fiKiii
IN MEMORIAM
FLORIAN CAJORI
Digitized by the Internet Arciiive
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ELEMENTS OF ALGEBRA
BY
WOOSTEE WOODRUFF BEMAN
Professok of Mathematics in the University of Michigan
DAVID EUGENE SMITH
Principal op the State Normal School at Brockport, N.Y.
BOSTON, U.S.A.
GINN & COMPANY, PUBLISHERS
1900
Copyright, 1900, by
WoosTEB Woodruff Beman and David Eugene Smith
ALL RIGHTS RESERVED
CAJQIU:
PREFACE.
In the preparation of this work the authors have followed their
usual plan of attempting to allow the light of modern mathe
matics to shine in upon the old, and to do this by means of a
textbook which shall be usable in American high schools, acade
mies, and normal schools.
In general, the beaten paths have been followed, experience
having developed these and having shoVgti their safety and value.
But where there is an unquestionable gain in departing from
these paths the step has been taken. For example, the subject
of factoring has recently attracted the attention it deserves ; in
fact, several writers have carried it to an unjustifiable extreme ;
but there are few textbooks that mention the subject after the
chapter is closed ; it is taught with no applications, and the stu
dent is usually left with the idea that it has none. The authors
have departed from this plan, and have followed the chapter with
certain elementary applications, using the method in solving easy
quadratic and higher equations, ^making' much use of it in frac
tions, and not ceasing to review it and its applications until it
has come to be a familiar and indispensable tool. By following
such a scheme the student knows much of quadratics before he
reaches the chapter on the subject, and he enters upon it with
increased intelligence and confidence.
The arrangement of chapters has been the subject of consider
able experiment of late. But the plan adopted in this work is,
in general, based upon the following :
1. The new should grow out of the old, as the expressions of
algebra out of those of arithmetic, the negative number out of
familiar concepts, factors out of elementary functions, quadratic
and higher equations out of factoring, the theory of indices out
of the three fundamental laws for positive integral indices, the
complex number out of the surd, and so on.
iii
iv PREFACE.
2. The student's interest should be excited as early as possible,
and it should be maintained by reviews and by applications to
modern concrete problems. To this end the equation has been
introduced in the first chapter, with simple applications, and
general review exercises have been inserted at frequent intervals.
3. The new should be introduced where it is needed. To put
the remainder theorem where it is usually placed, at the end of
the work, is entirely unwarranted ; it is needed just before fac
toring. To put complex numbers after quadratics is equally unsci
entific, for they are met on the very threshold of this subject.
Considerable attention has been given to the illustration of
algebraic laws by graphic forms. The value of this plan is evi
dent ; the picture method, the coordination of the concrete and
the abstract, the onetoone correspondence between thought and
thing — this has been recognized too long to require argument.
This method of making algebraic abstractions seem real is fol
lowed in the presentation of certain fundamental laws (p. 37),
in the study of certain common products (p. 51), but more espe
cially in the treatment of the complex number (p. 236) — a subject
usually passed with no understanding, — and (in the Appendix)
in the study of equations.
Where the time and the maturity of the class allow, the Appen
dix may profitably be studied in connection with the several chap
ters to which it refers. This arrangement allows the teacher to
cover the usual course, or to make it somewhat more elaborate if
desired.
It need hardly be said that no class is expected to solve more
than half of the exercises, the large number being inserted to
allow of a change from year to year.
It is the hope of the authors that their efforts to prepare a text
book adapted to American schools of the twentieth century may
meet the approval of teachers and students. It is believed that
they have lessened the general average of difficulty of the old
style textbook, while greatly adding to the mathematical spirit.
June 1. 1900.
W. W. BEMAN, Ann Arbor, Mich.
D. E. SMITH, Brockport, N. Y.
TABLE OF CONTEE^TS.
CHAPTER I.
INTRODUCTION TO ALGEBRA.
PAGE
I. Algebraic Expressions 1
II. The Equation 8
III. The Negative Number 17
IV. The Symbols of Algebra 21
V. Propositions of Algebra 25
CHAPTER II.
ADDITION AND SUBTRACTION.
T. Addition 27
II. Subtraction 31
III. Symbols of Aggregation . 35
IV. Fundamental Laws 37
CHAPTER III.
MULTIPLICATION.
I. Definitions and Fundamental Laws 40
II. Multiplication of a Polynomial by a Monomial . . 45
III. Multiplication of a Polynomial by a Polynomial . . 46
VI TABLE OF CONTENTS.
PAOE
IV. Special Products Frequently Met 51
V. Involution . . 53
CHAPTER IV.
DIVISION.
I. Definitions and Laws ....... 60
II. Division of a Polynomial by a Monomial .... 61
III. Division of a Polynomial by a Polynomial ... 63
CHAPTER V.
ELEMENTARY ALGEBRAIC FUNCTIONS.
I. Definitions 69
II. The Remainder Theorem 74
CHAPTER VI.
FACTORS.
I. Types '78
II. Application of Factoring to the Solution of Equations . 91
III. Evolution .92
CHAPTER VII.
HIGHEST COMMON FACTOR AND LOWEST COMMON
MULTIPLE.
I. Highest Common Factor 103
II. Lowest Common Multiple 110
CHAPTER VIII.
FRACTIONS.
Definitions . . . 114
I. Reduction of Fractions 115
TABLE OF CONTENTS. vil
PAOE
II. Addition and Subtraction ..... 123
III. Multiplication 126
IV. Division 132
V. Complex Fractions 135
VI, Fractions of the form ?; » t: ^ 35 140
Y
CHAPTER IX.
SIMPLE EQUATIONS INVOLVING ONE UNKNOWN
QUANTITY.
I. General Laws Governing the Solution .... 145
II. Simple Integral Equations 153
III. Simple Fractional Equations 154
IV. Irrational Equations Solved like Simple Equations . . 160
V. Application of Simple Equations 163
M CHAPTER X.
SIMPLE EQUATIONS INVOLVING TWO OR MORE
UNKNOWN QUANTITIES.
Definitions
I. Elimination by Addition or Subtraction .
II. Elimination by Substitution and Comparison
III. General Directions
IV. Applications, Tw^o Unknown Quantities .
V. Systems of Equations with Three or More
Quantities
VI. Applications, Three Unknown Quantities
Unknown
179
180
183
186
189
192
197
CHAPTER XI.
INDETERMINATE EQUATIONS. . . 201
VIU TABLE OF CONTENTS.
CHAPTER XII.
THE THEORY OP INDICES.
PAGK
I. The Three Fundamental Laws of Exponents . . 204
II. The Meaning of the Negative Integral Exponent . . 205
III. The Meaning of the Fractional Exponent . . , 208
IV. The Three Fundamental Laws for Fractional and Negative
Exponents 212
V. Problems Involving Fractional and Negative Exponents . 216
VI. Irrational Numbers. Surds ...... 220
VII. The Binomial Theorem 233
CHAPTER XIII.
COMPLEX NUMBERS.
I. Definitions 236
II. Operations with Complex Numbers 242
CHAPTER XIV.
QUADRATIC EQUATIONS INVOLVING ONE UNKNOWN
QUANTITY.
I. Methods of Solving 246
II. Discussion of Roots 260
III. Equations Reducible to Quadratics 266
IV. Problems Involving Quadratics . . . . . 276
CHAPTER XV.
SIMULTANEOUS QUADRATIC EQUATIONS.
I. Two Equations with Two Unknown Quantities . . 284
II. Three or More Unknown Quantities .... 298
III. Problems Involving Quadratics ..... 300
TABLE OF CONTENTS. IX
CHAPTER XVL
PAGK
INEQUALITIES. MAXIMA AND MINIMA. . 304
CHAPTER XVII.
RATIO, VARIATION, PROPORTION.
I. Ratio 310
II. Variation 319
III. Proportion . ... . . ^ • • • 326
CHAPTER XV III.
SERIES.
Definitions 334
I. Arithmetic Series 335
II. Geometric Series ........ 340
III. Miscellaneous Types 346
CHAPTER XIX.
LOGARITHMS 348
CHAPTER XX.
PERMUTATIONS AND COMBINATIONS. . 364
CHAPTER XXI.
THE BINOMIAL THEOREM. . .373
APPENDIX.
I. Proof of the Binomial Theorem for Positive Integral
Exponents (p. 57) 377
II. Synthetic Division (p. 67) 378
X TABLE OF CONTENTS.
PAOE
III. The Applications of Homogeneity, Symmetry, and Cyclo
Symmetry (p. 73) 380
IV. Application of the Laws of Symmetry and Homogeneity
to Factoring (p. 88) 387
V. General Laws Governing the Solution of Equations
(p. 152) 390
VI. Equivalent Systems of Equations (p. 185) . . . 394
VII. Determinants (p. 198) 395
VIII. Graphic Representation of Linear Equations (p. 202) . 408
IX. Graphs of Quadratic Equations (p. 296) .... 415
TABLES.
Table of Biographies 423
Table of Etymologies 427
ELEMENTS OF ALGEBRA.
CHAPTER I.
INTRODUCTION TO ALGEBRA.
I. ALGEBRAIC EXPRESSIONS.
1. There is no dividing line between the arithmetic
with which the student is familiar and the algebra which
he is about to study. Each employs the symbols of the
other, each deals with numbers, each employs expressions
of equality, and each uses letters to represent numbers.
In arithmetic the student has learned the meaning of 2^ ;
in algebra he will go farther and will learn the meaning of
2^. In arithmetic he has learned the meaning of 3 — 2 ; in
algebra he will go farther and will learn the meaning of
23.
In arithmetic he has said,
If 2 X some number equals 10,
the number must be ^ of 10, or 5.
In algebra he will express this more briefly, thus :
If 2a; = 10,
then X = 5;
indeed he may already have met this form in arithmetic.
By arithmetic he probably could not solve a problem of
this nature: The square of a certain number, added to 5
times that number, equals 50; to find the number. But
after studying algebra a short time, he will find the solu
tion quite simple.
1
2 .. , __ ^ ELEjMENTS OF ALGEBRA.
In arithmetic it is quite common to use a letter to repre
sent a number, as r to represent the rate of interest, i to
represent the interest itself, p the principal, etc. In
algebra this is much more common. In arithmetic it is
customary to denote multiplication by the symbol x , the
product of 5% and $100 being written 5% x $100, and
the product of r and phj rxp\ but in algebra the latter
product is represented by rp.
In expressing 5 times 2 we cannot write it 52, because that means
50 + 2. Bat where only letters are used, or one numeral and one or
more letters, we may define the absence of a sign to mean multiplica
tion. Thus, ah means a x 6, that is, the product of the numbers rep
resented by a and h; bah means 5 times this product.
EXERCISES. I.
If ct = 5, b = 7, c = 3, d = 1, e = 4, find the value of
each of the expressions in exs. 19.
^ J ^ o 7 21 ae
1. oabd. 2. facde. 3.
4. bed
nrr—^ — be — ad /.
35 ab
23c "■ ^3cde'
b cd
If a = 2, b = S, c = 4:, d = 5, find the value of each of
the expressions in exs. 1017.
10.
abe abc
bed acd
11.
a { d c — b
lb 3
12.
a b d
b c be
13.
4 6_8 10
abed
14.
c 3d 2b
a b 3a
15.
a ^ I c d
4~6"^8""10
16.
a { c 1 d — b
7 1
3.
17.
a + b ^c + d
5a — d
INTRODUCTION TO ALGEBRA. 3
2. A collection of letters, or of letters and other number
symbols, connected by any of the signs of operation (+, — ,
X , H, etc.) is called an algebraic expression.
E.g. , 3 X + 2 a is an algebraic expression, but 3 + 2 is an arithmet
ical expression. So 2 a is an algebraic expression, 2 and a being
connected by the (understood) sign of multiplication; also a, since
that means 1 a.
3. An algebraic expression containing neither the + nor
the — sign of operation is called a term or a monomial.
T^ c L r I — lOaftic . , ^ ,
E.g.., ^ao., 5 Vox, —z , are monomials. In the expression 2 aic
£i6 yz
+ Sby — ISy^, the expressions 2 ax, 3 by, and 5 y^, are the terms, and
each taken by itself is called a monomial. The broader use of the
word term is given in § 46.
4. An algebraic expression made up of several terms or
numbers connected by the sign + or — is called a polynomial.
The word means manytermed. On all such new words consult
the Table of Etymologies in the Appendix.
5. A polynomial of two terms is called a binomial, one
of three terms a trinomial. Special names are not given
to polynomials of more than three terms.
E.g., a^ —  is a binomial. 5 Va 1 ah^cd is a trinomial.
EXERCISES. II.
1. Select the algebraic expressions in the following list :
(a) Sa%c. (b) ^a'bcd.
(c) ^  c'd\ (d) x^ {!/ + z\
(e) 23V7H1. (f) 2x^?>x'^xV\.
2. Out of the algebraic expressions select the monomials.
3. Out of the polynomials select the binomials; tri
nomials.
4 ELEMENTS OF ALGEBRA. •
6. In the operation of multiplication expressed hy ax b
X c, or abc, the a, b, and c are called the factors of the
expression, and the expression is called a multiple of any
of its factors.
Factors should be carefully distinguished from terms. The
former are connected by signs of multiplication, expressed
or understood ; the latter by signs of addition or subtraction.
7. Any factor of an expression is called the coefficient of
the rest of the product. The word, however, is usually
applied only to some factor whose numerical value is ex
pressed or known and which appears first in the product.
E.g.^ in the expression 3 ax, 8 is the coefficient of ax, and 3 a is the
coefficient of x.
Since a = la, the coefficient 1 may be understood before
any letter.
8. As in arithmetic, the product of several equal factors
is called a power of one of them.
j&. ST., 2 X 2 X 2 is called the third power of 2 and is written 2^ ;
aaaaa is called the fifth power of a and is written a^.
9. The numbersymbol which shows how many equal
factors enter into a power is called an exponent.
E.g., in 2^, 3 is the exponent of 2 ; in a^, 5 is the exponent of a.
The exponent affects only the letter or number adjacent to which it
stands ; thus, ab^ means abbb.
The exponent should be carefully distinguished from the
coefficient. In the expression 2 ax^, 2 is the coefficient of
ax^, and 2 a oi x^; 3 is the exponent of x.
Since x may be considered as taken once as a factor to
make itself, x^ is defined as meaning x. Hence, any letter
may be considered as having an exponent 1.
There are other kinds of powers and exponents besides tliose which
have just been defined, and these will be discussed later in the work.
INTRODUCTION TO ALGEBRA. 6
10. The degree of a monomial is determined by the number
of its literal factors.
E.g.^ a^ is of the 5th degree, a^ft* of the 7th, 3 abc of the 3d., and 5 a
of the 1st. A number, like 5, is spoken of as of zero degree because it
has no literal factors.
11. The word degree is usually limited, however, by refer
ence to some particular letter.
Thus, while ^a^'x^ is of the 5th degree, it is said to be of the 3d
degree in x, or of the 2d degree in a, or of zero degree in other letters.
12. Terms of the same degree in any letter are called
like terms in that letter.
Thus, 3 ax2 and 5 ox^ are like terms, being of the same degree in
each letter. 3 ax^ and 5 ftx^ are like terms in x.
13. The degree of a polynomial is the highest degree of
any of its terms.
Thus, ax2 + 6x + c is of the second degree in x.
14. As in arithmetic, one of the two equal factors of a
second power is called the square (or second) root of that
power, one of the three equal factors of a third power the
cube (or third) root, one of the four equal factors of a fourth
power is called the fourth root, etc.
The word root has also a broader meaning, as in "the
square root of 2," an expression which is legitimate, although
2 is not a second power of any integral or fractional num
ber. This meaning will be discussed later.
The square root of a is indicated either by Va or by a^ the cube
root by Va or by a^, the fourth root by Va or by a^, etc. In a^, the
i is called a fractional exponent, and the term is read "a, exponent
i," or "the square root of a," or "a to the i power," a reading
which will be justified by the subsequent explanation of the word
power.
6 ELEMENTS Oi? ALGEBliA.
From what has been stated it will be seen that one of the
features of algebra is the representation of numbers by
letters. The advantages of this plan will soon appear.
Thus, if a number is represented by n, 5 times the square
of that number will be represented by 5 n^. If two num
bers are represented by a and b, 3 times the cube of the
first, divided by 5 times the square root of the second, will
3 a^ 3 a^
be represented by —  or by
15. Those terms of a polynomial which contain letters
constitute the literal part of the expression.
E.g. , the literal part of x^ + 2 a; + 1 is x2 + 2 x.
The expression is^ also used with respect to factors. Thus, the
literal part of ia^2 is a.
EXERCISES. III.
1. What is the numerical value of each term in the fol
lowing expressions, it a = 1, b = 2, c = 5, d = 3 ?
(a) ab^c'd\ (b) c^ + b^Sa.
(0)2^1002,. (d)« + A + ^ + ^.
2. In ex. 1, what is the numerical value of each poly
nomial ?
3. In 13 a%^x, what is the coefficient of £c ? of b^x ? of
a%^x ? What is the degree of the expression ? What is its
degree in ic ? What is the exponent oi a? ot b? oi x?
4. In the following monomials name the coefficients of
the various powers of x, and also the exponents of x :
(a) ^ (b) x^ (c) ^x».
(d) 23 a V. (e) ia^'cxK (f) ^a^Vbx.
INTKODUCTION TO ALGEBRA. 7
5. From ax^, 3 hx^, cx^, a^x, and 10 ahx^, select the like
terms in x or any of its powers.
6. From 3 ax"^, 9 mx, 14 ax^, ax^, 9 ax^, and 144 x, select
the like terms.
7. Express algebraically that if x^ + y^ + 2xy be divided
hj x { y the quotient \& x \ y. (Use fractional form.)
8. What is the degree of the polynomial ax^ [hx \ c?
What is its degree in ic ? What is its value if a = b = c — l,
and X = b?
9. Express algebraically that if the sum of a^, ab, and
b'^ be divided by the square of the binomial c — d, the quo
tient is X.
10. What is the meaning of the expression
(That from 4 times the square of a certain number there
has been subtracted, etc.)
11. Also of the following expressions :
(a) a' + 2ab + b\ (b) a^  b\
(c) ^a^4.b^ + a^. (d) a^ + 3 a^^* + 3 ab^ + b\
12. Eepresent algebraically the sum of 3 times the square
of a number, f the cube root of a second number, and 5
times the 5th power of a third number. What is the value
of the expression, if the three numbers are respectively 2,
8, 1 ?
13. Given a = 4, b = Q>, c = 9, d = 1Q>, e = 8, find the
value of each of the following, and designate the expres
sion as a monomial, binomial, etc. :
(a) 2 a'bc^. (b) d^e^  b.
/ \ i , z , J , 2 /JN 25 abcde
(c) ct* f & + ^ + e\ (d) —^
(e) 25#ja2^ + 5. (f) ^b^c^J^eK
o
8 ELEMENTS OF ALGEBRA.
II. THE EQUATION.
16. An equality which exists only for particular values of
certain letters representing the unknown quantities is called
an equation. These particular values are called the roots of
the equation.
Thus, X + 3 = 5 is an equation because the equaUty is true only for
a particular value of the unknown quantity x, that is, for x = 2. This
equation contains only one unknown quantity.
