. The nth term of an A. P.
Let a stand for the first term of _an A. p.,
/ for the nth term,
and d for the common difference.
Then, by the definition of an A.P.,
the second term = a-\-d,
the third term = a-\r'2id^
the fourth term = a + 3c?,
the fifth term — a + 4id^ etc.
It is evident from these expressions that the coefficient of d
in the expression for any term is less by one than the number
of the term.
Hence, the nth term = a-{'(n — l)d;
that is, /=a + (n—l)(f. Formula A.
This equation, or formula, will enable us to find the value of
any one of the four numbers, I, a, n, d, if the values of the
other three are known.
356 ALGEBRA
Example 1. Find the 20th term of the series 4 + 7 + 10 + 13 + • • • .
Herea=4, d=3, n=20.
Substituting in formula A, we have
Z=4 + (20-l)3=61.
Example 2. The 4th and 15th terms of an A. P., are 9 and 31,
respectively. What is the series ?
Here the 4th term is a + 3d=9, (1)
and the 15th term is a + 14S=i-25-| 2(-9) + (25-l)4 1=975.
Example 2. The first term of an A. P. is 2, and the sum of 20
terms is 135. Find the common difference.
Here, a=2, n=20, iS=135, d is unknown.
Froin formula C we have
135=10(4 + 19d).,
Solving this for d, we get d=^.
Example 3. How many terms of the series 3 + 1 — 1 — 3—5— • • •
must be taken to make —140?
Here, a=3, d=—2, S= — l'iO, n is unknown.
From formula C we have
-U0=^n{Q + {7i-l)(-2)}.
Simplifying this gives the equation
n^-4n-U0=0.
The solutions are 14 and —10. The negative solution has no
meaning in this problem, hence 14 terms must be taken.
358 ALGEBRA
Example 4. In a certain A. P. the common difference is 1^,
and the 12th term is 12^. Find the sum of the first 7 terms.
We first find a from formula A, then find S from formula C.
From formula A,
12^=a + lll|.
Hence, a=— 4.
Then from formula O,
Example 5. Find the sum of all the numbers between 100 and
500 which are multiples of 6.
The numbers between 100 and 500 which are multiples of 6 are
617, 6-18, 619, • • • • 6-83.
Hence, the sum is
6-17 + 618 + 6194- +6-83,
or 6(17+18 + 19+ • • • • • +83).
The sum of the A. P. enclosed in parentheses is obtained by
formulas A and C.
From formula A, 83=17+ (n-l)l,
whence 7i=67.
From formula O, >9=-y (34 + 66)=3350.
Hence, the required sum is 6-3350, or 20100.
Example 6. Show that the sum of r + 2*-^ + 3'+ n^
=inin + l)i2n + l).
Note. — While this series is not an A. P. it illustrates an application
of it.
Since (x-\-lf—ocr^=3x^ + Sx + l is true for all values of .t, we may-
give to £c a succession of values 1, 2, 3, etc., and by adding the n
identities, obtain the given series. That is, from
(ic + 1)*— a?^=3a?H3a?+l, we have
when x=l, 2'-l=»=3r + 31 + l,
when x=2, S^-2^=3-2^ + S2 + l,
when a?=3, 4^— 3^=3-32 + 3-3 + l,
when x=n, (ii + l)^— n''=3-nH3-n + l.
PROGRESSIONS 359
Now adding columns, and observing that the second term
of each left hand member cancels the first term of the member
above it, we have
(?i+l)^-P=3(P + 22 + 3H • • • • w'') + 3(l + 2 + 3+ • • • • n) +
(1 + 1 + 1+ • • • • to n terms)
=3(P + 2'^ + 3^+ • • • • n')+pi(n+l) + n. _
Solving for 12 + 2^ + 32+ • • • • »^^ we obtain |n(n + l)(2n + l).
(Let the pupil show how this solution was obtained.)
248. Arithmetical means. If thr^e numbers form an A. P.,
the middle term is called the arithmetical mean of the other
two terms.
Thus, in the series 3 + 8 + 13, 8 is the arithmetical mean of 3
and 13.
If ic is the arithmetical mean of a and b, then, by the defini-
tion of an arithmetical progression,
a + b
whence, x= ^ -
Hence, the arithmeiical tnean betioeen two numbers equals half
their sum.
249. In an A. P. all the terms between any two terms are
called the arithmetical means of those two terms.
Thus, since 2 + 4 + 6 + 8 + 10 is an A. P., 4, 6, and 8 are all
arithmetical means of 2 and 10.
Any number of arithmetical means may be inserted between
any two given numbers.
Example 1. Insert 7 arithmetical means between 10 and 30.
Since there are to be 7 arithmetical means, 30 must be the 9th
term of an A. P. of which 10 is the first term.
Hence, we have a=10, /=30, n=9.
360 ALGEBRA
Substituting in formula A, we have
30=10 + 8d.
Solving, d=2l.
Hence, the required series is
EXERCISE 100.
1. Find the 30th term -in 1 + 6 + 11 + 16+
2. Find the 16th term in -8-5—2 + 1 + 4+
3. Find the 23rd term in — |— 1- + ^ + |+
4. Find the 54th term in 11 + 17 + 23+ ••• •.
5. The 6th term of an A. P. is 17, and the 15tli term is 44
Pind the common difference.
6. The 3rd term of an A. P. is 0, and the 9th term is 22.
Find the common difference.
7. The fifth term of an A. P. is 21, and the 8th term is 33.
What is the 12th term ? The 20th ?
8. The 2nd term of an A.P. is 7, and the 11th term is 20^.
What is the 7th term? The 15th term?
9. Which term of the series 1 + 4 + 7 + 10+ • • • is 46 ?
10. Which term of the series 10 + 6^+3-1-
is -25?
11. In an A. P. whose common difference is 6, the 11th term
is 72. What is the first term ?
12. Find the sum of the first twenty terms of the series
42 + 39 + 36
13. Find the sum of the first thirteen terms of the series
8 + 12 + 16+
14. Find the sum of the first fifty odd numbers.
PROGRESSIONS 361
15. Find the sum of the first fifty even numbers.
16. Find the sum of the first ten terms of a series whose
first term is —6 and tenth term 25i.
17. How many terms of the series 15 + 12+9+ • • • • niust
be taken to make 45 ?
18. In an A. P. whose first term is 8, the sum of tlie first 9
terms is 324. What is the 9th term ?
19. Find the sum of all odd numbers of two digits.
20. Find the sum of all even numbers between 100 and 300.
21. Find the sum of all numbers between 50 and 250 which
are divisible by 4.
22. Show that the sum of the first n odd numbers is 71"^.
23. Show that r+2^ + 3^+ n'= ||(^ + 1) V
Suggestion. See Example 6, §247. Remember that (ic+1)*— £c*=
4a73+6a?2+4£t?+l.
24. Insert four arithmetical means betweeji 5 and 25.
25. Insert six arithmetical means between —10 and \.
26. Find three numbers which are in A. P., such that their
sum is 18, and such that the product of the first and last is
greater than the second by 14.
27. A man had a cistern dug 12 feet deep. The first foot
cost $1, the second 11.25, the third $1.50, and so on. What
did the digging cost ?
28. A body falls toward the earth at the rate of a feet the
first second, 3a feet the second second, 5a feet the third sec-
ond, and so on. How far will it fall in t seconds ?
29. A man pays $50 of a debt the first year, $75 the second
year, $100 the third year, and so on. In this way he pays the
whole debt of $1100. How many years does it require ?
362 ALGEBRA
GEOMETRICAL PROGRESSIONS.
250. A geometrical progression is a series of terms in Avhicli
the ratio of any term to the preceding term is the same for all
terms of the series. This ratio is called the common ratio,
and may be either positive or negative. The name geometrical
progression is usually abbreviated to G. P.
Thus, the series 1 + 2 + 4 + 8 + 16+ • • • ■ is a G. P. in which
the common ratio is 2.
The series 2— 1 + 1 — 1 + 1—3^4- • • • • is a G. P. in which the
common ratio is — |.
In either series, if any term be multiplied by the common ratio,
the product will be the next term.
251. The nth term of a G. P.
Let a stand for the first term of a G. P. ;
I for the nth term ;
and r for the common ratio.
Then, by definition of a G. P.,
the second term = ar,
• the^ third term — ar'^^
the fourth term = ar^^QiG.
It is evident from these expressions that the coefficient of a
in any term is r, with an exponent less by one than the num-
ber of the term.
Hence, the nth t€rm = ar''-^ \
that is, l=:ar"-^. Formula yl.
This equation, or formula, will enable us to find the value
of any one of the four numbers, Z, a, r, n, if the values of the
other three are known.
Note. — Since n is an exponent, its value cannot in general be found,
except by inspection, without the use of logarithms. See Chapter
XXVI. In all of the problems in this chapter n can be found by
inspection.
PROGRESSIONS 363
Example 1 . Find the nioth term of the series 2 + 6 + 18+
Here a=2, r=3, n=9.
Hence, Z=2-3«=13,122.
Example 2. The tenth term of a G. P. is ^-f^, and the first
term is 1. Find the common ratio.
Here a=l, n=10, 1=^\y-
Hence, -^-g =lr^j
and r=}.
Example 3. The fifth and eighth terms of a G. P. are ^f and
— Ill, respectively. Write the series.
Here the fifth term =ar*=^, (1)
and the eighth term =ar^=— |ff. (2)
Dividing (2) by (1), r'=-^;
whence *'=— I-
Replacing r by -| in (1), aff =ff ;
whence a =3.
Therefore, the series is
3 0i 4 8 I 1 6 33 I 64 128 i
^'^5^^fc
252. Sum of n terms of a G-. P. Let the sum of n terms of
a G. P. be represented by /S.
Then, jS=a + ar'\-ar'' + ar'-{- +ar"-2 + ar»-\
= a(l-^r + r' + r'+ • • • +r"-2+-^n-i^
=a(l:Z^\ §76.
1-r ^
Hence, 5=-^^—^. Formula^.
1— r
But, by formula ^, l=ar''-^.
Hence, from formula A and formula B, we get
A a—r/
1-r'
The five numbers, a, r, ?i, /, /^, of a G. P. we call the elements
364 ALGEBRA
If any three of the five elements of a G. P. be known, then
the other two may be found by the use of formulas A, B^
and C.
Example 1. Find the sum of the first six terms of the series
.4M-64-9 + -V-+
We have a=4, r=|, /i=6.
Hence, ^^ 4{l-(fr} ^
Example 2. The first term of a G. P. is 3, the sixth term is
9375, and the sum of the first six terms is 11,718. What is the
common ratio ?
By formula (7, •
•11,718 = 3=^^;
1— r '
whence r=5.
253. Sum of an infinite number of terms of a G. P.
From § 252 we have
\—r
This may be written in the form
S=-
1—r \—r
Now, if r he less than 1, r'* will be less still, and by increas-
ing n sufficiently, r" can be made less than any assigned value
which we may take as small as we please.
Thus, if r=i, t^^tV, ^''=^4, ^'^fK, '^'=-toW, ^"=1™, etc.
Now, at the same time that r^ becomes less than any as-
signed value, ^ will also become less than any assigned
value, however small. Hence, if n be taken sufficiently great,
5 will approach indefinitely near in value to = .
1—r 1—r 1—r
PROGRESSIONS
'6^)0
Consequently, in a G. P. where the common ratio is less than
1, by taking 71 sufficiently great, the sum of n terms, can be
made to differ from :^—-- by a number less than any assigned
value, which may be taken as small as we please.
Hence, the sum of an infinite number of terms of a G. P,
ichose common ratio is less than 1 is defined as the limit
^* That is, when n is infinite and r is less than 1,
1-r
S=A-' Formula J9.
1 — r
Example 1. Find the sum of the infinite series of terms
9 + 6 + 4+ • • •.
Here r=|; hence, the formula >S=T-— may be- applied. We
9
have ^=3— — g=27.
This formula can be used to find the value of a repeating
decimal.
Example 2. Find the value of .3333
This is a G. P. Avhose first term is y\, and common ratio ^V-
Hence, ^=-^=1=,].
Example 3. Find the value of .12232323
We have. 12232323 =TVo+To¥oT7+ro¥o^+To^¥oiroo + -
Hence, .12232323-
•Too T^-J-9 00— 9 900'
254. Geometrical means. If three numbers form a G. P.,
the middle term is called the geometrical mean of the other
two terms.
366 ALGEBRA
Thus, in the series 4 + 12 + 36, 12 is the geometrical mean of 4
and 36.
If X is the geometrical mean of a and ^, then by the defini-
tion of a geometrical progression, we have
h X
Solving, x=i/ab.
Hence, the geometrical mean of two numhers equals the square
root of their product.
255. In a G. P. all of the terms between any two terms are
called the geometrical means of those two terms.
Thus, since 1 + 3 + 9 + 27 + 81 is a G.P., 3, 9, and 27, are geomet-
rical means of 1 and 81.
Any number of geometrical means may be inserted between
any two given numbers.
Example 1. Insert three geometrical means between ^ and 32.
Since there are to be three geometrical means, | must be the
first term, and 32 the fifth term, of a G. P.
Hence, by formula A we have
ir*=32;
whence r=4.
Therefore, the required series is
1 + ^ + 2 + 8 + 32.
EXERCISE 101.
1. Find the ninth term of the series 1 + 6 + 36+
2. Findthetenth term of the series Jg—i + 1—4+
3. Find the eighth term of the series 2 + 3 + 4i+
PROGRESSIONS 367
4. Find the twelfth term of the series 1 2 + -^4~ * ' * ' •
X X
5. The second term of a G. P. is i, and tlie eighth term is
2^g. Find the tenth term.
6. Tlie first term of a G. P. is 3, and the tliird term is 6.
Find the sixth term.
7. The second term of a G. P. is 3, and the fifth term is ^-j-.
Find the fourth term.
8. The common ratio of a G. P. is 3, and the seventh term
is 81. Find the first term.
9. The third term of a G. P. is i, and the eighth term is 128.
Write the first eight terms.
10. Insert two geometrical means between 125 and —8.
11. Insert three geometrical means between 1 and 4.
12. Insert five geometrical means between 2 and ||-.
13. Find the geometrical mean of 6 and 96.
Find the sum of :
14. Five terms of i + i + |+
15. Twelve terms of 2—4 + 8-
16. Ten terms of If + 22+5/3
17. Six terms of 64-32 + 16- ..••..
18. Ten terms of -f +i.-5_|_ ......
19. In a G. P. whose first term is 1, and common ratio 4,
how many terms must be added to make 21 ?
20. In a G. P. whose first term is 1, and common ratio — i,
how many terms must be added to make |^ ?
Find the sum of an infinite number of terms of :
,21. 15 + 5 + 1+ ......
22. -3-^i-^V+ •
368 ALGEBRA
23. I. + 1 + -J+ ,
24. i + i + i+ • .
25. 5-3 + 1-
26. In a G. P. the common ratio is i. What must be the
first term in order that the sum of an infinite number of terms
may be 80 ?
27. In a G. P. the first term is 5. What must be the com-
mon ratio in order that the sum of an infinite number of terms
may be 4y-p=o,
or A=a, B=b, C=c,- P=p.
(2) One, or both, series infinite. In the convergent infinite
series a + bx-\-cx- + dx^ + ex*+fx^^gx^+ * ' * '? which is written
in ascending, positive, integral powers of x, any term whose
value is not zero may be made greater than the sum of all
terms that follow it by making x sufficiently small.
Let us choose the term clx^. Now let k be greater than any
coefficient following d. Then kx\l + x + x^ -}- x^ + • • • • ) is
greater than ex* + fx^ + gx^ -{- Or, since
l + x + x' + x'^ • • • • =Y37-,
X X
kx*YZ^ is greater than ex*+fx^+gx^+ *
1 kx
But dx^ is greater than kx*YZZ~'> i^ i _ is less than d.
X X X X
Now, since d is not zero, .. _ can be made less than d, for
by taking x sufficiently small, the numerator may be made less
than any assigned value, while the denominator will approach 1.
* When some of the terms are negative and the negative signs before
the terms are changed to positive signs, a new series greater than the
oUl arises; lience if the principle is proved for all terms positive, it is
evidently true when part of the terms are negative.
378 ALGEBRA
Hence, dx" can be made greater than ea;*+/£c^ + <7£«^-f
In like manner, ayiy term whose value is not zero in the
above series can be made greater than the sum of all the fol-
lowing terms.
If, now,
A^Bx^Cx^^Dx"-^ =a-\-hx-^cx^-^dx^\
be true independently of the value of a?, we have
{A-a)^{B-h)x-^{C-c)x^-V{D-d)x'^-\- • • • • • =0.
Now, A— a must be either zero or not zero; and if ^— a
were not zero, by taking x sufficiently smalL^— « could be
made greater than the sum of all terms that follow. But this
can not be possible, since the sum of the whole series is zero.
Therefore, ^— a=0, qy A=a.
Removing A— a, it can be sho^n in like manner that B=h\
thence removing B—h^ it can be shown that C=c ; and so on.
263. Expansion of fractions into series.
A fraction may be expanded into a series by the use of the
principle of undetermined coefficients.
Example 1. Expand ^ ■ in ascending powers of x.
Let us assume
=A + Bx+Cx' + Dx^-
1 + x
where A, B, C, etc., are independent of x.
Multiplying by 1+a?,
l + x'=A + {A + B)x+{B+C)x' + {C^-D)3(^ +
Comparing coefficients,
A=l.
A + B=0\ whence ^= — 1.
ji5+C=l; whence C=2.
O+i)=0; whence i)= -2.
UNDETERMINED COEFFICIENTS 379
Hence, the required series is
=l-x + 2x^-2^ +
1 + x
To determine what power of x shall occur in the first term
of the expansion, we arrange both numerator and denominator
in ascending powers of a?, and perform the first step of the
division.
Example 2. Expand » ~o' q 4 i^ ascending powers of x.
ZX — itr — oX
Dividing, the first term becomes ^xr^.
Hence, we assume
^~f^ ^ =^3(r^ + A + Bx-\-Cx'+Dx^+ ......
aX — X^ — oX
Multiplying by 2iP— x"^— 3a?*,
Comparing coefficients,
2A=0; whence A=0.
2^—1 = 0; whence B=l.
2C-A-|=-4; whence C=K^ + |-4)=i.
2D-B-^A=Q; whence D=^{B + ^A)=l.
Hence, the required series is
3_4a^ -1^1 + 3^+^^2+3^+ .
2£c-ar*-3a?*
EXERCISE 103.
Expand to five terms in ascending powers ofx:
r=^' l-x^' l + SiC + a;^
„ l-\-X K £C + ic' g ^X — X^-^-X"
"^^ T+^* • 2 + a;' ^' x'-x''
$80 ALGEBRA
10. 1;Z^. 12. f-^/-f.. 14. ^^+^
,, l-2a;^ + a^ ,o 1-2^' ik ^x'^x'
11. -5 — T-i — -^- 13. .^ — ZT-S- 15.
ar*— 4a;*— ic'^' ' cc + a;^ — 2£c^* * 1 + a;— aj^*
264. Expansion of surds into series, and extraction of roots.
The principle of undetermined coefficients may be used to
expand surds into series. The expansion is true only for
those values of x which make the series convergent.
Example 1. Expand yi + x in ascending powers of x.
Assume V'l + x=A + Bx + Cx^ + D,Tcr^ + Ex* + Fx^+
Squaring, by the rule for squaring a polynomial,
l+x=A' + 2ABx+{B' + 2AC)x^ + {2AD-h2BC)2(^ +
{C' + 2AE+2BD)x*-{-{2AF+2BE+2CD)x^+
Comparing coefficients,
A^=l; whence A =1.
2AB=1; whence B=A^=^.
2AC+B'=0', whence C=^^=-h
2A ^
2AD + 2BC=0; whence D=-^^=^t^,
2AE+2BD+C'=0- whence E=-^'^^±^'=-^j^.
ZA
. Hence, i/l + a?= 1 + \x—\^ + -^^x^'—^^^x" +
Note.— if, in the above example, we use ^=-1, different values for
B, C, etc., also will be obtained, and the resulting series will be the
other square root.
The same method may be used to find the square root of a
polynomial which is a perfect square. In this case the series
is finite.
UNDETERMINED COEFFICIENTS 381
Example 2. Find the square root of 4:—4:X + Qo(f—8a^ + 6x*
Evidently the highest and lowest terms of this expression must
be the squares of the highest and lowest terms of the root.
Hence, the highest term must contain x^' Accordingly, we as-
sume
V4—4x+Qx'—8x'-{-6x*—4x^+x''=A + Bx+Cx^+Dj(^.
Squaring,
4-4x+9x''—8x' + Qx*—4:X^ + x^=A^ + 2ABx+{B^ + 2AC)x^
+ {2BC+ 2AD)x' + {C' + 2BD)x' + 2CDx^ + D'x^,
Comparing coefficients,
2A5=-4;
J52 + 2AC=9;
2BC+2AD=-8',
C' + 2BD=6;
2CD=:-4:;
D'=l.
Each of these equations is satisfied when A =2, B= — 1, C=2
D=-l.
Hence, y4—'ix+9x'—8x^ + Qx'—4x^ + x^=2—x+2x^—x\
EXERCISE 104.
Expand to five terms in ascending powers of x :
1. y'l-x. 4. i/l-^x-x\ 7. i/l-2a;.
2. i/l + 9a;. 5. yl-x + x\ 8. i/l+2x-x\
3. i/l-4a;. 6. i/^ + x". 9. i/8 + a^.
382 ALGEBRA
Find the square root of :
10. l + 2x+Sx' + 2x' + x\
11. l + Ax + lOx'-i-Ux' + dx'.
12. lQx*-'60x-nx' + 2^x' + 2b.
13. U-4x'-2ix-^Ux'-22x'-j 17x' + 4x^.
14. l-4«-32a^ + 64a«-64a^ + 12a^ + 48a*.
265. Reversion of series.
The series i/=A + I^x+ Cx^-i-Dx^+ • • • • is said to be
reverted wlien x is expressed in tlie form of a series of terms
written in ascending powers of y.
A series may be reverted by use of the principle of unde-
termined coefficients.
Example 1. Revert the series y=x+x^ + x^ + x*+ • • • • .
Assume x= Ay + By^-{-Cy^+ Dy^+
Substituting,
x=A{x+0(y'-^x^ + x*+ • • ')+B{x^ + 2x^ + Sx^+ . . . .)
+ C{x^ + 3x'+ ' ■ .) + D{x'+ • • •)•
or, x=Ax+{A + B)x^ + {A + 2B+C)x^ + {A + 3B + SC+D)x*+ • • • •.
Comparing coefficients,
A=l.
A +1?=0; whence B=—A=—l.
A + 2B+C=0; whence C=-A-25=l.
A + 3B+3O+i)=0; whence D=-A-^B-SC=-1.
Hence, x=y—y'^-\-y^—y*+ • • • • .
EXERCISE 105.
Revert to four terms :
1. y=x—x'^-{-x^—x*-i- •
2, y=x + 2x' + ^x' + 4:X*-{-
UNDETERMINED COEFFICIENTS
383
^y»2 /y»3 ,-y,4
3. y = . + |+|-+|+ .
5. y=x—2x'^-^4x^—Sx*-
rv* ^7*2 ™3 ^y,4
-J rO tl/ *o *o
7. y=ic + £c^ + aj''^ + a;^+ •
8. y = a.-g- + ^-y-
266. Partial fractions. Two or more fractions whose sum
is a given fraction are called partial fractions.
The process of separating a fraction into partial fractions is
the opposite of addition of fractions. A fraction may be sep-
arated into partial fractions by use of the principle of undeter-
mined coefficients.
267. We consider first those cases in which the numerator
of the given fraction is of lower degree than the denominator.
(1) When each factor of the given denominator is of the first
degree^ and no two factors are equal.
Since the denominator of the given fraction must be the
common denominator of all of the partial fractions, assume
that the given fraction equals the sum of all the fractions
whose denominators are the factors of tlie given denominator,
and whose numerators are expressions independent of the
general number involved.
Example 1.
1 ft '¥•_!_ Q
Separate ^ ^^ \^ . into partial fractions.
4x^-9
Since 4ic2-9=(2ic + 3) (2aj -3), assume
18ir + 3 A B
4^_9-2iC + 3'''2a?-3'
334: ALGEBRA
Multiplying by 4a^—9,
■[8x + S=A(2x-S) + B{2x + S),
or i8x+S=(2A + 2B)x + 3B-SA.
Comparing coeflScients, -
2A + 2B =18, (1)
and SB-SA=3. (2)
Solving the system (1), (2), we get
J.=4, B=5.
Hence 1?^±-^=_A_ , _A_.
^®"^®' 4x^-9 2ic + 3 + 2x-3
Since the assumed equation is to be true for all finite values
of the general number, we may usually shorten the work by
assigning particular values to the general number that will make
certain undetermined coefficients vanish. This is shown in
the following example.
Example 2. Separate -5 — ^—- into partial fractions.
Since x^—25x=x{x—5){x + 5), assume
x'-75 _A jB_ C
x^—25x~ X a? + 5 X— 5*
Multiplying by x^—25x,
x^ — 75 = A(x + 5)(a?— 5) + Bx{x — 5) + Cx{x + 5) .
Now to make the terms containing B and C equal 0, let x=0.
Then — 75=— 25A; whence A=3.
To make the terms in A and C equal 0^ let x= — 5.
Then -50=50B; whence 5=-l.
To make the terms in A and B equal 0, let x=5.
Then -50=50C ; when C=-l.
Therefore, x^-75 ^3__1___1_
ir— 25a? x x + 5 x—5
(2) Wheyi one or more factors of the given denominator are
of higher degree than the flrst^ and no two factors are equal.
UNDETERMINED COEFFICIENTS 385
In this case evidently the denominators of the partial frac-
tions will be the factors of the given denominator as before ;
but since the assumed fraction must be general enough to in-
clude all fractions with the given denominator, the assumed
numerator whose denominator is of the nth. degree in the gen-
eral number involved must be the most general expression of
the degree n—1 in that general number. Thus, a partial
fraction whose denominator is x^ + 2 must be assumed in the
form — ,. : if the denominator is £c* + 3, the numerator
iC'^ + S '
must be assumed of the form Ax^ -{- Bx^ ^ Cx-\-D\ and so on.
Example 1. Separate ' .^ into partial fractions.
Since qc^-\-1 = {x-\-^){x'^—x-\-\), assume
5x^ + 1 A . Bx+C
+
x^ + i ~x 4- i x^—x + 1'
Multiplying by o?"^ + 1 ,
^x'' + l = A{x''-x^l) + {x + \){Bx-^C),
or ^x' + l = {A + B)x'' + {B-A + C)x + A + C.
Comparing coefficients,
A + B=5; B-A + C=0\ A + C=l.
Solving this system in A, B, and C, we get
A=2, B=S, C=-l.
nence, • ^^i-^^i + ^^_x+l'
(3) When ttco 0?' inore /actors of the gw en denominator are
the same^ i. e., ichen certain factors occur to a poicer.
In this case a repeated factor in the denominator may have
as many partial fractions corresponding to it as the number of
times the factor is repeated, i. e., as the power of the factor.
Their denominator will be this factor, raised, respectively, to
\hQ first power, second power, and so on, up to a power equal to
25
386
ALGEBRA
the number of times the factor is repeated. This is evident
S 4 2
since any such fractions as _— ^ + — -^2 + T^^rjya ^^^ ^^^
., , . , Sx'-2x + l
be united into — ^ ^^-^ — •
{x-iy
So(^ + Sx^ 18x 8
Example 1. Separate ^^+io^ " ^^*^ partial fractions.
Since ic* 4- 43?* = orXa? + 4) , assume
8.x^ + 8x''-18a;-8 _ A BCD
x* + 4:3(^ ~x + 4: X x^-af'
Multiplying by a?* + 40?^,
83(^ + 8x'-18x-S=Ax^ + Bx\x + 4:) + Cx{x+4:) + D{x+4),
or 8o(f + 8x'-lSx-S={A + B)x' + {4:B+C)x' + {4:C+D)x + 4:D.
Comparing coefficients,
A + B=8,4:B+C=8- 40+D=-18; 4D=-8.
Solving this system in A, J5, C, and D, we get
A=5, B=3, =0-4, D=-2.
Hence, x'-\-4.oe' ~x + 4:'^x x' x^' ' -
268. When the degree of the numerator is equal to, or
greater than that of the denominator, the fraction must first
be reduced to a mixed expression, then the fractional part
separated into partial fractions by the methods of § 267.
9a?^ + 9x^ 6
Example 1. Separate .. 2 , k^_2 ^^^^ partial fractions.
Reduced to a mixed expression, by division,
:3X — 2+,
Since 3icH5.T— 2=(a?+2)(3a?— 1), assume
16^-10 A B
'dx'-\-^x—2~x+2^2>x—l'
By the method of §267, we get A=6, B=—2.
„ 9£c' + 9i»2-6 006 2
UNDETERMINED COEFFICIENTS 387
EXERCISE 106.
Separate into partial fractions :
\.A,
2£c^-cc-l • a3^(£cH-l)
2 + 3a; - a;''-4a; + 3
5 + 38a; .. 2a;^- 1 3a;-12
^- 6^q^5^=^* 27-8af*
13a;-21 ^g 2a3^ + a;^-3a; + 4 _
^- (ir-l)(a;-2)(a^ + 3)' ' {x' -^-l^ix-X)
^x'-\-\hx 17 6^!z:4^i.
^- \x'^-^:x'-x-\ ' ^X^-l)'
7 -- + ^- + 1^ . 18. Si-SI-
'• (a;+l)(i« + 2)(a:+3) a, +a; +i
■ 1 19 -^ + 2«' + 5
8. ^j^- ^*'' (a?-l)(a;+l)
„ Sa^'-ia 20 2a;^-g«^-^
^- (2x-3)»- (a:» + 2)(x' + 2)
10. . r^ ,.. • 21- "'"'
23.
24.
8 + 12a;-2a;^-14a;=^-10a;*-2ar^
■ x\x+2y
CHAPTER XXVI,
LOGARITHMS.
269. Exponential equations. In all the equations which we
have discussed, the unknown numbers have appeared as bases,
with known coefficients and exponents. There are problems
which lead to equations in which the unknown numbers ap-
pear as exponents. Such equations are called exponential
equations.
Thus, 2^=16 is an exponential equation, the unknown number
appearing as an exponent. The solution of this equation is a? =4.
4^=8 is an exponential equation. Its solution is ir=|, because
4^=V?=8.
270. The solutions of some exponential equations can be
found easily by inspection. But in general this is impossible.
Thus, to solve 3^=243 is to find the power to which 3 must be
raised to give 243. This is seen by trial to be 5. Hence, x=5.
But 2^=12 cannot be solved by inspection, x is more than 3,
because 2-^=8; and x is less than 4, because 2*=16. Hence,
x=3+a fraction. In fact, x here is what is called an incommen-
surable number whose exact value cannot be found.
The general exponential equation can best be solved by the
aid of a set of numbers called logarithms.
271. Logarithms. T\\q logarithm of a number i'^ the expo
nent which indicates the power to which a given base must be
388
LOGARITHMS 389
raised to produce that number. That is, if a*=w, then x is
the logarithm of n to the base a ; and is written
Thus, since 2=^=8, log28=3.
Since 3*=81, log381=4. j-v r '- --£■ ^^'=' - ^-?
Since 5*= 625, log5625=4. '>*-^ ^- -^ ^ ^'
Since 4-2=i:,^-~, log4T'^=~2.
Since 9-^=~|=.^V,log9^V=-|. ^x^.^;^- ~%,
/ It follows that the exponential equation a^ = n and the
logarithmic equation aj=log,,?i are equivalent.
EXERCISE 107.
Express the followijig relations in terms of logarithms
1. 2^=^32. ^ 32-sr3. 7^=343. 5. 5« = 15625.
2. 3* = 81. '^ 4. 10^ = 10000. 6. 4-'=gL-.
7. 5-2 =J-^. 8. 10
2_ 1
2T- ^'^ -^^ — TO 0- ^ -^f^
Vo
I 60
^Express the following relations by means of exponents :
9. log39-2. "3'i;^ 11. log,16 = 2. 13. log,^V=-2. Z'^:^-^
10. log2l6=4. 12. logs4=f. 14. logio.001 = -3.
15. log,oo.001 = -f.
Find the values of the following logarithms:
16. logaS. 20. log,64. 24. log6216.
17. log,327. 21. log, .5. 25. logsyi^.
18. logiolOO. 22. log2.25. 26. log^ol.
19. logio-OOOl. 23. logger 27. log.l.
390 ALGEBRA
To the base 4 what numbers have the following logarithms ?
28. 1. U 30. f 32. -i.
29. 3. 31. -2. 33. 5.
34. -4.
272. Fuiidamental principles of logarithms.
Fi'om the definition of a logarithm it follows that any posi-
tive number, except 1, may be used as the base of the logarithm
of any arithmetical number. The following fundamental
principles apply to logarithms to any base.
(A) The logarithm of 1 to any base is ; that is,
This principle follows from «"=1.
{£) The logarithm of the base itself is 1 ; that is,
%„a = l.
This follows from a^ = a.
( C) The logarithm of a product equals the sum of the loga-
rithms of Its factors ; that is,
( loa^mn = loq.m + loa^n. \^
To pjrove this, let a^=m^ and ay = n. \
Then m)i = a''-ay=a^+y.
Hence, lQgam?i=a; + y = log„m + log„?i.
In like manner, this principle can be proved to hold for any
number of factors.
{!>) The logarithm of a quotient equals the logarithm of the
dividend minus the logarithm of the divisor; that is,
l^9a{z] =logam—log^n.
To prove this, let «^=m, and a«'=n.
Then -.=a^-^a?'=Q-c-y^
LOGARITHMS 391
Hence, log„ (^J =a;-y =log„m-log„w.
(^) The logarithm of a poicer of a number equals the loga-
rithm of the number, multiplied by the exponent of the power;
that is,
To prove this, let aJ^^n. —
Then 7i^ ={a''Y= ap*.
Hence, \og^{n^=px=p log^^w.
{F) The logarithm of a root of a number equals the logarithm
of the nwnber^ divided by the index of the root; that is,
To prove this, let
Then l/^=l/a^=a''. -
Hence, , log„i/ 7i=~ -^ , or - log„;i.
Note. —Principle {F) might be considered a special case of {E) ,
yn being written as the power n^.
By the use of the above principles we shall be able to replace
the operations of multiplication and division by those of addi-
tion and subtraction, and the operations of involution and
evolution by those of multiplication and division.
Example 1. Express loga"^ in terms of logaO?, loga?/,
log„2;, and lo^a^v.
\og~~= =\<^axy^-^o^aZ\/'^ By (D).
Zy tV /
=l0gaa? + l0ga2/'-l0ga2;-l0gal/w By (0).
=logaa; + 2 loga2/-loga2;-i log„?<^. By {E) and {F),
392 ALGEBRA
Example 2. Express 2 logaic— | logaV + h ^^SaZ as a single log-
arithm.
2 logaX-f loga2/ + i log„2;=log„ar^-log„2/^ + loga2^^ By (^).
=log„^ By(C)and(i)).
Examples. If logio2=.3010, logio3=.4771, find logio|/6T
logio|/6=^ Iogio6=|(logio2 + logio8)
=^(.3010 + . 4771)
= .3890.
Example 4. If logio2=.3010, logio3=.4771, logio5=.6990,
3y
find log,o^^.
1/800
Factoring, 360=23-3='-5; 800=2^-5^
I/' 360 3 .
' ^^^i«^*^^=logioi/360-logioT/800
=ilog,o2^-3=^-5-ilogio2^-5'
= K3 logio2 + 2 logio3 + logioS) -
I(5logio2 + 2logio5)
=K.9030 + .9542 + .6990)-i(1.5050 + 1.3980)
=.1263.
EXERCISE 108.
Express the following logarithms in terms of log„£c, log.y,
log„2j, log„^ .•
''^- ^ 6. log^-!^.
2. log^a^y. /_ ^_ 1/^1/^
K 1 1/^1/ y a;
LOGARITHMS 893
B.log.g. 10.1ogA-* ^^ ^
9 loff (^^V. Ill -" «5~*
2a;y''
13. l„g„/^+logy_^. 14. log„3^^^
Express as single logarithms :
15. log„a5+log„y + log„2-log„?^.
16. log„aj-log„y + log„^-log„^.
17. 2 log^a; + 2 log„y-2 log„2-2 log„«^.
18. i log„a;-i log„y.
19. }log„^() + |lbg„2.
20. 3 1og„a;-ilog„(y + ^).
21. 3log„g)+2 1og.(|
If logio2 = .3010, logio3 = . 4771, logio5 = . 6990, logio7 = .8451,
find to the base 10 the logarithms of the following numbers :
22. 18. 29. 70. l/98
3d. ■ .,
23. 15. 30. 210. y 144
24. 20. 31. 1000. . _ 3 ^ •
25. 50. 32. 6|. 37. ^;^l'g .
26. 24. 33. i/rs. 1^5 1 7
27. 21. 34. yW^ ^ (5^3
28. If. 35. 1^180. ''°- (5|)^-
273. Common logarithms. The logarithm of a number to
the base 10 is called a common logarithm.
Thus, logio6, \og^f,124:, logjo t\j logjo-lOOO, are common- loga-
rithms.
394: ALGEBRA
The logarithms of all arithmetical numbers to any one base
constitute a system of logarithms. The common logarithms of
all arithmetical numbers constitute what is called the common
system of logarithms.* The common system of logarithms is
used in practical calculations. This system is superior to
other systems for practical use because its base 10 is also the
base of our decimal system of numbers.
The rest of the discussion in this chapter will be confined to
commo?i logarithms ; and in the following sections the base
will be understood to be 10, and will not be written.
274. Characteristic and mantissa.
From the definition of a logarithm we have :
since 10" = 1, log l'=0 ;
since 10^ = 10, log 10 = 1;
since 10=^ = 100, log 100 = 2;
since 10=^=1000, log 1000 = 3; etc.;
and
since 10-^ = .l, log .1 = — 1 ;
since 10-^ = .01, log .01 = -2;
since 10-=' = .001, log .001 = -3 ; etc.
It Is evident from the above that the logarithm of a positive
integral power of 10 is a positive integer, and that the
* The system of common logarithms was introduced in the seven-
teenth century by Henry Briggs. Accordingly, logarithms to the base
10 are also called Briggs* logarithms. Another important system is
the Napierian system, named after Napier, a contemporary of Briggs.
Tlie base. of the Napierian system is the sum of the infinite series
. , 1 1111
fI~'~J2'*"p'^|4'^J5 • "^^'^ s""^ ^^ this series is approximately
2.7182818, and is represented by the letter e. Napier himself, however,
did not use the base e in his system.
While the common system is used in practical calculations, the
Napierian system is used in theoretical investigations.
LOGARITHMS 395
logarithm of a negative integral power of 10 is a negative
integer.
Moreover, since 65 is greater than 10^ and less than 10^ log
65 will be greater than 1 and less than 2. Hence, log 65 = l + rt
decimal. Also, since 382 is greater than 10^ and less than 10^,
log 382 will be greater than two and less than 3. Hence, log
Z%'± = 1-\r a decimal
Evidently, the logarithm of any positive number, except a
positive or negative integral power of 10, will consist of an
integral ami a decimal part.
Thus, log 825=2.9165, to four decimal places.*
The integral part of a logarithm is called its characteristic, and
the decimal part is called its mantissa.
275. Determination of tlie characteristic
A number having one figure in its integral part lies between
10" and 10\ Hence, its logarithm lies between and 1 ; i. e.,
it equals + « decimal. A number having two figures in its
integral part lies between 10^ and 101 Hence, its logarithm
lies between 1 and 2 ; i. 6., it equals 1 + a decimal. Similarly,
if a number has three figures in its integral part, its logarithm
lies between 2 and 3; i. e., it equals 24-« decimal^ and so on.
Therefore, if a number is greater than i, the characteristic of
its logarithm is positive, and is less by 1 than the number of
figures in the integral ]Kirt of the number.
Thus, in log 6841.27 the characteristic is 3. In log 362.781 the
characteristic is 2. .
Again, a number less than 1, and having no zero imme-
* It should be remembered that the decimal part is an incommen-
surable number, and may carried to any degree of accuracy required.
396 ALGEBRA
diately following the decimal point, lies between lO"" and 10-^
Hence, its logarithm lies between and— 1 ; i. e., it equals
— 1 + « decimal A number less than 1, and having one zero
immediately following the decimal point, lies between 10"^
and lO-l Hence its logarithm lies between — 1 and — 2 ; ^. e., it
equals — 2 + «<^ecma^. Similarly, if a number less than 1
has two zeros immediately following the decimal point, its
logarithm will lie between —2 and — 3 ; i. e., it equals — 3-f-a
decimal ; and so on.
Therefore, if a number is less than i, the characteristic of its
logarithm is negative^ and is greater by 1 than the number of
zeros immediately following the decimal point.
Thus, in log .0683 the characteristic is —2. In log. 00.08 the
characteristic is —4. In log .3974 the characteristic is —1.
In writing the logarithm of a number the mantissa is always
positive, and if the characteristic is negative, the minus sign
is written above the characteristic to signify that it applies to
the characteristic alone.
Tl^s, log .00357=3.5527. This means -3 + .5527.
276. IVie common logarithms of 7iumbers which do not differ^
except in the position of the decimal pointy have the same
mantissa.
This follows from the fact that changing the position of the
decimal point in a number is equivalent to multiplying or
dividing the number by some integral power of 10.
Thus, 3.261 X 10=32.61; 3.261 x 102=326.1 ; 3.261--102=. 03261.
But when a number is multiplied or divided by a power of
10, an integer is added to, or subtracted from, its logarithm;
hence the mantissa is not changed.
LOGARITHMS " 397
For example, it has been found that
log 2680=3.4281;
lience, log 268 =2.4281;
log 26.8=1.4281;
log 2.68 = . 4281;
log .268=1.4281.
277. Tables of Mantissas. The common logarithms of sets of
consecutive integers have been computed and tabulated. At
the end of this chapter is a table which contains the mantissas
of the logarithms of all integers from 1 to 1000.
Since the mantissa of the logarithm of a number depends
only upon the sequence of figures, and not upon the position of
the decimal point, only the mantissas of the logarithms of in-
tegers need be tabulated. Since the characteristics may be
found by the rules of § 276, they are left out of the table.
Logarithms may be computed to any number of decimal
places, the number depending upon the degree of accuracy re-
quired in their use. In the table in this book the manttesas
are computed to four decimal places. In this table the first
two figures of each number are found in the column headed JV,
and the third figure in the horizontal line at the top of the
table. The mantissas, with decimal points omitted, are found
in the columns headed 0, 1, 2, 3, etc.
In finding the logarithm of a number, find its characteristic
by § 275, and look in the tal)le for the mantissa. In looking-
for the mantissa of a number containing less than three figures,
annex ciphers until it has three figures.
Thus, to find the mantissa log 38, look for the mantissa of 380.
To find the mantissa of log 3, look for the mantissa of log 300.
398 ALGEBRA
278. Use of the table ; to find the logarithm of a given number.
{a) When the given number contains not more than three
figures, the mantissa of its logarithm is obtained directly from
the table.
Example 1. Find log 32.7.
By § 275 its characteristic is 1. By § 276 the required mantissa
is the mantissa of log 327.
Look for 32 in the column headed N. Looking along the hori-
zontal line of numbers opposite 32, to the column headed 7, we
find .5145, the required mantissa. Hence,
log 32.7=1.5145.
Example 2. Find log .91.
By § 275 the characteristic is —1. The required mantissa is
the mantissa of log 910, This is opposite 91 in the column headed
0, and is seen to be .9590. Hence,
log .91=1.9590.
{h) When the given number contains more than three figures,
use is made of the principle that when the difference of two
numbers is small compared with either of them, the difference
of the numbers is approximately proportional to the difference
of their logarithms.
Example 3. Find log 2.8465.
Shift the decimal point until it follows the third figure.
The required mantissa is that of log 284,65.
Now 284,65 is greater than 284 by .65.
The mantissa of log 284=, 4533.
, The mantissa of log 285 =,4548.
Subtracting, .4548— ,4533=. 0015.
Hence, an increase of 1 in 284 causes an increase of ,0015 in
the corresponding mantissa. Therefore, an increase of .65 will
cause an increase of .65 X -0015, or approximately .0010, in the
mantissa.
LOGARITHMS
Adding .0010 to the mantissa of log 284 gives .4543.
Attaching the characteristic, we have
log 2.8465=0.4543.
Example 4. Find log .008214.
The characteristic is —3.
The required mantissa is the mantissa of log 821.4.
Mantissa of log 821^.9143.
Mantissa of log 822 =.9 149.
.9149-.9143=.0006.
.4X.0006 = .0002.
.9143 + . 0002=. 9145.
Hence, log .008214=3.9145.
S99
EXERCISE 109.
Find the logarithms of:
1. 215.
11.
100.
21. .06843.
2. 673.
12.
900.
22. 4268.4.
3. 940.
13.
1.
23. 1.096.
4. 717.
14.
2.
24. .00012.
5. 4,62.
15.
1684.
25. 99.99.
6. 19.9.
16.
34.27.
26. 2031.7.
7. 830.
17.
100.5.
27. .0083326
8. 16.
18.
926.81.
28. ,50416.
9. 29.
19.
.8632.
29. 68593.
10. 8.5.
20.
.00315.
30. .074803.
279. To find a number whose logarithm is given.
(«) When the given mantissa is found in the table^ the
sequence of figures of the required number may be obtained
by reversing the process (a) of § 278. The position of the
decimal point is determined by reversing the rules in § 275.
400 ALGEBRA
Example 1. Find the number whose logarithm is 1.9325.
Looking in the table, we find the mantissa .9325 opposite 85
and in the column headed 6. Hence,
.9325= the mantissa of log 856.
Since the characteristic is 1, the number must contain two
figures to the left of the decimal point. Hence,
1.9325=log 85.6.
Example 2. Find the number whose logarithm is 2.5289.
From the table we have
.5289=mantissa of log 338.
Since the characteristic is —2, the number must be less than 1,
and must contain one zero immediately to the right of the decimal
point. Hence,
2.5289=log .0338.
(b) 'When the given mantissa is not found in the table^ the
required number is obtained by reversing the process of (i),
§ 278.
Example 3. Find the number whose logarithm is 3.6496.
The mantissa .6496 is not in the table, but the mantissas of the
table between which it lies in value are .6493 and .6503.
Now .6496 is greater than .6493 by .0003.
Also .6493= mantissa of log 446,
and .6 503= mantissa of log 447.
Subtracting, .6503-. 6493=. 0010.
Hence, an increase of .0010 in the mantissa .6493 causes an in-
crease of 1 in the corresponding number. Therefore, an increase
of .0003 will cause an increase of .0003-^.0010, or .3, in the
number.
Adding .3 to 446 gives 446.3.
Therefore, .6496=mantissa of log 446.3.
Since the characteristic is 3, we have
3.6496=log4463.
LOGARITHMS 401
Example 4. Find the number whose logarithm is ^3.8684.
This mantissa is not in the table, but the next less mantissa is
.8681, and the next greater is .8686.
Now .8681=mantissa of log 738,
and .8686= mantissa of log 739.
.8686-. 8681 = . 0005,
and .8684-. 8681 = . 0003.
.0003-^. 0005 = . 6.
■ 738 + . 6=738.6.
Hence, .8684= mantissa of log 738.6.
Locating the decimal point, we get
3.8684=log .007386.
EXERCISE 110.
Find the numbers whose logarithms are :
1.
1.7582.
6.
6.6884.
11.
4.0096.
16.
5.0220.
2.
3.8615.
7.
2.4786.
12.
1.4703.
17.
3.0392.
3.
.1847.
8.
3.6021.
13.
2.9765.
18.
4.4756.
4.
T.4609.
9.
3.7251.
14.
2.8460.
19.
.8735.
5.
2.6804.
10.
2.8976."^
15.
1.4072.
20.
1.5734.
280. Cologarithms. The cologarithm of ic is the logarithm of —
From this definition it follows that
colog x=/og(-J =log l-/og x=—log x;
that is, the cologarithm of a number may he obtained by changing
the sign of its logarithm.
Since the mantissa of a logarithm is alwaj^'s written positive,
to change the sign of the logarithm would give a negative man-
tissa. In order to avoid a negative mantissa in the cologarithm,
26
402 ALGEBRA
usually in place of —log £c, its equivalent, (10— log cc) — 10, is
used. Evidently, therefore, the cologarithm of a number may
be found by subtracting the logarithm of the number from 10,
and indicating the addition of —10 to the remainder.
Example 1. Find colog 485.
log 485=2.6857.
Hence, colog 485=(10-2.6857)-10=7.3143-10.
If log X lies in absolute value between 10 and 20, then in
order to make the mantissa positive we use for colog x the
form
(20-log£c)-20.
In general, if convenient, we may use the cologarithm in the
form
{a— log x)— a,
where a is any number that will make the mantissa positive.
Example 2. Find colog 267000000000. '
log 267000000000=11.4265.
Hence, colog 267000000000=(20-11.4265)~20=8.5735-20.
If the characteristic of the logarithm is negative, then the
— 10, or —20, will disappear from the value of the cologarithm.
Example 3. Find colog .00814.
log .00814=3.9106.
Hence, colog .00814= (10-3.9106) -10,
= (10 + 3-.9106)-10,
=12.0894-10,
=2.0894.
It has been shown that to obtain the logarithm of a quotient,
we subtract the logarithm of the divisor from the logarithm
of the dividend. Since colog x= —log «, instead of subtracting
the logarithm of the divisor ice may add its cologarithm.
Example 4. Find log
1 6827
^^^ 8iT6
Hence,
LOGARITHMS
6827
81.6'
log 6827 + colog 81.6.
log 6827= 3.8342.
colog81.6= 8.0883-10.
log ^=11.9225-10,
= 1.9225.
403
EXERCISE 111.
Find the cologarithm of :
1. 72.8. 6. 68.27. 11. .00321.
2. 691. 7. 1375. 12. .6847.
3. 4.56. 8. 261.3. 13. .0315.
4. 326.7. 9. 18329. 14. .0000623.
5. 12.34. 10. 43165. 15. .0004721.
16. .0005638.
281. Computation by logarithms. By the use of logarithms
long computations can be avoided. Multiplication^ division,
involution, and evolution, may be replaced by addition, sub-
traction, midtiplication, and division, respectively.
In using logarithms with negative characteristics it is some-
times convenient to add some number to the logarithms, in
order to make the characteristics positive, then to indicate the
subtraction of the number. Thus, 3 .6271=^7.6271-10.
Example 1. Find |^ ."00327.
We first find log i^. 00327.
log 1^.00327 =i log .00327 § 272, F.
=i (3.5145)
=^ (2.5145-5)
=^5029-1
=1.5029.
40^
ALGEBRA
]Sl4>w 1.5029=log .318555.
621.3X.03247
71.8
Hence, x7.00327=. 318555.
Example 2. Find the value of
Taking the logarithm,
^^^621.3X.03247_^^^ 621.3 + log.03247 + colog 71.8.
71.8
log 621.3=2.7933,
log. 03247 =2.5115=8.5115-10,
colog 71.8= 8.1439-10 ,
Sum=19.4487-20
^ =1.4487.
But 1.4487=log .281.
rru — ^e^^^ 621.3X03^47 ^q^ „^r^l„^^ir^n^■a^■^
xj.x\:/iCJLvi.c,
" 71 8 .«^*, «^|^a.v^^x*xxo*wv.u.j .
Example 3.
Find the sixth power of .428.
log .428«=6 log .428
=6(1.6314)
= 6(9.6314-10)
= 57.7884-60
•
=3.7884.
But
3.7884=log .00614.
Therefore,
.428«=.00614, approximately.
Example 4.
Find the value of 2.476 X (-1.724).
The sign of
the product must be determined by the laws of
signs. By logarithms we obtain merely the absolute value of the
product.
.
We have
log 2. 476 =.3938
log 1.742=. 2410
.6348
Now
.6348=log 4.313.
Therefore,
2.476 X (-1.742) = -4.313, approximately.
LOGARITHMS 405 >
5/- ^ ^ ^^H^f-J-y
Example 5. Find the value of ^ / • Q27^ X 32 . 6 X |/ 54^«-j:„ , ,or
/ .Q27^X32.6Xt/ 54Kc^. or ' •'
Eepresent this expression by x.
Then log a?=i(4 log .027 + log 32.6 + 1 log 542 + 1^ colog 4.12
+ 1 colog 7.14+^ colog 6.28)
4 log .027=7.7256=3.7256-10
log 32.6=1.5132
i log 542= .6835
i colog 4.12=^29. 3851-30) = 9. 7950-10
i colog 7.14=K49.1463-50) = 9.8292-10
^ colog 6. 28=i(59.2020-60) = 9. 8670-10
loga?=
5) 35.4135-40
7.0827-8
=
1.0827.
But
r.0827z
=log .12097.
Therefore,
x=
=.12097, approximately.
EXERCISE 112.
Find by logarithm
s the value of :
1.
71.6X.327.
8.
.1965--18.97.
2.
1.068 X. 0039.
9.
6.765-f-(-.01286)
3.
4.
5.
6.
681.7X4.235.
3.1416X1728.
•1.414x(--0632).
(-4617)X(-.03269).
10.
11.
12.
5334X.02374
-47.43X3.246*
2.476x73.81
.524x6184*
1657
7.
7.631--6214.
1.025X326.81*
13.
17.86^
16. 219.4^
17. V 9.268".
14.
.06814'".
16. (-
-.2596)*.
18. i>6278.i:
406 ALGEBRA
19. i> -20035. 21. .684*. ^^- ('^A)*-
20. .068^ 22. (^)«. 24. 3.69i-^.
25. (.6827X.114)^
26.
y 27* X -028^X16.75^
y i/629Xi>87I0Xt5/12C3*
5 /
27. 98.7i/ .068X1/28.59
' ^ i^6Xi>206:4Xi/:009l*
282. The solutions of exponential equations. The solution
of an exponential equation may be obtained by the aid of
logarithms.
Example 1. Solve 52^-11(5^) +24=0.
Factoring, (5^-8)(5^-3)=0.
■ Hence, 5^=8, (1)
or 5^=3. (2)
From (1), taking the logarithm of both members,
X log 5=log 8.
Therefore * a,-l2g_§_:i9031_
±nererore, ^~log 5~.6990--^''^^^^-
From (2), a;log5=log3.
Therefore r-^-^^-^i^- 882^
ineretore, x-^^^ .--^gg^-.6825.
Example 2. Solve 5*^-^=4^-1.
Taking the logarithms of both members,
(3^-l)log5=(.r-l)log4.
Removing parentheses,
Sx log 5— log 5=x log 4— log 4,
or ar(3 log 5 -log 4)= log 5— log 4.
Hence, ^^ Jog 5 -log 4
3 log 5-log 4
_ .6990-. 6021
■"2.0970 -.6021
=0648.
LOGARITHMS 407
EXERCISE 113.
Solve :
1. 3^=81. 6. 23^=25. 9. 4^+1 = 8-2^+2.
2. 5'--25 = 0. 6. 4^-1 = 3^+1. 10. '\/'2^^=i/¥=^.
3. 2-^ = 64. 7. 2^-1 = .32^-5. 11. 1/3^1=^ =3»-^.
4. 3^=15. 8. 52^—12^+1=0. 12. 2'««'^=8.
13. 5'«'^^ = 625. 14. 176.82 = 2.36.
15. 32-^-4(3-)-12 = 0.
16. 2*^-3(22-^) =4.
17. 2(4^)-4^-6-0.
EXERCISES FOR REVIEW (VIII).
1. Find the value of a in
a+2 4-ffl_7
a — 1 2a 3'
2. Find the value of t in
16
l/«^-l + 6=;
3. Solvea!(£c''-4) + (a3-2)=0. . .
4. Solve for x and y
^^^ + y^-*-
5. Solve (a;-l)(£c + 2)(aj^-6ic + 9)=0.
6. On how many nights may a different guard of 5 men be
taken from a body of 26 men ?
1\2«
7. What term in (x-] — j does not contain x ?
8. What is 'dn undetermined coefficient ?
408 ALGEBRA
9. What is the theorem of undetermined coefficients ? Upon
what principles did the proof in this book depend ?
10. What use can be made of the theorem of undetermined
coefficients ?
Make use of the method of undetermined coefficients in the
following :
11. Expand .. _o to five terms in ascending powers of x.
12. Expand ^_ _ i mto a series.
\—x
13. Expand .. . , ^ into a series.
14. Expand j/l—Sx to five terms in ascending powers of x,
in two Avays.
15. What is reversion of series ? Revert to four terms
y=x—Sx^-i-bx^—7x*+ • • • .
16. What are partial fractions ?
17. In what way can a given fraction be separated into
partial fractions ?
18. In what form must the partial fractions be assumed
when each factor of the -given denominator is of the first
degree, and two or more of the factors are equal ? Illustrate.
19. In what form must numerators be assumed when a
factor of the given denominator is of higher degree than the
first, and no two factors are equal ? Why is this ?
20. Separate into partial fractions :
(n\ ^ + ^^-^' r^^ 2^ + 2
^""^ {i-x){i-^xy' (^) ^(^17*
LOGARITHMS 409
21. What is an exponential equation ? Illustrate.
22. Solve 2^ = 128; 4- = G4; 9'^ = 27.
23. Give the integers between which the solutions of the
following equations lie : 2^ = 3; 7-^ = 100; 5^ = 600.
24. How can the solutions of exponential equations be
obtained when they can not be seen by inspection?
25. What is a logarithm 'i
26. To what exponential equation is logio65 = i:c equivalent?
27. Find the value of 10^^49 ; log g 216; loggyV^ ^^^a^^ ^
log„a^; log J; log^a.
28. What is a system of logarithms ?
29. What are common logarithms ?
30. Give in terms of log^aj and log„y, {a) log^icy ;
ih) iog„(^^) ; (c) logx; (^0 log^r X.
31. Prove the following identities :
(«) log,.(m-f-/i) —\og^m—\o^^n ;
{h) log6mP=^logftm; (c) logioeXlog,10 = l.
32. Give the equivalents of the following and show how you
get them :
{a) log„a; {h) logj; (c) lcg„(l-~a) ;
{d) log,o.001; (6) ilog^S.
33. What is the value of \ log39-2 log^S + log^a?
34. If a number is not an exact power of 10, what kind of
number is its common logarithm ?
36. Define characteristic ; mantissa.
36. How do you obtain the characteristics of the common
logarithms of numbers?
37. Give the characteristics of the following logarithms :
log 628.75; log 1.864; log .00031 ; log .681; log 6931.7.
410 ALGEBRA
38. Does a change in the position of the decimal point in
a number affect the value of the mantissa of its logarithm ?
What determines the value of the mantissa ? Why is this ?
39. Explain how to find the mantissa by using the table of
this book, when the given number has (a) less than three
digits; (b) three digits; (c) more than three digits. Upon
what principle does this last depend ? Is the result absolutely
correct ?
40. Find the logarithms of 61 ; 372 ; 4 ; 3180 ; 96.4 ; 132.67 ;
4166.8; 1.726; .065; .0002; .68532.
41. Explain how to find the number corresponding to a
given logarithm, {a) when the given mantissa is given in the
table, (b) when the given mantissa is not given in the table.