2 + 3 = 5 expresses an equality, but it is not an equation as the
word is used in algebra.
17. The discovery of the roots is called the solution of the
equation, and these roots are said to satisfy the equation.
Thus, if X + 5 = 9, the equation is solved when it is seen that x = 4.
This value of x satisfies the equation, for 4 + 5 = 9.
18. If two algebraic expressions have the same value
whatever numbers are substituted for the letters, they are
said to be identicaL
Thus, a^ _ is identical to a^ _f 5^ and a + 6 to 6 + a.
An identity is indicated by the symbol = , as in a^ + & = 6 + a^.
19. The part. of an equation to the left of the sign of
equality is called the first member, that to the right the
second member, and similarly for an identity.
The two members are often spoken of as " the left side " and " the
right side," respectively.
The extensive use of the equation is one of the character
istic features of algebra.
The importance and the treatment of the equation will
best be understood by considering a few problems.
In each case we say, " Let x = the number," meaning that x is to
represent the unknown quantity.
INTRODUCTION TO ALGEBRA. 9
1. Find the nuviber to twice which if 3 is added the
result is 11.
1. Let X = the number.
2. Then '2x = twice the number.
3. Hence, 2x + 3 = ll. (Why?)
4. Subtracting 3 from these equals, the results must be equal, and
2x=ll 3, or8.
5. Dividing these equals by 2, the results must be equal, and
x = 4.
Check. To see if this value of x satisfies the equation, substitute it
in step 3. Since 2x4 + 3 = 11, the result is correct. This is called
checking or verifying the result.
20. A check on an operation is another operation whose
result tends to verify the result of the first.
^.gr., if 11 — 7 = 4, then 4 + 7 should equal 11 ; this second result,
11, verifies the first result, 4.
The secret of accurate work in algebra and in arithmetic
lies largely in the continued use of proper checks.
21. A check on a solution of an equation is such a substitu
tion of the root as shows that it satisfies the given equation.
This substitution must always be inade in the original
equation or in the statement of the problem. Thus, in the
above solution it would not answer to substitute the root,
4, in step 4, because a mistake might have been made in
getting step 4 from step 3.
2. Twothirds of a certain number, added to 5, equals 17.
What is the number ?
1. Let X = the number.
2. Then x + 5 = 17, by the conditions of the problem.
3. Subtracting 5 from these equals, the results must be equal, and
lx = \2.
4. Therefore, x = 18.
Check. I of 18 = 12, and 12 + 5 = 17.
10 ELEMENTS OF ALGEBRA.
3. 72 divided by a certain number equals twice that num
ber. What is the number ?
1. Let X = the number.
72
2. Then — = twice the number, by the conditions
^ of the problem.
3. Therefore, — = 2x.
X
4. Multiplying these equals by x, the results must be equal, and
72 = 2x2.
5. Dividing these equals by 2,
36 = x2.
6. Extracting the square roots of these equals,
Q = x.
4. If from 35 a certain number is subtracted, the differ
ence equals the sum of twice that number and 20, What is
the number ?
.1.
Let
X = the number.
2.
Then
35  X =:= 2 X + 20.
(Why ?)
3.
Then
35 = 3x + 20, by adding x.
4.
Then
15 = 3x.
(Why ?)
5.
Then
5 = X.
(Why?)
Check. (What should it be ?)
From the preceding problems it will be seen that the two
members of an equation are like the weights in two pans of
a pair of scales which balance evenly ; if a weight is taken
from one pan, an equal weight must be taken from the other
if the even balance is preserved ; if a weight is added to one
pan, an equal weight must be added to the other ; and, in
general, any change made in one side requires a like change
in the other.
These facts are already known from arithmetic, where the equation
is frequently met. Even in primary grades problems are given like
2 X (?) = 12, this being merely an equation with the symbol (?) in place
of X.
INTRODUCTION TO ALGEBRA. 11
22. The axioms. There are several general statements
(of which a few have already been used) so obvious that
their truth may be taken for granted. Such statements
are called axioms.
The following are the axioms most frequently met in
elementary algebra.
1. Quantities which are equal to the same quantity, or to
equal quantities, are equal to each other.
That is, if 5 — X = 3, and 1 + x = 3, then 5 — x = 1 + x.
2. If equals are added to equals, the sums are equal.
That is, \tx = y, then x\2 = y \2.
3. If equals are subtracted from, equals, the remainders
are equal.
That Is, if X + 2 =: 9, then x = 9  2, or 7.
4. If equals are added to unequals, the sums are unequal
in the same sense.
" In the same sense " means that if the first was greater than the
second before the addition of the equals, it is after. Thus, if x is
greater than 8, x + 2 is also greater than 10.
5. If equals are subtracted from unequals, the remainders
are U7iequal iii the same sense.
That is, if x is less than 16, x — 3 is less than 13.
6. If equals are multiplied by equal numbers, the prod
ucts are equal.
That is, if  = 6, x = 3 x 6, or 18.
3
7. If equals are divided by equals, the quotients are equal.
That is, if 2 X = 6, x = 6 r 2, or 3.
8. Like powers of equal numbers are equal.
That is, if X = 5, x2 = 25. We here speak of x as a number because
it represents one.
12 ELEMENTS OF ALGEBRA.
9. Like roots of equal numbers are arithmetically equal.
That is, if a;2 = 36, X = 6. The axiom says " arithmetically equal,"
because it will soou be found that there is an algebraic sense in which
roots require special consideration.
These axioms should at once be learned by number.
23. Stating the equation. The greatest difficulty experi
enced by the student in the solution of problems is in the
statement of the conditions in algebraic language. After
the equation is formed the solution is usually simple.
While there is no method applicable to all cases, the fol
lowing questions usually lead the student to the statement :
1. What shall x represent? In general, x represents the
number in question.
E.g., in the problem, "Twothirds of a certain number, plus 10,
equals 30, what is the number ? " x represents the number.
2. For what number described in the problem may two
expressions be found?
Thus, in the above problem, 30 and "f of a certain number, plus
10," are two expressions for the same number.
3. How do you state the equality of these expressions in
algebraic language ?
fxH0 = 30.
EXERCISES. IV.
Form the equations for the following problems :
1. The difference of two numbers is 14 and the smaller
is 3. What is the larger ?
2. A's money is three times B's, and together they have
$364. How much has B?
3. The sum of two numbers is 60 and the difference is
40. What is the smaller number ?
INTRODUCTION TO ALGEBRA. 13
Typical solutions. In the solution of problems involving
equations, the axioms need not be stated in full except when
this is required by the teacher. The check (which is a
complete verification) should always be given in full, except
when the teacher directs to the contrary. The following
solutions may be taken as types :
1. What is that number to whose square root if 2 is added
the result is
7?
L Let
2. Then
3. ..
4. ..
X = the number.
Vx + 2 = 7, by the conditions.
x = 26.
Ax. 3
Ax. 8
Check. V25 + 2 = 5 + 2 = 7.
2. What is that number from twothirds of which if 5 is
subtracted the result is 10 ?
1. Let X = the number.
2. Then f x — 5 = 10, by the conditions.
3. .. x5 + 5 = 15, orfx= 15. Ax. (?)
4. .. x = 22^. Ax. (?)
Check, f of 22i = 15, and 15  5 = 10.
3. Find the value of x in the equation Vx + 1=^17.
1. \^ + 1 = ^ + 7. Given
2. .. Vi = 6h or V. Ax. 3
3. .. x = ^fi, or 40f Ax. 8
Check. (Give it.)
4. Find the value of x in the equation 5x — 3 = x + 7.
1. 5x — 3 = x + 7. Given
2. .. 5x = X + 10. (Why ? See ex. 2, step 3)
3. .. 4 X = 10, for 5 X — X means 5 x — 1 x.
4. .. x = 2i. (Why?)
Check. (Give it.)
14 • ELEMENTS OF ALGEBKA.
EXERCISES. V.
1. Find the value of x in the equation 2 £c f 2 = 30 + cc.
«Ai20^ ^ .. . 2 X
2. Also m — = 5. 3. Alsoin = 
X X Z
4. Also in a;2 + 7 = 88. 5. Also in a;^  1 = 35.
6. Also in f X + 5 ^ ^ ic + 20.
7. Alsoin.22cc + 30 = 17:z; + 70.
8. Also in 250 ic  20 = 20 :c + 440.
9. Also in 12.75 a; + 6.25 = 7.25 x + 17.25.
10. What number is that which divided by 3 equals ^ ?
11. What is the number whose half added to 16 equals
21?
12. What is the number whose tv/entieth part added to
10 equals 20 ?
13. What is that number to whose square if 5 is added
the result is 41 ?
14. What is that number to whbse square root if 5 is
added the result is 41 ?
15. What is that number from onethird of which if 27
is subtracted the result is 5 ?
16. There is a number by which if 9 is divided the quo
tient is that number. Find it.
17. The sum of a certain number and 9 is equal to
the sum of 1 and three times that number. Find the
number.
18. The sum of a certain number, twice that number,
and twice this second niunber, is 70. What is the first
number ?
19. The united ages of a father and son amount to 100
years, the father being 40 years older than the son. What
is the asre of the son ?
INTRODUCTION TO ALGEBRA. 16
Practical applications. The equation offers a valuable
method for solving many practical problems, of which a
few types will now be considered.
1. What sum of money placed at interest for 1 year at
4^% amounts to $836 ?
1. Let X = the number of dollars.
2. Then x + 0.04ix =: the number of dollars in the prin
• cipal + the interest.
3. But 836 = the number of dollars in the prin
cipal + the interest.
4. .. x + 0.0^x = 8S6.
6. Or 1.04^05 = 836.
6. .. x = 800. Ax. 7
7. .. the sum is 800. ■
Check. 800 + 0.04^ of 800 = 836.
It should be noticed that since x stands for the number
of dollars, when it is found that x = 800 it is known that
the result is $800.
In the applied problems of algebra, x is always taken to
represent an abstract number, and the first step should
always state definitely to what this abstract number is
to refer.
2. A commission merchant sold some produce on a com
mission of2^Q, and paid %%> for freight and cartage, remit
ting $117.50. For how much did he sell the produce ?
1. Let X = the number of dollars received.
2. Then x — 0.02 x = the number after deducting 2%.
3. And x — 0.02x — 5 = the number after deducting for
cartage also.
4. .. x0.02x5 = 117.50.
5. .. 0.98x = 122.60. (Why?)
6. .. X = 125. (Why ?)
Check. 1250.02of 1255 = 117.50.
16 ELEMENTS OF ALGEBRA.
3. After deducting jJ^ and then ^ from a certain sum
there remains $49.50. Required the sum.
1. Let X = the number of dollars.
2. Then x — j\j x = j% x, the number of dollars after
deducting jL.
3. From this ^^ x is to be taken \ of it,
x^^xiof x%x=ifx/^x
4. ., fx = 49.50.
5. .. x = 49.50 I
= m.
.. the sum is $66.
Check. 66  j\ of 66 = 59.40. 59.40  i of 59.40 = 49.50.
EXERCISES. VI.
1. In how many years will $100 double itself at 5%
interest ?
2. What sum of money put at interest for 2 years at 6%
amounts to $84 ?
3. In how many years will a sum of money double itself
at 6% simple interest ?
4. In how many yeai's will $80 amount to $200, at 6%
interest ? (80 + (k x 6% of 80 = 200.)
5. What is the rate per cent of premium for insuring a
house for $2000, when the premium is $30 ?
6. Taking, the number of units of area of a circle as
being 3^ times the square of the number of units of length
in the radius, find the radius of the circle whose area con
tains 77 units.
7. After selling some goods on 5% commission, a mer
chant remits, as the net proceeds, $79.80. How much is
his commission ? (Let x = the number of dollars for which
the goods were sold ; after finding x take 5% of it.)
INTRODUCTION TO ALGEBRA. 17
III. THE NEGATIVE NUMBER.
24. In remote times men could count only by what
are often called natural numbers, that is, 1, 2, 3, 4, 5, •••.
Such numbers suffice to solve an equation like a; — 3 = 0,
an equation in which x must evidently be 3.
Mankind then introduced the unit fraction, that is, a
fraction with the numerator 1. Such numbers are neces
sary in solving an equation like 2 cc — 1 = 0. (Solve it.)
Then came the common fraction with any numerator, as
§j fj TT' •••• Such numbers are necessary in solving an
equation like 3 cc — 2 = 0. (Solve it.)
The idea of number was then enlarged to cover the cases
of V2, Vt, v5, • • •, which are neither integers nor fractions
with integral terms. Such numbers are necessary in solv
ing an equation like £c^ — 2 = 0. (Solve it.)
25. Many centuries later the necessity was felt for fur
ther enlarging the idea of number in order to solve an
equation like x + l = 0, oric + 6t = 0, a being one of
the kinds of number above mentioned. This led to
the consideration of negative numbers, — 1, — 2, — 3, • • •,
and the meaning of these numbers will now be inves
tigated.
26. If the mercury in a thermometer stands at 5°
above a fixed point and then falls 1°, we say that it
stands at 4° above that point. If it falls another
degree, we say that it stands at 3° above that point,
and the next time at 2°, and the next time at 1°.
If the mercury then falls another degree, it becomes
necessary to name the point at which it stands, and we
call this point zero and designate it by the symbol 0.
If the mercury falls another degree, we must again name
the point at which it stands, and instead of calling this
18
ELEMENTS OF ALGEBRA.
point "1° below zero," we call it "minus 1°" or "negative
1°," and we designate it by the symbol — 1°. Likewise, if
the mercury falls 1° lower, we say that it stands at — 2°,
and so on.
27. Thus we find a new use for the word minus and the
symbol — . Heretofore both the word and the sign have
indicated an operation, subtraction ; they now indicate the
quality of a number, showing on which side of zero it
stands, and thus they are adjectives.
In speaking of "west longitude," "west" is an adjective modify
ing "longitude"; in speaking of "minus latitude," "minus" is an
adjective modifying "latitude"; so in "minus 2°," "minus" is an
adjective.
28. It thus appears that our idea of number can
be enlarged to include zero, and still further to
include the series of natural numbers extended
downward from zero.
If necessary to distinguish 1° above from 1°
below 0, the former is written +1° and called
either "plus 1°" or "positive 1°," and the latter
is written — 1°. But unless the contrary is stated,
a number with no sign before it is considered
positive.
29. It thus appears that positive numbers may
be represented as standing on one side of zero, and
numbers on the other.
Thus, if west longitude is called positive, east longitude
is called negative, and vice versa ; if north latitude is called
positive, south latitude is called negative ; if a man's capi
tal is called positive, his debts are called negative, etc.
E.g., if the longitude of New York is 73° 58' 25.5'' west and that of
Berlin is 13° 23' 43.5" east, the former may be designated as + 73° 58'
25.5" and the latter as  13° 23' 43.5", their difference being 87° 22' 9".
of 3 or
Of2 "
0+1 .'
INTRODUCTION TO ALGEBRA. 19
Similarly, if a man begins the year with $5000, and during the year
loses his capital and gets 1 2000 in debt, he is $ 7000 worse off than at
the beginning. It may then be said that he started with $ 5000 and
ends with — $2000, the difference being the $7000 which he lost.
30. Since two such expressions as + a and — a, or + 5°
and — 5", represent different directions, but equal measiu^es,
they are said to have the same absolute value.
The symbol  — a  is read, " the absolute value of — a."
Hence, I  5° I = 1 + 5° 1 , although  5° does not equal + 5°.
Since the difference between — 5° and + 5° on a ther
mometer is 10°, it appears that we sometimes find the dif
ference between two numbers by adding absolute values.
31. There are numerous signs used in algebra, as +, — ,
X, ^, V~, exponents, etc. But by the sign of a term is
always meant the { or — sign, which indicates the quality
of the term,, whether positive or negative.
Thus, in a^ 4 7 6, the sign of 7 6 is plus (understood), while in aV— 7 6
it is minus.
32. Positive and negative numbers, together with zero,
are often called algebraic numbers, positive numbers being
called arithmetical.
Zero is considered either as having no sign or as having both the
plus and the minus signs.
EXERCISES. VII.
These are intended for oral drill and should be supple
mented by many others of this type.
1. A ship in 8° west longitude (+ 8°) sails so as to lose
1° in longitude. On what meridian is it then ? Suppose
it loses 7° more ? 3° after that ?
^ 2. What is the difference in latitude between + 10° and
 20° ? between  90° and  90° ?
20 ELEMENTS OF ALGEBRA.
3. Show that I 5  7 I = I  10 + 12 I = I  22 + 20 I = 2.
4. What is meant by I — 4 1 ? by the absolute value of
 8 ? of  3 ?
5. What is the absolute value of 10  17 ? of 17  10 ?
off? of+l?
6. What other numbers have the same absolute value as
4 3,  5, + 10,  V2, ?
7. What is the difference in time between 50 years b.c.
and 50 years a.d. ? Indicate this by symbols.
8. Draw a line representing a thermometer scale ; mark
off 0°, 30°,  25°. What is the difference between 30° and
 25° ?
9. If the weight of a piece of iron is represented by
+ 10 lbs., what will represent the weight of a toy balloon
which pulls up with a force of 3 lbs. ?
10. Suppose the piece of iron and the balloon mentioned
in ex. 9 were fastened together. What would be their com
bined weight ?
11. If the upward pull of a toy balloon is represented by
+ 3 lbs., what will represent the upward pull of a piece of
iron weighing 10 lbs. ?
12. What is meant by saying that a person is worth
 $1000 ? Suppose $2000 is added to his capital. How
much is he then worth ?
13. Draw a circumference and show that the difference
between 50° and 10° equals 150°  +   10° , or 60°.
Also that the difference between 10° and  10° is 20°.
14. If the weights of two pieces of iron are respectively
100 lbs. and 300 lbs., and to these are attached a balloon
with an upward pull of 500 lbs., how shall the combined
weight be represented ?
INTRODUCTION TO ALGEBRA. 21
IV. THE SYMBOLS OF ALGEBRA.
33. As already seen, algebra employs the symbols of
arithmetic, often with a broader meaning, and introduces
new ones as occasion demands. The following classifica
tion will enable the student to review the symbols thus
far familiar to him, and may add a few new ones to his
list. Others will be considered from time to time as
needed.
1. Symbols of quantity.
a. Arithmetical numbei's, i.e., positive integers and frac
tions.
b. Algebraic numbers, the above with the addition of neg
ative numbers and zero. Others will be considered later.
c. Letters denoting algebraic numbers ; these are the
symbols of quantity chiefly used in algebra.
2. Symbols of quality.
a. The symbols + a7id — to indicate positive and nega
tive number, as in + a, — b, etc.
b. The absolute value symbol, as in 1 — 3  , indicating
that the arithmetical quality of — 3 is considered.
3. Symbols of operation.
a. Addition, +•
b. Subtraction, — .
c. Multiplicatio7i, X, ■ , and the absence of sign. Thus,
a xb, ab, and ab, all indicate the product of a and b. It
is quite customary in algebra to say " a into b " for
" a times b."
d. Division, h, /, :, and the fractional form. Thus,
a 7 b, a/b, a : b, and ? all mean the quotient of a divided
hyb.