42. Find the numbers whose logarithms are 2.3385 ; 1.8998 ;
3.7528 ; 3.8594 ; .6300 ; T.4835.
43. Define cologarithms. When would you use the colog-
arithm of a number? Are cologarithms necessary ?
44. Find colog 6.73.
45. What is the advantage in using logarithms in long
computations ?
"46. By use of logarithms find 32.61 X7.26--403.
47. By use of logarithms find i^671.4.
48. By use of logarithms solve
(«) 3*"— 13-3^ + 36 = 0. {b) 30^+145x^52^-23.^
49. Transform |/ ^-^ into a form adapted to computation
by tables whea a, h, and c are definite numbers.
/
TABLE OF MANTISSAS
411
N.
I
2
3
•4
5
6
7
8
9
10
0000
0043
0086
0138
0170
0212
0253
0294
03^
0374
ii
0414
0453
0492
0531
0569
0607
0645
0682
0719
0755
12
0793
0828
0864
0899
0934
0969
1004
1038
1073
1106
13
1189
1173
1206
1339
1371
1303
1335
1367
1399
1430
1 14
1461
1493
1523
1553
1584
1614
1644L
1673
1703
1733
1 15
1761
1790
1818
1847
1875
1903
1931
1959
1987
3014
i 16.
2041
2068
2095
3133
8148
2175
2201
2227
3353
3379
17
2304
3330
235o
3380
2405
2430
2455
2480
2504
8539
18
2553
2577
2601
3635
3648
2672
2695
2718
2T4^
2765
19
2788
2810
2833
2856
2878
2900
2923
2945
2967
2989
2iU
.3010
3033
3054
3075
3096
3118
3139
3160
3181
3201
21
3223
3>43
3363
3384
3304
3324
3345
3365
3385
3404.
22
3424
3444
3464
3483
3502
3522
3541
3560
3579
3598
23
3617
3636
3655
3674
3692
3711
3729
3747
3766
3784
24
3802
3830
3838
3856
3874
3892
3909
3927
3945
3962
25
3979
3997
4014
4031
4048
4065
4082
4099
4116
4133
26
4150
4166
4183
4300
4216
4233
4249
4265
4281
4298
27
4314
4330
4346
4363
4378
4393
4409
4425
4440
4456
28
4472
4487
4503
4518
4533
4548
4564
4579
4594
4609
29
4624
4639
4654
4669
4683
4698
4713
4728
4742
4757
30
4771
4786
4800
4814
4829
4843
4857
4871
4886
4900
31
4914
4928
4943
4955
4969
4983
4997
5011
5024
5038
32
5051
5065
5079
5093
5105
5119
5132
5145
5159
5172
iT
5185
5198
5311
5334
5237
5350
5263
5276
5289
5302
34
5315
5338
5340
5353
5366
5378
5391
5403
5416
5428
35
5441
5453
5465
5478
5490
5502
5514
5527
5539
5551
36
5563
5575
5587
5599
5611
5623
5635
5647
5658
5670
37
5682
5694
5705
5717
5729
5740
5752
5763
5775
5786
-sa-
5798
5809
5933
5831
5933
5833
5944
5843
5955
5855
5966
5866
5977
5877
5988
5888
5999
5899
6010
39
5911
40
6021
6031
6043
6053
6064
6075
6085
6096
6107
6117
41
6128
6138
6149
6160
6170
6180
6191
6201
6212
6333
42
6233
6343
6353
6363
6274
6284
6294
6304
0314
6335
43
6335
6345
6355
6365
6375
6385
6395
6405
6415
6425
44
6435
6444
6454
6464
6474
6484
6493
6503
6513
6533
45
6532
6543
6551
6561
6571
6580
6590
6599
6609
6618
46
6628
6637
6646
6656
6665
6675
6684
6693
6702
6713
47
6721
6730
6739
6749
6758
6767
6776
6785
6794
6808
48
6812
6831
6830
6839
6848
6857
6866
6875
6884
6893
49
6902
6911
6930
6938
6937
6946
6955
6964
6972
6981
50
6990
6998
7007
7016
7024
7033
7042
7050
7059
7067
51
7076
7084
7093.
7101
7110
7118
7136
7135
7143
7153
52
7160
7168^
7ir7
7185
7193
7202
7310
7218
7226
7335
53
7243
7351
7359
7267
7275
7284
7393
7300
7308
7316
54
7324
7332
7340
7348
7356
7364
7373
7380
7388
7396
'ii\
41^
ALGEBRA
N.
55
1
2
3
4-
5
6
7
8
9
7404
7412
7419
7427
7435
7443
7451
7459
7466
7474
56
7482
7490
7497
7505
7513
7520
7528
7536
7543
7551
57
7559
7566
7574
7582
7589
7597
7604
7612
7619
7627
E8
7G;J4
7642
7649
7657
7664
7672
7679
7686
7694
7701
59
7709
7716
7723
7731
7738
7745
7752
77G0
7767
7774
69
7783
7789
7796
7803
7810
7818
7825
7832
7839
7846
61
785a
7860
7868
7875
7882
7889
7896
7903
7910
7917
62
7924
7931
7938
7945
7952
7959
7966
79^3
7980
7987
63
7993
8000
8007
8014
8021
8028
STJ^
8041
8048
8055
64
8062
8069
8075
8082
8089
8096
8102
8109
81^6
8122
65
8129
8136
8142
8149
8156
8162
8169
8176
8182
8189
66
8195
8202
8209
8215
8222
8228
8235
8241
8248
8254
67
8261
8267
8274
8280
8287
8293
8299
8306
8312
8319
68
8325
8331
8338
8344
8351
8357
8363
8370
8376
8382
69
8388
8395
8401
8407
8414
8420
8426
8432
8439
8445
70
8451
8457
8463
8470
8476
8482
8488
8494
8500
8506
71
8513
8519
8525
8531
8537
8543
8549
855.';'
8561
8567
72
8573
8579
8585
8591
8597
8603
8609
8615
8621
8627
73
8633
8639
8645
8651
8657
8663
8669
8675
8681
8686
74
8692
8698
8704
8710
8716
87^2
8727
8733
8739
8745
75
8751
8756
8762
8768
8774
8779
8785
8791
8797
8802
76
8808
8814
8820
8825
8831
8837
8842
8848
8854
8859
77
8865
8871
8876
8882
8887
8893
8899
8904
8910
8915
78
8921
8927
8932
8938
8943
8949
8954
8960
8965
8971
79
8976
8982
8987
8993
8998
9004
9009
9015
9020
9C25
80
9031
9036
9042
9047
9053
9058
9063
9069
9074
9079
81
9085
9090
9096
9101
9106
9112
9117
9122
9128
9133
82
9138
9143
9149
9154
9159
9165
9170
9175
9180
9186
83
9191
9196
9201
9206
9212
9217
9222
9227
9232
9238
84
9243
9248
9253
9258
9263
9269
9274
9279
9284
9289
85
9294
9299
9304
9309
9315
9320
9325
9330
9335
9340
86
9345
9350
9355
9360
9365
9370
9375
9380
9385
9390
87
9395
9400
9405
9410
9415
9420
9425
9430
9435
9440
88
9445
9450
9455
9460
9465
9469
9474
9479
9484
9489
89
9494
9499
9504
9509
9513
9518
9523
9528
9533
9538
90
9542
9547
9552
9557
9562
9566
9571
9576
9581
9586
91
959a
9595
9600
9605
9609
9614
9619
9624
9628
9633
'92
9643
9647
9652
9657
9661
9066
9671
9675
9680
93
9685
9689
9694
9699
9703
9708
9713
9717
9722
9727
94
9731
9736
9741
9745
9750
9754
9759
9763
9768
9773
95
9777
9782
9786
9791
9795
9800
9805
9809
9814
9818
96
9823
9827
9832
9836
9841
9845
9850
9854
9859
9863
97
9868
9872
9877
9881
9886
9890
9894
9899
9903
9908
98
9912
9917
9921
9926
9930
9934
9939
9943
9948
9952
99
995C
9961
9965
9969
9974
9978
9983
9987
9991
9996
«'n
APPENDIX
A. SQUARE AND CUBE ROOTS BY FORMULA.
(SUPPLEMENTARY TO CHAPTER VII).
1. In Chapter VII the processes of extracting the square
and cube roots of expressions by inspection were discussed.
We shall here show how to extract the square root or the
cube root of a polynomial by use of a formula. This is a gen-
eral process which may be used in case the method by inspec-
tion fails. We may then show how to extract the square root
or cube root of any arithmetical number by use of a formula.
2. Square roots of polynomials. The square root of any pol-
ynoniial will he a polyno7mal.
If the root has only two terms, it will be of the form a-Vb.
In that case the rule for obtaining the root comes from the
identity
The work is usually arranged as follows :
a2 + 2aZ» + d' | a + ^, root
j^
la^h I 2ab-^h-
2ab-\-b'= 2ab + b''
When arranged in descending powers of a, the first term of the
root, a, is the square root of the first term of the polynomial, a^.
Subtracting a"^ from the polynomial, we get the remainder
2ab+b'^. Evidently the second term of the root, &, may be ob-
1
2 APPENDIX
tained by dividing the first term of this remainder, 2a&, by 2a,
OT twice the part of the root already found. The divisor 2a is
called the trial divisor. If to the trial divisor we add &, the new
term of the root, we get 2a + b ; and if this in turn be multiplied
by 6, the product, 2ab + b^, will be the above remainder. Hence,
2a + 6 is the whole or complete divisor. And a + ?> is the entire
square root.
The above process will determine the square root of any
polynomial whose square root is a binomial.
Example 1. Find the square root of 9a?^— 12xi/ + 4i/2.
Here, as in all cases, the given polynomial should first be ar-
ranged in descending or ascending powers of some letter. This
expression is already written in descending powers of x. The
work will be as follows :
a^ + 2ab + b^= 9x'-12xy-\-4y^
a'= 9x'
a + b
3x—2y, the root
2a + b=6x-2y
2ab + b'=
-12xy + 4y'
— 12xy + 4y^
Since 9x^—12xy + 4y^ is of the/or?7i a"^ + 2ab + b\ therefore a^=9x\
Hence, a=Sx, the first term of the root. Subtracting 9x^ we
have for remainder, —12xy + Ay'^.
Now 2a, the trial divisor, becomes 6ic. Dividing —12xy, the
first term of the remainder, by 6x, we obtain —2y, the value of
b of the formula. Adding this to 6x, the complete divisor be-
comes 6a;— 2^/. Multiplying this by —2y, we have —12xy + 4:yK
Subtracting this from the remainder —12xy + 4:y^ leaves no
remainder.
Hence, the entire root is 3x—2y.
If the root contains three terms, we have by grouping terms,
the identity
APPENDIX 3
Hence, when the root has three terms^ the first two may he
found as above ; then hy grouping terms these two tnay he used
as one, and the third term ohtained hy a repetition of the p>ro-
cess used to ohtain the second term,.
In like manner, when a root contains four terms the first three
may he used as one, and the fourth ohtained by another repetition
of the process used to ohtain the second. This process may he
extetided to any number of terms.
Note. — The student should take care first properly to arrange the
order of the terms in any case, either in ascending or descending powers
of some letter. Each remainder should also he arranged like the original
expression.
Example 2. Extract the square root of 16x* + 1 — 3a?+24a^ + Sa?^.
First arrange in descending powers of x. The work is as
follows :
a +b +c
4:X^ + SX — ^
a'=
16x* + 2^x^ + 5x'-
16x'
-3a? + i
2a + b=8x^ + Sx
2ab + ¥=
24x' + 5a?2-
-3x + i
2{a + b)+c=Sx^ + 6x-
2cla + b) + c^=
-\
-40^=^-
-4x2-
-3x + i
-3x + i
Since a^=\^x^, the first term is 4x^
Hence, the trial divisor, 2a, is 8x^.
Dividing 24ar* by this gives 3x, the second* term of the root.
Adding Sx to the trial divisor gives the complete divisor,
8x^ + 3x.
Multiplying the complete divisor by Sx gives 24:X^ + 9x^.
Subtracting this product from the first remainder leaves
— 4a?^— 3a?4-5, the second remainder.
Now using the root found, 4^x^ + 3x as one term, the new trial
divisor becomes 8x^ + Qx.
Dividing — 4x^ first term of the remainder, by 8x^, the first
term of the trial divisor, gives — |, the third term of the root.
4, APPENDIX
Then the complete divisor becomes Sx^ + Qx—^.
Multiplying this by — ^, and subtracting, the third remainder
becomes zero.
Hence, the entire root is 4a?^ + 3j?— ^.
Note. — Since every expression has two square roots, differing only
i!i sign, a second expression for the root in any case may be obtained
by changing tlie signs of the terms in the root found. Thus, in Ex-
ample 2, another root is ^—dx—ix^.
EXERCISE 1.
Extract the square roots of the following :
1. Ax' + 4x + l. 5. 8«-4a^ + a* + 4.
2. ic^ + 25y- — lOicy 6. l + %x—x'' + Sx' — 2x' + x\
3. ^x' + x'-2x' + 4:-4x. 7. 49aj« + 42£c«-19£c*- 12.^^ + 4.
4. 4a* + 49-3a'-70a + 20«l 8. x*-2x'i/-2xf + SxY + y\
9. x'-Qx^>/ + UxY-12xi/' + 4:i/\
10. y* + 4y + 4y^ + 2y + 4 + i,. 12. x^-^Sx + 12-^ + i,
11 o . «' I o A I A4 , 2^' I ^' 19 2 lla^^3«^^9a*
Find to three terms the approximate square roots of :
14. 1 + x. 15. 1-2CC. 16. a' + l. 17. m^ + 3.
3. Square roots of arithmetical numbers. A^ii/ arithmetical
number is in nature a polynomial.
Thus, 5263=5000 + 200 + 60 + 3
.= 5-1000 + 2-100 + 6-10 + 3
= 5-103 + 2-102 + 6-10 + 3.
APPENDIX 5
Hence, the square root of an arithmetical number will be ob-
tained in practically the same manner as the square root of a
polynomial.
The squares of the numbers 1, 2, 3, • • * * 9, 10, are 1, 4, 9, • • • •
81, 100, respectively. Hence, the square root of an integer of
one or tivo figures is a number of one figure.
The squares of the number 10, 11, • • • • 99, 100, are 100,
121, • • • • 9801, 10000, respectively. Hence, the square root of
an integer of three or /our figures is a number of two figures.
Likewise, the square root of an integer otfive or six figures is a
number of three figures. And so on.
Therefore, ^/* the figures of an integer are marked off from
right to left in groups of two^ the number of figures in the square
root will he equal to the number of groups^ any one figure
remaining on the left being counted as a group.
Thus, marked off into groups, 21904 becomes 2'19'04'. Since
there are three groups, the square root of 21904 will contain three
figures.
If the square root of a number be a number of two figures,
we may denote the tens of the root by a and the ones by b.
Then the root \fill be represented by a-^b. Hence, the given
number will be represented by a^-^^ab + b"^.
The use of the identity
\/a' + 'lab^b^ = a-\-b
to extract square roots of arithmetical numbers is best shown
by examples.
Example 1. Find the square root of 3969.
Pointing off, we have 39'69. Since there are two groups^ the
root must be a number of two figures. Since 60^ is less than
3969, and 70^ is greater than 3969, the root must lie between 60
and 70; i.e., the tens' figure of the root is 6, the square root of
Q APPENDIX
the greatest square in the left-hand group of the given number.
Therefore, in the above identity, a denotes 6 tens, or 60.
The work may then be performed as follows :
I a + b
39'69' | 60 + 3=63 , root
a''= 36 00
2a=120
2a + 6= 123
2ab+b'=
369
369
Since a=60, a^=SeOO.
Subtracting 3600, the remainder is 369.
The trial divisor^ 2a, becomes 120.
Dividing 369 by 120 gives approximately 3. Hence, &is proba-
bly 3.
Therefore, the complete divisor, 2a 4- 6, becomes 123.
Multiplying 123 by 3 gives 369,
Subtracting this from the first remainder leaves zero.
Hence, 60 + 3, or 63, is the required root.
If the square root be a number of three figures, we have, by
grouping terms,
(a + b + cy={a + by + 2(a^b)c + c\
Hence, when the root is a member of three figures., the first two
may he found as above ; then their sum may he used as one,
and the third term obtained by a repetition of the process used to
obtain the second.
In like maimer., when a root contains four figures., the sum of
the first three may be used as one, and the fourth obtained hy
another repetition of the process used to obtain the second. The
process may be extended to any number of figures.
APPENDIX 7
Example 2. Find the square root of 203401.
Pointing off we have 20'34'01'. Hence, there are three figures
in the root. The work is as follows :
20'34'01'
a"^ 16 00 00
a-\-h-\-c
400 + 50 + 1=451, root
2a = 800
2a + 6=850
2ah + h''=
43401
42500
2(a + 6)=900
2(a + 6) + c=901
2c{a + h) + c''=
901
901
The largest square in 20 is 16. Hence, a^= 160000. Therefore,
a = 400.
Subtracting 160000 leaves 43401 for remainder.
The first tiHal divisor, 2a, is 800.
Dividing 43401 by 800 gives approximately 50.
Hence, the complete divisor, 2a + b, is 850. *
Multiplying this by 50, we get 42500.
. Subtracting this from 43401 leaves 901 for second remainder.
Now adding 400 and 50 gives 450, a + 6.
The second trial divisor, 2{a + b), is 900.
Dividing 901 by 900 gives approximately 1, the third figure of
the root.
Hence, the complete divisor, 2{a + b) + c, is 901.
Multiplying this by 1, we get 901.
Subtracting this from the second remainder leaves 0.
Hence, the required root is 451.
When the square root of a number has decimal places^ the
number itself will have tioice as many.
Thus, (.23)2=. 0529.
8
APPENDIX
Therefore, to mark off a number which contains a decimal^
begin at the decimal point and mark to the left and to the rights
putting two figures in each group.
Thus, 723.618 will become 7'23'.61'80'.
Example 3. Find the square root of 50.9796.
a+ 6+ c
7.00 + .10 + .04=7. 14, root
50'.97'96
a^= 49.00 00
2a= 14. 00
2a + 6=:14.10
2ah + h''=
1.97 96
1.4100
2(a + 6) = 14.20
2(a + 6) + c=14.24
2c(a + 6) + c'=
.5696
.5696
An approximation to the value of the square root of a number
which is not a perfect square may be obtained to any degree of
accuracy desired.
Note. — For convenience in writing, each new trial divisor, which is
the sum of the parts or the root already found, may be denoted, respec-
tively, by a, a', a", etc., and each new term denoted by h, h', h", etc.,
as in the following example.
Example 4. Find to three decimal places the square root of 2.
Since we want three decimal places it is convenient to annex
6 ciphers. The work is as follows:
2.'00'00'00'
1.00 00 00
1 + .4 + . 01 + . 004=1. 414
2a=2.
2a + 6=2.4
2ah^h''=
1.00
.96
2a'=2.8
2a'+6'=2.81
2a'h' + h'^=
.04
.0281
2a"=2.S2
2a" + &"=2.824
2a"h" + b"'=
.0119
.011296
APPENDIX
EXERCISE II.
Find the square root o^ :
I. 7396. 2. 1849. 3. 26244. 4. 41209. 5. 12.25.
6. 146.41. 7. 125.44. 8. 4.6225. 9. .026244. 10. 2611.21.
Find to three decimal places the square root of :
II. 5. 12. 3. 13. .6. 14. 105. 15. 371. 16. .75. 17. 11.8.
4. Cube roots of polynomials. If the cube root of a polynomial
has only tico terms, it will take the form, a-\-b, and the rule for
extracting the cube root will come from, the identity __
The work is usually arranged as follows :
d^ + ?,ci^h-{-?>ab'^ + Jf \a + h, root
a^ ■ "■
3a2 + 3a6 + 62
^a^h + ?>ah^-^lf:
'Sa'b + 3ab' + ¥
'Sa'b + Sab' + h^
When arranged in descending powers of a, the first term of the
root, a, is the cube root of the ^rs^ term of the polynomial, a'\
Subtracting a^ from the polynomial leaves the remainder
Sa''b + 3ab'' + b\
The second term of the root, 6, may be obtained by dividing
the first term of this remainder, 3a^6, by 3a^, or three times the
square of the part of the root already found.
This divisor, 3a^ is the trial divisor.
If to the trial divisor we add 3a6, or three times the product of
the new term of the root by the old part of the root, and &^ or the
square of the neiv term of the root^ we obtain 3a^ + 3a6 + b^ ; and
if this in turn be multiplied by 6, the product, Sa^b + Sab^ + b^,
will be the above remainder.
10 APPENDIX
Hence, Sa^-\-^ab+b^ is the complete divisor.
If the complete divisor be multiplied by b and the product
subtracted from the first remainder, the second remainder becomes
zero.
The above process will determine the cube root of any
polynomial whose cube root is a binomial.
Example 1. Find the cube root of 8x^—S6x*y^ + 54x^y*—27y^.
This is arranged in descending powers of x. The work is as
follows :
I a + b
83C^-S6xY + UxY-27y^ \2x'-Sy\ root
a»= 8x^
Sa'=12x'
Sa' + Sab + b'=12x'-18xY + ^y'
3aH7+3ab'-\-b'=
-36xY + 54^V— 272/^
-36;rY + 54a;V-27?/«
Since a'=8a?*, then a=2x^ the first term of the root.
Subtracting 8a^ leaves —36xY + 54x^y*—27y^.
The trial divisor, 3a^, becomes 12x*.
Dividing —SQx^y^, the first term of the remainder, by 12a:*
gives — 32/^ the second term of the root.
3a6, three times the product of 2x^ and — 3?/^ is —18x^y\ The
b^ of the formula, the square of the new term —Sy^, is 9?/*.
Hence, the complete divisor becomes 12x*—18xY + 9y^.
Multiplying this by —3y^ gives — 36icV + 54xV— 272/« ; and
subtracting this from the first remainder leaves 0.
Hence, the cube root is 2x^—3y^.
When the root has three or more terms, by grouping the
terms of the root already found, these terms may be used as one,
and the next term found by a repetition of the process used to
obtain the second.
APPENDIX
Example 2. Find the cube root of
11
I a+b+c
x^-6x^+l5x*-20af + 15x^-6x-\-l \x'-2x+l
a'= x^
-63(^ + 15x*-20x' + 15x'-
-6x^ + 12x*- 8x^
-Qx + 1
a'=a + b=x^—2x
Sa'''=3x*-12.x^ + 12x^
3a''-\-t^a'c + c^=3x*—12x^ + 15x'—6x+l
3^*-12ie + 15x2-
3x'-12x^ + 15x'-
-6x+l
-607+1
EXERCISE III.
Find the cube root of
1. x' + Qx'i/ + 12x}/-i-Sy^.
2. x'-Sx' + Sx'-l.
3. x' + 9xy + 27xy^27i/',
4. 27a'-10Sa'b' + U4:Ctb*-Q4b\
5. l-6a + 12a^— 8a\
6. x'-Sx' + Qx'—7x' + Qx^-Sx-^l.
7. l~9a + SSa'-QSa' + Q6a'-S6a' + Sa\
8. S + x^^9x' + Sx' + Ux^ + 12x + lW.
9. a' + Ua + -,-112 + —— -12a\
10. 4ic=^-9ic^-6iB-l+ic«-6£c^ + 9a;*.
11. a;« + 3aj* - 54a; + 28ic'-9£c'- 6x^-27.
a;' ^y ,^y'' y^
^^' 27""6" • "T'S"-
19 APPENDIX
5. Cube roots of arithmetical numbers.
The cubes of the numbers 1, 2, 3 • • • • • 9, 10, are 1, 8, 27
729, 1000, respectively. Hence, the cube root of an in-
teger of one, two or three figures is a number of one figure.
The cubes of the numbers 10, 11 99, 100, are 1000,
1331 970290, 1000000, respectively. Hence, the cube
root of an integer of four^ five or six figures is a number of two
figures.
Likewise, the cube root of an integer of seven, eight or nine
figures is a number of three figures. And so on.
Therefore, if the figures of an integer are marked off from
right to left in groups of three^ the number of figures in the cube
root icill be equal to the number of groups^ one or two figures
remaining o?i the left being counted as a group.
Thus, marked off into groups, 2515456 becomes 2'515'456'.
Since there are three groups, the cube root of 2515456 will contain
three figures, i.e., ones, tens and hundreds.
If the cube root of a number be a number of two figures, we
may denote the tens of the root by a and the o?ies by ^, and
hence, represent the root by a + b. Hence, the given number
is represented by a^ + 3a^b + ^ab'' + b\
Therefore, the cube root of the number may be obtained by
use of the identity
^a' + Sa'b + ^ab' + b' = a + b.
The process is best shown by examples.
Example 1. Find the cube root of 39304.
Pointing off, we have 39'304'. Hence, the root will be a num-
ber of two figures. Since 30^ is less than 39304, and 40^ is greater
than 39304, the root lies between 30 and 40; i. e., the tens' figure
of the root is 3, the cube root of the greatest cube in the left-hand
group of the given number. Therefore, in the above identity a
denotes 3 tens, or 30.
APPENDIX
The work is then performed as follows:
39'304'
a'= 27 000
13
30 + 4=34,
root
3a2=2700
Sa' + 3ab + b' =307 Q
Sa''b + 3ab^ + b^=
12304
12304
Since a=30, a^=27000.
Subtracting 27000 leaves 12304.
The trial divisor^ 3a^ becomes 2700.
Dividing 12304 by 2700 gives approximately 4. Hence, b is
probably 4.
The complete divisor, 3a^ + 3a6 + 6^, becomes 3076.
Multiplying 3076 by 4 gives 12304.
Subtracting this from the first remainder leaves 0.
Hence, 30 + 4, or 34, is the required root.
When the root is a 72 umber of three figures the first tv^o may
he found as above ; then their sum may be used as one number^
and the third obtained by a repetition of the process used to
obtain the second. And so on.
Example 2. Find the cube root of 2515456.
Denoting the parts of the root already found by a, and a'
respectively and the new parts by b and b\ we have the following
work :
3a' =30000
3a2 + 3a& + ?>-^=39900
3d'b + ?>ab''-\-¥=
2'515'456'
1 000 000
11515456
1197000
! a + ?) +c
100 + 30 + 6 = 136, root
3a ■-'=50700
3a'2 + 3a'6' + 6''''= 53076
3a'26' + 3a'6'=' + 6'2=
318456
318456
There will be three figures in the root.
14: APPENDIX
The first number is found to be 100.
Then the first trial divisor is 30000.
Dividing 1515456 by 30000 gives approximately 50.
But it will be found that the resulting complete divisor^ when
multiplied by 50, will give a number greater than 1515456.
Hence, 50 is too large. It will be seen by trial also that 40 is too
large. Therefore we use 30.
Then the complete divisor becomes 39900.
Multiplying 39900 by 30 gives 1197000.
Subtracting gives a second remainder 318456.
Now letting a' be 130, the trial divisor, 3a'^ becomes 50700.
Dividing 318456 by 50700 gives approximately 6.
The second complete divisor, 3a'^ + 3a'6' + 6'^, becomes 53076.
Multiplying 53076 by 6 gives 318456.
Subtracting this from first remainder leaves 0.
Hence the cube root is 136.
For reasons similar to those given in § 3, to mark of a
miniber which contains a decimal^ begin at the decimal point
and mark to the left and to the rights putting three figures in each
group. If necessary, ciphers may be annexed to the right of
the decimal point.
An approximate value of the cube root of a number which is
not a perfect cube may be obtained to any required degree of
accuracy.
Example 3. Find the cube root of 1860.867.
1'860'.867' [ 10. + 2. + .3:^12.3, root
a^= 1 000 . :
3a^=300.
3a2 + 3a6 + 6'=364.
3a26 + 3a62 4-6'=
860.867
728.
3a'^=432.
3a'2 + 3a'6' + 6''=442.89
132.867
132.867
APPENDIX
15
EXERCISE IV.
Find the cube root of :
I. 103823. 2. 2744. 3. 15625.
5. 571787. 6. 340068392. 7. 523606616.
9. 328.509. 10. 41.063625.
Find to two decimal places the cube root of :
II. 2. 12. 10. 13. 106. 14. 3.7.
4. 148877.
8. 59.319.
15. .15
16
APPENDIX
B. HIGHEST COMMON FACTORS AND LOWEST COM-
MON MULTIPLES BY DIVISION.
(SUPPLEMENTARY TO CHAPTER X).
1. In Chapter X we showed how, by factoring the expres-
sions, to find the highest common factor^ {11. C. F.) or the
lowest common multiple (X. C. M.), of two or more expressions.
The highest common factor of two expressions may be ob-
tained also by a division process, known as the Euclidian Pro-
cess, which we shall here discuss. It may be used in those
cases where the factors are not readily found. •
2. H. C. F. by division. The process involves the following
principle :
If A = BQ^ R, ichere A, B, Q, and B are integral expres-
sions in the same general number, then the H. C. F. of A and B
is the same as the IL C. F. of B and R.
This principle is established by showing that every common
factor of A and 5 is a common factor of B and i2, and every
common factor of B and J2 is a common factor of A and B.
Since A=J5^+i2, it follows from the Distributive Law that
every common factor of B and jR is a factor of A, and hence is a
common factor of A and B.
Again, R—A—BQ, and therefore it follows as above that every
common factor of A and jB is a factor of R, and hence is a com-
mon factor of B and R. This proves the principle.
Now suppose that A and B are two integral expressions whose
highest common factor is required, the degree of JB being not
higher than that of A.
APPENDIX 17
Divide A by B, and call the quotient ^i, and the remainder R^.
Divide B by R^, and call the quotient Q2, and the remainder R^.
Divide Ri by R2, and call the quotient ^3, and tlie remainder
R^; and so on.
NoAV since the remainder must be of lower degree in the
letter of arrangement than the divisor, by this process the
successive remainders must diminish in degree. Hence, finally
a point will be reached when either the remainder is 0, or
does not contain the letter of arrangement.
Jf the remainder is 0, then the last climsor is the required
highest common factor.
For, since the dividend equals the product of the divisor and
quotient plus the remainder, we have from the above divisions
J3=R,Q, + R,,
i?„_2 — i?n-l Qri + ^n->
where i?,^ is the last remainder.
From these equations it follows by the preceding principle
that the H, C, F. of B and B^ is the same as the 11. G. F. of A
and i?; the H. C. F. of B^ and B^ is the H. C. F. of B and
i?„ and hence of A and B ; the II. C. F. of B., and B^ is the
II C. F. of 7?i and B^, and hence of A and B ; and so on.
That is, the II C. F. of A and B is the II C. F. of any two
consecutive remainders in this succession of divisions.
But if A\=0, the H. C. F. of B,,_^ and i?„ is B,,_, itself.
Therefore, the II. C. F. of A and B is B„_i the last divisor.
If the last remainder is not 0, but merely free of the letter of
;|^g APPENDIX
arrangement, then there is no common factor containing the
letter of arrangement.
For, if 7?„ is merely free of the letter of arrangement, then
J?„_i and i?„ can have no common factor which contains the
letter of arrangement. Therefore, A and B can have no com-
mon factor which contains the letter of arrangement.
*3. It is clear that the above process consists simply of re-
placing the two given polynomials by two new polynomials
which have the same H. C. F., then replacing these by two
other polynomials which have' the same IT. C. .F., and so on.
Hence, it is allowable at any stage of the work to multiply either
the dividend or divisor by any expression u^hich has not a factor
belonging to the other. Likewise, it is allovxible to remove from
either the dividend or divisor a factor 10 J lich is not common to
both. A factor common to both dividend and divisor may be
rernoved, provided that it is introduced into the H. C. F. as finally
found.
Example 1. Find the H. C. F. of Gx^ + llir + S and 2x' + 7x + Q.
6a^2 + lla;+ 3 | 2a?^ + 7a?+6 2x^ + 70.' + 6 | 2jg + 3, H. C. F.
6x"'^ + 21x-4-18 3 2x' + ^x x + 2
— 5 )-10.r-15 4x + 6
2x+ 3 4a^ + 6
In the first division we obtain the remainder —10a?— 15. From
this remainder we remove the factor —5, since —5 is not a factor
of the divisor 2x'^ + 7x+Q.
Now the H. C. F. of the given expressions is the H. C F. of
2ic' + 7x + 6 and 2x + 3. Dividing 2x'^ + 7x-\-Q by 2a? + 3 leaves no
remainder. Hence, their H. C. F. is 2a? + 3 itself.
Example 2. Find the H. C. F. of 2x' + 3(^-7x'-20x and
^af-12x^ + 5x.
APPENDIX 19
Removing the common factor x the work is as follows :
2ie + a?2-7a?-20
2
40?^-
o?
|4a?=^-
7
-12a?+5
-12a? + 5
4a?2-12a? + 5
4x=^-10a?
— 2a? + 5
— 2x+5
|2a?-5
|2x-l
4ar*+ 2a-''-14a?-40
4:.x^-12x'+ 6x
14ar^— 19a?— 40
2
28a?2-38a?-80
28x*2-84a? + 35
23) 46a?-115
2a?- 5
H. C. F=x{2x-5), ov2x^-5x.
Here, in order to avoid fractions, we multiply 2x^ + x'^—7x—2Q
by 2 before dividing.
We then seek the H. C. F. of 4a?2— 12a? + 5 and the remainder
14a?=^-19a?-40.
Again, to avoid fractions, we multiply 14a?^— 19a?— 40 by 2.
The second division gives the remainder 46a?— 115. From this
we remove the factor 23, since 23 is not a factor of 4a?^— 12a?+5.
We then seek the H. C. F. of 20-- 5 and 4a;=^ — 12a?+ 5. Dividing
leaves no remainder. Hence, their H. C. F. is 2a?— 5 itself.
Now to get the H. C. F. of the original expressions we must
multiply 2a;— 5 by the factor x which was removed in the
beginning.
Example 3. Find the H. C. F. of ^oc'y-^xy' + y^ and
3a?^ — Sx^y + xy"^ — y^.
Since y is a factor of the first expression, but not of the second,
it may be removed.
20
APPENDIX
We now seek the H. C. F. of 4:X^—5xy+y^ and
3ar^ — 3x^y + xy^ — y^. The work is as follows :
Sa^ — Sx^y + xy^ — y^
___^ 4
12a?' — 1 2x^y + 4xy^ — 4?/^
12a^-15x'y + Sxy^
3x^y-\-xy^—4:y'^
4
12a?^2/+ 4£C2/^— 16?/^
19^/'' ) 19xy'-19y'
x-y
4xl—5xy + y^
4iX^—4txy
4:X^—5xy + y^
3x
4:X^—5xy + y^
32/
-y, H. C. F.
4x-y
— xy + y'
— xy + y"^
4. The H. C. P. of three or more expressions. The 11. C. F. of
three or more expressions may be obtained by finding the
H. C. F. of any two of the expressions, then finding the 11. C. F.
of this IT. C. F. and the third expression ; and so on.
Example 1. Find the H. C. F. of a' + a'-Ua-24:, a^-Sa'
— 6a + 8, anda^ + 4aHa— 6.
Thei?. C.F.old' + a'-Ua-24:Siuda^-3a'-Qa + 8isa''-2a-8.
The H. C. F. of a'-2a-8 and d' + 4a' + a-
Hence, a + 2 is the H. C. F. required.
■6 is a + 2.
EXEBCISE
FindtheH. C. F. of:
1. a^-Sx-2, x^-x^-4:.
2. aj'^ + ic-e, ar'-2a;^-£c + 2.
3. ar'' + 7ic^ + 5a;-l, Zx^ -^ hx' + x-1.
4. 2a2-5a + 2, 4a^ + 12«2-a-3.
5. 4aj^-12a;-27, eaj'^-Slaj + lS.
APPENDIX 21
6. 2a^ + 9a'-6a-6, Sa' + 10a'-2Sa + 10.
7. Sx' + x-2, 4x' + 2x'-x + l.
8. 2a;'-7a'-2 + 7a, ba' + a'-Za'-A.a + A.
9. 2m'' — 3m' — 8?72 — 3, 3m* — 7m^ — Sm'- m— 6.
10. 9a=^^-22a5=^-85*, 3«=' + 13a'^> + 12a^»l
11. 2x''-^x''y-2xy\ 2x^ + lxhj-\-Zxy\
12. a^-^alf-2h\ 2a'-^a'b-ab' + Qb\
13. a;y-6ajy + 6ecy-3a;y + 2.^/, 6a!V-15xV + 21a;V-12a;y
14. 2lab- 17 ab'-bab'-\-ab\ baly'-SAab'-7ab.
15. ic* + 4£c^ + 4:x\ x'y + 5a;\y + ^x-y.
16. a;^ + 3a;' + 4, £c=' + 2£c'-4a; + 8.
17. ic=' + 3a; + 2, jc*-6a;=^-8aj-3.
18.
x'-x^^\,x'^x^-^\.
19.
x''-x'-%x^^\2, x''^4x'—Zx^-
20.
a?-\, 2a'-a-l, Sa'-a-2.
21.
2m' + 3m -5, 3m'- m- 2, 2m'
)n — S.
22. x' + x-Q,x' + Sx'-Qx-^, x'-2x'-x + 2.
23. 3«' + 5«' + a— 1, a' + 3a — 1— 3«^ «' — l + 7a' + 5a.
24. a'-9a-10, a'-30-7«, lO + a'-lla.
26. 2y' + Qy' + 4y\ 3?/=^ + 9y' + 9y + 6, 3^=* + 8y' + 5y + 2.
5. L. C. M. by use of H. C F.
The L. C. M. of tico exjyressions may be obtained by cUviduiy
their product by their H. C. F.
This may be proved as follows :
Let A and B be the expressions whose lowest common
midtiple is required, and let ^be their highest common factor.
Then A and B may be written
A = Hq,,
and B=IIq^^
in which q^ and q^ have no common factor. (Why ?)
22 APPENDIX
Hence, AJB = Hq^Hq^ = IT{Hq^q.^.
Now the L. C. M. must contain all factors common to A
and i?, all other factors of A^ and all other factors of B.
Hence, the L. C. M. equals Hq^q^.
Therefore, from the above equation,
AB==H(Hq,q,),
we see that the product of A and B equals the product of
their 11. C. F. and their L. C. M Therefore, their L. C. M.
ma (J he obtained hy dimding their product by their H. C. F.
6. L. C. M. of tliree or more expressions. The Z CM. of
three or more expressions may be obtained by first finding the
Z.C.M. of any two of the expressions, then finding the
If. CM of that result and a third expression ; and so on.
EXERCISE VI.
Find the Z. C M of :
1. 2a'-5a4-3, 2«=^— 7a + 5.
2. ^a' + Ua+lO, Qa' + ba-U.
3. a^-x'-Ux + 2i, x'-2x'-bx-}-e.
4. x'-12x+lQ, x'-4x'-x' + 20x—20.
5. 12a^ + 13a^ + 6a + l, lQa' + lQa' + 7a^l,
6. 4:rv'-12n' + bn, 2n* + n^-7n'—207i.
7. x'-Sx+2, x*-Qx' + Sx-Z.
8. 2x' + 6x-^, 2x'-{-x'-6x + 2.
9. 2a' + Sa-b,Sa'-a-2,2a'-Va-S.
10. a^+x-Q,x^-2x^-x + 2,x' + Sx'-Qx-S.
11. 2x' + Sx-2, 2£c'' + 15a;-8, iB^+10i« + 16.
12. 2a*+W + 4a\ 3«^+9a^+9a+6, Sa' + Sa' + ba+2.
APPENDIX
C. FUNDAMENTAL PROPERTIES OF NUMBERS.
1. In the following sections we shall establish some fun-
damental laws of numbers which are at the foundation of
arithmetic as well as of algebra. These laAvs were assumed in
Chapter II. They apply to any numbers. We shall now
establish these laws for commensurable mimhers. By the aid
of the theory of limits they can be shown to hold for incom-
rnensurahle numbers as well.
We shall first establish the laws for integers.
2. Law of order in addition. Numbers integers.
• Numbers to be added may be arranged in any order ; that is,
The proof of this law is based upon the meaning of an integer.
{a) When the numbers are positive integers. By definition,
positive integers are merely arithmetical numbers, each repre-
senting the number of units in a given magnitude.
Let us consider two magnitudes of the same kind, one con-
taining a units and the other b units. Then, a-\^b means the
number of units in the new magnitude formed by putting the
second magnitude with the first. And b-\-a means the num-
ber of units in the new magnitude formed by putting the
first magnitude with the second. But evidently the total
number of units in the two magnitudes is the same whether
the second be put with the first or the first put with the
second. Hence, a+ ^=5 + «.
By similar reasoning, this law can be shown to be true for
any number of positive integers.
24 APPENDIX
(b) When the numbers are negative mtegers. It has been
proved that the sum of any number of negative integers is
obtained by finding the sum of their absolute values, as in (a)
above, and attaching to this sum the negative sign. But by {a)
the sum is the same for any arrangement of their absolute
values. Hence, the law is true also for any number of 7iegative
integers.
(c) Wheii the numbers are integers^ some i^ositive and some
negative It has been proved that the sum of two numbers,
one positive and the other negative, is obtained by finding the
difference of their absolute values and attaching to this the
sign of the number having the greater absolute value. This
difference would be the same, whatever the arrangement of the
numbers. Hence, the law of order in addition liolds for two
integers^ one positive and the other negative.
By (a), (^), and (c), it follows that an integer, positive or
negative, may be brought to any position without changing the
value of the sum. Hence, the law is true for any number of
integers^ some positive and some negative.
Note.— This law is also called the commutative law of addition.
3. Law of grouping in addition. Numbers integers.
Numbers to be added may be grouped in any manner ; that is,
a-h6 + c = a-h(6 + c).
Since a, 5, c, are integers, we have by § 2,
a-\-b-^c=b + c + a
= (J + c) + a
Similar reasoning will apply in the case of any number of
integers.
Note.— This law is also called the associative law of addition.
APPENDIX 25
4. Law of order in multiplication. Numbers integers.
The product of two or more numhers is not changed by chang-
ing the order in ichich the midtiplications are j)erfornied ; that is,
abc = acb = bac, etc.
(«) When «, ^, c, are positive integers, place a objects in a
group, and form h rows of c groups to the row.
c columns
h rows
a a a a
a a a a
a a a a
Since there are a objects in each group and b groups in
each column, there must be ab objects in each column. And
since there are c columns, the total number of objects must be
abc.
In like manner, there are ac objects in each row, and b
rows. Hence, the total number of objects must be acb.
Hence, abc and acb represent the same number of objects.
Therefore, abc=acb.
If now a=l, this becomes bc=cb.
(b) If there be any number oi positive factors, by (a) any
two consecutive factors may be interchanged (considering the
product of all factors preceding these two as the lirst factor
a in (a) above) ; and by repetition of the process of intercliang-
ing two consecutive factors, all of the factors may be arranged
in any order without changing the value of the product.
(c) The law hokls if some of the factors are negative. For,
the absolute value of the product is the same whatever the
26 APPENDIX
signs of the factors, and the sign of the product has been
shown to depend only upon the number of negative factors.
Any change in the order of the factors could not, therefore,
change the sign or absolute value of the product.
Note.— This law is also called the commutative law of multiplica-
tion.
5. Law of grouping in multiplication. Numbers integers.
Factors may he grouped in any manner / that is,
ahc=a{hc).
When «, J, c, are integers, we have by § 4,
abc=hca
= {bc)a.
= a{bc).
Similar reasoning will apply in the case of any number of
factors.
Note.— This law is also called the associative law of multiplica-
tion.
6. Law of distribution. Numbers integers.
The product of tic o expressions equals the sum of the products
obtained by multiplying each term of either exj^ression by the
other / that is,
(a + b + c)x—ax + 6 jr + ex.
(a) When x is positive.
Since ic=l + 1 + 1+ to £c terms,
then (a + b + c)x={a + b + c)^{a^b + c) + {a + b + c)-\-
to X terms (Def. of multiplication.)
=<^4-a + «+ • • • • to a; terms
-]rb-\-b + b+ ••••tocc terms
+ c + c + c-{- ' ' ■ ' to ic terms §2
= ax + bx + cx.
APPENDIX 27
Similar reasoning will apply to any number of terms in the
multiplicand.
(b) When x is 7tegatwe.
Since — £c= — 1 — 1 — 1— .... to ic terms,
therefore (a + ^ + c) ( — a?) = — (a + ^ + c) — (a + ^» + c)
— (a + 6 + c) • • • • to a; terms (Def. of multiplication).
= —a — a—a— • • • to £c terms
—h—b—b— • • • to £c terms
— c—c—c— • • • to £c terms § 2.
= a{—x)^-b{—x)^-c{—x).
Similar reasoning will apply to any number of terms in the
multiplicand.
7. In order to prove the preceding laws when fractions are
involved, it is necessary first to establish the following prin-
ciples.
m
(1) a—=am-^n;
that is, to multiply any expression a by the fraction — multiply
a by m, then divide the product by n.
This follows immediately from the definition of multipli-
cation.
(2) To prove abc = acb
lohen a is a fraction, and b and c are integers.
If b and c are positive, form b rows of the fractions a with c
fractions in each row. Then there are c columns.
28
b rows ^
APPENDIX
c
columns
A
a
a
a
a
a
a
a
a
a
Now, by definition of multiplication, ab represents the sum
of the rt's in one column. And since there are c columns, abc
represent the sum of all of the «'s.
Again, ac represents the sum of the «'s in one row. And
since there are b rows, acb represents the sum of all of the a's.
Therefore, abc = acb.
If b or c, or both b and c, are negative, the proof folloAvs as
in (c), § 4.
(3) To prove
a m_m a
b 71 n b
This is established as follows :
a ni
~b n
By (1).
= (y-)
" I,
m-~-nnb
—nnb
(quotient X divisor = dividend.)
a
^-^■bm
., m a -
Also— T-no^
n b
(m a\
m a
n b
bn
By (2).
(quotient X divisor = dividend.)
By (2) and § 4.
APPENDIX
29
= — a-^hbn
n
By(i)
= (^.«)^6.6.«
m
(quotient X
divisor = dividend.)
= m,a
(quotient X
divisor = dividend.)
= €1-^1.
§4.
_, a tn ^ m a .
Hence, -r — 710= — -rno.
' n n
Axiom 7.
Dividing by 5, then by w,
a m
Tn
m a
-n'~b'
If n=l, this becomes
a
a
= m~^.
8. Fundamental laws when fractions are involved.
(1) The Imo of order in multiplication. In (3) § 7 we have
established the law of order in multiplication for two factors,
where one or both are fractions. By the same method the
law can be shown to hold for any number of factors.
(2) The law of grouping in multiplication., ^fhen. some of the
numbers are fractions, now easily follows.
Thus, abc=bca By (1).
= {bc)a
=a{bc).
The same reasoning would apply to any number of factors.
(3) The laic of distributioii. We are now able to complete the
proof of the law of distribution.
30 APPENDIX
We have x(a+b+c)—aia-\-xb-{-xc, whatever the expressions
represented by x, a, b, c, because the multiplier a + b + c is
obtained by first taking unity to form a, then to form b, then
to form c, and adding the three results ; and hence to obtain
the product x must be used in the same manner.
Now, since x(a + b + c)=xa + xb + xc,
(a + b + c)x=ax-{^bx + cx, by the laAV of order.
The same reasoning would apply to any number of terms.
(4) The law of order in addition noio follows. It has been
shown that integers or rational fractions can be reduced to
equivalent fractions having a common denominator, and such
that the resulting numerators and denominators are all
integers.
Suppose that when reduced the fractions are
etc.
X
Then, f+|+f+ .... = " + ^ + f • • ■ • §50, Chapt. VI.
But the value oia + b + c • • • • is not changed by changing
the order of the terms. Hence the value of - + ^+-+
Jb Jb tJu
is not changed by changing the order of its terms.
(5) The laid of grouping in addition, when fractions are in-
volved, can now be established by the same reasoning that was
used in § 3.
INDEX
[Numbers refer to pages.]
Abscissa, 193.
Absolute term, 243.
Addition, 1, 35, 46, 47, 146.
of negative numbers, 35.
of fractions, 146.
elimination by, 183.
Algebraic expressions, 5.
numbers, 32.
sum, 35.
Antecedent, 298.
Arithmetical numbers, 33.
means, 359.
progressions, 354, 355.
Arrangement of expressions,
Axioms, 21.
60.
Cologarithms, 401.
Combinations, 342.
Commensurable numbers, 299.
Common difference, 355.
Completing the square, 235.
Complex fractions, 155.
Complex numbers, 223.
Conditional equations, 20.
Conjugate surds and imaginaries,
219, 226.
Consequent, 298.
Constants, 310.
Continuation, symbol of, 56.
Coordinates, 192.
Continued proportion, 304.
Base of power, 12.
Binomial, 15.
square of, 77.
Binomial theorem, 81.
extraction of roots by, 353.
general term of, 349.
proof of, 346.
Brace, bracket, etc., 6.
Character of roots, 341.
Checks, 48.
Clearing of fractions, 165.
Coefficients, 13.
numerical, 12.
undetermined, 375.
31
Degree, of terms, 131.
of an equation, 161.
Difference, 1.
Discriminant, 241.
Distributive law, 18.
Division, 4, 64, 66, 153.
Divisor, dividend, 4.
Elimination, 181, 182, 184, 186.
Equations, 19.
equivalent, 179.
exponential, 388, 406.
fractional, 165.
graphic representation, 192, 373.
graphic solution, 197,375,378,279.
32
INDEX
inconsistent, 179.
independent, 179.
indeterminate, 178.
ill quadratic forms, 253.
integral, 163.
irrational and radical, 161.
linear, 164.
quadratic, 230.
simultaneous, 179.
solutions of, 178.
symmetrical, 267.
systems of, 180.
Equivalent equations, 179.
Evolution, 84.
Exponents, 12.
laws of, 56, 64, 75, 324.
Expressions, 5.
literal, 13.
Extremes and means, 300.
Evaluation of an expression, 6.
Factoring, 103.
equations solved by, 233.
Factorial -n, 340.
Factors, 12.
Formula, the, 168, 169.^
Formula for
solving quadratics, 238.
Fourth proportional, 300.
Fractional equations, 165.
exponents, 324.
Fractions, 71.
complex, 155.
partial, 383.
Geometrical progression, 354, 362.
means, 365.
Graphic, representation of and
solution of equations, 192, 197,
273, 275, 278, 279.
representation of imaginary
numbers, 228.
Harmon ical progression, 354, 370.
Highest common factor, 131.
Homogeneous equations, 264.
expressions, 131.
Identical equations, identities,
20.
Imaginary numbers, 223.
Incommensurable numbers, 299.
Inconsistent equations, 179.
Independent equations, 179.
Indeterminate equations, 178.
fractions, 321, 322.
Inequalities, 291-295.
Inserting signs of grouping, 53.
Integral equations, 161.
expressions, 130.
Involution, 75.
Irrational numbers, 86, 212.
Known and unknown numbers,
24.
Law of order, 16.
of exponents, 56, 64.
of grouping, 16.
of signs, 41.
expressed by an equation, 214.
Letter of arrangement, 60.
Like and unlike terms, 15.
Linear equations, 163.
Literal numbers, 9, 11.
advantage of, 9, 11.
Logarithms, 388.
characteristic of, 395.
computations by, 403.
common, 393.
INDEX
33
Napierian, 393.
Lowest common multiple, 135.
Mantissa, 395, 397.
Mean i)ioportional, 304.
Means, arithmetical, 359.
geometrical, 365.
liarmonical, 370.
Means and extremes, 300.
Minuend, 1.
Monomials, 15.
Multiples, L. C. M. 135.
Multiplicand, multiplier, 3.
Multiplication, 3, 39, 57, 58, 150.
Negative exponents, 326.
numbers, 30-38.
Numbers, definite, 9.
algebraic, 33.
commensurable, 299.
constants and variables, 310.
finite and infinite, 319, 320.
general, 8, 9.
imaginary and complex, 223.
known and unknown, 24.
natural, 9.
opposite, 33.
rational and irrational, 163, 212.
real, 223.
Opposite numbers, 33.
Ordinate, 192.
Parentheses, 6.
Partial fractions, 383.
Permutations, 337.
Polynomials, 15.
square of, 79,
Powers, 12.
Prime factors, 104.
Progressions, 354.
arithmetical, 354, 355.
geometrical, 354, 362.
liarmonical, 354, 370.
Problems, directions for solving.
173.
Products, 4.
special, 92.
Proportion, principles, 299.
Quadratic equations, 230.
graphs of, 273, 275, 277.
principles, 242.
systems of, 261.
Quality, signs of, 32.
Quotients, 5.
special, 96.
Radicals, 83, 86, 212.
Ratio, 298.
Rational expressions, 6, 86.
Rationalizing factor, 220.
Real numbers, 223.
Remainder theorem, 123.
Removal of si^ns of grouping, 51.
Roots, 83.
of a polynomial, 90.
of an equation, 21.
character of roots, 241.
sum and product of, 242.
Series, 254, 378.
convergent, 354.
divergent, 354.
finite, 354.
fractions expanded into, 378.
oscillating, 354.
reversion of, 382.
Signs of grouping, 6.
Simultaneous equations, 179.
34
INDEX
Solution of equations, 21, 178.
graphic method, 275.
by special devices, 270.
Square of binomial, 77.
of polynomial, 79.
Square roots, 79.
Subtraction, 2, 37, 49.
Subtrahend, 2.
Surds, 86, 212.
Symbol of continuation, 56.
Symmetrical equations, 267.
Synthetic division, 127.
Systems of equations, 180.
consistent or determinate, 180.
defective, 269.
equivalent, 180.
impossible, 180, 190.
indeterminate, 190.
solution of, 180.
Theorem, trinomial, 81.
of undetermined coefficients,
375.
remainder, 123.
Third proportional, 304.
Transposition ,165.
Trinomial, 15.
Undetermined coefficient^, S!7b.
theorem of, 376. j
Unknown numbers, 24.
Variables, 310.
limit of, 318.
Variation, 310.
direct, 311.
indirect, 311.
joint, 312.
Vinculum. 6.
Terms, 14.
Theorem, binomial, 81, 346.
Zero exponent, 64.
operations with, 321.
ANSWERS
Exercise 1.
8.
8. 11. 33.
14. i. 17.
44.
20. 13.
^.
15. 12. 0.
15. 38. 18.
16.
21. 24.
10.
2^ 13. 48.
16. 4. 19.
6.
22. 0.
23. 216.
24.
1. 25. 18.
Exercise 2.
26.
25.
1.
5. 4. 27.
7.
230. 10.. ^V-
13.
23.
16. 2, r
2.
11. 5. 40.
8.
37. 11. 28.
14.
h
17. 17..
3.
270. 6. C6.
9.
24. 12. 49.
15.
liV
18. 576.
19. 20tli; X taken as a factor 20 times.
20. 4; 5; 1. 21. 7; x\ 3.
Exercise 4.
5. 35a6a?2^.
12. 14a'.
19. ll^-'^a?.
6. 30a&c.
13, ^\x^y.
20. 484.
7. 2xyz.
14. 'S2x2j^.
21. 693.
8. 14abcxy.
15. mah. ,
22. 24.8.
9. 15a^bx^y.
16. 28irV-
23. x; x^y; y\
10. iabcxyz.
17. lOax.
24. x2; a; 6.
11. 15a.
18. loab^
Exercise 5.
25. 7?i2; 3m; 21.
1. 3. 3. 2.
5.
3.
7. 2. 9. 2.
11.
2. 13. I. 15. 1.
2. 7. 4. 4.
6.
3.
8. 1. 10. 1.
Exercise 6.
12.
3. 14. 5. 16. 2.
1. 8 boys.
3. 10 yrs. and 30 yrs.
5. 16, 48.
2. $3.
4. 45, 80.
6. 8, 15.
2 ANSWERS [Ex. 6-9
7. B, $250. A, $500. C, $800. 8. 4 cows; 12 hogs.
9. 112 yds. by 224 yds. 10. 8 hrs. H- H ^^^' Per hr.
12. 64.
16.
65,72
20. 18 mi.
24. 12, 24.
13. 9.
17.
32,48
21. 12 hrs.
25. $24,000.
14. 48 lbs.
18.
57,58
, 59. 22. If hrs.
26. 7.
15. 24, 39.
, 19.
6 da.
23. 15, 20, :
25.
27. 5 yrs., 35 yrs.
28.
12 mi.
29. 3 lbs.
30.
A, 25; B, 27.
Exercise 7.
1. 3.
4. -27.
7. -3V 10.
-9.
13. 14.
2. 7.
5. 2i.
8. -.24.-^ 11.
-2.
14. 4.
3. -22.
6. 0.
9. 9. 12.
— 5.
15. 14.
16. -3.
18.
6.
19.-39.
23.
-200 lbs; -50 lbs.
24. 25 ft ;
+5;
+25;
-5; +15. 25. $210; -
-$160.
Exercise 8.
1. -14.
7.
540.
13.
81. 19. 24.
25.
-3. 31. f.
2. -14.
8.
-280
14.
-64. 20. 108.
26.
-xV 32. -54.-^
3. 14.
9.
ri
15.
-1. 21. 576.
27.
-2.5. 33. -5i.
4. 14.
10.
1.
16.
216. 22. -162.
28.
-12. 34. 8.
5. 30.
11.
0.
17.
.432. 23. -9.-
.29.
-i. 35. -192.
6. -12.
12.
32.
18.
72. 24. 9.
Exercise 9.
30.
18. 36. -28f.
1. 3a?.
3.
-4c3.
5. IQax^.
7. 10F<^.
2. -4a26.
4.
-3a5cd. 6. lA.
8. -B^.
9. 2lpq.
13.
-23m27i2.
17.
— 5ahc,
10. 86f.4C. 14. -X. 18. ^yz.
11. -5ofiy\ 15. -18^3. 19. 12a-9x.
12. -Slpqr. 16. 5a^l^c. 20. Qah-ldxy.
21.^ -2a5c2+9a26c+7a52c.
28. 3a2+262. 24. {2+a+h)xyz.
23. (3a+56-7c)a;2, 25. (-7c+2+Sa)y\
Ex. 10-13]
ANSWERS
1. -dx-^y+5z.
2. 7a— c.
7. 2oc^+2x.
8. 4a2.
9. 2i)d^+6x^y—Qxy^—2y^.
Exercise 10.
3. -3P+dQ+4R-5S. 5. 15p+6(/-S
4. dac—ixy. 6. Sab+Qbc.
10. 23Vi-|6^ic.
11. 5x^+i^x-2.
12. -9a?4+4a?34-2£c2+7a7+4.
9. 7x-10y.
10. -a75-ic3 + .x2.
n. 4a26+2a62+268.
12. 2AB+5xy-'7PQ.
Exercise 11.
1. 6a%\ 5. 136?/2.
2. — CoTi/. 6. 24x^y'^z.
3. aic2. 7. 2x-7y+10z.
4. -lOabc. 8. 4a+14a6— 7c.
13. cc2-ll£c+13. . 14. x''+x^+oc^+x*+2o(^+x+l
15. a*4-3;rS+?-7+3c2s5.
16. a;3-2a;2+^-l; a_2^24.^_l. _2x^+x-l.
17. _as-a862+4a253_2ic. 18. a.'3_^7^2^_^,^_l.
19. 9.125a-7.6a;2-6.25?n3-5??i2-l.
20. —x^+x^+x—2. 24. —x^+Qx^—Qx—4. 27. —0734.407—6.
22.