22 ELEMENTS OF ALGEBRA.
In arithmetic the symbol : is used only between num
bers of the same denominations ; but in algebra, where the
letters represent abstract nmnbers, this distinction does
not enter. For ease in typesetting the symbol / is often
used in. print ; in writing, the fraction is usually employed.
e. Involution and evolution are indicated by exponents.
Evolution is also indicated, as in arithmetic, by the symbol
V? a contraction of r, the initial of radix (Latin, root).
Thus, ^3 i^^eans aaa,
8^ means one of the three equal factors of 8, or 2.
4. Symbols of relation.
a. Equality, =.
b. Identity, = ; thus, a = a, read " a is identical to a."
Also read " stands for," as in r = rate, F = x^ \ 2 xy, etc.
.c. Inequality : > greater than, < less than, ^ not equal
to, > not greater than, < not less than.
5. Symbols of aggregation.
The expression m (a \ b) means that « + ^ is to be mul
tiplied by 7n. The parenthesis about a { b is called a
symbol of aggregation.
The bar, brackets, and braces are also used, as in
a
m \a — [b \ X (a — b — c) { xa'] — d\, and in
x'^ + 2a x + c^ = {a + b)x'^ + {2ab + G)x + G'^
b
+ c
but the term parenthesis is often employed to mean any
symbol of aggregation. The subject is more fully dis
cussed on p. 35.
6. Symbols of deduction.
•.', since.
."., therefore.
INTRODUCTION TO ALGEBRA. 23
7. Symbol of continuation.
• • • , meaning " and so on," as in the sentence, " consider
the quantities a, a^, a^, • • • ."
34. Conventional order. Mathematicians have established
a custom as to the order in which these signs shall be con
sidered when several are involved, as in an expression like
a ■\ h ^ c ^ d ^ ef^  g ^ hk^ ^'
In the above expression six operations are involved, as
follows :
Direct. Interse.
Class I. Addition.
Class II. Multiplication.
Class III. Involution (Powers).
Subtraction.
Division.
Evolution (Roots).
The mathematical custom is expressed in the following
conventions :
1. If tivo or more operations of the same class come
together (^without symbols of aggregation), the operations
are to he performed in the order indicated.
E.g., 2 + 34 + 1=2, and 2x8=4x2 = 8.
2. If two or more operations of different classes come
together (without symbols of aggregation), the operations
of the higher class are to be performed first.
I.e., involution and evolution precede multiplication and division,
and these precede addition and subtraction.
^.t/., 5 + 2 X 8  22  Vs = 7.
This conventional order can, of course, be varied by the
use of symbols of aggregation.
E.g., 2 + 3 X 5 = 17, but (2 + 3) X 5 = 25.
24 ELEMENTS OF ALGEBRA.
There are also certain exceptions to this conventional
order, but they are not of a natui'e to cause any confusion.
E.g.,, ah r cd means {ah) i (cd) and not , and similarly in other
cases of the absence of sign where division is involved.
Similarly, when the sign of ratio (:) appears in a proportion it has
not the same weight as the symbol ^ . Thus, 2 + 3:12 — 2 = 1:2
means (2 + 3) : (12  2) = 1 : 2.
EXERCISES. VIII.
1. li a = 1, h = 2j c = ^, d = 4:, find the value of each
of the following expressions :
(a) {a + b'^f. (b) b(c\dy.
(c) 5d/bc a. (d) (#  c? = b) c\
{q) 3a + b X cd. (f ) (^ + *) (p + d).
(g) 2 + a^H^ Ha + 5. (h)2axb^dxca.
2. Read the following expressions :
(a) a + a^ = a { a^.
(b) alb'ip>a\lb> 1.
(c) a^ ^L a \ a^ — a, .' . a"^ <. a \ aP'.
.(d) ••• a = 2, .'. a" = 4., a^ = S, a^ = 16, • • •.
(e) a^ { a^ =^ a^, and a^ + a^ ^ a^, if a is positive.
3. Show that the following are equal when a = 2 and
b = 8. That is, substitute 2 for a and 3 for b in each
member.
(a) (a + by==a^ + 2ab + b\
,(b) (bay = b^2ba + a^
(c) (b^a^)/(ba)=b + a.
(d) (a + b) (a^  ab + b^) = a^ \ b\
(e) {b""  a"") / (b  a) = b'' ■} ba + a\
(f ) {a + by = a^ + 8 a% + 3 ah^ + b\
INTRODUCTION TO ALGEBRA. 25
V. PROPOSITIONS OF ALGEBRA.
35. A proposition is a statement of either a truth to be
demonstrated or something to be done.
E.g., algebra investigates this proposition : The product of a"* and
Qn ig dm + n_ j^ also considers sucli statements as this : Required the
product of a \ b and a — b.
36. Propositions are divided into two classes, theorems
and problems.
A theorem is a statement of a truth to be demonstrated.
E.g., The product of a"* and a" is «"» + ".
A problem is a statement of something to be done.
E.g., Required the product of a + b and a — b.
A corollary is a proposition so connected with another as
not to require separate treatment.
The proof is usually substantially included in that of the proposition
with which it is connected.
REVIEW EXERCISES. IX.
1. What is the degree of the expression 3 ax^y^ ? What
is its degree in ic ? in y? in ic and y? in z?
2. Distinguish between coefficient and exponent. What
is the coefficient of x in the expression  ? the exponent ?
3. What is the meaning of the expression ab ? of 26 ?
of a I ? of 2f ? What is the value oi ab it a = 2, b = 6?
ofafiia = 2,x = 3,y = 4:?
4. What is meant by the etymology of a word ? What
is the etymological meaning of binomial? of trinomial?
of monomial ? of aggregation ? of theorem ? (See Table
of Etymologies.)
26 ELEMENTS OF ALGEBRA.
5. Show that if a = 7 and h = 5,
(a) (a + b)(ab) = a^P.
(b) (aby = a'' + P2ab.
(c) (a  by = a^  b^  3ab(a  b).
6. Show that a a = 3, b = 2, c = 1,
(a) (a]byc'^ = (a + b + c)(a + b  c).
(b) la + b + cy = a^ + b^{c^ + 2ab\2bc + 2 ca.
7. What meaning has the number " minus 2 " to you ?
8. What is the value of 8^ ? of 9^ of 16^ • 32^  16' ?
9. Show by substitution that 1 is a root of the equation
in ex. 10.
10. How many terms in the equation 2x^\3x — 4^ = 1?
How many members ?
11. Draw a diagram illustrating the fact that the abso
lute value of the difference between — 5 and 10 is 15.
12. What is the degree of the polynomial x^ + 3x^y^ +
3 xy^ \ 5y + 6 ? What is the degree in a? ? my? in z?
13. Write the following in algebraic language : The sum
of the square of a number, 3 times the number, and 5, is
equal to 9.
14. Represent algebraically the sum of the cube of a
number, 5 times the square of the number, and 6, less half
the number.
15. What is meant by solving an equation ? by a root of
an equation ? by checking a solution ? Illustrate with the
equation ic — 2 = 0.
16. What is the number from which if 5% be taken, and
10% from the remainder, and 20% from that remainder,
the result is 41.04?
17. Write ou^ thiee problems which you can now solve,
but which you could not solve when you began to study
algebra.
CHAPTER II.
ADDITION AND SUBTRACTION.
I. ADDITION.
37. In elementary arithmetic the word number includes
only positive integers and fractions, or at most a few indi
cated roots like V2, VS, • • • . Hence, the word sum, as there
used, applies only to the result of adding two positive num
bers.
In algebra the word sum has a broader meaning, and
includes the results of adding negative numbers and num
bers some of which are positive and others negative.
E.g.^ consider the combined weight of these three articles: a 211).
weight, a 4lb. weight, and a balloon which weighs — 5 lbs. (i.e., pulls
upward with a force of 5 lbs.). Together they would evidently weigh
1 lb. Hence 1 lb. is said to be the sum of 2 lbs., 4 lbs., and — 5 lbs.
So the result of adding a debt of $100 to a capital of $300 is a capital
of $200 ; hence, $200 is said to be the sum of $300 and  $100.
38. In this broader view of addition two cases evidently
arise :
1. Numbers with like signs.
2 lbs. + 3 lbs. = 5 lbs.
A balloon pulling up 5 lbs. and one pulling up 8 lbs. together pull
up 13 lbs., or ( 5 lbs.) + ( 8 lbs.) =  13 lbs.
2. Numbers with unlike signs.
A balloon pulling up 5 lbs. and a weight of 2 lbs. together pull up
3 lbs., or  5 lbs. + 2 lbs. =  3 lbs.
27
28 ELEMENTS OF ALGEBRA.
39. From considerations like these we are led to define
the sum of two algebraic numbers as follows :
1. If two numbers have the same sign, their algebraic sum
is the sum of their absolute values, preceded by their common
sign.
Thus, to add — 3 and — 2 means to add 3 and 2 and to place the
sign — before the result.
2. If they have not the saTue sign^ their algebraic sum is
the difference of their absolute values, preceded by the sign
of the one which has the greater absolute value.
Thus, to add — 3 and 2 means to find the difference between 3 and
2 and to place the sign — before the result, since I — 3 I > I 2 I.
3. In the special case where the two numbers have the
same absolute value (i.e., where they are equal and of oppo
site signs), the sum is zero.
E.g.,2 + {2) = 0.
4. If one of two numbers is zero, their algebraic sum is
the other number.
Thus, 3 + means  3.
40. The algebraic sum of several numbers is defined as
the sum of the first two plus the third, that sum plus the
fourth, ■ • • .
Thus, a + h + c + d means a + & with c added, and that sum with
d added. I.e., a + 6 + c + d means [(a + 6) + c] + d.
EXERCISES. X.
1. Find the sum of  20, + 3,  47, + 80.
2. Also of + 2,  3, + 5,  4, + 9,  3,  6.
3. Also of 2 x% 5 x2,  6 x^ Sx^?
4. Also of 127 mn, 62 mn, — 93 mn,  17 mn?
ADDITION AND SUBTRACTION. 29
5. $50 + 17 +($21) + (130)=?
6. 5 + 219 + ( 376) + ( 40) + 10 + ( 37) = ?
7. ( 7) + 4 + ( 2) + 18 + 13 + ( 20) + ( 6) = ?
8. Sa{(2a) + (5a) + 8a + 6a + (10a) = ?
9. What is the sum oi 3 a, 5 a, —6 a, Sa, 10 a, — 3 a,
17a?
10. 12 x^^ + 4 ccV + ( 16 x^y) + ( 3 xhj) + 10 xhj
= (?)x'y?
11. 5 lbs. + 55 lbs. + ( 40 lbs.) + ( 27 lbs.) + 121 lbs.
+ ( 19 lbs.) + ( 5 lbs.) = (?) lbs ?
12. What is the combined weight of two balloons weigh
ing, respectively, — 10 lbs. and — 18 lbs., and thiee pieces
of iron weighing, respectively, 6 lbs., 12 lbs., and 14 lbs. ?
13. On seven consecutive midnights in January, in Mon
treal, the temperature was 30°, 18°, 10°, 4°, 0°,  7°,  20°.
What was the average midnight temperature for the week ?
14. What is the combined weight, under water, of a piece
of cork weighing — 2 oz., a stone weighing 3 lbs., a piece of
wood weighing — 1 lb. 3 oz., and a piece of iron weighing
5 lbs. ?
15. A merchant finds that he has cash in bank $575.50,
stock worth $4875, due from customers $1121.50, that he
owes a note and interest amounting to $350.25 and bills
amounting to $827, and that he owns a bond and mortgage
of $1000. Express his capital as the sum of these various
items with their proper signs.
16. A ship sailing up a river would go at the rate of 15
miles an hour if it were not for the current ; the current
averages 5 miles an hour for the first 3 hours of the ship's
progress, and 4 miles an hour for the next 2 hours. How
far has the ship gone at the end of 5 hours ? Express this
as the sum of several algebraic numbers.
2a+ 6  3c
46+ c
 (5 a  6 + c
4a + 46  c
4 + i15=
2+ 5 =
 12  i + 5 = 
lOi
7
30 ELEMENTS OF ALGEBRA.
41. To add several literal expressions, called the addends,
is to find a single expression called the sum, such that what
ever values are substituted for the letters the value of the
sum shall equal the sum of the values of the addends.
E.g., the sum of a, 2 a, 7 a, — 4 a, is 6 a ; for suppose 1 is substi
tuted for a, we have 1+2 + 7 — 4, which is 6 ; and if 2 is substituted
for a we have 2 + 4 + 14 — 8, which is 12,
and so for any other values.
Similarly, the sum of the addends in the
annexed problem is — 4 a + 4 6 — c ; for if
a = 2, 6 = i, c = 5, we have — 10^ + 7 — 7i
= — 11, and similarly for any other values
of a, 6, c.
Since these values are entirely
arbitraiy, they are usually called _ 8 + 2 5=ll
arbitrary values.
42. Hence, it appears that to add like terms is to add the
coefficients, and to add polynomials is to add their like
terms, the literal parts being properly inserted in the sum.
The sum is supposed to be simplified as much as possible. Thus,
the sum of 4 a — 6 and 6 + a is 5 a, not 4 a + a.
EXERCISES. XI.
1. Add ^x'^^2xy\4.y% ^x^ ?>xy 'ly'', ^x'' + xy.
2. Add 6Vm { X, dVm — X — 3y, SVm — 2y,Siiid 8x.
Check the work by letting m = 4:, x = 1, y = 1.
3. Add 2a + Sb — c, — 4c, 7a, —6b\Sc, and — a\b
— c. Check the work by letting a = 1, b = 1, c = 1.
4. Add 17 X — 9y, Sz + 14:X, y — 3x, x — 17 z, and x —
3y \ 4:Z. Check the work by letting x = 1, y = 2, z = 3.
5. Add 16 m ^ 3 n — p, p \ 4: q, — q \ 7 m — 3 n, n — q,
and 3n \ 2 p. Check the work by letting m = l, n = l,
p z=z 2, q = 4:, 01 hy assigning any other arbitrary values.
ADDITION AND SUBTRACTION. 31
II. SUBTRACTION.
43. Subtraction is the operation which has for its object,
given the sum of two expressions and one of them, to find
the other.
The given sum is called the minuend, the given addend is
called the subtrahend, and the addend to be found is called
the difference or the remainder.
That is, the difference is the number which added to the
subtrahend produces the minuend. In other words,
difference + subtrahend = minuend.
E.g., •.•4 + 5 =9, .. 4 = 95;
v4 + (3) =1, .. 4 = 1 (3);
v4 + (5) =1, .. 4=l(5);
...  4 + { 3) =  7, .. _ 4 =  7  ( 3).
These results are illustrated as follows: the difference
between the temperature of 9° and that of 5° is 4°; that
between 1° and 3° (i.e., 1° above and 3° below 0) is 4°;
that between  1° and — 5° {i.e., 1° below and 5° below 0)
is 4° ; that between — 7° and — 3° is — 4°, that is, the mer
cury must fall 4° from  3° to reach  7°.
We may, therefore, think of subtraction as the inverse of
addition, or the process which undoes addition.
Example. What is the remainder after subtracting
3 a2 + 4 a6  5 62 from 4 a^  6 a6 + 2 62 ?
What term added to 3 a2 makes .9 k ». , o 7,9
4 a'' — 5 a6 + 2 6^
4a2? Evidently a2. 3a2 + 4a6562
What term added to 4 ah makes — ; — TnrvWTi
a^ — 9ao + 1 0^
5 ah? Evidently  9 a6 ; for the
addition of — 4 a6 makes 0, and the further addition of — 5 a6 makes
— 5a6.
Similarly, 7 62 is the term v^^hich added to — 5 62 makes 2 62.
.. 4 a2  5 a6 + 2 62  (3 a2 + 4 a6  5 62) = a2  9 a6 + 7 62.
32 ELEMENTS OF ALGEBRA.
Check. Let a = 1, 6 = 2. Then
4a25a6 + 262 4 _ lo + 8 = 2
3a2 + 4ab5b2 3+ 8 20 =9
a2  9 a& + 7 62 1  18 + 28 = 11
Since this is an identity, it is true for any values of a and h. Hence,
the work may be checked by letting a = 1, 6 = 2. The minuend then
becomes 2, and the subtrahend — 9, and the remainder 11, which is
2 (9).
44. Theorem. The subtraction of a negative number should
be interpreted as the addition of its absolute value.
Given a and — b.
To prove that a —{—b) equals a plus the absolute value
of — 5 ; that is,
that a—(^—b)=a\\ — b\oYa + b.
Proof. 1. a — {—b) must be such a number that
a{b) + {b)= a. Def. of subt., § 43
2. Adding b to both members, and remembering
tliat (_ J) + ^ = 0, § 39, 3
and that a \ = a, § 39, 4
we have a — (— ^) = <x + &,
which we were to prove.
Corollary.. *.• a — {—b) = a\b, .' . — (— b) = b. This
is usually expressed by the phrase, Minus a minus is plus.
EXERCISES. XII.
1. From 3a — 4:b \c subtract 2a — 5b — c. Check the
work by letting a = 3, b = 1, c = 2, or by assigning any
other arbitrary values ; also by adding the remainder and
subtrahend.
2. Prom 3a — 5cc + fm subtract 4 a — m. Check by
letting a = 5, x = 1, m = 3, or by assigning any other
A
ADDITION AND SUBTRACTION. 33
arbitrary values ; also by adding the remainder and sub
trahend.
3. From 13x + y — 3z subtract 5 x — 7 1/ + z. Check
as in exs. 1, 2.
4. From Ta"" \3ab  6b^ subtract a'' + 3 ab  2 b\
Check as in exs. 1, 2.
5. What expression added to 2 x^ — 3 x?/ — 15 y^ makes
7x'^3xij + z? Check.
6. Perform the following subtractions, checking each as
in exs. 1, 2.
(a) 3ic37a;2 + 2ic13 (b) l.Sjy^  2p^r + O.Sr^
(c) 2a^3ab{ b  c
V 17 ai 13b {12c
(d) 2a^3a''b\2 a^b^  3 a%^ + 15 ab''  b'
6a^ f 3 a^b^  a%'' jJ^
(e) 18^2^,^ + 4a 3Z>« (f) 2xyy''^3x'
1 a^ 2a + 4^>^ 6^ x''\y''3xy
7. What is the difference between the capital of a man
who has a stock of goods worth $5000, $750 in the bank,
and owes $1000 on a mortgage, and that of one who has
a stock of goods worth $6000, has overdrawn his bank
account $275, and owns a $500 mortgage ?
8. If P = a2 + 2 a6 + 52^ g = 2a2 + aZ» f 6^, and B =
— 4:ab — 7 b^, find the values of the following expressions.
Check in each case by assigning arbitrary values to a and b.
(a) FQ. (b) P  P. (c) QE.
(d) QP. .  ^  .„. _ _
(g) FQE.
(b) FE. (c) QE.
(e) F{QE. {i) F + EQ.
(h) EQF. (i) QFE.
34 ELEMENTS OF ALGEBRA.
45. Detached coefficients. Additions and subtractions may
evidently be performed without the labor of writing down
all of the letters. Since the coefficients of like terms are
added, these coefficients may be detached and added sepa
rately, the coefficients of like terms being placed under one
another. Missing terms are indicated by zeros.
Thus, the second of the following additions is the simpler :
(1) (2) Check.
a2 + 2a6+&2 1 + 2 + I = 4
_3a2 a6+ h^
31+1
= 3
4a23a&362
433
= 2
2a2_2a6 &2
221
=  1
2a2.
2 ah 62.
Since, if the arbitrary value 1 is assigned to each letter, the value
of each term is its numerical coefficient, the check requires merely the
addition of the coeflBcients.
EXERCISES. XIII.
Perform the following operations by using detached
coefficients, checking the results by the above method.
1. Add a% + a%^  4 a^^ 3 a«^>  h\  a%^ + h\ 4 ab^.
2. Add 5 a;* — 2 x^if' + y^, x^y + xy^, x^ — xy^, — x^y + y\
3. Add x^ — x^y + xy^ — y^, 2x^ + 3 x^y — 4 xy^ + y^,
x^ — y^.
4. Add p^ 4 Sp^ + 4j9  6, y _ 2j9 + 1, p3 _ ^^ 3^
+ 2p + 3.
5. From a'' + 2ab + b^ subtract a^  2 ab + b'i
6. From x^ + x^y + xy^ + y^ subtract x^ — x^y + xy^ — y^
7. Given P = x'' + Sx'y + 3xy' + f, Q=  3x'y +
3 xy^ — 3y^,E = x^ — y^, find by using detached coefficients
the values of the following, checking as above :
(a)Pg. (h)QR. (G)RP. (d)QF.
(e)IiQ. (i)FR. (g)F^Q + B. (h)P{.QE.
ADDITION AND SUBTRACTION. 35
III. SYMBOLS OF AGGREGATION.
4^. Symbols of aggregation, preceded by the symbols +
and — , may be removed by considering the principles of
addition and subtraction already learned.
Since a {(b — c) = a { b — c,
and a — (b — c) = a — b \ c,
therefore, a symbol of aggregation preceded by \ may be
neglected; if preceded by — it may be removed by changing
the sig7i of each term within.
E.g., 2a + (3&c + a) = 2a\Sb c + a = Sa \ Sb  c.
2a(3 6c + a) = 2a3& + ca = a36 + c.
For the same reasons, any terms of a polynomial may be
enclosed in a symbol of aggregation preceded by + ; also in
a symbol of aggregation preceded by — provided the sign
of each term, within is changed.
E.g. ,a + & — c + (Z = a + (6 — c + d) = a + & — (c — d).
The word term now takes on a broader meaning than that
given in § 3. E.g., in the expression a — b(G — d), b(c — d)
is often considered as a term. So in general, where no con
fusion will arise, polynomials enclosed in symbols of aggre
gation, with or without coefficients, are often called terms.
E.g., (a — 6)x2 + (a + 6)x I (a^ — h^) may be considered as a tri
nomial.
EXERCISES. XIV.
Kemove the symbols of aggregation in the following :
1. p^ + 2pq + q'(q'p^).
2. a''3b''h(2a^ + 7b''c^).
3. a^  (3 a'^b Jra^b^)b^ + 3 a%.
4. 2x'^3xy + y^{2x'' + 3xyy^).
5. 5m^  (3 m^ f 1)  (4 7/2.^ + m^  3) + (m^ + 1).
36 ELEMENTS OF ALGEBRA.
47. Several symbols of aggregation, one within another,
may be removed by keeping in mind the principles already
mentioned.
The order in which these symbols are removed cannot
affect the result, but the simplest plan will be discovered
by considering the following solution.
Simplify a — \^a \ b — (c — d — e) \ c"], (1) beginning
with the inner symbol, (2) beginning with the outer symbol.
(1) (2)
1. a — [a + h — (c —d — e) + c] 1. a— [a + b — {c —d — e) \ c]
2. =a — [a + 5 — (c — d + e) + c] 2. =a — a — b + (c—d — e)—c
3. =a — [a\h— c + d — e + c] 3. =a — a — h+ c—d — e — c
4. ^a — ab+c — d\e— c 4. =a — a — 6+ c —d + e — c
5. = — b — d \ e 6. = —b —d + e
How many changes of signs were made throughout solution (1) ?
how many in solution (2) ? Hence, which solution is the better ?
From the second step of solution (2) could you have written down
step 5 at once ? Could you have done this from step 2 of solution (1) ?
On this account which is the better solution ?
From the above solution it appears that it is better to
rem,ove the outer parentheses first. A little practice will
enable the student to remove them all at sight if this plan
is followed.
EXERCISES. XV.
Remove the symbols of aggregation in the following
expressions, uniting like terms in each result.
1.  [a^ {2abb'' w") + b'^y
2. 4.a''\5b'' + a[^&a'3a{b''a)\\,
3. a^x — [_ax^ \ a^ — (a^x — a^) + x^~\ — ax"^ + x^.
4. 10 m^ + 5 mn — [6 m^ + ti^ — (2 mn — m^ \ n'^)'] — n^.
5. —(—(—(••• — (— 1) • • •))), an even number of sets
of parentheses ; an odd number of sets.
I
ADDITION AND SUBTRACTION. 37
IV. FUNDAMENTAL LAWS.
48. The following laws have thus far been assumed :
I. That a + b = b \ a, a and b being positive or negative
integers, just as in arithmetic 3 + 4 = 4 + 3. This is
called the Commutative Law of Addition, because the order
of the addends is changed (Latin com, intensive, + mutare,
to change).
II. That a\b\c^a{(b\c), the letters represent
ing positive or negative integers, or both, just as in arith
metic 3 + 4 + 5 = 3 + 9. This is called the Associative
Law of Addition, because b and c are associated in a group.
III. That ab = ba, a and b being positive integers, just
as in arithmetic 2 • 3 = 3 • 2. This is called the Commuta
tive Law of Multiplication.
That these laws are valid for the kinds of numbers indi
cated will now be proved, although the proof may be
omitted by beginners if desired.
49. I. The Commutative Law of Addition".
1. If 3 marbles lie on a table, and 4 more are placed with
them, the result is indicated by the symbols 3+4.
2. If the original 3 marbles be removed, 4 will remain ;
and if the 3 be then replaced, the result will be indicated
by the symbols 4 + 3.
3. But the number of marbles has not been changed.
..3 + 4 = 4 + 3.
4. But this proof is independent of the particular nmn
bers 3 and 4, and hence, a and b being any positive integers,
a \ b ^^ b { a.
5. The proof is evidently substantially the same for sev
eral groups. Hence,
a + 6 + c + " = a + c + 6H — • = b { c >r a \ , etc.
38 ELEMENTS OF ALGEBRA.
6. And since, if some of the terms are negative, we deal
with their absolute values, adding or subtracting as indi
cated, and prefix the proper sign to the result, therefore
the above proof is sufficiently general.
I.e., a + b — c = a — c + b, because in any case we are to take the
difference between the absolute values of a + 6 and c, and prefix the
proper sign.
50. II. The Associative Law of Addition.
To prove that a \ b + c = a \ (b + c), the letters repre
senting positive or negative integers, or both.
1. '.•c + b + a = (c + b) + a. Def., § 40
2. = a+(c + b). ■ Com. law, § 49
3. .. a + b + G= a +(b \c). Com. law, § 49
The proof is evidently similar, however many terms are
involved or however the grouping is made.
51. III. The Commutative Law of Multiplication.
To prove that ab = ba, the letters representing only posi
tive integers.
*****... a in a row
b rows.
1. Suppose a collection of objects arranged in b rows, a
in a row, or, what is the same thing, in a columns, 6 in a
column.
2. •.' there are b in one colmnn, in a columns there are
ab objects.
3. *.• there are a in one row, in b rows there are ba objects.
4. But the collection being the same, ab = ba.
ADDITION AND SUBTRACTION. 39
REVIEW EXERCISES. XVI.
1. Distinguish between an equation and an identity, illus
trating each.
2. Show that 2 — 3 = 3 — 2, and state a proposition
covering such cases.
3. What is the etymological meaning of coefficient ? of
subtraction ? of literal ? of minuend ?
4. Why is not the arithmetic definition of sum sufficient
for algebra ? What do you mean by sum in algebra ?
5. What is the advantage in using detached coefficients
in addition ? Make up an example illustrating this.
6. What is the number which added to — 5 equals ?
equals 2 ? Hence, what is the difference — (— 5) ?
2(5)?
7. Eemove the symbols of aggregation in the following
expressions. By beginning at the outside you can usually
write the result at sight, except for simplifying.
(a) [x+{x + ^j {xy) x'].
(b) a — \a —\_a — a — {b — a)~\\.
(c) Z a [b + c  (a  b) { a']— h.
(d) x''[2x^ + y'(x^  2/2  ^?^=^0 + 2/'] 2/'.
8. Enclose any two terms (after the first) in parenthe
ses:
(a) «;2_^2_2c2_3^»c. (b) Zp''4.pq2q'' + r\
(c) 4 ic^  2 x2  7 ic + 1. (d) m'^ — m^ + m^  m + 1.
9. What is meant by the Commutative Law of Addi
tion ? Have you proved it for all kinds of numbers ? If
not, name a kind for which it has not yet been proved by
you. Similarly for the Associative Law of Addition.
CHAPTER III.
MULTIPLICATION.
I. DEFINITIONS AND FUNDAMENTAL LAWS.
52. Multiplication originally had reference to positive
integers and was a short form of addition. It was, for
this case, defined as the operation of taking a number
called the multiplicand as many times as an addend as there
are units in an abstract number called the multiplier, the
result being called the product.
E.g., in this limited sense, to multiply .$2 by 3 is to take $2 3 times,
thus, 3 X $2 = $2 + $2 + $2 = $6.
But as mathematics progressed it became necessary to
multiply by simple fractions, and hence to enlarge the defi
nition to include this case.
By the primitive meaning of the word times it is impossible to take
2 X $2
$2 f o/ a time. But the product of $2 by f may be defined as
o
So the product of c by  may be defined as the product
of a and c, divided by b, c being either integral or fractional.
As mathematics further progressed it became necessary
to multiply by negative numbers, and hence to enlarge the
definition to include this case. The natural definition will
appear from a simple illustration.
Suppose 5 men move into a town, each paying $ 1 a week in taxes.
They are worth 5 x $l = $5a week to the town.
Suppose 5 such men move out. This may be represented by saying
that the town gains — 5 men, or, in money, — $5.
40
MULTIPLICATION. 41
Suppose 5 vagrants move in, each being a charge of $1 a week.
They are worth 5x(— $1) = — fSa week to the town.
Suppose 5 such vagrants move out. This may be represented by
saying that the town gains — 5 vagrants, or, in money, $ 5.
Hence, it is reasonable to say that
$ 1 multiplied by 5 = $ 5, for the first case ;
$1 " " —5 =—$5, " second "
.fl " " 5= 15, " third "
$1 " " 5= $5, " fourth "
53. From such considerations multiplication by a negative
number is defined as multiplication by the absolute value of
the multiplier, the sign of the product being changed.
E.g., allowing the word times to indicate multiplication in general,
 2 times 3 means  ( 1  2 I x 3), or  (2 x 3), or  6 ;
_2 " 3 " _ [I 21 X (3)] " [2x(3)] " (6),
or + 6.
54. General definition of multiplication. The above partial
definitions may now be put into one general definition :
To multiply a number (the multiplicand) by an abstract
numher (the multiplier) is to do to the former what is done
to unity to obtain the latter.
The result of multiplication is called the product, and the
product of two abstract numbers is called a multiple of either.
E.g.., consider the meaning of 3 x $2. Since 3 = 1 + 1+1, there
fore, 3 X $2 means $2 + $2 + $2 = $6.
Consider also f x f . Since  = (1 + 1) 4 3, therefore, f x f means
(f + f)  3, or V ^ 3, or if
Consider also (—2) x (—3). Since — 2 = — (1 + 1), therefore,
( 2) X ( 3) means  [(  3) + ( 3)], or  ( 6), or 6.
55. The expression a • is defined to mean 0.
This is the natural definition, because 2x0 must mean + 0.
And since it will be shown that the order of factors
can generally be changed without altering the product, the
product 0Si is defined to be the same as a • 0, or 0.
42 ELEMENTS OF ALGEBRA.
56. The product of three abstract numbers is defined to be
the product of the second and third multiplied by the fiist.
J.e., abc means a{bc), the product of 6 and c multiplied by a.
The product of four or more abstract numbers may be understood
from the above definition for three, ^.g.i abed means cd multiplied
by b, and that product by a.
57. Law of signs. From the definition it appears that
like signs produce plus, and unlike signs minus.
I.e., + X + = +
+ X  = 
 X 4 = 
 X  = +
58. Reading of products. As already stated, the original
meaning of the word times referred to positive integers.
The expressions f times, ^ of a time, and — 2 times are
meaningless in the original sense of the word. But with
the extension of the definition of multiplication has come
an extension of the meaning of the word times, so that it
is now generally used for all products, as in § 53.
Thus, the expression 2^ times as much is generally used, although it
is impossible to pick up a book 2^ times. So (— 2) x (— 3) is read,
" minus 2 times minus 3," although we cannot look out of a window
— 2 times.
As already stated, the word into is sometimes used in
algebra to indicate the product of two or more factors,
(— a) (— b) being read " — a into — Z»."
The parentheses about negative factors are omitted when
no misunderstanding is probable. Thus, (— a) • (— b) may
be written — a X — b, or even —a — b. But — a^ and
(— ay are not the same, the former meaning — aa and the
latter — a — a, oi 4 a^.
MULTIPLICATION.
43
59. The Associative and Commutative Laws of Multiplica
tion. Before we are able to proceed with certainty in mul
tiplication, it is necessary to show that we can change the
order and grouping of the factors to suit our convenience.
For example, to prove that a6c, which by definition means a(6c),
= {ab)c = {ac)b = b{ac) • • • .
Proof. 1. Suppose this solid to be composed of inch
cubes, and to have the dimensions 4 in., 5 in., 6 in.
2. Then, since there are 4 cubes in
the row OA, and there are 5 such
rows in the layer CA, there are (5 • 4)
cubes in that layer. And since there
are 6 such layers, there are 6 • (5 • 4)
cubes in all.
3. Similarly, since there are 6 in
column OB, and there are 4 such
columns in layer BA, there are (4 • 6)
cubes in that layer. And since there
are 5 such layers, there are 5 • (4 • 6) cubes in all.
4. Similarly, there are 4 • (5 • 6) cubes.
5. But the total number is the same,
..6 (5 4)= 5 (4. 6) = 4. (5. 6).
6. And since the proof is independent of the numbers,
.' . a ■ (b • c) = b • (g ' a) ^ c • (b • a) = (a • b) • c = • • • .
7. By taking d such solids it could be proved that
a (b •cd) = (ab) ■ (cd) = (a'bcy'd = b a (dc)= • • • ,
and similarly for any number of letters.
8. And since in multiplications involving negative num
bers we proceed as if the numbers were positive, prefixing
the proper sign, therefore the proof is general for all
integers.
44 ELEMENTS OF ALGEBRA.
EXERCISES. XVII.
Perform the multiplications indicated :
1.
2.7.
2.
4.3.
3.
72.^.3.
4.
h'hh
5.
(2)^. (3)1
6.
(iy^.(2y.
7.
4.5.3.2.1.1.
8.
l._2.3.4.5.6.
9.
1.2. 3. 4.
10.
5.3.1._1._3._5.
1.
l.(_ 2)2.33. (4)^
12.
(_ 1)100. (_l)99.^_ 2)5.
3.
4. 3. 2. 1.0.1.2.
3
4.
60. The index law. Since a^ = aa, and a^ = aaa, there
fore a^ ■ a^ = aa • aaa = a^. Similarly, if m and n are posi
tive integers,
a™ = aaa ■ ■ • to m factors,
and a" = aaa ... to n "
a"* . a** = aaaa • ■ • to m ^ n "
This is known as the index law of multiplication.
Hence, 2 a^b^^c^ ■ 5 a^ftV = 10 a^b^c^.
The cases in which m and n are negative, zero, or fractional are
considered later.
EXERCISES. XVIII.
Perform the multiplications indicated :
1. a^.(^ay.
2. 25ab^G^d^'2a^bh''d.
3. — a ' — a^ ■ — a^ ■ — a^ ' — a^.
4. a(ay(ay(~ay(ay.
5. x'^ x^ . X.
6. £c"*?/" • ic"?/'" . x^y"^.
7. x^y^z^ . £cy2^3 • xHj^z^ • xyz.
8. a:^'^ . a:;^ . x^  y^ ■ y^ • y'^ • z^  z^ • z.
MULTIPLICATION. 45
II. MULTIPLICATION OF A POLYNOMIAL BY A MONOMIAL.
61. I. When the monomial is a positive integer, as in the
case of. a(b — c).
1. '.' a = l{l\l\toa terms,
2. .'. a(b — c) = (b — c)H(6 — c)\(b — c)\ to a terms,
Def. mult. § 54
3. = b ^ b \ b { •■• to a terms,
— c — G — c — "toa terms, § 49
4. = ab — ac. Def. mult. § 54
E.g., 2{xiy) = {x + y) + {x + y) = 2x + 2y.
II. When the monomial is a positive fraction, as in the
0
X l + lHl + toa! terms
case of  (b — c).
1. •.•
y
, ic _ (b — c) \(b  g)\ ■ ■■ to X terms
Zi. • . (o c ) ^^ ■ >
^ ^ Def. mult. § 54
xb — XG . _
6. = } as m I
y
4. = } because xb ?/ths minus xc yths
y y ./ J
is the same as (xb — xc) ^/ths.
III. When the monomial is negative, and either integral
or fractional, as in the case of (— m) (b — c).
1. '.' — m = m • 1, preceded by the sign — ,
2. .'. (— m)(b — c) = m (b — c) preceded by the sign — ,
Def. mult. § 53
3. = (mb — mc) preceded by the sign — ,
I and II
4. = — m^ + mc. § 46
46 ELEMENTS OF ALGEBRA.
62. From the results of these three cases it appears that :
To multiply a polynomial by a monomial is to multiply
each term, of the polynomial by the monomial and to add
the products.
Since the multiplier is distributed among the terms of
the multiplicand, this statement is known as the distribu
tive law of multiplication.
E.g., 3 a2 (a*  6) = 3 ««  3 oPh. This can be checked by letting
a = 1, 6 = 2. Then 3 a2 («! _ &) = 3 . _ 1 ^ _ 3^ and 3 a^  3 a26 =
3 _ 6 =  8.
EXERCISES. XIX.
Perform the following multiplications, checking the re
sults by assigning arbitrary values to the letters :
1. a''{a^\b^G').
2. 5 m^xy {xz^ — S z^x — 4).
3. 7x''y(Sxy^ + 2xy).
4. 25ab^c^d\2a'b^i2c''d).
5. 5a[3a + 2(a2)].
6.7 m^n^ (2 m — 3 ?i — 4 mn + 6 m^n).
III. MULTIPLICATION OF A POLYNOMIAL BY A
POLYNOMIAL.