2aH5-l|a:2+10
25.
a,'8_2a;2+8a7-2.
28. 36.
23.
7a3+2a2-3a-
-7. 2S.
a;3-4a7+6.
29. -9.
30.
3.
32. -3.
34. 16.
36. 16.
31.
-18.
33. 8.
35. 26.
Exercise 12.
37. 8.
1.
3a- 6. _
2.
507+5?/.
7. -2b.
3. 52+4^(
4.
4a-2b.
5.
3aj3-7£c2+l.
8. -2a?3-
-4x^+x-4:.
6.
-3(K8-a;2?/+6a;2/2+22/8.
9. 7x'+Sxy-5y^-2y».
10.
a-b.
12. 5-2/.
14. -4a;+3a. 16. 2.
11.
15a-26.
13. e+ic.
15. 1.
17. x:
1. x^-oc^+oc^-ix^-x+l),
2. ax—{by+cz—dw).
Exercise 13.
3. a-(-25+3o-4d).
4. -(10e-5/+9r).
ANSWERS
[Ex. 13-16
5. (2+a)c(^+{b-S)x^+(6-c)x.
6. 7+{5-2a)x+{l+Qb)x^+{d-4a)x^.
7. (a-'d)x*+ (2+a)c(^+ {l-b)x^+ {d-c)x-l,
8. _(5_2)7/-(l-a)?/+5.
9. -(-p+q-r)y-{dq-2p)y^-sy*.
10. -(6x+2)y^-{d-x^)y^-(dx^-5)y.
11. 10+(2+b+c)x-(a+2)x'^.
12. (a+Q)x^-5x. 13. 5ii;2-3a7-l.
14. {l+a)x^+ (a.-b)x^+ (b+c+l)x+4:.
15. (a-3)ic-^+(a2+a-3)a'2+2aa;. 17. (a-b)x'2+ib-c)x+c+d.
16. .Tio+(2-a+&)ic5+?>+c-d. 18. (b+2)x^-{c+S)x+5-a,
19. pa:;3_p^2_(gjfr7')a;_^_g.
Exercise 14.
6. 4a^b^c^.
7. -^^5^4.
8. -P^Q^.
12
1. -«7;,i2.
2. 60a^66.
3. 210a;V-
5. -3a2364c7c?5.
y. —pn^'n
10. 14.21875a;62,3.
48pV'-
2i^22^11.
2(a+b)5.
-15a3(6+c)*.
11.
11.
12.
13.
14.
16. x'^\
17. Qa\ 19. (a3)io.
Exercise 15.
5 . da^y^zw — Sxf^j/zHv + Soc^yzw^.
6. 10p2g2,._15pg2^2+20p2gr2.
7. ^35-^53.
8. -80ir8+12ar5-8^.
10. 1 5a363c3ic22/2_ Qa^b'2c^x'^y'^+da^bcx^^+ dal^cx^y^+dabc^x^y^z.
11 . — 5a%22/4 + a^x'^y^— -ja^xy* + ^'^x^yK
-14?/. 16. -6ir-12?/.
1. a^b^-2a^b^+a^b\
2. ic'^— a^+.r^— a;4.
3. -6a6624.5a5j>3_2a364.
4. Ax'^y^+Qx^y^-Wx'Y-
9. -Ix^y^z^-lx^y^z^
15. -26c.
1. a2+2a5+52.
17. 2ic3+3ic^?/-5ir2/2.
20. 24ic2-30iC2/.
Exercise 16.
2. a2-62.
18. 2ab^-2a^b.
19. 7a?4-42/*.
3. 6a;24-iC2/-22/2.
Ex. 16J ANSWERS 5
5. 307*4-072-4. 7. 20p2g2-14pV-6p2?-2.
8. .t2-h5o;4-6. 10. 16a2-49. 12. 12x^y^ -Sc-^d^.
9. 4a;2+4a;-35. 11. ic2-l00. 13. ia^_|f£c2+^.
14. ^3ga2_|a5-^\62. 17. 0.94a2-5.55a6-3.64&2.
15. ^od»-^x^y+^\xy^-j%y^. 18. a^x^-b^if-bcyz+acxz.
16. 5.625372+ 15.375^^-1 1.257/2. 19. p^q^+2pq^r-p^qr+q^r^-pqr\
20. ait;4-5a;-f-ca;+a?/+b?/+e2/+a-2;+62;+c2;.
21. 2a7*+irS-8.T2+23;^7_i2. 22. a;5+£c4+ic3_a;2-a;-l.
23. 4a77-2a76+7£)75— 7a74+7a;3-7j;2+3o?-5. .
24. x8+;;c4^_i, 27. a4+a252+54. W/^^^
25.^-a^2+<^23^_^^ ^ 28. ic^-l. "— «^~ '
2a^2«®-«'^'^-14«i^*+19«I'^^-^^ 23. ^2^2_4^8C_ 4^3(7+ 16^502.
30. — 2p2c2+3gr2(^2_2gc?re-r2e2+pgcd+3jpcre.
31. -2a;8+3a;c6+2aj6-6a2a;2-3«ic2+4a£c4+2a2x4.
• 32. a4-4a36+6a262-_4a53+54,
33. £c«4-aa7^— 4a2£t^— 46t3a'3+4a4o72+3a5o7.
34. a262_ 52c24_26c2fZ + 2b2cd-c^d^-2bc(P- bM^.
35. ai4-2aio6"3+2a666-a2&9." 41. |a3-ia2&+i.|a52- ^^^63.
36. a;"+i-f-a?«?/+a7y«+2/«+^- 40 a?* llx^a^ a*
37. a^2«_2^2n, ■ 30 900 30'
38. a6<^+a*^'^62«+a2o53._,.55c, 43. a73-a72+|£c-|.
39. x^"+^-x^''+^y^+x^y»-^-y^. U. Sx^+8x^-\^x^+%^x^-\^x+3.
40. a;» + 2_£pn+l4.2it:n_^^«_l + ^H-2^ 45. ^9^;;p4_4 3£p2_|„^_
46. 2.8x'*-7.36;r3+5.7a;2+4.16a'-10.24.
47. (r2«— 07*2/*— ^"Z/'+Z/*^'- ^- ^^^4- 1007^4-35072+5007 +24.
48. 07»+'" + 07'«2/" + 07"2/'" + 2/"+'". 55. 1 07^ + ^0727/+ ^077/2 +^',^3,
49. x^+2x^—x—2. 56. 07*"— 2/*". ^
50. o?8-l. 57. a6— &6.
51. 6a3+a25-lla62_663. 68. 4a6.
52. x^-1. 69. 072-907+6.
53. ai2_5i2, 60. 2x^-2xy-2y^.
61. 6£c8+12a;2-14a;-4. 62. 5a'^-oa^b-2a^b^-ab»+b*+a^b'^+a*b^-ab\
5
ANSWERS
[Ex. 17-19
•
Exercise 17.
1.
aK
6.
-Aa^bc^. 11 -is^t.
16. -2st.
2.
a^l^.
7.
ip\ 12. 5a7.
17. 9r+2.
3.
-2a863.
8.
-12ia4. 13. a^.
18. ^fz.
4.
-^xy^.
9.
7a262.. 14. -V^»r
'.
19. |s».
5.
465c.
10.
-|??i%p3, 15. £c<-?/«-i.
Exercise 18.
20. Vs2M.
1.
x-^l.
5. 10a4+964.
9.
2x'^y^+ix^y—%x^y^.
2.
xy-l+4x\
6. a2-2a&+3b2.
10.
2x-|2/+if^2/2.
3.
-x^+5-dx^.
7. —a+b+c.
11.
a— 6+c.
4.
-3x^+2x^
8. |j;7-2^'V. •
12.
a2+a6+62.
13. —la^—'^-ixy'^+'^-ix^y.
15. 5a7«-2a;»+i. le, a«_a2«.
14. 13£c4-20.8a?3+39.
17. a7«+2/". 18. l+£f2+ic4.
1. 3a+l. 4.
2. 5.2^+1. 5.
3. y+1. 6.
10. s(^-\-xhj+xy'^+y^.
11. x^—xy+y^.
14. a^-a262_^54. 19.
15. 1207-1. 20.
16. x+1. 21.
17. x^—2xy+y^. 22.
18. .r4+£c2+i. 23.
29. —d-dx—x"^.
30. a2_,_2a6+a+4?>2_25+l
31. a;3_4^2+ii£t;_24.
32. x—y+z.
33. £C3_,_^2y_|_a72/24-2/8.
34. a4+3a3+9a2+27a+81.
35. 2a2+9a-5.
36. x^+ocf^y+x^y^+y^.
Exercise 19.
5a+l.
3a2-5a-2.
x^+2xy+y^.
7. c-4.
8. a-5.
9. a2+a5+62.
12. x^—x^y+x'^y^—xy^+y*.
13. a4+a252_^54.
it'4+a73-|-.x2+a;H-l, 24. a;*— 072+!.
2ic4+aic2_3a2, 25. ^y-L
2a2-5a+7. 26. o^— 2a:2rt+2a?a2_a3.
£C^+£c2//2+i/4. 27. l-3£c+2a;2-a^.
a;2— 6^7+9. 28. a;2+2ir2/+%^-
37. a2+5a+6.
38. x'^—xy—xz+y^+z'^—yz.
39. it'2+2a??/+2/2— 0^2;— ?/2;+;2;2.
40. ^a2--Ja6+|62.
41. ?i+^+a^62+4a253+i664.
lb 4
42. |£C2— ^.T2/+2/2_
43. ^x^-lx^y+^xy^.
Ex. 19-21]
ANSWERS
44. a2'»+2a'"6'"— 62'n.
46. 2a;«-r3a;»-i— 5ic»-2.
1. a"
2. a28.
3. -aiofois.
4. a?6y/8.
5. x^hf^z-^.
6. a6a?3o?/i8.
7. Oojio.
9. 343aV-
10. ^2x^^y^.
11. -32a20o;^.
12. aj2o?/i2.
45. a^« — £c2a^a_j_^^2_f_9^. 29. a^-20.r2+100.
x^y^+iaxy+ia^. 30. a,'8— 1007*4-25.
in^-Qm+9. 31. x^-2x^+x*.
i;j4_6„j,2_|_9. 32. 4a7*+4r8H-ic2.
m^n^—imny+iy^. 33. 25ir*— 100:^+076,
36. 4a;2-126a?7/4-962^2.
37. 4a262+16a62c+1662c2.
ir2o_|_2icio+l. 42. £c2«_^2a;»+l.
ic24_2£pi2+l. 43. a;2»-2a7»+l.
45. 4a;2»+i2a;«a»+9a2'«.
8 ANSWERS [Ex. 21-23
46. a2n+6_3rt«+8|ia+8+^2<.+6. a* _2a^a^ ofi
47. 4a:2'»+2+4?iar2»+J+n2aj2».
it'i'> , .T^ , a?*
49. a2»62n-2_2a2n-l&2»-l4.a2«-252».
53. 4a2+4G(b+?>2+4ac+26c+c2.
a2 2ac cf
°"- b^ bd d^' 54. 9a2+6a6-6ac+6^-26c+ca-
55. 16-16a-86+4a2+4a&+62.
56. 16-246+ 9&2_24c+186c+9c2.
67. if-i^+c^+ab-^-^+'^.
Exercise 22.
1. l4-2a;+3ic2+2a^+a;*. 2. a8+2a6+5a*+4a2+4.
3. Ax^+12x^+2oa^+2ix^+iexf^.
4. a;W+2a;8+3^+2a^+2cc4+2a?3+a;2+2a7+l.
5. icc»-12x^+25x^-2^x^+16.
6. a2+4624-9c24-i6d24.25e2-4a&+6ac-8ac?+10ae-126c+16&d-206e-
24c-
-6. 32. a^+2a?3-2a;2_3a;+2.
26. x^—2xy+y^+10x-
■10^/+21. 33. x^-
Aa^x^+Sa*.
27. p2_,_2pg+^2_6p_|
6g-160. 34. a;2n_
-^nyn_2\y^\
28. x*+2x^+ix^+'6x-
■18. 35. a;2'»+24-a7n+i2/»-i-62/2''-2.
Exercise 29.
1. a;+3.
7. 5ab-i.
13. l+6m2n8.
2. £c-3.
8. l+7aa;.
14. 13a?t/3-12a*.
3. a2+5.
9. l+4a:4.
15. 5a +1562.
4. 63-4.
10. t^-1.
16. a»+6».
5. 2;z;+l.
11. 8a?2+92/2.
17. a?»+i-a»-i.
6. 3a-l.
12. lOa-lla^.
18. a+6+c.
19. a+b+x+y.
20. {a+b)^-{x-y)\
10. m8+m2n,+mn2+w8.
Exercise 30.
11. l-2/+2/2-2/8+2/*-2^.
12
ANSWERS
[Ex. 30-33
14. a^-a^+a^-a+l. 17. 8a^-12a^h+18a%^-27b^.
15. a^+x'+x^+x+l. 18. 27a3-45a2Z>2+75a64- 12566.
16. a2+62. 19. a^-cc^y^+x^y^—x^y^+y^.
22. a^+a^2/+^2/'^+^2/'+^^2/*+^2/^+2/^-
23. x'^—a^y+x^y^—x^y^—x^y^—x^y^+xy^—y'^.
24. a^2/6+a;22/4a8+a^2a6_,_a». 25. 25a-862.
26. 64p6 + 32p5g8^_16p4g,6_,_8p3^_|.4p2gl2_f_2pgl5_^gl8^
27. a^»— a;2"2/"+a7''2/*«— ^3». 31. a^"— a2»+a»— 1.
Exercise 31.
4. ar4(a^+5).
5. x^y^(x^+y*).
6. 5a2(a-26).
1. a;(aj2— 2^+1).
2. ir2/(a;+3a?2!/— 5?/).
3. a?(a;2-3).
10. Sx'^{2x^+Sy^+xy).
11. 12a*5'»(262-3a2).
12. 4x2(ic2+l)i
13. 2a8(4a2+2a5+62).
18. ^^(l+a).
1. 5(a;-4)2.
2. 4(a-b)2.
3. a(.T8+3)2.
4. Mxy+h)^.
5. 5a(2a-3)2.
1. {x+Q){x+7).
2. (x+8){x-Q).
3. (a?-5)(a;-4).
4. {x-7)ix+A).
6. (07+8) (a?+9).
6. (a;-|-10)(aT-5).
7. (a;-14)(a?+4).
7. 7a;2(4£c2+?/2).
8. 9x^(2x^-1).
9. a*(a2-6a6+262).
14. a262c2(a253 + 52c4 + (j3c2).
15. 6a782/(4a722/2_2+7a?32/).
16. 9m?i(3m2+4mn+9?i2).
17. 14a2?/2(4?/2-a2/+2a2).
19. a»(a7«— 1). 20. 5a''-'^y''-'^{a+2y)
Exercise 32.
6. -l(3a;-l)2.
7. -l(a2-4)2.
8. h^{5a-x^y)^'
9. Sxy{x-ly)^.
10. 8(6a2+56)'^.
16. 3a(a-6)6.
Exercise 33.
8. (£c-4)(a?-8).
9. (a+15)(a-12).
10. (m-16)(m+15).
11. (^-21)(f4-20).
12. (a+l)(a+i).
13. (c+10)(c-7).
14. (6+14) (6+6).
11. 7ix+y)^
12. 5a(a-6)8.
13. 2xy(x^+y)».
14. -5aa?(a-36)8.
15. 2(2a-6)*.
15. (c+21)(c-4).
16. (d-ll)(d+5).
17. (2+r)(l+r).
18. (6+s)(4-s).
19. (3+^) (5-2/).
20. (2+a)(l-a).
21. (2*^-2) (22+1).
Ex. 33-34]
ANSWERS
13
30.
49.
50.
51,
52.
53.
54.
55.
56.
57.
58.
59.
60.
1.
2.
3.
4.
5.
16.
17.
18.
23.
23.
24.
{x''-2)(x^-5).
(x^-8)ix^+2).
ix'2+18){x'^-10).
(x^+l){x^+2).
(x^-S)(x^~2).
(0^+13) (a^-3).
(£C«-16)(a^+14).
(ax+ll)(ax+2).
(a+136)(85-a).
(xy—Sah) (xy—2ab).
{xy^+9ab) {xy^-2ah),
(xy-5c){xy+2c).
(x^y^+7a)(x^y^+2a).
{x^y^-M^h) (ic3?/3-4a86).
(am2+5c2)(am^+6c2).
(l_2a)(l-a).
(l+9.T)(l-3a;).
(a+6-f-2)(a+6+5).
{x-y-\Q){x-y+^).
{{x+yy+W} {{x+y)^+^}.
31. (b7/-10)(6?/+3). 40. (a66-7)(a66+5).
32. (ab+10)(«6+20). 41. {xy^+n){xy^-4:).
33. {xy-8){xy-2(i). 42. (p5Jg4_l)(p2g4_2).
34. (mri+10)(mn— 6). 43. {x+2y){x+y).
35. (abcH-15)(a6c-2). 44. {x-^ij){x-2y).
36. (j72/2;-10)(a7^2f-9). 45. (a7+72/)(ic+107/).
37. (x2j/2-l)(a;22/2_2). 46. (a+136)(a-36).
38. (a^.v8+3)(aV+ll).47. (a-206) (a+26).
39. (aj42/*-14) (07*2/4+9). 48. (5+3a)(6-a).
61. {a+h—x—y){a+h-2x-2y).
62. 2(a;-12)(iC+7).
63. 3(£C+3)(a;-2).
64. 5(a?+4)(a;+5).
65. 2(a;-25)(a;+8).
66. a{x+1[){x—2).
67. a2(a;+7)(£C-5).
68. x{x+la){x—Qa).
69. 307(07— 24^) (aj+ 42/).
70. 0722/2(0^+72/) (07+22/).
71. 2o7(o72/+26)(o72/+5).
72. 2(10-'a;)(ll+a;).
(2ir+l)(oj+2).
(307+1) (0^+2).
(2o7+5)(o;+2).
(2o;-l)(cc+3).
(2o;-3)(o;-l).
(7a3-l)(2a3+l).
(3a2ic24.3)(2a2.:c2+3)
(a6c+4)(5a&c— 1).
3a (207+3) (07+1).
22/(2oj+7)(9oj-l).
Exercise 34.
6. (5oj+l)(o;-2).
7. (3o;+2)(4a?-l).
8. (6o;-5)(o;-l).
9. {2t-l){t+\).-
10. (o;+2)(8o;-l).
11. (4c-3)(3c-4f.
12. (ft-16)(56+l).
13. (a+5)(7a+l).
14. (2/+3)(102/-9).
15. (r2-2)(2r2+3).
19. (ar2y'-3)(2a722/+5).
20. (2a263_5)(3(i253_4),
21. 3(a?+2)(4o;-l).
25. {x+y){2x+y). 28. {x-iy){^x+y).
26. (4o;-2/)(3.T-2/). 29. (2«i-3a)(6m+a).
26(a-9)(3a+4). 27. {2x+y){x-2
30. (20^2/— 5)(i>^+3).
14
ANSWERS
[Ex. 34-35
31. (a;2+ 42/2) (3£c2- 2^2).
32. Sa(x-2y){9x—y).
33. 2('Sah-2c)(ab+c).
34. {ax+b)(x-l).
35. (y+2a)(2y+h).
36. {2z-h)(z-a).
37. (aa?+l)(a?+5).
38. 2cj(a;+a)(a?— &).
1.
2.
3.
4.
13.
14.
15.
16.
17.
18.
19.
20.
21.
24.
25.
26.
27.
(.17+6) (07-6).
(a;+10)(a;-10).
(a;4-7)(a;-7).
(x+8){x-8).
{10y-7b){10y+7b).
(9a-85)(9a+85).
(25a;- 155) (250?+ 155).
(3aa7-l)(3aa7+l).
{ixy+3){ixy-d).
(2ay-b)(2ay+5).
(5a7a— 4)(5a?a+4).
(10a^-95)(10£n/+95).
(4072;— 5c) (4a72;+5c) .
(13071/2;- 5a5c) (13o7i/2;+5a5c).
(la-lb) (ia+ib).
{lxy-ia)i^xy+^a).
{i\ab-i^x) {^\ab+^%x).
(f-fa^2/)(l+l^^).
(I_i^0(j5c)(l+-ijpa5c).
Exercise 35.
5. (o;+12)(o;-12).
6. (a;-14)(o;+14).
7. (07+a)(07— a).
8. (x+n)(x—n).
9. (2a;+a)(2o7-a).
10. (3o7-a)(3ir+a).
11. (2o7-3a)(2o;+3a),
12. (4o7— 5a)(4o7+5a).
31 /-E — ^^\ (^ -I. ^^\
' \4:b yfUb'^yl'
32. l^-l\l^+l\
''- {TWb-''y)(mb-'y)'
34 i^-W^+^\
\9y 2xJh)y^2xJ'
35. /
^_ 10xyz \L^ -JOxyz y
c /\ c
36. (a+5-l)(a+5+l).
37. {a+b—c—d)(a+b+c+d).
38. (x+y—a—b){x+y+a+b).
39. (o7— 2/— a+5)(07— 2/+a— 5).
40. (2o;4-22/-3a-35)(2o;+22/+3a+35).
42 l <^+^ _ c+d \( a+b j^ c+d\
' \a-b c-dl\a-b^c-df'
43 i 5(x+y) ijx-y) ) ( 5{x+y) 4(x-y) )
' i 7(a+5) 9(a-5) J | 7(a+5) "^9(a-5) f '
Ex. 36-38J ANSWERS
Exercise 86.
1. (a-b-x){a-b+x). 7. (a+5&-l)(a+55+l).
2. {{+x-y){l + x+y), 8. (a-h-2)(a-b+2).
3. (x-y-l)(x-y+l). 9. (a-4b-2c)(a-ib+2c).
4. (« + 3b-3c)(a+36+2c). 10. {l-a+dx)(l-\-aSx).
5. (l_aT-42/)(l+a;+%). 11. (3£c-2a+3c)(3a?+2a-3c).
6. (x+2a-y)(x+2a+y). 12. (a+6-c4-d)(a+6+c-d).
13. {x-2y-a-3b){x-2y+a+Sb).
14. (a-56— 2?w-f-3n) (a-56+2m-3n).
15. {x-Qa-2y-b) (x—Qa+2y+b).
16. (32/-a?)(2/+3a?). 19. (lla-46)(-a-26). •
17. {Sx-4y)(x-2y). 20. (-Sy) (2£c). ,
18. (-2x-y){ix~9y). 21. (9a+36)(lla-6).
Exercise 37.
1. ix^+x+l){x^-x+l). 2. (l-aj+2ic2)(l+a;+2a;2).
3. (x^-x^+l){x^-x+l){x'^+x+l).
4. (a2+2a6+362)(a2_2a5+362). 5. (a2+3a+3)(a2-3a+3).
6. (2^2+207?/+ 5?/2)(2a;2-2.T2/+ 5^2),
7. (2a2+562+4aZ>)(2a2+562-4a6).
8. (x^-nxy-dy^){x^^'6xy-dy^).
9. (6a2-4a?/-5^2)(6a2+4ay-52/2).
10. (7a4-3a2?>+262) (7a*+3a26+252).
11. (237*- 6a;22/3+ 52/6) (2a^+6a;2?/3+52/«).
• 12. {8x^y^-2xy+l){8x^y^-\-2xy+l).
13. (a*-4a2a72/2_3a.22^)(a4+4a2iC7/2-3ic2^4).
14. (2ni^ii*—5ab»mn^-6a%^) (2m^u'^+r)ab^mn^-5a^b^).
15. (.T6-cT8i/r3+4)(a7«+a;Vl34-4).
16. (oa^-2ab»c+2b*c^) (5a^+2ab^e+2b*c^),
Exercise 38.
1. (x+y){x^—xy+y^).
2. {x+y)(a^—oc^y+x^y^—xy^+y*).
3. (x+y)(x^-a^y+x*if-x^y^+x'^y*-xy^+y^).
15
l^ ANSWERS [Ex. 38-39
5. {x+y) {x^o-x^y+x^y^-x'^y^+afiy*-x^y^4x*y^—x^y''+x^y^-xy^+y^oy
6. (dx+a) {9x^-'6xa+a^). 9. 5(2a+36)(4a2-6a6+962).
7. {4x+^y) {Ux^-12xy+9y^). 10. 2(Q+x)(S6-6x+x^).
8. (5a+6&)(35fx2-30ab+3662). 11. 3(3a;+l)(9a;2~3a?-hl).
12. (b+i)(b*-b^+h^-b+l).
13. (i+x){i—x+x'^—x^+a^—x^+x^).
14. (2+!c)(16-8ic+4a;2-2ic3+a?*).
15. (l+5a)(l-5a+25a2-125a3+625a*).
16. {ah+l){a^t^-ah+l).
# 17. (007+2) (a4a?4-2a3ic3+4a2x2-8aa;+16).
18. {a+5xy)(a^—5axy+25x^y^),
21. 0a+5c)(^a4-Ja36c+^a252c2-ia&3^34.54c4j.
22. (4a+i&)(256a4-16a36-}-a262-3-Va63+^^e&*).
23. (2x+j^y) (Q'ix^-lQa^y+ix*y^-x^y^+ix^y*-j\xy^+-i^y^).
24. (a7+2/'^)(a72— ic?/2_,_2/4),
25. (3a;2+22/)(9a?4-6a;22/+42/2).
26. (x+l){x^-x+l)(x^-x^+l).
27. (a+6)(a*-a36+a262-a53+b4)(ai'>_a555+5io),
28. (1+a) (l-a+a2-a3+a4) (I_a6+ai0-ai5+a20).
Exercise 39.
1. (x-y)(x2+xy+y^). 2. {a-h){a^+ab+l^).
3- (a;-^)(a74+a?82/+a?2^2+a,^_^2/4).
4. (ic— y) (x^+afiy+x^y^+y^x^+x^y^+xy^+y^).
6. (x-y) {x^^+x^y+x»y'^+x^y^+x^y*+x^>y^+x^y6+o(fiy'i+x^?f+xy^+y^^).
7. (2a-y)(Aa^+2ay+y^). 10. {2d^-b)(ia'^+2a%+b^).
8. (3a;-4a)(9x2+12a7a + 16a2). n. (i_|;^)(i+|^+^a;2).
9. (5-7a)(25+35a+49a2). 12. {x-l)(x*+x^+x^+x+l).
Ex. 39-40] ANSWERS I7
13. (2-y)(lQ+8y+4y^+2y^+y*). 14. (3-a)(81+27a+9a2+3a8+a4).
15. {x—2a)(x'^+2x^a+4x'^a^+8xa^+lQa'^),
16. (x^l) {x^+x^+x^+oc^+x^+x+l).
17. (l-a)(l+a+a2+a8+a4+a5+a6).
18. {2-xy) {Q4:+d2xy+16x^y^+8x^y^+4x^y^+2xY^+xf^y^).
19. 3(aT-l)(9a?2+3aj+l). 20. 3(6-a)(36+6a+a2).
21. ^(x-y)(x2+xi/+7j'^).
«« / a\ / . a^a x^a^ xa^ a\
22. (^-g|(a^+-^-+^+^+_).
23. 2(^x-y) (^i^x^+la^y+ix'y^+lxy^+y^).
24. 2a?(fa;-2/)(fa?2+3a.^+2/'^).
25. a8(^_(|r2/)(a;2+a?a?/+a22/2).
29. (a-b-c+d) {(a-by+ (a-b) (c-d) + (c-d)^} .
30. (x-2y-2a-b) {{x-22j)^+ {x-2y}^(2a+b) + {x-2y)^{2a+by^i-
(x-2y){2a+b)^+{2a+by}.
Exercise 40.
1. (a2+62+a6i/2)(a2+62-a6i/2).
2. (a2+62)(a2+52+a&i/3)(a2+62_a&i/3).
3. (a4+64+a262|/2) (a4+54_a262|/2).
4. (a2+62) (a8_a652_^aV-a266+68).
5. (a4+54+a262|/3)(a4+54_a252|/3)
(a2+62+(^;^y^2) (a2+62-a5i/2).
6. (a^.+l)(a2+l+a\/d)(a^+l-ai/3).
7. (ic2+l)(ir;i2-;t;io+x8-ar6+£c4-a;2+l).
8. (x^+l+x\/2)(x^+l—x\/2).
9. (a;24-l)(a;8-aj«+a3*-aj2-|-i);
10. (l+a:*+a'2|/2)(l+aj*-ic2|/2).
11. {4x*+l+2x^\/2)(4x*+l-2x^l/2),
18 ANSWERS [Ex. 40-41
12. (l+9ic8+3ir*i/2)(l+9a38-3a^]/2).
13. (4^2+92/2) (ix^+gy+mxyy'S) (4a;^+9y^-Qxy\/3).
Exercise 41.
1. (a2+62)(a+5)(a-6).
2. (a+6) (a-6) (a2+a6+52) (a2_a?>+52).
3. (a-f-6) (a-&) (a2+?)2) (a2+52+ct?,^/2) (a2+52_ot5-j/2).
4. (ci+b) (a-b) (a'^-a^+a^b^-ah^+b^) (a^+a^b+a^b^+ab^+b*),
5. (a+b) (a-b) (a^+b^) (a^+ab+b'^) {a^-ab+b'^) (a^+b^+ab\/d)
(a2+52_c(5-^/3),
6. (a+b) (a-b) (a^+a^b+a'^b^+a^b^+a^^+ab^+b^)
(a^-a^b+a'^b^-a^^+a^^-ab^+b^).
7. (a-^b^)(a^+b^). 9. {a^-'l^){a^+b^){a^+b^).
8. (a3-64)(a3+64). 10. (a5-62) (a5+62.). ,
11. (a?2+l)(a;+l)(£c-l).
yi2. (a;+l)(a?-l)(aj2_^a;+l)(a-2-a;+l).
13. (l+x)(l-x){i+x2){l+x'^+x\/2)(l+x'2-xi/2).
14. (l+a?)(l— a?)(l+a;+a?2-|-£c3_i_^)(l_a;_^^.2_£t.3_^^)^
^15. (1-0?) (i+x) (l+a-2j (l+a?+a-2) ( 1 -xi-x^) (l+x^+xi/d)
(l+x^-x\/d).
16. (2+a^2) (2-0^2).
17. (2+a;2)(2-a;2)(2+2iC+£c2)(2-2aj+a?2).
18. (9-073) (9+0^). 19^ (10_a8)(10+a8).
20. (2a-l)(2a+l)(4a2+l)(4a2-2]/'2a+l)(4a2+2i/2a + l).
21. {Sa^+2x»){Sa^-2a^)(9a^+Axfi).
22. (H-2a2) (l-2a2) (l+2a+2a2) (l_2a+2a2) (l+4a*+2a2|/2)
(l+4a*-2a2]/2).
23. (K2ic)(l-2aj)(i-a;+4a?2)(iH-a;H-4a?2).
-^24. {xy-ab){x'Y+abxy+a^b»){xy+ab)(x'^y^-abxy+a^b^).
25. (2+3a6)(2-3a5)(4+9a262).
26. (2-a) (2+a) (16+8a+4a2+2tt3+a4) (16-8a+4a2-2a3+a*).
Ex. 41-43] ANSWERS. 19
29. (Factors of x^'+a'^) (factors of aj^—a"). /
30. (£c»-3— 2/»+2)(cc"-8+2/»+2).
31. {(a+6)2+c2}(a-f-&+c)(a+6-c).
32. {(j;_2/)2+ (a-6)2; {x-y+a-h) (x-y-a+b).
33. {(a+b)2+c2}(a+6+c)(a+5-c).
34. [(2x—yy^+ (a+4b)2} (2x—y+a+ih) (2x-y-a-4b).
Exercise 42.
1. (a+b)(c-d). 5. (a2+i)(5_|_c). 9. (x-l)(y-l).
2. (a7-7/)(a-5). 6. (a+b)(x^+x+l). 10. (a-2)(6-3).
3. (a;+2)(a+6). 7. (2+d)(a-6+c). 11. (£C+b)(£c+a).
4. (x-y)(a+4). 8. (a;-l)(«-6). 12. (a?-2)(a?+a).
13. {x—a){x+a+c). 16. (a+6)(a7+l)(a7— 1).
14. (x-a)(l+x-a). 17. (a7+l)(a:-l)(a;+3)(ic2+l).
15. (a7+2/)(« + l)(«^-«+l). 18. (x-y)(x+y-'i).
19. (a-6)(£tf-l)(a.^+(r3+ic2+.T+l).
20. (a+1) (a-1) (6+1) (5-1) (a2+l) (b^+1).
21. (07— l)(acc+5). ' ' 24. (mn+l) (pq+2).
22. (2o^+2/)(^+l)(^^-^+l). 25. {2a-b) (2x-y).
23. 2fe2(a+6)(a-b). 26. (6+(i)(6-d)(a-c)(a2+ac+c2).
27. {a-l)(a+l)(x-l)(x^+x^+x^+x+l).
28. (aH^5)(£C+l)(ic+2). 29. {x-y)(x+2y\
30. {m—q)\m+q) (n+p) (n'^—np+p^) (n—p) (n'^^+np+p^).
31. (a+&) (a-J) (x+y) (x-y) (x^+y^) {x^+y^+xy\/2) {x^+y^-xy\/2).
32. (a-b){a+b)(a^+b^){x-y)(x2+xy+y% ^
33. a(a+6)(a;+l)(a;-l).
34. {x+y—z—w){x—y—z+w).
35. (a:+2/+'2^)(^+2/-^)(^-2/+^)(^-^+2/)-
36. {a-b)ia+b+cy 39. (ic2_2/24-i)(aj2+2/2).
37. (a+6+2c)(a+6+3c). 40. (x-y+z)(x^+xy-{-y^).
38. (a-5+c)(a-5-c). 41. {3-x~x^)(l+x).
42. (a+6)(aa;+62/+c).
20
ANSWERS
[Ex. 43
1.
2.
3.
4.
13.
14.
15.
20.
21.
22.
23.
24.
28.
(2a;+l)(ir-l).
(Sx+2){x+l).
(2a;-l)(aJ-2).
(4x-l){x-h2).
(x-i){2x-l)(x+e).
(x-l)inx^+x+2).
{a+2) (a^+a+l).
Exercise 43.
5. •(2a;-3) (x-d). 9. ^(x+2) (x^-x-1).
6. (3a;+5)(a;+3). 10. (x-l){x+S){x+5).
7. (Sx+2)(x-n). 11. (x+l)(x^-2).
8. (x-l){x-2){x-d). 12. (a7_l)(£c4-l)(a?2+4).
16. (a+3)(a2-a-l).
17. (a+l)(a+4)(a-5).
18. (a+4)(a+3)(a3+l).
19. (a+1) (a-1) (a2-ai/3+l) (a^+a\/2+l).
(2x-y) (x-y).
{^x-y)(x+y).
ix+y)(x-y)(x+2y).
(x-dy){x+Sy)(x~2y).
(a+db){a-h){a+2h).
a(x^-Sa^) (x^+Sa%
- (^-irnr-
47.
48.
49.
50.
51.
52.
53.
54.
55.
56.
ah(2x-l){x+1),
(2a+35)(2a+13b).
(ar-7)(ar+4).
a{x—a)(x—5a).
35. {y-z){x-z).
36. (x+'S) (x^+6).
.37. (2a-2b-y-z) (2a-2b+y+z).
38. (3£C-52/)(3a;-42/).
39. (a2-a54-62)(a2+a6+52).
40. {x^—ixy—y^){x^-{-4:xy-y^).
41. (a^3_6)(a?3+7).
42. {x^-5y^) (x^+ly"^).
44. £c(it?-3)(l-ir)(l+a!?).
45. (by+l+xy){l-xy).
46. (x-y-z) (x-y+z) (x+y-z) (x+y+z).
(x+y)^(x-y).
x(y-d){y+10).
(c—b)(a+d).
(x+d){x-2)(x^-x+Q).
(8f+9)(9f-5).
(a+1) (a-1) (5+1) (5-1).
(2z-M){z+U).
-a2(3a-£c)2.
x{x^+c)(b+ax).
(a2_b2_c2)2.
57. (ic-4a4) (a;-a252).
58. (£C— a— 5)(ic+2a).
59. (2x''-dxy-\-Sy^) (2x2+dxy+Sy^).
60. (a— c)(a+c)(5+c).
61. (x-S) {x^+2x+2).
62. (a+3)(a2-2a-l).
63. {x-l){x-Z)(x+b).
64. (a;-l)(a;-3)(ar-7).
65. (a;-l)(a;-2)(aj-3).
66.(a;+2)(ic+3)(a;-3).
Ex. 43-45]
ANSWERS
21
67. (y-5x-3z)(y-5x+dz).
68. (3a;»+52/")(2a7''— 32/").
69. (3a-36-3)(a-6-4).
70. (7a + 76+3c)(a+6-2c).
71. (x+3)(x-i)(x^-5).
72. (a-1) (a+1) (a-h).
73. (a?-a)(a;-3).
74. aa7(a?— 2a) (a?— a).
75. {x-2)i2x^+x-7).
76. (5a;-7)(7ar-5).
77. x»(x-7){x+S).
78. (a7-3b-2)(ar-3[>4-2).
79. (.x-3_5)(a;24.7).^^
80. (x-l){x-i-2)(x+S).
81. (a;-3)(a;2+7).
82. (x+l)(x^+24x-16).
83. (2a;+22/-2;)(5a7+52/4-6«).
84. (8a+8?)+9c)(3a+36-4c).
1.
2a262.
2.
3a363c2.
3.
15a^2/3.
4
8xyz^.
5.
a%.
6.
tzK
7.
icV-
8.
7a53.^2.
9.
Ixyz^
10.
dabd^.
11.
xy^z.
12.
2a^W.
13.
d6xyz.
14.
35a3&.
15.
6ma2.
16.
5a2a?.
17.
7a2Z>2.
18.
7CC2/2.
1.
6a268.
2.
30a36*.
3.
8a762c8.
85. (a;+2-2a+y)(a;+2+2a
-y\
Exercise 44.
19. ar32/5.
37. 2a;-3.
20. 2(a+6)2.
38. ax{x—a).
21. 7{x-yy.
39. a2_^5.
22. 4(a+£c)2.
40. a(a+2a;).
23. x-1.
41. 1.
24. 5(a?+l).
42. 7m2+5m+5.
25. VS(x--i)^x+l).
43. a+h.
26. a;+2.
44. 1-a;.
27. a:-l.
45. a-&.
28. 2a;+l.
46.* m-2.
29. 4.
47. a?— 2y.
30. x-1.
48. a2+a5i/2+52.
31. a;-l.
49. (a+l)2.
32. a-&.
50. l+ir+a;2.
33. a-b.
51. a—x.
34. ic.
52. a;-5.
35. a;-3.
53. 2{x-\-y).
36. a;+l.
54. a-1.
Exercise 45.
4. lQ2x^y^.
7. 100a2m2n«.
5. 42ic8?/822.
8. 60^6^8.
6. Sia'b^c^.
9. 144aW66.
^2 ANSWERS [Ex. 45-46
10. 108x^y\ 12. 3(a-6)8(a+?>)2. 14. oc^(l+x)^l-x).
11. 2(x+y)^ 13. 2£c2(a7-l)(ii7+l). 15. 30 (a+6)2(c-d)8.
16. 235a8a;(a-b)(2a+5)3. 23. (2x-l)(x+2)iSx+l).
17. (a;+l)(£t?-l)2. 24. (2x^-x-\0)(2x^+x-d).
18. (a?+l)(a?-l)(a;-2). 25. 2a7(a7+l) (072+50:4-6).
19. x{l-x){l+x)(l+2x). 26. 2a;2(a?+l)(a;-l)(£c2+a:+l).
20. ia-b)(a-\-h){2a+h). 27. (ic+l)(a;-l)(aj+2) (i»2+i).
21. (aH-2)(a+3)(a+4). 28. (l-aT^Xl+aj+icO-
22. (a;+6)(a;-5)(a7-l). 29. 2x^{l+x)(l-x){x-2){x^+l).
30. 12a;2(a7+7)(a;_2)2.
31. (a-&)(a+6)(a2+&2)(c[24.of5+52).
32. l-ic8. 37. 07(1-0?) (l+x)(l+a;2)(10-a;).
33. 076-1. 38. (07-l)(a;+l)(jr+2)(a;+3).
34. 6o;2(£c+3)(o;-l)(o;2+l). 39. 3o74(o7-l)(ic2+ir+l).
35. (a4-64)2. 40. (a+b)^a-hf.
36. (3a-2)2(3a+2)(9a2+6a+4). 41. (a;-3)(.T-12)(o;2-2).
42. (a+l)(a4-2)(a-l)2(a2-2a+3).
43. (6a3+a2-5a-^)(3a2+a-2).
44. (6o78-7o72z/-2a;y2)(3^2_,_^2/-4^2), ,
45. (07+1) (07-1) (x+2) (x-2){x-r^) {x-S).
46. (07+2) (207-1) (307+1).
47. (2a+l)2(2a;-l)2. 49. (dx-^)(2x+7)(ix-n).
48. (a-b)(x-y)K 50. (4a;+l)(2a;+7)(3a?-8).
Exercise 46.
1 « '7x^ ab-2
' b' 6- 2p-
2 2a2 5o^
^' 362- 7- "7^-
3 528 __3266c
4 — . 9 -^^
6. f . 10. ^|.
» 07+1
Ex. 46-48]
ANSWERS
23
16.
17.
18.
19.
20.
21.
x+2
x^ '
x-y
xy '
x+1
x^+x+1'
x—d
X + 4:'
X'
■x+1
X^ — 0(^+X^—X + l'
3.r+
x^'
1
2
3.
10. 3a +86+
2
X—1 +
ix
6x2'
x-V
1062
3a -26*
_3_
6£C2'
22.
23.
24.
27.
x+2
3x+i'
1+x+x^
x+x^+a^+oc^'
x'^+2xy+y ^
x'^+xy-\-2y^'
1+x^
1-x^'
l-g
2+a*
x+1
x+2'
Exercise 47.
28
4 — 3?
2 + 07'
x^+Zxy+y^
^^- x^-2xy-iy^'
b+c—a
30
31.
b—c+a
m-2
m+9'
c—d
c+d'
x+2 .
x-d'
4. £c2+2a?+l.
7. 2a;+l +
2a:2_i
5. x+y+
x-y
6. x^-dx—2+
8X + 4:
2v^
^ x+y
a'2-2.r+4+
x+2
11. x^+2x:^+'lx+8+
32
a;-2*
12. x^—xy+y'^
hx
6a;2-
2y3
i»+2/'
Exercise 48.
8a*
9a3
20a8a;
24a3ic3' 24a3.T3'
.r2— ,r ic^+a?
1'
1'
6a?3
24a«a^*
1
a;2-l-
3.
4.
18&4
3c3
12a262c'
xyz'
6£C+3
5a? +5
12a262c' I2a262c*
£C2 ?/2
(ri/2;' a?^/^;'
707+14
(a;+l)(a;+2)(2a;+l)' (a7+l)(a;+2)(2ic+l)' (a?+l)(a;+2)(2a:+l)
(2a;+l)(a;-5) (l-a?)(a;-1) x (x+l)
' {x+l){x-l){x-5y {x+l){x-l){x-5y {x+l){x-i){x-5)'
(a+b)(a'^+b-2)
a^-b^
2(a;+l)a
(a;+l)*'
2(a-b)(ag+b2)
2a;2
{a—b){a+b)
x{x+l)
{x+1)*^ {x+iy'
24
10.
11.
12.
13.
14.
15.
16.
1.
2.
3.
4.
5.
17.
18.
19.
26.
28.
ANSWERS
[Ex. 48-49
xir
a;(a;2_j,y2) .r^faT+y) ^^___
y{x^-y^y y{x^-y^y y{x'-y'^)'
6a?(a?+?/) -3?/(a?+y) x^
x(y-b)(z-c) y(x-a)(z-c) z(x-a)(y-b)
-2?/2
{x-a}{y-b){z-cy {x-a){y-h){z-cy {x-a){y-b){z-cy
x^-9
x^-1
(x-l)(x-2)(x-dy (x-l){x-2)(x-'dy
2a^+ia 4ac+12c
■(a+2)(a-2)(a+3)' (tt+2)(a-2)(a+3)*
b—c c—a
{a—b){a—c){b—c)' {a—b){a—c){b—c)' (a—b)(a—c)(b—c)'
n-b
y'^-z,^
z'^-x'^
x^-y"^
{x-y){x-z){y-zy {x-y){x-z)\y-zy {x-y){x-z){y-zy
Exercise 49.
- 4.^2+37 + 1
7.-r-2
- 2a;2 •
abc
Ux-j-9
lOa^ *
x+y+z
xyz
Ax
x^-1'
a+2h
10^2+20?
x'^-\ '
1
2+y'
7.
8.
9.
10.
a^—x
3a;«-2a;2+8a;-2
2oc^—5x^+2x '
4a2
a2-a;2*
x+1
X
2x+x^
1+x
11.
12.
13.
14.
15.
16.
20.
21.
(a;+4)(2a;-l)(3a;+l)'
a;2_i-
2&2
22.
4bcd + 6acd —Sabd— 2abc
ASabcd
y^ + xy^ + Sxh/ — x*
x^^ • ^^•
2xy—2y
x^—xy^'
27. 0.
oc^—x'^+dx+l
x^~l
24.
25.
30.
1-x
—x^+x^—2x
1-£C2 •
5— a;— 5a?2
l-X'2 *
4ab
cC^-b^'
-4
{x+\){x~'d){x+by
2a?2— 6a?— 14
a;2-2a?-8 *
a^
x-r
(a—b){a—c)
4ax+4bx—4:ah
452- a;2 •
Ex.
31.
35.
37.
38.
49-51]
x^-l
32
ANSWERS
26
24-8a;2
34.
1.
5.
9.
10.
11.
12.
13.
14.
15.
16.
1.
2.
3.
4.
14a;2+130
£C4-26£C2+25"
4.T3+2a;2+4a;-5
ad
x+y
x+2y
x^+xy+y^
x^+ocy+y^
x^—xy+y^'
a+1
a+o
x+y
x^
h^ '
^x'^-\-^xy
b-2+3ab+9a2
68
3a^
bxy
1
3a -36*
x^—xy+y^
x^—5x'^y^+4y^'
2bc^-2bd^+2a^-2bm
a^c^-b^c'^-a'^d'i+b^d^ '
^^ 2iic*-2a*+2a^xi-2a^
39.
40.
y ^
a^xz'
2y+S
x-1'
17.
2x^-x^
x^-1 '
x(x+y)
x-y
Exercise 50.
3. 1.
7. 07-1,
x^—x^a^—x^+a'^
a^+a%+ab^+b^
1
x-1
18. a+b.
19. x+y.
20. x^—x^y+xy^.
a
21.
22.
23.
24.
5.
6.
7.
a—b
2(x-y)
x+y
x+y-S
X
x-y
x+y
Exercise 51.
x'^+4xy+4y^
2x'^+xy—'Sy^'
ab^-a^
x+\
x+^'
8. 1.
41.
42.
a2-62
25.
26.
27.
1
2+x
4. a«6.
8. a^-x'^+x-\.
2ofi—x'^+Ax-2
a;2-4
225a2c2-16a?*
25c2
x^
x+2'
oc^—Ax
a?^- 9
29. 2x.
30. 6.
31. 6a;+9.
32. a;2+a;-4.
33. 4a2.
x—2
10.
a;'^+9a;+14'
(a-6)2(CT2+6g )
(rt-6)2+l •
^^' 6xy^ '
12.
x+x^
1-x'
26
ANSWERS
[Ex. 52-55
Exercise 52.
U
X
x-\
x+V
1+237
l+a7+a;2*
13.
a
6
a^+4a72+7a7+6
.T3+5x2+9a7+5"
7.
8.
a2+62
a2-62-
a2-a+l
10.
CT-4
a— 5*
?l2
11. 3a72-ir2/-32/2.
2a- 1
14.
307+3*
Exercise
53.
1. 1.
6.
8.
11. ^.
16. I,
21. 3.
2. -2.
7.
-6.
12. 15^.
17. 16.
22. If.
3. A.
8.
0.
13. 8.
18. -3
23. If.
4. 2.
9.
1-
14. -|.
19. V.
24. 12.
5. 2.
10.
i.
15. -37.
Exercise
54.
20. f.
25. 2.
1. 6.
7.
I
13. 4.
-
19. 4.
25. 8.
2. 6.
8.
h
14. -H.
20. 1.
26. 4.
3. -f.
9.
4.
15. 19.
21. |.
27. -2.
4. 5.
10.
6f.
16. f.
22. \.
28. -10.
5. -f.
11.
1.
17. 13.
23. ^\.
29. -S^.
6. 1.
12.
15. .
31. 3
18. -7.
Exercise
32
55.
24. 13.
30. ^.
1 ^+cy
' 2a '
6.
7.
1.
2a- 1
- 11.
a+b—c '
2. y-2.
3. 1.
8.
4-a*
—a.
12.
a(x+S)+h(x-S)
x-1
4 c(a;-l)-
07-
-a(a;+3)
9.
2ax-h
c
13.
4t+l
t+2'
5. 3a;.
,^
10.
0.
14.
-t.
Ex. 55-56] ANSWERS 2Y
15. 1 19. ?f. 23. ^.
r ' v^'
c
16.^. 20. f?. 24. £^'.
17. ^. 21. ?^. 25. |.
18. i!:. 22. E±D:, 26 .:^
27. f(F_32). 28.
rr
r+r'
Exercise 56.
1. $10. 6. 16 and 20. 11. 96.
2. 12 and 38. 7. -24. 12. 26.
3. -7 and 8. 8. 3 and 25. 13. 17 and 18.
4.-3 and 12. 9. 25 and 36. 14. j%.
5. 24. 10. 27. 15. 9.
jg 16 yrs.=: John's age. 21. 293^ mi. from station.
' 10 yrs.= James' age.
.« .« , .. 22. g^Vmin.
17. 10 yrs. and 40 yrs. ^
18. $125. ^^' ^^'
19. $72000. 24. 10 rods by 16 rods.
20. 45 dimes ; 6 quarters. 25. 6.
26. 1 J mi. per hr. 10 mi. 3^ mi. per hr.
27. -36. 31. $350.
28. A, $120; B, $160; C, $80. 32. 8%.
29. 11 and 12. 33. Silk $1.10; linen $.55.
30. -2. 34. 18 ft. by 20 ft.
35. In 8| hrs. ; 21 J mi. from starting point of first pedestrian.
36. 21 y\ min. after 4 o'clock. 41. 60.
37. 5y\ min. before 5 o'clock. 42. $3750 and $2500.
38. 15 min. 43. 12 lbs. iron ; 60 lbs. lead.
39. 8f da. 44. 25 oz.
40. 6 da. 45. 246. "
28. ANSWERS [Ex. 57-60
Exercise 57.
1. x=2, y=S. 9. a=5, x-Q. 17. x=7, y=10.
2. x=-l,y=^. 10.y=5,b=7. 16. x=^2, y=16.
Z. x=5,y=L 11. a=rf|,5=-if. 19. x=l, y=i.
4. x=-2, y=-S. 12. 07=36, y=d6. 20. x=-d, y=ll.
5. x=-h y=h 13. a=^, b=-'^. 21. x=iO, y=lb,
6. a- -2, b=-d. 14. ir=fff, 2/=fA. 22. x-1, y=A.
7. a7=4, a=ii. 15. jp=2, g=5. 23. x=a—b, y=a—b.
8. a=3, 6=7. 16. .t=12, ^=11. 24. aj=:a, y=b.
25. a;=a, 2/=|-
1. a;=13, y=17. 6. a=35, 5=20. 11. a=12, 6=-13.
2. x=4, y=-5. 7. a=7, 6=9. 12. 6=6, 2^=18.
Z. x=il,y=^j\\ 8. a;=15, a=8. 13. m=Yi^, ?i=ff.
4. a;=8, 2/=-2. 9. a=-l, 2/-3. 14. a;=7, ?/=-2.
5. x=2, y=l. 10. aj=14, y=zU. 15. a?=1.8, 2/=1.4.
16. 07=6, 2/=12,
17. a=3, 6=-7.
18 x=^^^-tl^ V-
2n+m' ^ 2n+m
Exercise 59.
1. a=17, 6=13. 7. a;=10, y=2A, 13. x=7, y=-2.
2. a=-3, £c=-7. 8'. aj=l, 2/=-f. 14. a? =5, 2/= 7.
3. a7=9, 2/=2. -^. a;=H, 2,=,?,. 15. a?=2, 2^=3.
4. m= — 11, n=7. 10. x=l,y=—\. 16. a;=6, 2/=4.
5. £c=8, 2/=l. 11. a=5, 2/=-3. 17. a;=9, 2/=-3.
6. a;=-V/-, 2/=/t- 12. a7=2, 2/=8. 18. a7=12, 2/= -3.
19. 0?=^, 2/=tV 20, aj=12, 2/=3.
Exercise 60.
1. a?=7, 2^=3. 4. a?=i, y=l. 7. a?=20, 2/=32.
2. a?=7, 2/=4. 5. a=4, 2/=3. 8. £c=-l, 2/=3.
3. a;=5, 2/=7. 6. a=-2, 6=-3. 9. a?=ll, y=-4.
Ex
. 60-64]
ANSWERS
(
10.
11.
x=5, y=-5.
x=8, y=l.
14.
15.
a?=20, y=12.
a;=12, y=10.
18. x- ^ ^,y- ^ ^.
a+b '^ a+b
12.
a=—%\h=—'l.
16.
x=67, y=10d.
19. x=a^+b^, y=ab.
13.
00 = 4:, 2/=f.
17.
a=10, x=8.
20. x=a^, y=b.
21. x=2p, y=iq.
22. x
Exercise 61.
=18a-246, y=SQb-2ia.
1.
x=u y=2.
4.
a=3, h=4.
7. x=2, y=7.
2.
x=-2, 2/-3.
5.
x=S, y-2.
8. a=l b=h
3.
x=h y=h
6.
a;=:5, 2/=2.
9.x=^^^,x=-^,
10. x=l,
2/=l.
11. 07=10, y=5.
29
Exercise 62.
1. x=l, 2/=3, z=^. 7. 07=4, 2/=5, 5r=6.
2. ic:=6, 2/=l, z=2. 8. a;=20, 2/=10, ;2;=30.
3. x=l, y=h z=h 9- a?=l, 2/-7, 2;= -4.
4. 0?=^, y=l z=i. 10. p=2, 3=3, r=4.
5. .T=-5, 2/=8, 2;=-9. 11. x=l, y=2, z=S, w=:4.
6. x^-l, 2/=8, ;s=l. 12. p--=2, q=-l, r=-3, s=5.
Exercise 64.
1. 21 and 8. 9. Father 35 ; son 10.
2. 15 and -6. 10. 36 and 27.
3. 3i ft. and 8i ft. 11. 22 and 16.
4. 96 and 24. 12. 26 and 8.
5. Orange 4 cents ; peach 1 cent. 13. |.
6. A's age 36 ; B's age 21. 14. j\,
7. 4 cows ; 24 hogs. 15. 18 and 40.
8. 5 dimes ; 20 nickels. 16. 48 and 8.
17. 10 persons ; $16.
18. A, $125 ; B, $250 ; C, $200 ; D, $450.
19. 54.
20. Carriage, $175 ; harness, $25.
21. 12 doz. at 12 cents per doz. ; 8 doz. at 4 for 5 cents.
22. $575, $1250, $1825.
30 ANSWERS [Ex. 64-65
23. 6 men ; $2.
24. $750 ; 6%.
25. 8.5 in. first year ; 10 in. second year.
26. A*s, 2| mi. per hr. ; B's, 3 mi. per hr.
27. 24 mi.
28. 45 mi. per hr. ; 55 mi. per hr.
29. A's, 6 yds. per sec. ; B's, 4 yds. per sec.
30. 8 hrs.
31. 2 mi. per hr.
32. Current 2| mi. per hr. ; crew 4 mi. per hr.
33. 90 ft. per sec, and 60 ft. per sec.
34. 360 revolutions per min., and 540 revolutions per min.
.35. A, 30 days ; B, 24 days.
36. A, 68f days ; B, 160 days ; C, 240 days.
37. Man $2.75 per day ; boy $1.25 per day.
38. A, $23,040 ; B, $7,680.
39. 35 hogs ; 60 days.
40. Alt., 9 in. ; base, 12 in.
41. Velocity in still air 350 yds. per sec. ; velocity of wind 6.5 yds. per
sec.
42. One in 16 min. ; other in 20 min.
43. $1.25 and $1.65.
44. 6yx from first ; d^^ from second.
45. 4 from first ; 3 from second.
46. 42 lbs. tin ; 14 lbs. zinc.
47. 8j% oz. first ; 4|^ oz. second ; 2}f oz. third.
48. 432.
49. 24, 60, 120.
50. Length 30 rods ; width 20 rods ; area 600 sq. rods.
51. 384.
Exercise 65.
1- 2l/2. 3. _2^;3; 5. 6l/£
2. 51^5. 4. 3i>?. 6. 6i/3.
Ex. 65-67]
ANSWERS
7. 5i/4.
8. 42i/2.
9. soil's.
10. 21^102.
18. 2ah-\/2a-h. 28
19. a2v^2.
20. iy^.
21. ii/6. 3°-
^V9c.,2.
{a—x)\/ a—x.
(^+2/)l/4a.
11. 91^5.
12. Sxyi/Sy.
13. ia^b^i/dab.
22. ii/is".
23. ^^i^ii:
24. i^/^.
25. .-^al^^-
31.
32.
33.
3 / — r
■1
14. 2axy\/2xy^.
2^>/^-4^C.
15. -5ir22/4y^5a.2.
16. 2x»y^y.
17. a'^y^i/Zay'^.
2/^
26. ^V'a^bcK 34.
27. 4^^'2'- 35.
-i
|l^l+2a;+2a^.
Exercise 66.
1. 5i/2.
3. 1/5.
6.
4V3.
2. 21/3:
4. 5l/6.
6.
-4/5.
7. Ii'/e:
14. fl/2-|/3
8. (3b2+2a6-a2)^^
15. (3-2a+4a7)i^^.
9. (3a-2a25+262)|/26r
16. VV3.
10. 51/ 3a2.
17. -I1/5:
11. {x+2y=2)\/y.
18. -/^v^i:
12. (2a+3&)i/a6c-a6|/c.
19. 2|/5H-6-S
ii;^4-:
13. {x'^-xy)y'x^-^yH/xy\
20. 3v 2+11/2:
Exercise 67.
1. 1^08, v^a*r
4. v^343, ^/144, 1^64.
2. y^^ k'^, I^^
5. i/a8>6i«, y'a^bis, j/ a^b*.
3. F 5^ |^2«; J^P5;
6. 1/5.
31
32
ANSWERS
[Ex. 67-
7. i/48.
- ^i-
19. 51^25920.
8. 1/ iol
20. i/a^-b^.
9. 1/192.
- i/i-
21. (a;-?/) i/a?+2/-
^
22. a+b.
10. y'567.
16. 21^3.
23. -1.
11. l/f.
17. 61^500.
24. 21/2;
12. i/|.
25. 4.
13. \/h'
18. 1^55296.
26. 5.
27. 6+2i/2+2i/3+2l/6!
28. ai/2a6.
31. c2i^3a*62.