63. Eequired the product of (a j 6) (c + ^).
1. Let m = (a + b).
2. Then m{G + d) = mG \ md, § 61
3. = (a + b)c +{a \ b)d, • .• m = {a + b)
4. = ac + be \ ad + bd. §§51,61
From this it appears that to multiply one polynomial by
another is to multiply each term of the first by each term of
the second and to add the products.
This is the general form of the distributive law of multiplication.
MULTIPLICATION. 47
The following example illustrates the process :
x^ \2xy +y^
^ + y
Product by x, x^ + 2 x^y + xy^
Product by y, x^y + 2 xy^ + y^
Sum of products, cc^ + 3 x^y + 3 xy^ + y^
.'. (X + 7j) (x' + 2xy + 7/2) = x' + 3x'y + 3xy'\ y\
Check. Let x = 1, y = 1. Then
1+2+1=4
1 + 1 = 2
1 + 3 + 3 + 1 = 8, or 2. 4.
Since the identity liolds true for any values of x and ?/, it holds true
\l X = y = \^ as in the above check. It is evident, however, that the
value 1 does not check tlie exponents. Where there is any doubt as
to these, other values must be substituted.
EXERCISES. XX.
Perform the following multiplications, checking the re
sults by assigning arbitrary values to the letters.
1. {a + h){x + y). 2. (x + y){xy).
3. {x^  y^) {x + y). 4. {p^ + q'') {x^  3 /).
7. (a3 + a2 + a + l)(al).
8. {bx^x + l){3x^x2).
9. {2x + 3y z){2x3y + z).
10. (ic* + £c3 + x^ + a; + 1) (x — 1).
11. (x + y){x^ + 3xhj + 3xy'' + y^).
12. (3 ^2 _ 2 a) (5 a«  2 ^2 _ 3 ^ _p 4)_
13. {a b) (a' + a^b + a^b"" + a^b^ + a^b'' + a^'b^ + ab^ + ^^).
48 ELEMENTS OF ALGEBRA.
64. A polynomial is said to be arranged according to the
powers of some letter when the exponents of that letter in
the successive terms either increase or decrease continually.
In the former case the polynomial is said to be arranged
according to ascending powers, in the latter according to
descending poivers of the letter.
E.g.^ x^ + 3x3 + a;2 4 1 is arranged according to descending powers
of X. If it is desired to have all of the powers represented, it is written
x5 + 0x4 + 3x3+x2 + 0x + 1.
The polynomial x^ — 3 x^y + 3 xy^ — y^ is arranged according to
descending powers of x and ascending powers of y.
There is evidently an advantage in arranging both multi
plicand and multiplier according to the powers of some
letter, as shown by the following example :
Not Abkanged. Arranged,
2/2 + x2 + 2 x?/ x2 + 2 x?/ + 2/^
x + y x\y
xyi + x3 4 2 x2?/ x3 + 2 x^y + xy^
+y3 + x2y + 2x?/2 x2y + 2 X?/2 t y3
X2/2 + X^ + 2 X^Z/ + y3 ^ X2y + 2 X?/2 X3 + 3 X2?/ + 3 X?/2 f 2/3
= x3 + 3x2?/ + 3a;y2 _!_ yz Check. Let x = 1, ?/ = 1. Then
24 = 8.
The method at the right is evidently much simpler.
65. It is also evident that the product of the terms of
highest degree in any letter in the factors is the term of
highest degree in that letter in the product. Also that the
product of the terms of lowest degree in any letter in the
factors is the term of lowest degree in that letter in the
product.
Hence, if the factors are both arranged according to the descending
(or ascending) powers of some letter, the first term of the product will
be the product of the first terms, and the last term will be the product
of the last terms.
MULTIPLICATION. 49
EXERCISES. XXI.
Perform the following multiplications, checking the re'
suits by assigning arbitrary values to the letters :
1. x^ — if by x^ + if.
2. a?x + x^a by x^a — a^x.
3. x^y'^ — x^ — y^ by y — x.
4. X \ y \ z hj x \ y — z.
5. 1  a2 + a*  a^ by 1 + a'^
6. x'^ + xhj \ y^ by cc^ — 3 a? + ?/.
7. i 2/'  ^ 2/^ + i ^' by ^ 7/  ^ ;?;.
8. xyz — x^ — y"^ — z^ by a? + ?/ + ^.
9. p'^ — 2pq + ^^ by ^^ + 2^2' + q"^. •
10.  ^2 + 3 a^* + 52 by ^ahh'' + a\
11. «,^ — a* + a^ — a^ ^ (X — 1 by a + 1.
12. ic^ — 3 ic^?/ \ 3 o:;?/^ — 2/^ by x^ — 2xy \ y^.
13. cc?/ + 2 cc^ — 3 2/^ + ic^ + ?/^ + 4 ^^ by x — y — 2 z.
66. Detached coefficients may be employed in multiplica
tion whenever it is apparent what the literal part of the
product will be.
E.g.^ in multiplying x^ \ pz ■\ q hy x^ — x \ pq the coefficients
cannot be detached to advantage.
But in multiplying x^ + 2 xy + 2/^ hy x + 8 y, it is apparent that the
exponents of x decrease by 1 while those of y increase by 1 in each
factor, and that this law will also hold in the product. Hence, when
the coefficients are known the product is known also, and the multi
plication may be performed as follows :
Check.
1+2+1 =4
1 +3 =4
1 + 2 + 1
3 + 6 + 3
1 + 5 + 7 + 3 =.16
.. (X + Zy) (x2 + 2x2/ + 2/^) = x3 + 5x2y + 7x?/2 + 3y3.
50 ELEMENTS OE ALGEBRA.
67. If the coefficient of the first term of the multiplier is
1, as is frequently the case, the work can be materially
simplified by the following arrangement:
The problem is the same as the preceding one.
1
+ 3
1+2 + 1 Check. 4 • 4 = 16.
3 + 6 + 3
1 + 5 + 7 + 3 x3 + 5x22/ + 7ay2 + 3i
68. In case any powers are lacking in the arrangement of
the polynomial, zeros should be inserted to represent the
coefficients of the missing terms.
E.g.., to multiply x"^ {■ xy \ ip by x^ + i/^, either of the following
arrangements may be used :
1 + 1 + 1 Check. 23 = 6.
1+1+1 1
1+0+1 +0
1+1+1 +1
1 + 1 + 1
1+1+2+1+1
1 + 1 + 1
1 + 1+2 + 1 + 1 X* + x^?/ + 2 x2?/2 + x?/3 + 2^
EXERCISES. XXII.
Perform the multiplications indicated in exs. 113, by
detached coefficients, checking the results as usual.
P = cc^ — x^y ■\ xy^ — 2/^ Q ^ ^ — V) R ^ x'^ — xy { y^.
1. PQ, 2. PR. 3. QR. 4. P\
5. Q'R. 6. R\ 7. QR\ 8. Q^R\
9. {x + yf. 10. {xyy. 11. {x { yf.
12. (xijy. 13. (x + y)(xy).
14. Verify the following identities, (1) by substituting
arbitrary values, (2) by expanding both sides of the iden
tity, using detached coefficients or not as seems best :
(a) (x + y + zy (x^ + 2/' + ^') = 3(a5+7/) (y + z) (z + x).
(b) (x\yy\(y + zy+(z + xy(x' + y' + z')=(x+y+zy.
MULTIPLICATION.
51
IV. SPECIAL PRODUCTS FREQUENTLY MET.
69. In exs. 913 on p. 50 five products were found which
are so frequently used as to require memorizing. They are
as follows :
1. (cc + 2/)^ = ^^ + 2 ic?/ 4 ?/^ Hence, the square of the
sum of two numbers equals the sum of
their squares plus twice their product.
This theorem may be illustrated by a figure.
Here the square ^.0 = (x + ?/)2, the square
AP = x^, the square PC = y^, and there are two
rectangles equal to EF = xy. And
:• AC = AP + 2EF+PC,
.. (a; + 2/)2 = x2+2xy + ?/2.
n
G
H
y
xy
y
r
X
p y
X
x"
X
X
xy
y
B
xy
^<
2. (x — yy = x'^ — 2xy\ y^. Hence, the square of the
difference of two numbers equals the
sum of their squares minus twice their
product.
In the figure, AP^ = x% BH = y^, AC =
{X  y)2, and DP= CH = xy. And
:• AC = AP2DP\BH,
.. {X y)^^x^ 2xy + y^.
Expressions of the form x \ y, x — y, are
called conjugates of each other.
(xy)'
y.
H
3. (x + y) {x — y) = x^ — y\ Hence, the product of the
two conjugate binomials equals the differ
ence of their squares (i.e., the square of
the minuend minus the square of the sub
trahend).
In the figure, AG = x^, AF = y^, and GG \ FB
= x{xy) +y{xy) = {x + y){x y). And
•.• GG + FB = AC AF,
.. (.T + 2/) (X  ?/) = X2  2/2.
xy ^^"y^
F
y y^ y
y
52 ELEMENTS OF ALGEBRA.
4. (x \ yy = x^ \ 3 x^y + 3 xy'^ + y^. Hence, the cube of
the sum of two numbers equals the sum of their cubes plus
three times the square of the first into the second plus three
times the square of the second into the first.
5. (x — yy = x^ — 3 x^y + dxy"^ — y^. (State the theorem.)
EXERCISES. XXIII.
By the help of the theorems of § 69 expand the expres
sions in exs. 118.
1. 42 X 38, i.e., (40 + 2) (40  2).
2. 23 X 17. 3. 95 X 85.
4. (^2 + 3)2. 5. (a^2)2.
6. (2^ + 1)'. 7. (2cc2l)3.
8. (2 x""  yf. ■ 9. [a(b + c)^.
10. (2cc2 + l)(2cc2_l). 11. (a2 + 3)(a2_3).
12. (a]bab)(a + b^ ab). 13. [(a + b) {a  b)J.
14. {a''\2ab{b'){a''2ab + b''). 15. (x'' + y') (x^  y^) .
16. 42^. 17. 4951. 18. 492.
19. Verify the following identities :
(a) (^2 + ^,2 _^ c2 + ^2) (^2 __ ^2 _^ ^2 _^ ^2^ _ ^^^ + &ic +
c?/ + cZ«)^ = {ax — bwy + {cz — dyy + («?/ — cw)^ + {dx —
bzy + {az  dwy + (6?/  cxy.
(b) (a? + 2/)^ — £c^ — ?/^ = 3 xy {x + ?/) (a?^ \ xy \ yy.
(c) (x + ij)^ x^ i/ = 5 xyix + ^) {f + xy + y^.
(d) (a; + yy x' y' = lxy{x + y) (rr^ + a:^/ + y^.
20. Expand the following expressions by the help of the
theorems of § 69, checking by arbitrary values :
(a) {x^ + yy. (b) {x' + yy.
MULTIPLICATION. 63
V. INVOLUTION.
70. The product of several equal factors is called a power
of one of them (§8).
The broader meaniug of the word power is discussed later (§ 130).
At present the term will be restricted to positive integral power.
71. The operation of finding a power of a number or of
an algebraic expression is called involution.
The student has already proved one important proposition in in
volution, viz., that a"* • a'* = a"* + »*, where the exponents are positive
integers (§ 60).
He has also learned how to raise the binomial x ±y to the second
and third powers {§ 69).
It now becomes necessary to consider certain other theorems.
72. Notation. If m and n are positive integers,
(aJ^y means a^ • a"^ • a™' ■ • • to n factors, each a^ ;
«"•" " aaa • to m" " " a.
E.g., (a3)2 means a^ • a^ = a^ + s = aS .
of' " a • a • a • • • to 32 factors, = a^ ;
of " a a a ■■ to 2^ " =aP.
73. d^ has already been defined to equal a.
74. The expression a^, a being either positive or negative,
is defined to equal 1, for reasons hereafter set forth (§ 214).
75. Theorem. The nth power of the m.th power of an
algebraic expression equals the math power of the expression.
Given an algebraic expression a, and m and n positive
integers.
To prove that (a/^y = a""^.
Proof. 1. (a™)" means a"" • a"" ■ a"" • • ■ to n factors, each a"*,
2_ = Q^m + m + m+ . • .to7i terms, each m ff QQ
3. = a'"".
64 ELEMENTS OF ALGEBRA.
76. Theorem. The mth power of the product of several
algebraic expressions equals the product of the jnth powers
of the expressions.
Given the expressions a, b, c,  •, and m an integer.
To prove that (abc •••)"• = a"'b"'c"' • • • .
Proof. 1. (abc • • •)'" means (abc • • •) • (abc • • •) • (abc •••)•••,
to m groups, each (abc  ■ •)
2. = (aaa • • • to m factors) • (bbb • • • to m factors) • (ccc
■ •• to m factors) • • • § 59
3. = a'^b'^c"*.  Def. of power
77. Law of signs. Since
j a \ a = \ a^,
and — a ■ — a = \ a"^,
but — a — a — a ^i — a^,
it is easily seen that
1. Powers of positive expressions are positive ;
2. Even powers of negative expressions are positive ;
3. Odd powers of negative expressions are negative.
EXERCISES. XX JV.
Express without parentheses exs. 112.
1. (a'^x'^y. 2. (aH"^)"". 3. (a%^c^d^y.
4. (a'^b^'cy. 5. (ab'^cy. 6. (a^^^V)*.
7. (ay, (ay. 8. (a^, (a^. 9. ( a'«^.«)2'»«.
10. (^a^'^y^. 11. [(ayy. 12. (a^^b^cy.
13. Prove that (a'^y ^ (a'^y.
14. Is it true that a""" = a"'" ? Proof.
15. Also that (a'^b'^y = (a"Z>'»)'«« ? Proof.
MULTIPLICATION. 55
78. Powers of polynomials. A polynomial can be raised
to any power by ordinary multiplication.
But in raising to the 4tli power it is easier to square and
then to square again, since (cl'^Y = a^.
E.g., to expand {x — 2yy.
1. (x2y)2 = x24x2/ + 4y2. §69
2. (x2  4 X2/ + 4 2/2)2 = [(x2 _ 4 xy) + 4 ?y2]2 § 46
3. = (x2 _ 4 x?/)2 + 2 (x2  4 x?/) • 4 2/2 + 16 y^
4. =x4  Sx^y + 16x2?/2 + 8x2y2 _ S2xy^ + IB?/*
5. =x''8x32/ + 24x2?/232x?/3 + 16y4.
Check. ( 1)4 = 1  8 + 24  32 + 16 = 1.
Similarly, to raise to the 6th power first cube and then
square, since (a^y = a^.
But to raise to the 5th, 7th, 11th, or other powers of
prime degree, multiply out by detached coefficients.
EXERCISES. XXV.
Expand the expressions in exs. 120.
1. (20 + 1)2. 2. (x'3 7/y.
3. (x + 3yy. 4. (2x7yy.
5. (x"" + 2/")*. 6. (a + b + cy.
7. (^x^yy. 8. (x3yy.
9. (2x^3yy. 10. (a h 2 b hey.
11. (abcy. 12. (a^h2ab + by.
15. (a^ob'^ay. 16. (x^ + xY + y^.
17. (ia + 2b + cy. 18. (31 m^  20 7^2)2.
19. (a  2 ^ 4 3 c)2. 20. (a — b + c dy.
56 ELEMENTS OF ALGEBRA.
79. The Binomial Theorem. It frequently becomes neces
sary to raise binomials to various powers. There is a simple
law for effecting this, known as the Binomial Theorem.
The student will discover most of this law in answering the following
questions :
Expand (a + by, (a + hf, {a + by, {a + by.
(a) How does the number of terms in each expansion
compare with the degree of the binomial ?
(b) How do the exponents of a change in the successive
terms ?
(c) How do the exponents of b change in the successive
terms ?
(d) In each case, what is the first coefficient ? How does
the second coefficient compare with the exponent of the
binomial ?
(e) In the case of the 4th power does the third coefficient
equal — r ? In the 5th power is it —^ ? What will it
probably be in the 6th power ? in the 7th ? in the 7ith ?
(f ) In the case of the 4th power does the fourth coeffi
4 • 3 • 2 5 • 4 • 3
cient equal ? In the 5th power is it ^ ? What
will it probably be in the 6th power ? in the 7th ? in the
nth?
(g) In the case of the 4th power does the fifth coefficient
equal „ ' ? In the 5th power is it ? What
will it probably be in the 6th power ? in the 7th ? in the
nth. ?
( h) In expanding (a + by, what will be the coefficient of
a% ? of a^b^ ? (The student should now be able to answer
without actual multiplication.)
MULTIPLICATION. . 57
80. Theorem. If the binomial a \ b is raised to the nth.
power, 71, integral and positive, the result is expressed by
the formula
(a + by = a" + na^''^b + ^^^^^~ ^ a"^b^
where :
2
23
^ .(. !)(. 2) ^^„_3^3^
1. The number of terms in the second member ^s n + 1 ;
2. The exponents of a decrease from n to 0, while those of
b increase from to n;
3. The first coefficient is 1, the second is n, and any other
is formed by multiplying the coefficient of the preceding
term by the exponent of a in that term and dividing by 1
more than the exponent of b.
The proof of this theorem, which has already been found inductively
on p. 56, may be taken now or it may be postponed until later in the
course. The proof is given in Appendix I.
81. Pascal's Triangle. The coefficients of the various
powers of the binomial f+n are easily found by a simple
arrangement known as PascaVs Triangle, from the famous
mathematician who made some study of its properties.
Coefficients for 1st power 1 1
" 2d
2 1
" 3d
3 3 1
" . 4th
4 6 4 1
" 5th
5 10 10 6
" 6th
6 15 20 15
Each number is easily seen to be the sum of the number above and
the number to the left of the latter.
Write down the coefficients for the 7th, 8th, 9th, and 10th powers,
thus enlarging Pascal's triangle.
For note on Pascal, see the Table of Biography.
58 ElyEMENTS OF ALGEBRA,
82. Various powers of f + n. These are needed in the
extraction of roots (§§ 128133) and should be verified by
the student.
(/ + nf ^fJ^^fH + Zfu" + n\
(/ + ny =r + 4.fn + e/^Ti^ + 4>^ + n\
(/+ ny = (Expand it.)
(f^ny= " "
{f+ny= " «
Illustrative problems. 1. Expand (2 a — 3 b'^y.
1. (2a  362)3 ^ (2a)3 + 3 (2a)2( 362) + 3 (2a) ( 352)2 + (_ 352)3
2. = 8 a3  36 a262 + 54 a64  27 6^.
Check. (l)3.= 836 + 6427=: 1.
In cases like this it is better to indicate the work in the first step
and then simplify.
2. Expand (  2/ + ^')''
2. =(2/)%2(2/);22+(2;2)2
3. ■ = xy + ?/2 + X2;2  2 2/z2 + z*.
4
EXEBCISES. XXVI.
Expand the following expressions :
1. (x + yy. 2. (1ay.
3. (x^ — yy. 4. (x — yy\
5. (a2Z')2. 6. (2CC + 2/')'
7. {x^yZy^y. 8. (a; + 3/  «)2.
MULTIPLICATION. 59
11. {ahcy. 12. (2a ^ly.
13. {ab + cy. ^ 14. {^x + 2yy.
15. {^^''iy'^Y' 16. (3a2_2a6 + 62)8.