32. ^]/a6.
aa'2 /
1
29. xy\/^'
OK ^ ^ / ,0 J
^^- (a+6)2l/"^-^--
30. a86ci/3c.
33- SaW^b^'
Exercise 68.
36. ^i^2a2^2.
1. fl/3.
3. Uj/U-l/2i).
5. 4|/3.
2. ii/l5.
4. 3v/2.
6. 5i/3-5i/2.
7. 3+1/6-1/15-1/10.
8. 3+i/T4-v^6-fi/31.
9. 1/15-4.
13. l+|ir— |i/a;'^+3£C.
14. l+il/2-il/6"-Ji/3:
15- A-Ai/iH-tVi/S-tWS.
16. |l/3+tl/'2-il/7-il/42.
17. i+ll/6-^l/l5.
10. b-i/b2-a2.
11. l+2a;2+2a7i/l+a;2,
12. ^+ii/;i^z^
13 (x+ l/xy+ -\/xz) (x+y—z—2\/xy y
x^+y^+z^—2xy—2xz—2yz
19 n—b+j/ac— }/hc )(a+ h—c—2\/ob)
1. 21/2.
3. a7i/4a72/^.
a2+6'2+c2-2a6-2ac-2bc
Exercise 69.
3. 125a?V8^.
4. 64a666.
5. a«.
6. a^b^y'a^.
Ex. 69-71] ANSWERS
7. 64a66?/5. 5 __ ^ 3/-
•^ 12. 1/4. 16. 2x2 1/2.
8. 243ari'2/24. . V '*. ^ V
9. 256(a2-62)2. 13. i>5^. 1^' l^"'-^.
10- V^S. 14. v^3H^2.
sa
18. |>2a;2.
19. 7.T3aV2a.
11. 1^2. 15. 2. „p-
•^ 20. \/a.
Exercise 70.
'• '^^ '' ^Y^ 13. (a+6)^—
2. 10i/-l. 8. 7i/2l/-l.
3. 25,/=T. 9. aV:^ ''• (^-3^2/)>/-l.
4 ii/ZT 10- 4a?3?/|/_i. 15. 7|/-1.
5. ii/-i. ^^- 7;^i/-i- 16. 121/-1.
6. fi/3T. 12. 9(a;-2/)2|/ZT. 17. 2y^,
18. (2i/2+|/10+i/7)|/-:^. 22.' 16+3i/=T.
19. 5a2|/ZT. 23. 2a+26.
20. (2x^y-h6x^y^)i/~. 24. 2|/::6.
21. Sx+y}/'^. 25. 2a;-5?/i/^.
Exercise 71.
1- -^^- 4. -24|/'^. 7. a363.
2. -12. 5. 420. 8. -2x^y2,
3. -70. 6. lOSOj/^. 9. 5.
10. 6+i/6+2i/^-3i/^.
11- 1- 12. 0. 13. -10i/l0-6i/5+5i/6+3i/3.
^^* ^~"' 19. -7|/^. 24. i]/IT.
20. -i/H^. 25. ^.
21. 2. 26. |,
15. y^-x^.
16. -5v^^.
17. -fi/'^. 22. |. 27. fic8
18. -5i/^. 23. 2. 28. -|v/^.
29. *-fl/^. 30. i+iv/^s+ii/iTs-ii/e.
34
ANSWERS
1
81. f-
-fv/-5.
32. f +1]/
-3.
33.
i-
tW-5'
34.
. -^■^/To-il/15-il/i4-il/21.
35.
x^-
-y+2xi/-y
x^+y.
^
Exercise 72.
1. ±2.
11.
±2i/-l.
20.
±5.
2. ±5.
3. ±13.
4. ±7.
12.
13.
21.
22.
±il/30.
±ii/39:
5. ±|l/30.
14.
±iV2.
23.
±ll/55.
6. ±il/30.
15.
±Ai/-^09.
24.
0.
7. ±25.
16.
±l/-3.
25.
±1.
8. ±|.
17.
±3.
26.
±V/-14.
9. ±>/7.
18.
±7.
27.
±f.
10. ±2.
19.
±1.
Exercise 73.
28.
±2.
1. 3, 4.
14.
-hh
27.
3,^.
2. -5,2.
15.
11,11.
28.
-2,5.
3. -1, -7.
16.
-h -h
29.
11, 12.
4. -6,9.
17.
11,-7.
30.
-10, f..
5. 2, i.
6. -3, i.
7. -2,1.
8. 1, -h
18.
19.
20.
21.
5,6.
-2,5.
3, -V-.
\S -2.
31.
32.
33.
3,-|.
4,9.
9. h -f.
10. 2, -i.
22.
23.
2, -V-.
34.
35.
— 1, 6.
-1, -9.
11. -2, i.
24.
4,^.
36.
28, -20.
12. -2, 1.
25.
i,i.
37.
h -v-
13. 4, 7.
26.
6, -V-.
38.
5,-f.
[Ex. 71-73
Ex. 74-75]
ANSWERS
Exercise 74.
1. 2, 4.
17.
1, -f.
33. 7, f.
2. -6,2.
18.
tV±tVv"141.
34. ±5.
3. -4, -10.
19.
-i±i|/-71.
35. 4, -W-.
4. -3,2.
20.
tV±tVi/13.
36. -i±il/5.
5. 6, 5.
21.
_4±|/-5.
37. -|±^|/185.
6. -7, 2.
22.
-3±i/-2.
38. i±ii/37;
7. 9, -1.
23.
i±T/3.
39. 2±ii/3.
8. 11, -2.
^^24.
-3±4i/-l.
40. 3, -|.
9. -3, 18.
25.
14, -8.
41. 4, 11.
10. -2, -10.
26.
3, 5.
42. 3, -V.
11. h -1.
27.
-5, -13.
43. -3, 5.
12. 3, -h
28.
±4.
44. 0, 4.
13. h f.
29.
4,-3.
45. |f±xVl/273.
14. -i, 3.
30.
±1.
46. ±4i/2.
15. f. |.
31.
5,7.
47. 2±ii/3:
16. -i 7.
32.
7, -V-.
Exercise 75.
48. W±^Vl/209.
1. -2,5.
16.
f±il/-ll.
31. 6, -1.
2. -3, -7.
17.
tV±tV1/337.
32. 4, -|.
3. -2, f.
18.
f±il/57.
33. 0, 1.
4. i, -2.
19.
if, -1-
34. 0, h
"^5. -l,f.
20.
-f±il/l3.
35. l±3l/^l. .
6. 7, |.
21.
-i±il/l7.
36. 7, -V-.
7. 1, -f.
^22.
l±3i/-l.
37- 1, -h
8. f , |.
23.
-i±i|/129.
38. 3, -V-.
9- I |.
24.
2±i/5.
39. 1, 1.
10. 8, h
25.
f±il/-7.
40. -i±il/5.
11. -|±il/-23.
26.
-l±2|/2.
41. 0, 1.
12. i±i^-34.
27.
3, -i.
42. 7, V-.
13. A±3W-1'^9.
28.
-i±il/-3.
43. 0, -5.
14. 4±2i/3
29.
3, -V-.
44. h -2.
15. 2, f.
30.
l|±iV|/l33.
45. 3, |.
35
36
ANSWERS
[Ex. 75-78
46. 3, |. 48.
3,
-5. 50. 4, -1.
52. 4, -|.
47. -¥, h 49.
4,
- 4. 51. 2, -h
Exercise 76.
53. 5, -|.
1. ic2_^_3o^o.
4. 2r)x^+30x+Q=0.
7.
x''+10x+24:=0.
2. x^-5x-U=0.
5. Gx'^+x-2=0.
8.
12a?2+25a?+12r=0.
3. 16ir2_s.^+i::^0.
6. 6ir2-7a?-10rr0.
9.
4a?2_7a;4-3z=0.
10. 8^2+6j;c4.1-0. 11. Positive.
12. One positive and one negative in each.
13. -A
±|/^.
±a.
14. ± 16. 15. (a) c not greater than
Exercise 77.
(^)c=i.
3. i&± 11/62+16.
4. a7=:3a or —^a ; a=^a7or —3a;.
5. x=a or 2a ; a=x or 4^x.
6. ?>=:£c or — iT— 2a ; x=b or —5— 2a.
7. — ir±i-i/r2— 4.9.
^ 1_
■2m'^2?^T
8- -9^±9^l/^'-4mf.
). X,
10.
11.
12.
13.
14.
15.
ID
mna
16. ±
17.
■v±\/v^+2c
i8.5=±V|?^;'=±i|/|';-4|/f.
Exercise 78.
1. 15 and 22.
3. 42 and 8.
2. 10 and 21.
4. 17 and 33.
Ex. 78-79] ANSWERS 37
5. 76 and 77 ; or -77 and -76. 8. /^.
6. 12 and 13 ; or -13 and -12. 9. 16 or -8.
7. 4 and 14 ; or —14 and —4. 10. Either 3 or 4.
11. 8 or - V-.
12. 2 in. and 5 in.; or 4 in. and 7 in.
13. 500 ft. by 596 ft. 18. 1 inch.
14. 12 in. and 16 in. 19. 2 ft.
15. 5 ft. 20. 6 in. by 12 in.
16. 25 yds. and 39 yds. 21. 32 rods by 60 rods.
17. 36 sq. in. 22. 5 hrs.
23. 30 mi. per hr. and 40 mi. per hr.
24. 8 and 12; or -12 and -8. 32. 100 ft.
25. 9 and 24. 33. 35 feet.
26. 6. 34. 10 in., 8 in., 6 in.
27. 2 mi. per hr. 35. 24 min. and 18 min.
36. 15f min.
37. 12 days.
38. 15 min.
39. A, 8 days ; B, 10 days.
40. 10 and 14, or -60 and 84.
41. 45 mi. per hr.; and 30 mi. per hr.
45. 64 sq. in. 48. 12 inches.
46. 4 inches. 49. 84.
47. 14 inches. 50. 4 ft. by 8 ft.
51. 20 A. 52. 529 sq. yards ; 4 yards.
Exercise 79.
1. 2, -2, 2i/^, -2i/~. 7. i/S, -1/^, 1/2, -1/2.
2. 1/5, -V/'S, 1/^, -l/^. 8- 1» -1. l/2, -V2.
. 9. 2, -2, 3, -3.
3. 2, -2, il/-6, -il/-6. ^0 ^^ _^; ^^^ _^
4. 1,-1,-2. jj -3±i/l0, -f±ii/29.
^' ^' ^' -^' 12 -^±y^ -i±i/5
6. 1, 3, -2. • 2 ' 2" •
28. 15doz.;
18 cents.
29. 12.
30. $500.
31. 42.
40.
4.1
42. 2ihrs.
*i,
43. 12.
44. 1.
38 ANSWERS [Ex. 79-80
13 ^±V~ 1±V~ 17. 1, ZllAl^.
14. 1 ± 1/2, 1 ± 1/2. ^«- 3' -^' ^^ ^' -^^' _
15. 1, -1, l/-3, -l/-3. 19- 1» -1> 2 ' T) .
16. 1, -2, zl±V^. 20. l/^±/-^ , -V2±i/-2
21. ±2, ±2i/~ l/2±l/^, -l/2±i/^.
22. ± i/2, ± |/^, 1 ± l/^, -1 ± l/^.
23.
24.
±l/-l, '^'%^
'-1 ~
-l/3±i/-
2
-1^
-|±iT^34+2i/-
-15; -
|±^l/34-
•21/
-n
1.
25.
l±l/-7 l±3i/-
2 ' 2
EI.
26.
2, 4, n!
r±v^l7
2 •
28.
-l±l/^
- -
-i±
2
VI
27.
1, 1, —
3±l/5
3 •
29.
1, 1, 1±
^
zl.
Exercise 80.
1.
3.
13. 144.
25.
None.
37.
9.
2.
3.
4.
5.
6.
7.
5.
1.
4.
7.
-5.
4, -1.
14. 16.
15. 9.
16. -|.
17. 7.
18. -12.
19. 4.
26.
27.
28.
29.
30.
31.
8.
12.
4.
-1.
I
38.
39.
40.
41.
42.
43
«'-^*
^' 2a2 •
13.
1-
16.
None.
0.
8.
-2.
20. 49.
32.
None.
44.
2.
9.
10.
11.
-8, 40.
-4.
3,2.
21. 0.
22. None.
23. 81.
33.
34.
35.
-h
7.
i.
45.
46.
47.
0,-3.
7.
1.
12.
5, A.
24. 4.
36.
i.
48.
1.
Ex. 80-83J ANSWERS 39
49. 6,2. 51. None. 53. i^+^f ,
.0. 2^^ 52. OiV^. 54. 3, -2, ±2^ZT.
55. 1, -1, -'Y^ , i±J^-
Exercise 81.
1. a;-l, y=:2 ; 0?=^^ 2/=- !• H- ^=^, 2/=2 ; a?=2, ^=3.
2. .r=3, y=—2 ; a7=— 3, 2/=2. 12. x—4:, y=l ; aj=2, 2/=:3.
3. x=l, y=i ; .T=4, y=^\. 13. 07=7, ?/=4 ; a;=— 4, y=—7.
4. ic=— 5, 2/=2 ; a7=-10, 2/=-8. 14. a?=l, i/=5 ; ir=-3, 2/=-3.
5. x=5, y=7 ; a7=-ff, 2/=-ff. 15. x=^, y=l ; ic=|, 2/=-|.
6. ir=2, 2/=l ; x=-l, y--=-2. 16. ir--=i, 2/=| ; x=i, y=l
7. £c=2, y=^ ; a^^-H, 2/=-2. 17. a;=ll, y=-8 ; a;=8, 7/=-ll.
8. x=^, y=2 ; £P=2, y=i. 18. a=3, 6=1 ; a=l, 6=3.
9. a;=-4, y=-3 ; ic=f, y=—\^-. 19. ^=4, 2i7=12 ; t=—^^, w=-\K
10. a?=6, ?/=l ; a:=l, ?/=6. 20. m =— 8,?i=— 3; m=-V, n=-V^.
21. A=5, B=z4:; A=z4, ©=5.
Exercise 82.
1. 07=5," t/=3 ; 07=— 5, y='S ; a;=5, ?/=— 3 ; x=—5, y——S.
2. 07=6, ?/=l ; a?=-6, y=l ; .t=6, y=-l ; a7=-6, ?/=-l.
3. x=5, y=2 ; a7=— 5, 7/=2 ; a;=5, y——2 ; a;=— 5, y=—2.
4. a7=ii/2, 2/=il/2"; x=-U/2, y=iV2; x=n'^, y=r-ii/2;
x=-^y2,y=-^l/2.
5. 07=2, 2/=3 ; a;=2, 2^=-3 ; £C=-2, ?/=3 ; 07=-2, y=-S.
6. 07=1, 2/=5 ; x=l, y=-o ; 07=-|, 2/=r) ; 07=— |, ?/=-5.
7. 07=6, y=i ; 0^=6, 2/= — | ; a-=— 6, 2/=| ; o?=— 6, o;=— f.
8. 07=i, y=i ; 07=i, 2/=-i ; x=-i, y=i ; 0?=-^ y=-h
9. £c=6, 2/=9 ; o;=6, y-—d ; 07=-6, jf=9 ; 07=-6, y=-9.
10. o;=4, y=2 ; a;=4, 2/=— 2 ; a;=— 4, y=2 ; 07=— 4, y=—2.
11. a?=4, 2/=3 ; a;=-4, ?/=3 ; a?=4, y=-3 ; a;=-4, 2/=-3.
Exercise 83.
1. x=2, y=l', x=^2, y=-l; a;=||/2, y=iy2; a;=-fi/2; 2/=-|l/2l
40 ANSWERS [Ex. 83-85
2. x=0, y=yl9 ; x=0, y=-yid ; x=S, y--^ ; a?=:-3, y=2.
3. x=i/% y--0 ; x=-y5, y=0 ; a7=5, 2/=-3 ; x=-5, y=S.
4. a;=3, t/=4; ic=-3, 2/=-4; x=lV% 2/=|t/3; a7rr-||/3, ^/zzr— fi/i".
5. ic=2, 2/=4 ; x--'Z, y--^ ; a;=i/2, y=^\/2; .t~-|/ 2, 2/=-3i/2.
6. a;=il, 2/=2;.rr=-l, 2/=-2 ; a7=|/3; 2/=0 ; a;=:-|/3, ?/=0.
7. a?— 1, 2/=5 ; a7=: — 1, 2/=— 5 ; £c=14, 2/=— 8 ; 0?= — 14, 2/=8.
8. x—2, y=5 ; a;=— 2, y=—5 ; ar=4|/3, ?/= — 1/3 ; ic=— 4i/3, y=V'd.
9. ic=2, 2/==5 ; a;=:— 2, y——5. (Defective system).
10. 07=3, 2/=5 ; a?=: — 3, y=—5 ; x=S, y=—5 ; x=—S, y=5.
11. (j?=:4, y=5 ; x=—4:, y=—^ ; x=Si/'S, y=l/'S ; 0?=— 3y 3, ?/=: — ]/3.
12. 07=5, y=l ; a7=-5, 2/=-l ; x=iy -10, 2/=-il/ -10 ;
Exercise 84.
1. x=~l, y=2 ; 07=2, 2/=-l. 2. a7=-2, ?/=5 ; £r=:5, y=-2.
3. 07=6, ?/=2 ; 07=— 2, ?/= — 6.
4. 07=5, 2^=-l ; 07=-5, y=l ; 07=-l, ?/=5 ; 07=1, ?/=-5.
5. 07=5, 2/=— 3 ; 07=3, 2/=— 5. 6. 07=6, y=z5 ; 07=5, y=G:
7 o^-l£l±i ^- 1^5-1 . ^_ t/5-1 l/5+l .
\-Vl -l-j/s ". l-T/5 ^_ l-|/5
^- 2 ' ^- 2 ' ^ 2 ' ^ 2—
8. 07=5, ^=4 ; 07=4, 2/=5 ; 07=— 5+i/— 14, y=— 5— l/— 14;
07=— 5— |/— 14, 2/=— 5+l/— 14.
9. 07=7, ^=-1 ; 07=-7, 2/=l ; a7=l, 2/=-7 ; 07=-l, 2/=7.
10. 07=6, y-^ ; 07=-9, 2/=-6.
11. a;=5, 2/=-7 ; o;=-7, 2/=5. 12. a;=4, 2/=5 ; o;=5, y^L
Exercise 85.
1. a;=3, 2/=2 ; o?=2, 2/=3. 2. o;=0, 2/=-3 ; aj=3, 2/=0.
3. a?=5, 2/=-2 ; o;=2, 2/=-5.
Ex. 85-86]
ANSWERS
41
4. x=l, y=2 ; x=2, y=l ; a;=:f +il/-55, 2/=^-il/-55 ;
5. irrz— 2, 2^=:8 ; x=8, y-—2.
6. 37=4, y=l ; a7=-l, ^=-4 ; a;=f+il/-79, 2/=-|+il/-79 ;
7. a;— 5, 2/=3 ; a;=5, 1/=— 3 ; 0?=:— 5, y=d ; a;=:— 5, y=~d.
8. 07=3, 2/=^3 ; x=3, y=d ; a?=— 3, 2/=— 3 ; x=—d, y=—S.
9. 37=4, y=2 ; a7rr-2, 2/=-4. 10. x=2, y--S ; a7=3, 2/=2.
11. x=S, y=l ; a;=l, y=S ; a7=2+5|/^, 2/=2-5i/^ a;=:2-5v/-l',
12. x=S,y=2;x--2,'y=-S.
Exercise 86.
1. x=25, y=9 ; x=9, ?/=25 ; ir=-81+8v ^97, 2/=-81-8]/^^97';
.T=-81-8l/^7, 2/:rr-81+8l/^I^.
2. x=l, y=4: ; a7=4, ^=1 ; x=-S-i ^, y=-S+y^;
3. x=4:, y='d ; 07=3, y=4: ; a?=— 3, 2/=— 4 ; a7=— 4, y=—d.
4. .T=i, 2/==i; 07=-^,?/=-^.
5. x=2, 2/=7 ; ^7=7, y=2 ; a7=f , 2/=! . ^^,-7^ 2/=|.
6. 37=2, 2/=l ; 37=1, 2^=2 ; 0;= — 1, y=—2 ; a7=:— 2, y=—l ;
V-i-yn.
^- -V-^-yn -yz-i+yii
^- 2 ' 2^ 2 •
7. 07=8, 2/=2 ; a;=2, y=8.
8. a;=3, y=l ; 07=-3, y= — l ; 07=1, y=S ; a7=-l, 2/=-3.
9. x=-i, 2/^796. 10. a;=27, 2/=8 ; x=-8, y=-27,
11. ^=6, y=n ; a.=3, 2/^6 ; a.=Z:lW57, ^..ziW:^.
^_ -19-3v57 .._ -19+3v^57
42 ANSWERS [Ex. 86-89
12. x=l, y=—5 ; a?=5, y=-l ; x=l, y=4: ; x=-i, y=—l,
13. Impossible system. 14. x=W, y=Q.
15. x—4:, y=6 ; x=5, y=4: ;
^_ -9-l/T6i ^^ -9+1/161 . .^._ -9+ 1/161 .,_ -9-i/167 .
2'^ 2' 2'^ 2
16. a;=0, ^=0; x^i, y-\ ; a?-!, 2/=-2 ; x=—\, y= — h
Exercise 88.
1. 4 and 9. 2. 5 and 11 ; 5 and — 11 ; —5 and 11 ; or —5 and —11.
3. 3 and 5. 4. 6 and 2 ; or -2 and -6.
5. 10 and 6 ; 10 and -6 ; -8 and 0. 6. 2 and 8. 7. 49 and 9.
8. 5 and 6 ; -2 and -1 ; or 2|/^ and -2]/^.
9. 1 and 4 ; 1 and —4 ; —1 and 4 ; or —1 and —4.
10. 62. 11. 36. 12. 10 and 4.
13. 4 in. and 9 in.; or 6 in. and 6 in.
14. 2 in. to width and 1 in. from length ; or 3 in. to width and 2 in.
from length. 15. 6 and 8. 16. $96.
17. Derrick 40 ft.; guy-rope 50 ft. 18. 40 feet and 25 ft.
19. f . 20. $3.25 ; 10 days. 21. 6 men ; 8 days.
22. $32 for apples ; $22 for potatoes.
23. $250 ; 6%. 24. 8 days, and 12 days.
Exercise 89.
7. a?>5. 9. a!<6. 11. a?>3or <~1.
8. x^\K 10. ic>ll. 12. a;>5or <-3.
13. X between Y- and 7.
14. X between a and -, when a is not 1.
a
15. X between 2 and 6. 18. Impossible system.
16. X between —14 and —3. 19. x between 4 and 12.
17. x5 and ^<5 that satisfy equation.
22. Values of a7>3 and y<^5 that satisfy equation.
23. Values of a;— H that satisfy equation.
Ex. 89-92] ANSWERS 43
24. Values of a;^ — |f and y^rii t^** satisfy equation.
25. Values of ic^-— ^ and y^:^r\y that satisfy equation.
1.
-V-.
2.
3.
f.
1
3i-
4.
h
a*
5.
._-_,
6.
T^.
7.
11, li f.
8.
/..
9.
|.
10.
3
2(a-5)'
11.
4.
52. 320 yds ; 4*
54.
|. 55.
1.
x=^y.
2.
42.
9.
V=:32t.
12.
16 cu. in.
13.
256 ft., 113 ft.
Exercise 90.
12.
4.
24.
pq\
13.
15 and 20.
25.
tV.
14.
15 and 21.
26.
^
15.
4 and 18.
a '
16.
48.
27.
X '
17,
150^.
28.
4.
18.
i'2.
29.
45.
19.
TO.
30.
3iu*.
20.
he
a'
31.
3v^6.
21.
y_z
32.
Vab.
X
33.
Vxy.
22.
4.
34.
2ad
23.
9a;6.
"3c"'
is; 480 yds. 53. 3 or 4.
55. 8 and 12 ; or -12 and -8. 56. 64 ft.
Exercise 91.
3. xy=24:. 5. x=8yz. 7. 7.
4. 25. 6. 2. 8. x^=4y^.
10. 30sq.in. 11. 201.0624 sq. ft.
14. 2.44 approximately.
15. 7^^^ ft. from end of heavier.
Exercise 92. '
1. F—km, where -F= force, ?7i=mass, and ifcrra constant.
2. a— ^2 , wherea=attraction, d=distance between tliem, m=rmass
of one, m'=:mass of other, and k=?L constant.
3. t= — j — , where f=tension, m^mass^ v=velocity, ?=length, and
fc— a constant.
44 ANSWERS [Ex. 92-94
4. t=1c\/ 1, where t=time, l=length, k=£L constant.
5. p—khh, where p=pressure, 7i=depth, &=:area of bottom, ^-—a
constant.
6. H8^=k, where H=hent, s— distance, A;=a constant.
7. R=k-, where 72= resistance, Z=: length, a = cross-sectional area.
a
and k=R constant.
Exercise 93.
"• ^- 12 ^ 20. A-
_ "■ ac2- «+b 28. 8i/a.
A 3— 13. -o^- " — ^ 1
5. ^gi, 14. M. '^- SMi^)- ^l/^
V. ^x2. ~ 24. -^•
16. |«V. a;")/^ «.-
52 31 y if .
17 '^^?/'^ 25. 1- • ,V-
9. y 27. 1
10. |/(|7. IS- 56^- ^^; "^4 32. ^>(^:^2
33 V^^+b . 34. 27. 36. i.
l/a^' 35. 4. 37. ^k-
38. |. 39. V. 40. ^i^. 41. ^j%^, 42. /^. 43. |. 44. 24.
Exercise 94.
1. 2a;2. 6. a^,
2. 1.
10. i/m\
3. a. '• ^ -• 12. ab.
4. 4^.
61/c
& ^ 15b
5. -5-^. 9. -^ 14. -rz::.
l/a8 V a46** l/aii
Ex. 94-95]
ANSWERS
45
15. ?5.
a
le. a46i2.
17. ^^^j^^
22.
23.
24.
25.
a2aj.
xyVx^^yK
30.
31.
32.
3
X+y
V^-y'
— s.
18. la'y'x.
X
26.
27.
28.
1
33.
34.
35.
a7
a;3+a;^2/^+2/«. '
a2-l. ,
^ifiy^
20. a.
xyi/xy^'
21. a.
29.
-1.
36.
x-y. j
37. 8a^2_i8ar-i-
38. a2-62.
39. x+y.
47-
-1507.
44.
45.
xi-
+ 1965+36-115-5-6&-10. i
40. ^+27.
46.
x-^
^+2a;-*
+3+2a!^+a;*.
41. 2a-20a^+18a~*.
47.
a4-
-1+a-*
42. a;^— £c32/^+2/^
48.
aJ+2aW
^6-1.
43. a;"^+a;~^+a7'
-^+1.
49.
a^i
—2a? V+y-
50. X-
i+2a;-^+3aj~^+4a;"^+5+4a^+ ^x'
^+2a;*+a?.
51. a^-a?~i
52. 9a;-' -162/5.
56.
57.
a-3+a-2+a-i+l. 1
53. -\/a+2i/h.
58.
x^+x^y^+x^y^+y^. \
54. ar-2-3ar-i+9.
59.
h
'
55. a-»+3.
60.
18.
V
Exercise 95.
V
■^v,,^^^^
1. 24.
6.
336.
11.
30.
16. 240. ^ 1
2. 720.
7.
60.
12.
180.
17. 12.
3. 720.
8.
TliuU
13.
24.
18. 56.
4. 48.
9.
A.
14.
1814400.
19. 56.
5. 7.
10.
8|.
15.
380.
20. 16380.
40 ANSWERS [Ex. 95-1
24. do.
25. 16.
ANSWERS
[E
26. 650.
28. 56.
30.
!l£:
27. 120.
29. 103^800.
32. 6 ; 10080.
Exercise 96.
31.
90.
6. 56.
11. 66i,2^.0 i
;22a.
16.
56.
7. 9.
12. 10.
17.
270.
8. 1140.
13. 10.
18.
28. '
9. 53130.
14. 325.
i
19.
560.
10. 2.
15. 4.
20.
53130.
21. 21.
• 22. 45.
Exercise' 97.
1. 4.
2. 210.
3. 82160.
4. 455.
5. 220.
1. oc^+5x^y+10x^^+10ai^+5xy^-]-y^,
2. 64ic«-576a76y-i-2160a-V~4320a^+4860a;V-2916a;^+T292/«.
"' H-8a;2+24.r*+32if«4-16a^.
•: l28ai*-h448ai2&-l-672ai062+560a823»»+280a6&*+84a4664.i4a266+57,
1 ~ 15a«+ 90a«-270a»+405ai2~ 343«i5.
x'^-6aTlOa24- 15a^a*_20ic«a«4- 15a?*a8 -6a'2«io-+- a^^.
1 -f 16.T+112a?2+448a^+1120a^+1792a--5+1792cc«+1024a77+256a^.
ar-*+4ar-2+6+4r2-fa?*.
32^10 ^_ 80ar-«-+-S0;r-«+ 40ar-4+ lOar-2-i- 1 .
a«- 6a65^+15a4b- 20a«6^+ 15a262- 6aZ)^+fe8.
,. 5a^h 5am 20a2?)8 40a6* . 326*
"^- 32^24*^ T"+ 27~~+~8r"^"2ir'
• r f-9a W+ 36a ^b^+8ia 64-12('. '^^ " ^'^ ■ ~ ' ' -
♦-36a"'^&^+9«''V^+68.
'. 18. -8064a;iV- 19- -36a?-ia '.
Ex. 97-98J ANSWERS ^^
20 ^*^^?^ oo ^•'>''>04
81 ' "* — — isT""
21. ^%%ayc-^. 24 _,^^.22
22- 210a;i 25. l+dx-5o^+Safi-afi.