"■(0' ■ ■•(9'
REVIEW EXERCISES. XXVII.
1. Solve the equation 184 — cc^ = 40. Check.
2. What is the etymological meaning of multiply? of
abstract ? of ascending ? of descending ? of coTnmutative ?
3. Show that the arithmetic definition of multiplication
is not broad enough for algebra. Explain the de^nition
in § 54.
4. What is the broader meaning of the word times in
algebra ? Illustrate.
5. What is the Index Law of multiplication ? Has it
been proved by you for all kinds of exponents ? If not,
for what kind ? Prove it.
6. What is meant by the Distributive Law of multipli
cation ? Prove the law.
7. Make up an example illustrating the advantage of
arranging the terms according to the powers of some letter
in multiplication.
8. What are the advantages in using detached coefficients
in multiplication ? Illustrate by solving a problem.
CHAPTER IV.
DIVISION.
I. DEFINITIONS AND LAWS.
83. Division is the operation by which, having the product
of two expressions and one of them (not zero) given, the
other is found.
Thus, 6 is the product of 2 and 3 ; given 6 and 2, 3 can be found.
The given product is called the dividend, the given expres
sion is called the divisor, and the required expression is
called the quotient.
84. Since = a • (§ 55), it follows that  should be
defined to mean 0.
85. Law of signs. Since
\ a ■ \ b ^ + ab,
\ a • — b ^ — ab,
— a ■ } b = — ab,
and — a — b = \ ab,
it therefore follows, from the definition of division, that
■i ab T + a = { b,
— ab i { a = — b,
— ab . a = + 5,
and \ ab . a = — b.
That is, like signs in dividend and divisor produce +,
and unlike signs — , in the quotient.
DIVISION. • 61
86. Index law. Since o^"*~'' • a" = o^"», by the index law of
multiplication (§ 60), therefore, — = a™""", by the definition
of division.
Hence, 10 a%H^ f 5 oFbH'^ = 2 a^¥c^.
The above proof is based on the supposition that m>n, and that
both are positive integers. The cases in which m and n are zero,
negative, and fractional, and in which m<n, are considered later.
EXERCISES. XXVIII.
Perform the following divisions :
1.  125   25. 2. 80   16.
3 a%e 25 a'bc'
3. — 4. 3T7
— oc — o a^OG
 10 xy^z^ 49 a;^yQ^V
^* 5 2/V ' ^' Iwz^
56a^'b^'c 27 (a b)
— 8 a^%G ' a — b
II. DIVISION OF A POLYNOMIAL BY A MONOMIAL.
87. 1. •.' ma \ mb ^ mo = m (a \ b + c). § 61
„ ma { mb \ mc , , , r. .c ^ j • ■ •
2. .*. = a \ b + s. Del oi division
m
3. Hence, to divide a polynomyial by a monomial is to
divide each term, of the polynoinial by the monoinial and to
add the quotients.
Thus, ^^!^ ^^^^ = 2ah. Check. Let a = 2, 6 = 3. Then
 17 a62
17 ■ 2 . 27  34 • 4 ■ 9 „ „ „  306 ,
= 2 • 2 — 3, or = 1.
 17 • 2 • 9  306
62 * ELEMENTS OF ALGEBRA.
EXERCISES. XXIX.
Perform the divisions indicated; check by assigning
arbitrary values.
27 x^y  27 V 121m%»110mV
^ 21xy ' ' llmhi'
x^ + Sx^y + 3xh/ + xy^  S a'b  12 a'P + 9 a^b^
^ X ' ^ 3a'b
a^^3a^ + 3 a^b''  7 ab^
5. ~ ~*
— a
34 a%^c  17 ab'^c^ + 51 a^b^c
^' liable
200 xY  "^^ ^V + 1^5 ^V .
'^' 25icV
^'  13 ic'"^/'
5j9^  15 j9^g + lOj^'g'  20^^
2(a + bY3(a + by + 2(a + bf
(a + by
48 a;^V  36 x^hj + 72 x^^  108 xy
^^' 12 xy
(x' {2xy + yy +{x^ + 2xy^ y'^y
(x''Jr2xy + y')
(2xiy {■^{2xiy{2xiy
^^' (2xiy
 52 a^%'  78 a'b^'  26 a^'^b^^  130 a'b^
^^'  26 a%^
DIVISION. 63
III. DIVISION OF A POLYNOMIAL BY A POLYNOMIAL.
88. As a preliminary to the explanation of this foriu of
division it is necessary to observe the following important
points :
1. In division, if dividend, divisor, and quotient are
arranged according to the descending powers of some letter,
then the first term of the quotient is the quotient of the
first terms of dividend and divisor.
That is, in dividing x^ + Sx^y + 3 xy^ + y^hy x + y, the first term
of the quotient is x^. For it has been shown (§ 66) that the term of
highest degree in any letter in the product (dividend) equals the prod
uct of the terms of highest degree in that letter in the multiplicand
(divisor or quotient) and multiplier (quotient or divisor).
E.g., in dividing x"^ + x^y + 2 x^y'^ — 2xhj^ — xy^ — y"^ hy x* + 2x^y
+ 3 x2?/2 4 2 xy^ + y^, the first term of the quotient is x^. If the terms
in each polynomial were written in reverse order, the first term of the
quotient would evidently be — y^.
2. If the product of the divisor and the first term of the
quotient is subtracted from the dividend, a partial dividend
is obtained which is the product of the divisor by the other
terms of the quotient.
That is, in dividing x^ + Sx^y + Sxy'^ + y^hj x + y,we know (by 1)
that x'^ is the first term of the quotient. Now if
from x^ + Sx^y + 3 xij^ + y^
we take x^ (x + y) or x^ + x^y
the remainder . 2 xy + 3 xy^ + y^
is a partial dividend and is the product of the divisor, x ^y, by the
other terms which follow in the quotient.
This is evident because the whole dividend is the product of x{ y
by the quotient ; hence, the 2 x^y + 3 xy"^ + y^ is the product oi x + y
by the other terms of the quotient.
It will be noticed that this is similar to the division with which the
student has become familiar in arithmetic ; each remainder is the prod
uct of the divisor and the rest of the quotient.
64 ELEMENTS OF ALGEBEA.
89. The operation of division can now be explained. Let
it be required to divide 3 x^y \ if \ x^ ^ ?> xif by y \ x.
It has been shown (§ 88) that, if the expressions are
arranged according to the descending powers of x, the
first term of the quotient is x^.
x^ ^ 2xy { y^ = quotient.
Divisor = x + y ) x^ + Sx^y + Sxy^ + y^ = dividend.
If x^ {x + y) or x^ + x^y is subtracted,
the remainder 2 x^y + 3 xy^ \ y^ is a, partial dividend,
tlie prodiict of x \ y
by the rest of the quo
tient. . •. the next term
of the quotient is 2 xy.
Subtracting 2xy{x + y) or 2 x'^y 4 2 xy^
the remainder xy^ + i/^ is also a partial divi
dend, the product of
« 4 y by the rest of
the quotient. .. the
next term of the quo
tient is 2/'^.
Subtracting y"^ (x + y) or xy" + y^
there is no remainder, and the division is complete.
90. Exact division. If one of the partial dividends be
comes identically 0, the division is said to be exact. If not,
the degree of some partial dividend will be less than that of
the divisor ; such a partial dividend is called the remainder.
This subject will be further considered in the chapter on fractions.
If D ^ dividend, d = divisor, q = quotient, and r = remainder, then
D — r = dq ;
that is, if the remainder v^^ere subtracted from the dividend the result
would be the product of the quotient and the divisor.
91. Checks. 1. Since the dividend is the product of the
quotient and the divisor, one check is by multiplication.
.• D — r = dq, any remainder should first be subtracted.
2. The work may be checked by arbitrary values.
DIVISION. 65
92. Arrangement of work in division. The full form of
the work is as follows :
x^ + 2xy +2y^ = quotient.
Divisor = x + y) x^ ^ S x^y + 4:xy^ i 5 y^ = dividend.
x3 I x^y =x'^{x + y).
2 x^y + 4 xy'^ + 5y^ = 1st partial dividend.
2x^y + 2 xy^ =2xy{x + y).
2xy^ + 5y^ — 2d partial dividend.
2xy2 + 2y3 = 2 ys (x + y).
(See check below.) Sy^ = remainder.
It is better in practice to abridge this work as follows :
x2 + 2 x?/ + 2 ?/2
X + 2/ ) x^ + 3 x2y + 4 xy2 ^ 5 ^3
x« + x2y
2x^y
2 x2y + 2 xy2
2 X2/2 + 5 y^
2 xi/2 f 2 y^
Sy^
It is still better to detach the coefficients if possible.
1 + 2 + 2
1 + 1)1+3 + 4 + 5
1 + 1 Check. Let X = y = 1.
(1 + 1) (1 + 2 + 2) = 1 + 3 + 4 + 5
or 2 • 5 = 10.
2
2 + 2
2 + 5
2 + 2
3 x^ + 2xy + 2y'^, and 3 y^ remainder.
Similarly, to divide x^ — 1 by x — 1.
1 + 1 + 1
11)1+0 + 01
1 — 1 Check. Let x = 2.
1 (2  1) (8  1) = 4 + 2 + 1
^^ or 1 . 7 = 7.
1
11 X2 + X + 1.
66 ELEMENTS OF ALGEBRA.
EXERCISES. XXX.
Perform the divisions indicated in exs. 114. Check
the results by substituting such arbitrary values as shall
not make the divisor zero.
1. x^ — y^ loy X — y.
2. x^^ — a^'^ by x^ — a^.
3. 32 a^ 7/ by 2 a  b.
4. x^ + x'^y^ + y^ by x^ \ xy \ y^.
5. a^ { h^ \ e^  3 abc hy a \ b + c.
6. a^ + 3a^ + 3a\l hy a^ + 2a + l.
7. x^ — 3x^ {3x ^y^ — 1 by x + y — 1.
8. x'^ — 2 ax^ + 2 rt^x — a^ by x'^ — a^.
9. 1 by 1 — ic, carrying the quotient to 6 terms.
10. a^2a'' + 2a^ + 6a'' + a~l by  a^ + a + 1.
11. _(t6 48aS614a4Z>2 4.Q,3^3 + 6a2^,4by a33«^Z' + 2a^'l
12. a^5a^b + 10a^b^10a'b^\5ab^b^hya^2ab\b^
13. a;^ + 2/^ 4 «^ — 3 0??/^ by x^ \ y^ \ z'^ — xy — yz — zx.
14. a3*4xy + y^ by ?/^ — cc?/ + cc^. (Rearrange the divisor.)
Perform the divisions indicated in exs. 1531 by using
detached coefficients, checking as above.
15. £cs  5 £c2 _ 3000 by xh.
16. 16x^81?/* by 2x43?/.
17. 3 ic^  7 X  2  2 £t;2 by 1 4 if;.
18. a* 4 24 a + 55 by fi^2  4 tt 4 11.
19. x^ 2 a^x^ + a* by x""  2 ax ^ a\
20. x«3x« + 6x^7.t2 43 by x^2.t2 + 1.
21. p^ +jj^ 4 4y^ 9^9 + 3 by p^ \ p'^  3^ 4 1
22. x^ — ic^ + 2 x^ 4 4 x^ — 7 ^2 4 4 a? — 1 hy x^ + x — 1.
DIVISION. 67
23. x^ + 7 xhf — 5 xV — ic V + 2 ?/^ — 4 ay'' by {x — yf.
24. 26 rx2 _^ 4 ^3 _ 3 ^4 ^ ^5 _ 92 ^_^ 55 ^^ «^3a4ll.
25. 24 7M*14m39m284 + 43m by 7  3 /?^ + 4 7/2,2.
26. ic«3a;^5cc5 + 2£c^ + 5a;^ + 4x2 + 2 by .t« + 2x1.
27. 3 m« + 7 m^  12 «^^ + 2 m^  3 wt^ __ 13 ^^^  6 by m^
+ 3 m  2.
28. a;^x^2x^ + 5iz;^5iz;^ + 8x2 + 6ic12 by £c«
2iz;2 + 3.
29. :r« + 2 a;^ + 3 ic« + 4 (cc^ + 1) + 5 iz;* + 6 ic^ + 7 ic^ + 8 ic
by (X + V)\
30. 10 m«  11 iw'  3 m^ + 20 wt^ + 10 1)1" + 2 by 5 m^
 3 ??i2 + 2 m  2.
31. x^ + 2a;« + 3a;^43ic^ + 3x8 + 3x2 + 2a34l by x^
+ x^ + x^ + x^ + X + 1.
32. Divide the product of {x V){x 2){x  3)(x  4)
by 2(43x) + x2.
33. Divide q^ {1 by 2' + 1, and hence tell the quotient
of 100001 by 11 {q = 10).
34. Divide 4:^^ { 2t^ + 5t^ + St + 1 by 2^ + 1, and hence
tell the quotient of 42581 by 11.
35. Divide the sum of  x^ + 4 x* + 7^ x^ + 11 x'^ + 7 x
+ 4 and  x^ + 4 x^ + 6^ x^ + 9 x^ + 4 x by their difference.
36. Divide 1 + x^ by 1 { x carrying the quotient to 5
terms. From the form of this quotient tell what the next
5 terms will be.
37. The product of two polynomials is 2w* — 13 m^n +
31 m%^ — 38 nm^ + 24 n^. If one of them is m^ — 5 mn +
6 n^, what is the other ?
Where the time allows, the work in Synthetic Division
(Appendix II) should be taken at this point.
68 ELEMENTS OF ALGEBRA.
REVIEW EXERCISES. XXXI.
1. Solve the equation 2 — (3 — 4 — a?) = 3.
2. Solve the equation — 2 ic + 4 = — 12. Check.
3. Solve the equation f£c + 4 = Jaj + 4. Check.
4. What are the advantages in detaching the coefficients
when practicable ?
5. What is the etymological meaning of quotient ? of
coefficient ? of associative ?
6. The cube of a certain number, subtracted from 1,
equals 9. Find the number.
7. What is the sign of the product of an odd number of
negative numbers ? Prove it.
8. If from twice a certain number we subtract 7 the
result is 15. Find the number.
9. Three times a certain number, subtracted from 5,
equals — 10. Find the number.
10. Why do you arrange both dividend and divisor
according to the powers of some letter ?
11. Why do you avoid using such an arbitrary value in
checking division as shall make the divisor zero ?
12. If to three times a certain number we add 2 the
result is five times the number. Find the number.
13. What is the value of
a\a  h^a''  2 cQ)''  a  b ^ c) \ b~\ c\
when ^ = 3, 5 = 1, c = 2?
14. What is the Index Law of Division ? Have you
proved it for all values of the indices ? If not, for what
kinds of indices ?
CHAPTER V.
ELEMENTARY ALGEBRAIC FUNCTIONS.
I. DEFINITIONS.
93. Every quantity which is regarded as depending upon
another for its value is called a function of that other.
E.g., with a given principal and rate, the interest depends upon the
time ; hence in this case the interest is called a function of the time.
Similarly, the expression ic^ _ 3^ + 21 is a function of x, etc.
94. A function of x is usually indicated by some such
symbol as f(x), F{x), f^{x), P{x), ■••.
Thus, if the expression x^ — x \ I is being considered, it may be
designated by /(x), read "function of x," or simply "function x."
If some other function of x, as x* — x^ + 2 x^ — x + 4, is also being
considered, it may be distinguished from the first one by designating
it by F{x). read "/major of x," or "/major function x."
P (x), fx (x), ■ . . are read " P function ic," "/one function
ic," • • • . The Greek letter <^ (phi) is also very often used in
this connection, </> (x) being read " phi function x.^'
95. If f(x) is known in any discussion, /(a) means that
function with a put in place of x.
E.g., if
/(x)=x2 + 2x + 1,
then
/(a) = a2 + 2a + l,
/(2) = 22 + 22 + 1 =
and
/(O) = + 0+1=1.
60
70 lOLEMENTS OF ALGEBRA.
96. A quantity whose value is not fixed is called a vari
able ; if the value is fixed, it is called a constant.
E.g., in the expression y'^\2y + ^, y may have any value, and
hence y is a variable. But when it is said that y  2 = 3, the value
of y is fixed, and hence ?/ is a constant, 5.
I
97. Every algebraic expression which, in its simplest
form, contains several variables is called a function of those
variables.
E.g. , x'^ ^ 2xy {■ y^ is a function of x and ?/, and may be designated
by/(x, 2/), read "function of x and ?/," or simply "/of a; and y.""
But x + y + a — y — xis not a function of x and y.
EXERCISES. XXXII.
1. If f(x) = x'x^ + xl, what are f(a), f(a^), /( 2),
/(l),/(0)?
2. If f(x) = x'^\x{l, and F(x)=x — 1, find the value
of /(ic) • F(x). Check by letting x =^ 2, i.e., by finding the
valueof/(2).i^(2).
3. If f(x) = x^ + 3x^ ^ 3 X + 1, and <ty(x) = x"" + 2x + 1,
find the value of f(x) ^ 4, (x). Check by using /(I) ^ </> (1).
4. It f(x, y) = x^ Sxhj + Sxf y^, Sind fi(x, y) = xy,
find the value oif(x, y)fi(x,y); also oif{x,y)^f^(x,y).
Check by using /(2, 1) and/i(2, 1).
5. If F{x, y,z) = x^ + if + z^ 3 xyz, and/(ir, y, z) =
X \ y \ z, find the value of F(x, y, z) if{x, y, z). Check
by letting x = y = z = 1.
6.. li f^{x)= x^ + 2x + 1, f^(x)= x" 2x + 1, and/8(a;)
= x'^l, find the value of /i {x) f^ (x) f^ {x). Check by
letting cc = 2.
7. If f{x) = a;*  10 a;« + 35 a^2 _ 5Q ^ _^ 24, find the values
of/(l),/(2),/(3),/(4).
I^H98. An algebraic expression is said to be rational with
respect to any letter when it contains no indicated root of
that letter. In the contrary case it is said to be irrational
with respect to that letter.
E.g.^ 4a + v2 is rational with respect to a,
but 2 + 4 \a is irrational witli respect to a.
So x2 — X Va + va is an irrational function of a, but it is a
rational function of x.
99. A rational algebraic expression is said to be integral
with respect to a letter when this letter does not appear
in any denominator. In the contrary case it is said to be
fractional.
E.g., is an integral algebraic expression, with respect to a,
but 2 — is a fractional expression, with respect to a.
W
X 1
x2 1 — is an integral function of x,
a a
but x2 — Vx is not, because it is not rational,
1 2
is not, because it has x in both denominators.
X x2
«lOO. An algebraic expression is said to be homogeneous
en all of its terms are of the same degree.
E.g., 7 a^x + 4 a^ + x'^ is homogeneous, but 3 a^x + 4 ax^ is not.
So ax'^y + Jfixy'^ + chj^ is homogeneous as to x and y, but not as to
^alone, nor as to y alone, nor as to a, x, and y.
101. An algebraic expression is said to be symmetric with
respect to certain letters when those letters can be inter
changed without changing the form of the expression.
E.g., x'^\2xy + y^ is symmetric as to x and y, because if x and y
are interchanged it becomes y2 ^ 2 yx + x^ which is the same as the
original expression. Similarly, x^ + y^ 4 z^ + axyz is symmetric as
to X, y, and z, but not as to a, x, y, and z.
72 ELEMENTS OF ALGEBRA.
102. An algebraic expression is said to be cyclic with
respect to certain letters in a given order when its value
is not changed by substituting the second for the first, the
third for the second, and so on to the first for the last.
E.g., a (a — 6) + 6 (6 — c) + c (c — a) is cyclic as to a, &, and c ; for
if 6 is substituted for a, c for 6, and a for c, it becomes 6 (6 — c) +
c {c — a) ^ a {a — b), which is the same as the original expression.
It will be noticed that if an expression is symmetric it
must be cyclic, for a cyclic change of letters is a special
case of the general interchange of symmetry. But the con
verse is not true, for the special case does not include the
general one.
E.g., x'^ + y^ \ z"^ — x{x \ y'^) — y {y + z^) — z{z \ x^) is cyclic but
not symmetric ; but x^ + y^ + z'^ — xy — yz — zx is symmetric and
hence also cyclic.
The theory of cyclic functions is often called cy dosym
metry, or, where no misunderstanding will result, simply
symmetry.
EXERCISES. XXXIII.
Select from exs. 113 those expressions that are (1) homo
geneous, (2) symmetric, (3) cyclic, as to any or all of the
letters involved :
1. a^x — b^x + c*ic. 2. ic^ 4 2 x^y^ + y^.
3. x^z — 3 xyz^ + y^z^. 4. ab { be \ ca \ abc.
6. a'' + b^]c^Sabc.
6. abc  3 ac^ ^ bc^ — c\
7. x^ — x^y + ccy — xy^ + ?/*.
8. a\be) + b\ca)hc\ab).
9. a^ (b  ef + b\c ay \ c\a  bf.
10. x"^ ^y^ ^z^ + ax \by + ez^ mxyz.
11. be (b \ c)\ ca (c + a) + a^> (a + ^) + 2 abc.
ALGEBRAIC FUNCTIONS. 73
12. a^ (b^  c2) + b^ (c^  a^) + c« (a^  b^).
13. {a + b) {a^ + b^ G^) + {b + c) (b^ + c^  a^)
+ (c + a) (c^ + a^ b^.
Select from exs. 1420 those functions of x, of y, and of a
that are (1) rational, (2) integral functions of those letters.
14. ^x — X wa.
15. x^ \x^ + 1.
16. x'^/a — x/a^.
17. ic^ + ic^ + ic + y/x.
18. a;« + 3 a;^;/ + 3 ic?/2 + 7/8.
19. cc"* + x"'^ + x'»2 H Yx"^ + x + 1.
m
20. x"* — ic^, (1) when m is even, (2) when m is odd.
I
The applications of homogeneity and symmetry are numerous
and valuable. If the time allows, they should be taken at
this point. They are set forth at some length in Appendix
II.
It should, however, be said that symmetry and homo
geneity form two valuable checks, especially in multiplica
tion. If two expressions are homogeneous their product is
evidently homogeneous.
E.g.^ the product of x^ + 2x?/ — y^ and x — y cannot be x^ + x'^y^
3 x?/ + 2/^, because the factors being homogeneous the product must
be so.
Likewise, if two expressions are symmetric as to two or
lore letters, their product must be symmetric as to those
letters.
E.g.^ the product oix'^ — 2xy \ y^ and x \ y cannot be x^ — x'^y +
+ 2/8, because this is not symmetric as to x and y.
A knowledge of symmetry and homogeneity is of great
lue in factoring.
74 ELEMENTS OF ALGEBRA.
II. THE REMAINDER THEOREM.
103. If we consider the remainder arising from dividing
a function of x, say x^ {■ px + q,\}j x — a, we find an inter
esting law.
X + p \ a = quotient
X — a ) x2 + px + 5
x^ — ax
(p + a) X + 5
[p \ a) X — pa — a'^
a^ + _pa + g = remainder.
That is, the remainder is the same as the dividend with
a substituted for x.
Hence, if this law is general, we may find the remainder arising
from dividing x^ + 2 x — 3 by x — 2 by simply substituting 2 for x in
the dividend. This gives 2^ + 2 • 2 — 3 = 5, the remainder.
Similarly, it is at once seen that, if this law is general, x^''' + 2 x^ — 3
is exactly divisible by x — 1. (Why ?)
That the law is general is proved on p. 75.
EXERCISES. XXXIV.
Assuming that the remainder can always be found as
above stated, find the remainders arising from the follow
ing divisions :
1. cc" — 1 by a? — 1.
2. x'^^ — y^ by X — y.
3. 2 cc^  64 by ic  2.
4. 32 a' 1 by 2 a  1.
5. (7cc)i° + l by 7x + l.
6. x^ — x^ ^x^ — X + 1 by X — 3.
7. x^ { x^ — x^ — X + 1 hy X — 1.
8. 3x^ + 4.x^2x36 hj x2.
9. £c" + 1 by a; + 1, i.e., by x  ( 1).
ALGEBRAIC FUNCTIONS. 75
104. The Remainder Theorem. If f (x) is a rational inte
gral algehraic function of x, then the remainder arising
from dividing f (x) hg x — Si is f (a).
Proof. 1. Let q be the quotient and r the remainder.
2. Then f(x) = q(x — a)\ r. Def . of division
(I.e., the dividend equals the product of the quotient and the divisor,
plus the remainder, and this is true whatever the value of x.)
3. Step 2 is true it x = a, it being an identity.
4. But r does not contain x. (Why ?)
5. . • . f(a) = q (a — a) + r = \ r = r, from step 3.
6. I.e., the remainder equals /(a), or the dividend
with a substituted for x.
105. CoROLLAKiES. 1. If f (x) is a rational integral
algehraic function of x, then the remainder arising from
dividing f (x) by x + a is f (— a).
For X \ a = x — {— a) ] hence, — a would merely replace a in the
above proof.
[2. If f (a) =: 0, then f (x) is divisible bg x — a.
Tor the remainder equals / (a) , and this being the division is exact.
{. If n is a positive i?iteger.
(a) x° 4 y° is divisible bij x + y when n is odd.
Tor, putting — y for x, x» + 2/" becomes ( — y)" + y^^ which equals
"when n is odd, and not otherwise.
L^^ (b) x° + y*^ is never divisible by x — j.
I^KFor, putting y for x, x« + 2/" becomes 2/» + 2/", which is not 0.
IB (c) x" — y° is divisible by x + j when n is even.
For, putting — y for x, x" — ?/" becomes (—?/)" — y^, which equals
1 when n is even, and not otherwise.
IB (d) x" — y"^ is always divisible by x — y.
mK'PoY, i^uttiug y for x, x** — ?/" becomes 0.
B
76 ELEMENTS OF ALGEBKA.
Illustrative problems. 1. Pind the remainder arising from
dividing (x \ ly — x^ — 1 by x + 1.
Substitute — 1 for x, and f{x) becomes (— 1 + 1)^ — (— 1)^ — 1,
which equals + 1 — 1, or 0.
2. Also when (x — my + (x — ny \ (m \ ny is divided
by ic + m.
Substitute — m for z, and f{x) becomes (— m — m)^  (_ m — n)^
+ (m + n)3, which equals — 8 m^ — (m + n)^ f (m + n)^, or — 8 m^.
3. Also when nx'"'^^ — (ti + 1) ic" + 1 is divided by x — 1.
Substitute 1 for x, and w — (n + 1) + 1 = 0.
4. Find the remainder arising from dividing x^ { 5x*
Zx^2x + l by x + 1.
Here it is rather tedious to substitute — 7 for x. If the student
miderstands synthetic division (Appendix II) it is better to resort to
it, as follows :
11 + 53+02+ 7
7 17 1477 539 3759
12 1177 537 ;  3752 remainder.
Check. [8  ( 3752)]  8 = 470.
EXERCISES. XXXV.
Find the remainders in the following divisions :
1. ic^"' + ?/2'» by x + tj.
2. a;^ — 4 £c2 + 3 by £c + 4.
3. aj^^' + i + 2/2m + i ^j x + y.
4. 32a;i033a;« + l by a;  1.
5. a;* + 2a;23a;7 by a;  2.
6. aj^"^ + a;io  2 by ic  1 ; by cc + 1.
7. i«« + 3 cc2 + 50 by a; + 5 ; by cc  5.
8. x^ + y^ by a;2 + y\ (Substitute  y" for x\)
9. x^^ + 2/15 by ic^ + y^.
10. iC^O + 2/20 l3y ^4 __ ^4_
M
ALGEBRAIC FUNCTIONS. 77
REVIEW EXERCISES. XXXVI.
1. Solve the equation f(x)=f{2).
2. If f(x) ~x — l, solve the equation f(x) /(S) = 0.
3. If f{x) = x — 1, solve the equation [/(^)]^= a;'^ — 3.
4. If F(x) = x^ — 5x \ 1, solve the equation
F(x) = F(x) + 5x.
5. Is ax^ + bxy + a?/^ symmetric as to x and y? as to
a and b? as to a and x ?
6. Is this a rational function of x :
^x^x^V^ + 3xy/ai^?
Is it an integral function of a? ? Is it a rational function
of a?
7. If f(x, y) is symmetric as to x and ?/, is [/(ic, y)]^
also symmetric as to x and 3/ ? Illustrate by letting
(a;, y) = x + y.
8. May /(ic, ^) be not symmetric as to x and y, and
[/(^j 2/)]^ be symmetric ? Illustrate by letting f(x, y) =
xy.
9. Do you see any advantage in having a function sym
bol, as f{x), in the way of brevity ?
10. Multiply x^ \3x^y + 4. x'^y^ [^xtf + y^ by ic^ — xy
y'^, checking the result (1) by symmetry, (2) by homo
geneity.
11. Multiply x^ — S xhj {■ S xy^ — y^ by x'^ + 2xyy^
d check by symmetry or by homogeneity according to
hich one applies.
12. Divide x^ — y^ hj x — y, checking the quotient by
mogeneity.
13. Divide x^ + y^ hj x \ y, checking the quotient by
mmetry.
CHAPTER YI.
FACTORS.
I. TYPES.
106. Tlie factors of a rational integral algebraic expression
are the rational integral algebraic expressions which multi
plied together produce it.
In the expression 3x(x + l)(x'^ + x \ 1) (x^ + 2)
3 is called a numerical factor,
X " " monomial algebraic factor of the first degree,
ic + 1 " " linear binomial factor,
aj2 _j_ a;  1 " " quadratic trinomial factor, the term " quad
ratic " being applied to integral algebraic expressions of the
second degree in some letter or letters.
ic^ f 2 is called a cubic binomial factor, the term " cubic "
being applied to integral algebraic expressions of the third
degree in some letter or letters.
E.g., in the expression x^{x \ y + z) {x^ + y^), x^ is a monomial
cubic factor, x \ y + z is a. linear trinomial factor, and x^ + y^ is a.
quadratic binomial factor.
107. Rational integral algebraic expressions which in
volve only rational numbers are said to exist in the domain
of rationality.
JS.gr,, x2 + 2ic + i, but not x2 — V2. The former has no algebraic
fraction, and the latter involves an irrational number.
78
FACTORS. 79
108. The product of two integral expressions in the
domain of rationality is evidently another integral expres
sion in that domain. We say that an expression is reducible
in the domain of rationality if it is the product of several
integral expressions in that domain, and irreducible in the
contrary case.
E.g.^ 4x2 _ 9 is reducible, because it equals (2x + 3) (2 x — 3), but
x^ — 3 is not reducible, the word " reducible " alone meaning " reduci
ble in the domain of rationality. ' '
109. A rational integral algebraic expression is said to be
factored when its irreducible factors are discovered.
E.g., the factors of x* — 1 are x^ + 1, x + 1, and x — 1. When
X* — 1 is written in the form (x^ + 1) (x f 1) (x — 1), it is said to be
factored, because x2 + l,x+l,x — 1 are irreducible.
The expression x — 1 is irreducible, although it has the factors
Vx + 1 and Vx — 1, because these are not rational.
The term " factorable " is applied only to rational inte
gral expressions. 1^g, while ("V^ 4 l)(v^ — 1)^ "^ — 1?
expressions like Vx — 1 are not spoken of as factorable.
110. Factoring is the inverse of multiplication, and like all
inverse processes it 'depends on a knowledge of the direct
process and of certain type forms already known.
E.g.., because we know that
Ik {X + ?/)2 = x2 + 2 x?/ + 2/2,
r ifierefore we know that the factors of
x2 + 2 xy + ?/2 are x + y and x + y,
and those of m2 + 2 m + 1 " m t 1 " m + 1.
I^Bll. Although all cases of factoring give rise to identi
IBs, the symbol = is usually employed instead of = as
80
ELEMENTS OF ALGEBRA.
112. The tjrpe xy + xz, or the case of a monomial factor.
Since x {y \ z) = xy \ xz, it follows that expressions in
the form of xy + xz can be factored.
^.^., 4a;2 + 2x = 2x(2x + l). Check. 6 = 23.
A polynomial may often be treated as a monomial, as in
the second step of the following :
2/2 — my + ny — mn = y {y — m) \ n{y — m)
= iy + n){y  m).
Check. Let 2/ = 2, m = n = 1. Then 3 = 3.1.
It must be remembered that an expression is not factored
unless it is written as a single product, not as the sum of
several products.
E.g., the preceding expression is not factored in the first step ; ouly_,
some of its terms are factored.
EXERCISES. XXXVII
Factor the following expressions :
1. ic^ + x^y + ic*.
3. x^ — x^ — x^{ X.
7. m^ + 3 m^Ti + 3 m7i2.
2. a'^ \ 2 ab { 3 ac.
4. 3x^ — 4.ax^ + x^.
6. ahy — ay { y^ — hy.
8. w'^ — wy \ ivx — wxy.
113. The type x^ ± 2 xy + y^, or the square of a binomialj
Since {x ±yy=x^±2xy + y^ (§ 69, 1, 2), it follows that
expressions in the form oi x^±2xy [ y'^ can be factored.
E.g., a;2 + 4x + 4 = (x + 2)2. Check. 9 = 32.
x2 _ 6 x?/ + 9 y2 ^ (aj  3 vY. Check. 4 = ( 2)2.
EXERCISES. XXXVIII.
Factor the following expressions :
1. a;2 + 10cc + 25. 2. 4. x^ + 4. xy \ y\
3. 25 + a;210a;. 4. m« + 14m8 + 49.
FACTORS. 81
5. 121ic222£c + l. 6. 4a;2 442/(2/ 2a;).
7. 9x^2^xy + 16y^ 8. 81 a;^ + 72 ^y + 16 3/^
9. 4:9z^ + Slw^126zw. 10. (x + yy + 2(x + y)\l.
11. 169a^\169b^S3Sab.
12. a2^4«^ + 4 + 2(a + 2) + l.
113. «2_^2a5 + ^»2 + 2(a + ^)2/ + 2/''.
14. x^ \2xy + y^ + 2xzj2yz{ z\
15. m^ + 71^ + p2 + 2 mn — 2 mp — 2 np.
14. The type x^ — y^, or the difference of two squares,
ince {x + y)(xy) = x^ y" (§ 69, 3), it follows that
expressions in the form of x^ — y"^ can be factored.
5 •  3.
53 1.
3 1.
E.g., x'216 = (x + 4)(x4).
Check.
15
x*  16 = (x2 + 4) (x2  4)
= (x2 + 4) (X + 2) (X  2).
Check.
15 =
X4 4 x22/2 + y* = X* + 2 x22/2 + 2/^  x22/2
= (X2 + 2/2)2 _ r^2y2
= (X2 + 2/2 4. a;y) (a;2 4. 2/2 .
xy).
Check
EXERCISES. XXXIX.
Factor the following expressions :
1. ^^162/'. 2. ici6_l.
3. a" + a^^^ _^ ^>*. 4. 36a;292/'.
5. 16a;* + 4a;22^2_^^4 6 81 ic* + 9 a;^ + 1.
7. a;2 42a;2/ + 2/'^'. 8. (cc + y)^  (i»  2/)^
9. a'' + h''  x"" 1+2 ah + 2x.
I 10. a22a^> + &2_(a;2_2iC2/ + 2/^).
n 11. 4a2f4a3(=4a2_^4a + l4).
82 ELEMENTS OF ALGEBRA.
115. Forms of the factors. Although a rational integral
algebraic expression admits of only one distinct set of
irreducible factors, the forms of these factors may often
appear to differ.
E.g., since {x  2y) {2x  y) = 2x'^  5xy ] 2y^,
and {2y x){y2x) = 2x^5xy + 2y2,
it might seem that 2x'^ — 5xy + 2y^ has two distinct pairs of factors.
This arises from the fact that the second pair is the same as the
first, except that the signs are changed, each factor having been multi
plied by — 1. But this merely multiplies the whole expression by
 1 •  1, that is, by + I.
Hence, the signs of any even number of factors may he
changed without changing the product.
E.g., jc2  5 X + 6 = (x  2) (x  3), or (2  x) (3  x).
Check. 2 =  1 •  2, or 1 • 2.
X*  1 = (x2 + 1) (X + 1) (X  1)
zz:(x2+l)(Xl)(lX)
= (x2l)(X+l)(lx).
Check. •Letx = 2. Then
16l=5.31 = 53l=53 1.
EXERCISES. XL.
Factor the following, giving the various forms of the
results and checking each.
1. la\ 2. x^1.
3. 16 £C*. 4. a^ — h\
5. lQx^^ly\ 6. 2 + ^* 2^2^.
7. 121f £c222£c. 8. z^ + 22z^V2.
9. a;i<>26a;« + 168. 10. a''  c'' ^r b"" + 2 ah.
11. 16£c*f8a;2 + i_252/^ 12. x''l^x^{2^x^lx.
13. 121 x^ f 121 2/2  9  242 xy.
14. 4.x''\lt/2ijzz^ + 4.x.
FACTORS.
83
116. The type x^ ± 3 x^y f 3 xy^ ± y^, or the cube of a
binomial.
Since (x ±tjy = x^±3 xhj + 3xi/± if (§ 69, 4, 5), it
follows that expressions in the form of ic^ ± 3 x^y + 3 xif
± y^ can be factored.
E.g., 8x3 + 12x2 + 6x + 1 = (2x)3 + 3(2x)2 + 3 • 2x + 1
= (2 X + 1)3. Cheok. 27 = 33.
27 x6  54 x% + 36 xV 8^/3 =
(3x2)3  3 (3x2)2 . 27/ + 3 • 3x2(2 2/)2  (2?/)3
= (3x22?/)3. C/iecA;. 1 = 18.
x8 x2?/2 xy* y« /x\3 o{^V{y'^\ .o(^\{y'^\^ /v'^V
S~'~T^~6rY7 = \2)~^\2)\'3)'^^\2)\B) ~\s)
_/X_^\3
~ V2 3/
Check. Let X = 2, y = 3. Then l_9 + 2727=8 = (l 3)3.
EXERCISES. XLI.
Factor the following expressions :
1. lSx + Sx'^x^ 2. a^3a^ + 3al.
x^^ 3x^ + 3x^1. 4. 27x327a;2 + 9ic 1;
6. a^3a%^ + 3a^b'b\
6. 27 0^9  27 a« + 9 a^  1.