27. 16-82a+24a2-8a8+a4+3262-48a62+24a262-4a862-f.'M64-24a6*
+6a264+86«-4a&6+58.
>?8. l+3a!+6a;2+4ic3-6a^-2ic6+3a*-.'r9.
29. «3-53+c8-#-3a26+3a52+3a2c-3c2d+3cd2+362c-3a2d!-3t2d+3ae2
+3ad2_36c2-35d2_6a6c+6a&d-6acdH-66cd.
30. 1 6a4- 32a864- 24a262-8a68+ 64+ 96a8c2- 1 44a26c2+ TZab^c^- 12b»c^
+216a2c*-216a6c*+5462c4+216ac«-1086c6+81c8.
Exercise 98.
1. l+2a7+3a;2+4a^+
2. i-ix^+ix^-j\afi+ -'
3. a-8+4a-i0a;2+l0a-i2a^+20«-Wa^+ ,
"*• 32a:^ ^64a^^ 128.x' '^25()a,-8'^
5. l-^x^+Gx*-10j(fi+ •- • .
6. l-lx^-lx*-^\afi-
7. a*+fa"^62_^^^-f54_^^.5^^-l56.
8. ^+-J_+ 3,5
l/2£P 4i/2£cs 32i/2ic6 1281/ 2a-7
9. v^-iv^a;+iv^2if2-j|v 2arV
1.0. a;~*-|a;~^+||a;~~'^*_ma;~'oV •
11. i/3+ — 7^- -+ -(-)' ' ' •
^ '^2]/3 241/3 1441/ 3^ ^
12. i%>---37:~^ — H7:- 5---
^ ~ 3i/4 18|/'4 3241/4
13. ^V?^'- . 15. 5e-5.
14. 5ligf||T«~'^'ci 16. — s^Vj-^i^".
48
ANSWERS
[Ex. 99-102
1.
2.
3.
4.
5.
21.
26.
1.
2.
10.
11.
12.
13.
18.
19.
1.96794-..
2.9925+.
146.
37.
20i.
329.
3.
7500.
10, 6, 2...
1679616.
-16384.
Exercise 09.
3. 6.0276+. 5. 4.0193+. 7. 4.9984+.
4. 2.0800+. 6. 1.9947+. 8. 5.0990+.
3.0006+ . 10. 2.7589+.
Exercise 100.
6. 3^.
7. 49 ; 81.
8. 14i ; 26i.
9. 16th.
10. 11th.
24. 9, 13, 17, 21.
27, $28.50.
11. 12.
12. 270.
13. 416.
14. 2500.
16. 2550.
16. 97i.
17. 5 or 6.
18. 64.
19. 2475.
20. 19800.
25.' -8h -7, -H, -4, -2i, -1.
28. at^
Exercise 101.
5- jAi-
3. 34U.
^' -^- 6. ±12i/2.
9. T25+5V+i-l-i+2+8+32+128.
29. 8.
7. -8^.
8. i.
-50, 20.
1/2, 2, 2i/2.
24.
Iflll^. 20. 4.
3. 21. 22i.
26. 60.
28. |+|+^«^ + 3-V5+
36-12+4-1+
A%' 34. IH.
I "35. I
14. Hi
15. -2730.
16. 971.2+,
17. 42.
24. 1.
25. 3i.
27. -j\.
or ^+l+^%+jh+ ' '
31. A
30. |.'
37. 56^«^ ft.
38. 4 years.
39. $1.21i.
40. 149||ft.
Exercise 102.
1. A. 2. ;,V 3. ^%. 4. ^V
6. -i'^-HKA + tV+3\ + iV+-. 6. 0+4+2+|+l + t+ •
Ex. 102-105] ANSWERS 49
7. i. 8. 3. 9. -2, -6, 6, 2.
10. hh ^% 1%, ^» if. 11- 80+41/899, 80-4V/399.
Exercise 103.
1. l+x+x^+x^+oc^+ ' ' ' . 2. l+2a:+2a;2+2ir8+2ii;4+ ,
3. 2x—Bx^+dx^-dx^+Sx^-
4. l+x'^+x^+oc^+x^+ 5. x+x^+oc^+oc^+x^+ . . . . ,
6. i+^x-^x^+j%x^-^%x^+
7. l-x+Sx^-5x^+7x*-
8. 2x+5x^+Ux»+S'7x*+97x^+ • • • .
9. x-2+2x-^+2+2x+2x'^+
10. ocr-^—Qc—^—x-'^—l—x—x^—
11. x-^-h4x-2+17x-^+7S+S09x+ ■ • • • .
12. x^-x^-2x^+7x^-Sx^-
13. x-^—l+x—dx^+5x^-
14. |aj-2-fa^i+i/-ifa;+|fic2_ • • . . .
15. 2xi^-2x^+5x*-7x^+12x^-
Exercise 104.
1. l-ix-ix^-j^s^-j^^x^-
2. l+|a7—V-i»2+W^--rlF^+ • • • •
3. l-2£C-2ic2-4a^-10ii^- •••..•.
4. l+^x-^x'^+j\x^-j%%x^+
5. l-j^x+^x2+j%x^+j^^x^-
6. 2+ix^—,\x^+^\-^a^-j^^^^x^+ .
7. 1—x—ix^-ix^-^x*- • • • • .
8. l+x~x^+x^—^x^+
9 2i/34.j/L8_l/2^ . 1^3 9__5i/2 ,2 , . . .
10. l+x+x-\ 11. l+2a;+3a;2. 12. 4x^+3x-5.
13. 4.-dx+x^-2x\ 14. l-2a+4a2-8a8.
Exercise 105.
1. x=y+y^+y^+y^+ . 2. x=y-2y^+5y^-Uy*+
3. x=y-iy2+ly»-^\y^+
50 ANSWERS [Ex. 105-107
4. a;r=2/-32/2+13^3-672/4+
5. x=y+2y^+4y^+8y'^+
6. 07=3?/+ V2/2+ W/2/^+-¥tV2/'+ .
7. x=y—y^+2y^—ny^+
8. x=y+ly^+j%y^+^\\7f+
Exercise 106.
1 1 _ 1 12 1 ?_
• 1-x T+x' (a?-l)3 x-l'
2 -J_+_A_ 13. l--i-
3 8 _ 1 5__ 14 3__2 8
3(a?-2) 6(07+1) 2{x-iy ' x x+l (x+l)'^'
4 7 8 _ 1 __ 1 + 3^
• 3a;-2^2ii;+3' 2^-3 9+6^'+4i^'2-
5 _J_ + _:L__1_. 16 ^ 1 I -^'-1
' a?-l a?-2 £c+3 ' ic-1 a'4-1 (ic^— ar+l)*
A. ^5 35 10_ 17 A-1+I-_J_+ 3
4(2a?+l)^12(2a;-l) 3(a;+l)' ar a;2^a?3 a;-l^(x-l;2'
^' x+l a?+2 a;+3' 2{x-^+x+l) 2(x^-x+iy
8 _J x-2 19 _J 2 2
3(07+1) 3(.T2-a;-+-l)' ' 07-l a?+l 072+0;+ 1*
2 12 1 20 2a?-7 _ 2x'^+2x-4:
^- 207-3"^ (207-3)2 {2x-d)^' ' 3(x2+2) 3(0^3+2) '
ID 1 I 2 , 1_ 21 16 16 4
o;-!"^ (07-1)2 a;-2* ' {x+2y^ (x+2)^ x+2'
11 _i __i i_ 22 3 - 4 2
• 4(07-1) 4(07+1) 2(^24.1)- • (i_a;)3 (i_a;)2^i_aj-
23 1.1, 1 1
07+1 0;— 1 a;2+a7+l 072— 0?+l'
24. l-i + — 1 _L_
078 07^(a;+2)3 o;+2'
Exercise 107.
1. log232=5. 4. Iogiol0000=4. 7. logg^V^-^
2. log381=4. 5. log5l5625=6. 8. logiojU = -'^'
3. log7343=3. 6. log4^\ = -3. 9. 32^9.
Ex. 107-109J
ANSWERS
10. 2*= 16.
16.
3.
22. -2.
28.
4.
11. 42=16.
17.
18.
19.
3.
2.
-4.
23. 2.
24. 3.
25. -3.
29.
64.
12. 8^=4.
^0.
3)
13. 7-2-^V
31.
!»..
14. 10-3=.001.
20.
3.
26. 0.
32.
h
15. 100~^=.001.
21.
-1.
34.
27. 0.
33.
1034.
51
Exercise 108.
1. logaOC+\ogay+\ogaZ+\ogaU^'- 3. log„2+ ]ogaX + 2 logaV + S loga
2. 2 \oga00+2 logaV. 4. 2 logaX+logay-^ logaZ.
5. i logaX+^ \0gay-l0gaZ-i logatV.
6. i \0gaX+t loga?/— i logaZ—^ logaW.
7. logaX-logay-logaZ-logaW.
8. l0gaa;+2 logay-^OgaZ-2 ]ogaW.
9. 2(l0gaa7 + l0ga2/-l0ga;^-l0gaU'). H- i \ogaX+^ ]ogay-i logaZ.
10. I ]0gaa?-f logay. 12. i I0ga^+l0ga2/ + i log„a?.
13. j logaX-i logay +i logaZ-i \ogaW.
14. l0ga2 + logaa;+2 loga?/-loga3— loga^:-^ logaW.
15. log„^^.
21.
>og»|
30.
31.
2.3222.
3.
^«- ^^^«^-
22.
1.2552.
32.
-.8239.
23.
1.1761.
n. iog«,^:.
24.
1.3010.
33.
.6276.
18. logj/--.
1/2/
25.
26.
1.6990.
1.3801.
34.
35.
.7385.
.7517.
19. )logal/ ff^ '
^loga "^
^^.
27.
.3680.
36.
.2762.
28.
.2219.
37.
-.3460.
Vy+z
29.
1.8451.
38.
-.0537.
Exercise !
109.
1. 2.3324.
4.
2.8555.
7.
2.9191.
10. .9294.
2. 2.8280.
5.
.6646.
8.
1.2041.
11. 2.
3. 2.9731.
* 6.
1.2989.
9.
1.4624.
12. 2.9542.
52
ANSWERS
[Ex. 109-113
13.
0.
17.
2.0021.
21. 2.8353.
25.
1.99996.
14.
.3010.
18.
2.9670.
22. 3.6302.
26.
3.3079.
15.
3.2263.
19.
1.9361.
23. .0398.
27.
3.9208.
16.
1.5349.
20.
29.
3^4983.
4.8363.
24. 4.0792.
30. 2.8739.
28.
r.7025.
Exercise 110.
1.
57.3.
6.
488000.
11. 10223.8.
16.
105195.12.
2.
7270.
7.
301.
12. 29.53.
17.
1094.5.
3.
1.53.
8.
4000.
13. .09474.
18.
.0002989.
4.
.289.
9.
.00531.
14. 701.5.
19.
7.4733.
5.
.0427.
10.
790.
15. .2554.
20.
37.445.
Exercise 111.
1.
8.1379-
-10.
5.
8.9087-
-10.
9. 5.7368-10.
13.
1.5017.
2.
7.1605-
-10.
6.
8.1658-
-10.
10. 5.3649-10.
14.
4.2055.
3.
9.3410-
-10.
7.
6.8617-
-10.
11. 2.4935.
15.
3.3260.
4.
7.4859-
-10.
8.
7.5829-
-10.
12. .1645.
16.
3.2489.
Exercise 112.
1.
23.41.
4.
5428.6.
7. .001228.
10.
.8224.
2.
.004165
5.
-.08936.
8. 103.58.
11.
.0564.
B. 150.9. 9. -526.05. 12. 4.947.
13. 1817500. 14. .000000000002, approx.
15. 14.81.
18. 4.295.
19. -4.1166.
16. .004541. 17. 1.561, approx.
20. .001204. 22. .04327. 24. 5.1811.
21. .387. 23. 3025.7.
25. .00000285.
26. 14.53.
27. 85.6, approx.
Exercise 113.
1.
4.
4. 2.4651. 7. 2.165. ^
10. 13.
2.
2.
5. 1.5481. 8. 3.3852.
11. 5.
3.
3.
6. 8.6336. 9. 2.998.
12. 1000.
13.
10000.
14. 2.209. 15. 1.631. 16. 1.
17. i.
ANSWERS TO REVIEW EXERCISES
Exercises for Review (I).
5. 16|. 21. 45 ft. and 15 ft. 27. -$500.
8. 8, 16, 32. 22. 105 and 21. 28. +50« and -15°.
11. (a) 13f ; (&) 230f. 25. +75 and -15. 30. -3.
12. (a) 46; (5)9. 26. -15. 60 lbs. 32. 32.
34. 64 ; -27 ; ^V ; tV- 36. -4 ; 4 ; -27 ; |.
Exercises for Review (II).
1. -y^-xhj. . 5. lla;2-3a;+2.
2. 2ab-3a;-2c+5a;2. 7. a^+2a'^y +2x^+0^-^1/^,
3. 9a;2-92/+2a. 8. 2«8-6a2+a-5.
4. (3?/2+2a4-3cy)a?. 11. 2a-4&-2c.
13. (7-a)a.^+(7-a5).T8+(5-3c2)a;.
14. - {h-^+c)o(fi- (a+26-2)a;2- (4a-b)a;.
15. a2c^+{a-^)x^+{h-2)x+a^+h-^2.
16. a?; a"; x^K 17. -3a8aV-
19. 2x'5— 4a^+6.T3-10a;2.
21. _3a*+5a46-5a362+4a268-2a6*+266.
22. 40a2_35a5. 23. a^ ; a; a^ . ^8,
25. Quotient-a4+2a364-3a262^-4ab3-2a66+55*— 467 .
Remainder=6a5S-6a68— 466+469.
29. |a;4^. 30. a^ ; a""* ; b'^'m^" ; S^.
31. £c5//5 ; £ci22/i8 ; 64a96a>.
32. -8a^2/i%i2 . 81a2464ci2ci8 . -^^.
33. a2+2a6-t-62; Ax^-\2x^j^+^if ; ^a*+2a26+962 ; 0^-4^2+4.
34. rt2_|_52_,_c2_|_2a5+2cic+26c; Ax'^-\-^y'^-\-z^—\2ocy+ixz—Qyz.
53
64 ANSWERS [Ex. for Rev. II-III
36. ±2a^; Sa% ; -2a^b^ ; {a—hY\ imaginary.
38. Ax'^-a or a-4x2 ; 9+5n3 or -9-5n3 ; 2x'^-^y\
39. a2-b2; l6a^_9?/4; ^-86076.
40. x'^+{a+h)x+ah; a^+2a;2-24; 4a2624.20a6+21 ; i^c'^x'^+^^cx-^Z.
41. ic2_f.a:^^2/2 ; £t7^+£c2^2_|_|^4 . a:^_a:22/2_|_2/4 ; a^—a^+a^—1.
42. When ?i is even. 43. When n is odd.
Exercises for Review (III).
3. (a) 2x{Zx^-2xhj+'f); (b) a^b(da-2h+l); (c) 2x'^(x^-3);
(d) ab{a-b){a^+b^)', (e) x-y«-Hx^+y); (/) a-b-ia^+b').
4. (a) {xy^-l)(xy^+l)', (6) ia;(a?-62/) (a!+6?/);
(c) (a-6+2c3)(a-5-2c3); (d) (a+26+c) (a+26-c);
(e) (ir+2)(ic-2) (0^2+4). (^) (l_£c2+22/2)(l+a;2-22/2).
5. (a) (a:4-2)(ir-2)(x2-2£c+4)(a;2+2a;+4); (6) (a2+9)(a4-9a2+81);
(c) (l+ar2^2+a^)(l-a;2|/2+a^); .
(d) (x-l-y)(x^-2x+l+xy—y+y^).
6. (a) (a;-6)(a;+5); (6) (Sx-2)(x+l); (c) (3+4a:)(l-2ir);
(d) (5a?2-2/)(a^2+22/); (e) (2a:+32/)2; (/) (5-a^)(3+a;);
(r/) (a4+a2+l)(a4-a2+l); (/i) (6-2a) (&+4a).
7. (a) (a+6)(a+l)(a-l); (6) (£c-a)(a;+l) (ct2-£c+1);
(c) (a?— 3+a— b)(a'— 3— a+6); (d) (x+l)(a+b)(a+c).
8. (rt) (a;-l)(a;+2)(£t?-3); (6) (.t+2) (2a'-l)(a;+4);
(c) (a-3)(a-3)(a-5); (d) (6+3) (d^.,, 4).
9. (a) 2aa;(2aa;+3a2^-2/); (6) 5a3a^(a+a;)2 ; (c) (2a:-y)2;
(d) (a;i/+16)2; (e) (2a;2-1.5mn)2;
(/) (2a7- 152/2) (207+151/2); (l-14ir2/=^)(l + 14cc2/2); (gr) (x^+y2)(x+y);
(h) {x+2)(x-y); (i) (a-b){a^+b^); (j) {dn-4m)(2-7m'^);
(k) {x-y+2)(x-y-2); (I) (a-b+5)(a-5-5);
(m) (d+3c+l)(d+3c-l); (n) (5.T+72/2f)2 ; (o) (iri/3+3)(a:2/3+9);
(1)) (a72/-llz)(aj?/-13^); (g) (x-b)(x-24)',
(r) (itV+i2)(a-22,3._io); (s) (ir2+i)(a;+l); {t) (a-b)(a^+b^);
(u) {a+b+2ab){a+b-2ab); (v) {x^-\-y^){a+b){,a-b);
{w) (2a;- 3) (07+4); {x) (a+b)(c-d).
Ex. FOR Rev. III-IV] ANSWERS 55
10. (a) (3a8+l)2;(6) (4a+56)(3a-26); (c) (a+5)(a-b)(a+25)(a-2?>);
(d) {2x^-4xy+dy^)(2x^+ixy+2y^); (e) (8a+96)(2a-6);
(/) (a-i-h+c-d)(a+h-c+d); {g) (x-+,l)^; (h) (a+h){a-h-l)',
(i) {x^+xy+y^)(x^-xy+y^); (j) (Sx-2y)(2x-Sy);
(k) {7x-Sy)ix+4ty)', (l) (4a-7)(4a-5); (m) (a2'»-?>'»)(a2'»-hi!>");
(n) (fl+6)(a2-6a+36;; (o) (x^+2xy+y^+l)(x+y+l)(x+y-l);
(p) (a2+a+l)(a2-a+l); (g) 3a;2^(a;2+(CT/+2/2)(a;2-a;2/+i/2);
(r) a(a-l-a5-5); (s) (l+rt)(l-b); (f) (a+5+c)(a+6-c)
(c+«— b)(c— a+6); (w) {2a+b){dc—2d)', (v) {ax+hy){bx+ay);
(M^) (a+cc)(a-a;)(b2+a:2); (a?) (3a-l)(2a+36).
11. (a) aj-2; (6) a?— 5 ; (c) 2ic-l ; (d) £t'+3.
12. (a) 6a;2(i»-3)2; (5) 10a26(a-5)(a+l); (c) a?2(a;-3)2(a;+2);
(d) {dx+5){2x-l)(x^+^).
iPi /^\ 4ic2-19a;+24 . /m 18-6^_ . /^x 22a?+42
art-6x'3+7a;2+6a;-8 ' ' ' a?2-2ic-48 ' ' ' a?2_,_i0ic+21*
26-2a
18. (a)x+7; {b)dx+Q; (c) 3a?+30.
19. rJr-::,. 20. € 21. a. - 1-
x2^yr «"• a;2 • ""' 2ic2_i'
no / XA /^,^^ , . a^+h ,,.a^—Qax+x^ . . x^z+x—y^z—z ^..
23. (a) ; (6) 1 ; (c) ^:p^, ; (d) ^,^_^^, ; (e) ;^3|^— ; (/) 1.
Exercises for Review (IV).
5. (a) ^', (6)4, (c)i.
^ , , 1+x 1 ... 2ci . a+2d , 2at—a
6. (a) a=-2^, a;^^^^^ ; (5) a=^^-^, i^-^ST^ ^=—2" '
... . i * J * J a d . d
(c) t=pr^, i)=-^, r=-^, ^=- ; d=vt, v=-^, t=-.
7-2 8 — 9 ^^
^- "• ^- bx ^' 2 '
,n f—M- r,-f^ n-J^ 11 7«-:ST „-V^ r-—
12. 1. 18. (a)x=S,y=-4; {b)x=hy=^.
19. (r=— 1, 2/:=a+6;a=^, 6=r^.
56 ANSWERS Ex. for Rev. IV-V]
20. x=z4, y='S. 21. x=a-\-h, y—a—b.
Exercises for Review (V).
6. (l-6a2+126)x?^3a.
7. (aic2?/+aaj22;+3a2a;+ 12?/+ 12;2;) 1/^+7.
8. 61/ 7. 13. ^/ 635000.
8/ 6,—— 6/--
10. 1/125, 1/121, l7l3. 15. 3j;i/243£rii.
11. i^8a^ 16. les+isiy'Io+ssi^m
12. v^fO. 17. 2-1/5"
19. f{+|fi/3. 20. Z2aWi/ab'', 50a^i^3aj; 2yy'3x^y.
21. i/3a;22/3; i7a;+3 ; dx^\/x-y.
22. (a) 5+1/6 ; (6) 15+4i/l5; (c) 111/? ; (fi) x'ul/loT
27
(e) i/a?s^io ; (/) Aa^ ; (gr) 37-2+^ ; (h)i/2.
2b'. 2ic3|/Zi. 27. (a) 8ai/^; (&) -24^^. ; (c) 3 ; (d) 2-i/5.
28. 3-1/^. 30. -8+21/15.
31. M-^Si/35+Ai/^+At/^.
32. -^1/3+1^:^-1+11/31/-^.
36. (a) ±il/3; (b) ±3; (c) ± l/3.
38. (a) i or -5 ; (b) | or -| ; (c) i or -|.
39. (a) 8 or -4 ; (5) i or -5 ; (c) a or |.
40. (a) i or -f ; (b) i±il/l3 ; (c) 5 or | ; (cl) f or -5.
41. Cor 3. 42. 13 or -J/. 43. l±v^^^^ . 47. ^.
6
49. 7a;2-19ar-6=0 : x'^-2x-ji=0.
52. (a) 1/3, -1/3, 1/2, -1/2 ; (5) -1, -1, ^±|i/IT;
(c) 1, 1, -3±2i/2.
Ex. FOR Rev. V-VII] ANSWERS 57
58. (a) i+ll/19; (6) |; (c) 0.
59. Cube roots cf 1: 1, -i±il/^;cube roots of 8: 2, -lij/^ ;
cube roots of 27: 3, -f ±fl/^.
63. x-1, y=^2 ; x=2, y-\ ; £»7=-3+i/"^, 2/^-3-]/^ ;
.T=-3-i/^, ^=-3+1/ -2^
64. A-TTi'^ 65. F=|7rr3; Fi^^ttDS.
Exercises for Review (VI).
6. No. 22. 3.
9. 4£n2 ; i2cc^y2 . 6(a+6)2 ; |/^. 37. «i5 ; «* . ^,15 ; a'^n . «^.
10. 64; 256«4; 8?/3 ; (a+6)5. ' ' ' ' ^^ ' ^^'*
11. t ; 2.T22/ ; GOZ>a; ; a^+2/. ^9. a^ ; 0-2 ; 64aV?^-
17. |. 30. 8; 4; 5; ^
18. .^=10^. „„ b2 . 2a253 ^ 2
19. 60^=2/. • a2 ' 31/ ' (a-6)2'
21. 1. 33. i; 5;3V;i; -32.
3 — 4 — ?/^ ^^
34. .aio ; i/a2 ; Va^ ; ^/-, ; g^^'
35. £c"3+2a;~^2/-i+?/-2 ; 4a-2-12a-i?>~^+95-i ;
x-'^+^x'^y^ ^^x'^y^ +y ; a;"^-2/~^ ; a/^+h^ ; a^+a^b^+b^.
38. a greater than 2.
39. a? greater than V, 2/ greater than ^-f.
40. 19. 41. Between 8 and 15.
Exercises for Review (VII).
3. n{n-\) ; 60; 5040; ?i(n-l)(n-2) (?i-r+l).
4. 16. 5. 120; 720.
6. 6 ; 24 ; 720 ; n{n-\) (?i-2) 3-2 1.
7. 19958400. 8. 50400.
9. 35;190;l; ^^V^) ; ^^^^-^>-V^^^-^^+^^ 1.
10. 4950. 11. 25. 12. 66. 13. 103740. 14. 581400.
19. l-2a^+|i»«-t?a?9+/Ta?i2_^\a7i5+,t^a7i8.
58 ANSWERS [Ex. for Rev. VII-VllI
20. l+^x+-g\x-2+-sUj^+ • 3*- '^''h
21. „Cr-ia»-'+i6'"-i. ' 36. 4+8+12+16+20.
22 — £c20 38. ?^.
23. --'|Ja'2o. 39. a-i;ri=— 4.
24. l+JJ.r-5.r3+3a^-£C«. 41. f||Mf-
25. 2.828 ; 2.924 ; 2.024. 42. 11|.
28. 39. 43. l+3i+6+8i+ll + J3i+16.
29. —V. ^' ^'
30. '^"^ ; 50. 45. 4i/3.
31. 11. . 46. 7 ±41/3.
33. 345. 47. 3+6+18+54+162.
Exercises for Review (VIII).
1. 3or-t. 2. ±1/5. 3.-1,-1,2.
4 a+V2b-a'^ a-y2h-^\ ^_a- V2h-a'^
7/=r«+l/26— a2
^ (i;^=6
5. 1, -2, 3, 3. 6. 65780. 7. (?i+l)th.
11. l+6a7+18x2+54.T3+162.T*+ • • • ■ .
12. l+3a?+-4ic2+7ic3+lla?4+
13. l-2a;+a?2+a?3-2a74+
14. l-|ic-V-£c2— V/^-W/^^- •
15. a;=2/+37/2+13^3+672/*+
on (f,\ 1 I " 1 . /i)\ q 1 I " ,
• ^^ 2(1-.^) "^2(1+0;) (1+07)2'.^'^' 2(a+l)'^2(rt-l) '
(o\ 3 2 1. ,,'22 1
^^ a;+l 2.'r+l"^2a;-l' ^ ^ a; a;+l'^(a?+l)2 '
(p\ ^-1 5 _ 4 . , -. 3 1_ 3 5a?-3
'"^ a? a:2 a;+i (a?+l)2' ^-^ ^ ir2 a; 2(a?-l)"^2(.T2+a;+i)*
22. 7 ; 3 ; f. 26. 10^=65.
23. 1 and 2 ; 2 and 3 ; 3 and 4. 27. 2 ; 3 ; -4 ; 2 ; a? ; ; 1.
30. (a) logaOJ+loga?/; (5) logaa;-loga^ ; (c) nlogaO?; (cZ) i- logaO?.
32. (a) 1 ; (5) ; (c) -1 ; (d) -3 ; (e) 1.
33. 1. 34. An integer plus a fraction.
ANSWERS TO EXERCISES IN APPENDIX
Exercise I.
1. ^x+1.
2. x^-5y.
6.
7.
1+x—x'^+x^.
1x^+8x^-2.
11. «+62+|.
3. x^-x+2.
8.
x^-xy+y^.
12. X+4:--^.
4. 2a2+5a-7
9.
x^-Sxy+2y^.
X
13. .^-"-f ^
2 2x
5. a2-2a-2.
10.
y^+^y+l-
14. l+4^x-ix^+
• • .
16. l4-ia2.
-ia4+ •••••.
15. 1-x—^x^-
- •
• •
17. m^+l,m-^-^^m-^+ • • • • .
Exercise II.
1. 86.
5.
3.5.
9. .162.
13. .774 + .
2, 43.
6.
12.1.
10. 51.1.
14. 10.246+.
3. 162.
7.
11.2.
11. 2.236+
15. 19.261 + .
4. 203.
,8.
2.15.
12. 1.732+
17. 3.435+.
Exercise III.
16. .866+.
1. x+2y.
6.
X-2-X+1.
10. ar2-2;r-l.
2. 0^2-1.^
3. a;2+32/2.
7.
8.
l-3a+2a2.
2+i!t;+£t'2.
11. a;2-2a;-3.
4. 3a -462.
5. l-2a.
9.
a-4+2.
a
Exercise IV.
''■ M
1. 47.
4.
53.
7. 806. 10.
3.45. 13. 4.73+
2. 14.
5.
83.
8. 3.9. 11
1.25+. 14. 1.54+,
3. 25.
6.
698
9. 6.9. 12.
59
2.15+. 15. .53+.
30
ANSWERS
[Ex.
IN App
Exercise V.
1.
a;-2.
7.
x+1.
13. x^y—xy.
19.
x^+x^-
2.
x-2.
8.
a-2.
14. 1ab-ab^.
20.
a-1.
3.
x+1.
9.
m^-2m-S.
15. £c3+2a;2.
21.
m-1.
4.
2a-l.
10.
da+4b.
16. None.
22.
x-2.
5.
2x-9.
11.
2x^+xy.
17. x+1.
23.
a+1.
6.
a2+4a-5.
12.
a2_a6-2b2.
18. None.
24.
a -10.
25. y+2.
Exercise VI.
6. 4n5-15n3-33w2+20n.
7. x^-2x^-Qx^+20x'^-ldx+e.
8. 2a?4+7a^-2a;2-13.r+6.
9. 12a4+44a8+21a2-47a-30.
10. x^+4x^-Sx^-29x^+2x+24.
11. 2x^+2Sx^+Q0x^+28x-'S2.
12. 18a8+84a7+162a6+186a5+132a4+54a8+12a2.
1. 4a8-20a2+31a-15.
2. 24a8+50a2_31a-70.
3. ;r6-17a?3+12£c2+52.T-48.
4. x^-'i7x^+lQx^+Q0x-80.
5. 48a4+64a3+37a2+10a+l
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