7. 8 a;^  12 £c2?/ + 6 a;y2 _ ^3^
8. 54.x^27x + Sx\36x^
9. 1.331 ic^  7.26 ^2 _^ 6.6 X 8.
10. 64 xY  48 xy + 12 x'Y  1.
11. xY^^ + 6 xy^2 _^ ;l2 a;y« + 8.
12. 0.125 x« 0.75 x* + 0.15 cc^l.
13. (a + ^')« + 3(a + ^)2 + 3(a + ^>)Hl.
84 ELEMENTS OF ALGEBRA.
117. The type x° ± y". It has been shown (§ 105, Re
mainder Theorem, cor. 3) that
£c" __ y» contains the factor x {■ y when n is odd,
" " " ^^ X — y never,
^«. _ yn u a u rf^ j^ y ^Jien n is even,
" " " " cc — ?/ always.
Hence, it follows that expressions in the form of x"" ± y^
can often be factored.
E.g.^ x^ + y^ contains the factor x ■\y. The other factor can be
determined by division. It may also be determined by noticing that
a;3 ^ yZ ig symmetric and homogeneous, and that its factors must
therefore be x + ?/ and x^ _. ]^xy + ?/2, where k is to be determined.
Letting x = y = 1,
x3 + 2/3 = (X + y) (x2 + kxy + ?/2)
becomes 2 =2(2 + A;),
and therefore, A; = — 1,
whence x^ + t/^ = (^c + 2/) (a?^ — a??/ + y^).
This type occurs so often that the forms of the quotients
should be memorized :
signs alternating.
the
03" — y
signs alternating.
3. ~^ = cc^i + a;"2y + a;"y + x^^^f + • • •, the
signs being all +.
We are thus able to write out the quotient of (xi^ + ?/i5) ^ (x + y)
at sight, and so for other similar cases.
The integral parts of the quotients in 1 and 2 are the same, but
the remainders are different. E.g.^ if n is odd there is no remainder
in 1, but in 2 there is a remainder — 2 ?/».
FACTORS. 85
When the exponent n exceeds 3 it is better to separate
into two factors as nearly of the same degree as possible,
and then to factor each separately.
E.g., x8  ?/8 = {x^ + 2/*) (x*  y^)
= {X^ + 2/*) (X2 + 2/2) (X2  2/2)
= (X* + y^) {x^ + y^) (X \y){x y),
or the same with certain signs changed (§ 115).
This is better than to take out the linear binomial x \ y ov x — y
first, which would give
x8 — 2/^ = (x + y) (ic^ — x^y + x^y"^ — x^y^ + x^y^ — x^y^ + xy^ — y"^),
or (x — y) (x^ + x^y + x^y^ + ic*2/^ + x^y^ + x^y^ + xi^ + y^),
in which cases it would be difficult to discover the factors of the two
expressions of the seventh degree.
So x2«  2/2« = (x« + y^) (x«  ?/").
118. Binomials of the form cc" ± ?/" which have not the
factor x±y may contain ic"* ± t/"*.
X6 + y6 ^ (X2)3 + (2/2)8 = (a;2 + ?/2) (x^  X'^y'^ + 2/4).
EXERCISES. XLII.
'actor the following expressions :
■^t.
x' + 1.
2.
x'  16.
3.
x^  if.
4.
1  x^\
5.
x^^%y\
6.
x^ + f.
li
32a^s + l.
8.
£C2+1 + 1.
i
a;i2 __ 4096.
10.
^2^,4 _ ^,2^4
'^1
729^36 + 2/'.
12.
216a«Z»«.
13.
{x + 2/)« + 1.
14.
125a« + 27.
15.
64cc«7292/^ .
16.
27a» + 64^«.
17.
125^^27cc/.
18.
a^ + a + b^ + b. .
19.
{a  by (a + by.
20.
m^n^ + 2nl.
86 ELEMENTS OF ALGEBRA.
119. Thetypex^ + ax + b. 'Letx^\ax + b = (x\m)(x\n),
in which m and n are to be determined. Then
x^ \ ax + b = x^ \ (m \ n) X { mn.
It therefore appears that if two numbers, m and n, can be
found such that their sum, m + n, is a, and their product,
mriy is h, the expression can be factored.
E.g., consider a:2 + lOx + 21.
Here 10 = 3 + 7,
and 21 = 3 • 7,
x2 + 10 X + 21 = (x + 3) {X + 7). Check. 32 = 4 • 8.
Consider also x^ — 3 x — 40.
Here 3 = 58,
and  40 = 5 •  8,
x23x40 = (x + 5)(x8). Check. 42 = 6.7.
EXERCISES. XLIII.
Factor the following expressions :
1. x^ + Zx + 2. 2. x''x2.
3. ic^ + a;2_12. 4. x^^x + Q.
5. cc24iz;165. 6. j^^j^GOO.
7. a2aa130. 8. a;2_4^_21.
9. ^2 11 a 60. 10. x'4.x^4.5.
11. 4a;2__8£c45. 12. a^ + 17 a + 66.
13. ic2 + 41 £c + 420. 14. ic^ + 16a;2 + 55.
15. a2_24a + 135. 16. icy + 4a;V + 3.
17. x^  15 cc2 _ 100. 18. ^2  16 a  225.
19. a'^x^ + 5 a^x"" + 6. 20. ic^ ^ 7 a??/ + 10 1/.
21. 4a2 + 2aZ.2&2. 22. a'x""  5 a^x  U.
23. m2 38 m 4 165. 24. cc^ + lla^y  26/.
25. mV _ 7 ^ic  18. 26. m^ic* + 12 m,x^ + 35.
FACTORS. 87
'^" ax^ ^ bx \ G= (mx + 71) {px + q),
in which m, 71, p, and q are to be determined. Then
ax^ \hx { c^i mpx^ + (viq + pn) x + qn.
It therefore appears that the coefficient of x, mq + pn, is
the sum of two numbers ivhose product, mqpn, is the product
of the coefficient of x^, mp, and the last term, qn. Hence,
if these numbers can be detected, the expression can be
factored.
E.g. , consider 6 x2 + 17 x + 12.
Here 17 = 9 + 8,
and 6 • 12 = 72 = 9 • 8.
6ic2 + 17x + 12 = 6x2 + 9x + 8x + 12
= 3x(2x + 3) + 4(2x + 3)
= (3x + 4)(2x + 3). Check. .35 = 75.
Consider also 8 x^ + 7 x — 3.
Here 7 = 92,
and 6  3 =  18 = 9 •  2.
6x2 47x3 = 6x2 + 9x2x3
' =3x(2x + 3) (2x + 3)
= (3 X  1) (2 X f 3). Check. 10 = 2 • 5.
EXERCISES. XLIV.
factor the following expressions :
6,^2 + ^12.
2.
12PP1,
4a;2_4cc3.
4.
3 ^2 + 8 a + 4.
600 a'^ — a 1.
6.
9cc217cc2.
Ux^5xl.
8.
Sa^ + 22a{ 12.
12p''  7p + 1.
10.
6^ + 25 0^ + 1).
1 11
^12 _ 7 ^6^6 _ S ^12_
12.
16x''62x + 2T.
16 a^ + 43 a^* + 27 b\
14.
4.0 a^{ 61 abS4:b\
15
16xyz^{S9xi/z27.
16.
30x^4:1 xz15z\
88 ELEMENTS OF ALGEBRA.
121. Application of the Remainder Theorem. The presence
of a binomial factor is usually detected very readily by the
use of this theorem (§ 104).
E.g., x^ — 4x43 evidently contains the factor (x — 1), and the
other factor, x^ + x — 3, can be found by division.
Similarly, consider x^ — 2 x — 21.
Trying x — 1 we have
/(I) = 1  2  21 ?i ; .. X  1 is not a factor.
Trying x f 1 we have
/( 1) =  1 + 2  21 ;zi ; .. X + 1 is not a factor.
Trying x — 3 we have /(3) = ; .. x — 3 is a factor.
If the student understands Synthetic Division (Appendix
II), the test of divisibility is easily made by that process,
thus:
11 2 21
3  3 9 21
13 7 ; remainder.
Hence the factors are x — 3 and x^ + 3x + 7.
Check.  22 =  2 . 11.
Since the factors of — 21 are ± 1 and qp 21, ±3 and =F 7, the
number of trials necessary is very limited.
EXERCISES. XLV.
Factor the following expressions :
1. x819i»30. 2. x^Sx2.
3. m* — 2 mn^ + n^. 4. a^ — a^ — a — 2.
5. a^a2 \2a^ 6. a;« + 9 x^ + 20 cc + 12.
7. a^6a^{lla6. 8. a^ + 8a^ 112 a + 256.
9. a^'a^lBah 12.
For those who have studied symmetry as set forth in
Appendix III, the cases of factoring given in Appendix IV
are recommended at this point.
FACTORS.
89
MISCELLANEOUS EXERCISES. XLVI.
'122. General directions.
.. First remove all monomial factors.
}. Then see if the expression can be brought under some
of the simple types given on pp. 8187. This can probably
always be done in cases of binomials and quadratic trino
mials, and often in other cases.
3. If unsuccessful in this, the Eemainder Theorem may
tried, especially with polynomials of the form
£c" + ax'^~^y + hx'^~^y^ + • • •.
4. Always check the results, and be sure that the factors
are irreducible.
™
I. cc* + 4.
^3. x^ + y^.
^5. x^ — a;*?/*.
7. a;^ + x2 + ^.
9. a%^c^a%^G\
11. a^a^ 110.
L3. £C*  11 ic2 _^ 1.
15. 6a;2_23x420.
17. ccy 4 2 ccy + xy.
L9. a^lba%'' + ^b\
II. ah At y'^ — ay — hy.
13. £c*  8 icy + 16 ^/.
\h. (a + by + (aby.
27. 2/« + 37/^ + 62/ + 18.
29. 21 ^2 + 26 aZ>  15 h''.
2. rz;* + 4 /.
4. 1 + x^ + cc^
6. ic^ + 2/^ + icy.
8. x''2xSf + 7/.
10. £C2 (a;2 + 7/2^ + /.
12. ic'' + ic^^ + 2 icY
14. 2a;2 + lla;412.
16. 2/2^2^2^1.
18. (x + 2/)^ — ir'^ — 2/"^.
20. ax"^ + {a + b)x + b.
22. 12 x22/2  17 a;?/ + 6.
24. (x + l)25cc29.
26. 16 cc*  28 ic2y2 __ ^4^
28. 7x3 + 96a;2103£c.
30. a;4((fc2 4^,2^x2 + a262^
90 ELEMENTS OF ALGEBRA.
31. m«w« + l. 32. a^ ^a^b^ + b^
33. 9a;2_igy2 34, a;2"lla;" + 28.
35. a^ j^a2 a». 36. 9 a^"*  5  4 a"*.
37. 10a2_3606*. 38. a2(a2_24) + 63.
39. cc'*"' + x^"* + 1. 40. a^ — ac — bc — b'^.
41. ic* + «» + a;%*. 42. x" + 12 £C2/ + 36 t/I
43. a;2 ^ 16 ic + 63. 44. m^ — ^^ _ ^2 ^ 2 ^m;.
45. a;214a; + 49. 46. (a^ + 1)^ _ (^,2 + i)3.
47. a^ (a^ — 1)  ^6. 48. {x + y)^ + 4 (w + ^)^
49. 6 + 15 a^ _ 19 a. 50. 5 aZ> — Z'C + cc? — 5 ac?.
51. S —(x + y + zy. 52. cc^ + ?/^ — 4 cc^^ — 4 ic?/^.
53. a%ab^ + a% + ab''.
54. a2(^_^i)_52(^_l_l),
55. Sxy{x + y) + x^ + y\
56. 4 ccy  (a:2 + 2/2 _ ^y .
57. 2ic+(a;24)?/2a^y/2_
58. 121 a*  795 ^2^2 _^ 9 Z^
59. (a4)24(a4) + 4.
60. (a; 5)2 8(0; 5)+ 12.
61. x^{x — 2y) — y^{ij — 2x).
62. l(aZ.)110(aJ)2.
63. 10 + 16(a4&) + 6(a + ^>)2.
64. (m + ny + 10 (m + w) + 24.
65. 2 ic2  x^y + (2/  2) (a;y  a;)^.
66. x^ + y^ (w^ + g;2) _ 2 (xi/ + wz).
67. (a + by (a + by  (a + ^')' + 1
FACTORS. 91
II. APPLICATION OF FACTORING TO THE SOLUTION OF
EQUATIONS.
123. To solve an equation is to find the value of the
unknown quantity which shall make the first member equal
to the second. Such a value is said to satisfy the equation
(§ 17).
E.g., if cc2 = 4,
then x2  4 = 0, or (x + 2) (x  2) = ;
.. X = + 2 or — 2. That is, either + 2 or — 2 will satisfy the equa
tion ; for if X == + 2, then (2 + 2) (2  2) == ; and if x =  2, then
(2 + 2)(2 2) = 0.
If x2 + X := 6,
then x2 + X  6 = 0,
whence (x + 3) (x — 2) = 0. This equation is evidently satisfied if
either factor of the first member is 0. (Why ?)
If X + 3 = 0, then x = — 3, because —3 + 3 = 0;
and if x  2 = 0, " x = 2, " 22 = 0.
If x4  6x3 + 11x2 6x = 0,
then X (x — 1) (x — 2) (x — 3) = 0. This equation is evidently satisfied
any factor of the first member equals 0. (Why ?)
Hence, x may equalO, as one value ;
if X — 1 = 0, then x = 1, because 1 — 1 = 0;
Id if x2 = 0, " x = 2, " 22 = 0;
id if X  3 = 0, " X = 3, " 33 = 0.
EXERCISES. XLVII.
Solve the following equations :
1. a;2l = 0. 2. ic2 + 287 =48a;.
3. 2x^ + 2 = Bx. 4. 6a;213x + 6 = 0.
5. cc2 = 2x + 143. 6. x^10x^ + 21=0.
7. x^ {'ix'' + x = Q>. 8. ic6  14^4.^49 ^2^ 3g_
9. ic^13£c2 + 36 = 0. 10. 2ic367cc2 + 371rr = 0.
11. 2x^lx'' + ^x = 0. 12. .x^15a;2 + 10j; + 24=:0.
92 ELEMENTS OF ALGEBRA.
III. EVOLUTION.
124. If an algebraic expression is the product of two
equal factors, one of those factors is called the square root
of the expression. Similarly, one of three equal factors is
called the cube root, one of four equal factors the 4th root, • • •
one of n equal factors the nth root.
The broader meaning of the word root is discussed later (§ 130).
The process of finding a root of an algebraic quantity is
called evolution.
Evolution is, therefore, a particular case of factoring.
It is evidently the inverse of Involution, as Eoot is the
inverse of Power.
125. Symbolism. Square root is indicated either by the
fractional exponent ^ or by the old radical sign V~, a form
of the letter r, the initial of the Latin radix (root).
Similarly, a^ or V a means the cube root of a,
and, in general, a" " Va " " nth " "
For present purposes it is immaterial which set of symbols is used.
The student should, however, accustom himself to the fractional
exponent, which, while a little more difficult to write, has many
advantages over the older radical sign as will be seen later.
126. Law of signs. Since any power of a positive quantity
is positive, but even powers of a negative quantity are posi
tive while odd powers are negative (§ 77), therefore,
1. An even root of a positive quantity is either positive or
E.g., 4^ = ±2, 81* = ±3.
2. An odd root of any quantity has the same sign as the
quantity itself.
E.g., 8* = 2, (8)* =2.
FACTORS. 93
3. An even root of a negative quantity is neither a posi
tive nor a negative quantity.
E.g., V— 1 is neither + 1 nor — 1.
An even root of a negative quantity is said to be imagi
nary, and imaginary quantities are discussed later (Chap.
XIII).
127. The root of a monomial power is easily found by
inspection.
E.g.,. 4a^¥ = 22aab'bbb,
V4a2&4^ V(2 ■abb){2abb) = ±2abb
= ±2 ab\
3 y
Similarly, v 64 x^y^ = 4 xy'^,
V32 xi5y3o = 2 xh
6 ,
'64x12= ±2x2.
EXERCISES. XL VIII.
Simplify the following expressions :
1. V4V^. 2. ^ 8 ^6^,12.
3. ^3^ (a 2 by. 4. ^16a''"'s/a''^b''^\
5. ^2a^\^2b^^I?.
6. Vl6a;V°^^ V32icV'^'''
2x/ 2x+l/ 2x+\/
9. V729^W, V729<^i«^«, V729^i«^.
10. VV«i^3^^ ^^^ being even ; m being odd.
94 ELEMENTS OF ALGEBRA.
128. Roots extracted by inspection. The roots of the mono
mials given on p. 93 were extracted by inspection. Simi
larly, the square root of a square polynomial, the cube root
of a cube polynomial, etc., can often be found by inspection.
Illustrative problems. 1. What is the square root of
ic* + 4 x^y + 4 2/^.
1. ••• [± (/ + w)]2 =p 4 2/?i + w2, § 82
2. and •.• this polynomial can be arranged in a similar form, mz. ,
(x2)2 + 2x2(2 2/) + (2?/)2,
3. .. it is evidently the square of ± («2 f 2 y).
Check. (±3)2 = 1+4 + 4 = 9.
2. rind the cube root of x^ + 6^;^ + 12 x^ + Sy^
1. •.• (/ + ri)8 =p + 3/2n + 3/7i2 + n^ § 82
2. and •.• this polynomial can be arranged in a similar form, viz.,
(x2)3 + 3 (x2)2 . 2 y + 3 a;2 (2 2/)2 + (2 y)^
3. .. it is evidently the cube of x2 + 2 y.
Check. 33 = 1 + 6 + 12 + 8 = 27.
3. Find the square root of
a'' + 4.P + 9 c^ + 4: ab  6 ac  12 be.
1 '■' [±{x + y + z)]2 = x^\y^ + z^ + 2xy + 2yz +2zx,
2. and •.• this polynomial can be arranged in a similar form, viz.,
a2 + (2 6)2 + (_ 3 c)2 + 2 a (2 6) + 2 a( 3 c) + 2 (2 6) ( 3 c),
3. .. it is evidently the square of ± (a + 2 6  3 c).
Check. 02 = 1+4 + 9 + 46 12 = 0.
4. Find the fifth root of
^10 5a'b\ 10 a%'  10 a*b^ + 5 a^b^  b^
1. •.• there are 6 terms, and the polynomial is arranged according
to the powers of a and 6, it is the 5th power of a binomial (§ 82) whose
first term is a^ and whose second term is — 6, if it is a 5th power.
2. But (a2  6)5 equals the given polynomial. (Expand it.)
FACTORS.
95
EXERCISES. XLIX.
Extract the square roots of exs. 16.
1. ^x'ix' + ^^.
2. 4: a 12 Vab \9b.
3. Am^ — 12mx^ + 9x\
4. 9 a'b'  30 a'b'c' + 25 a'c^'.
5. 4 m^ + 4 mn + 12 mp \ n^ \ 6 np { 9p\
6. 4 ic*  12 a^V + 16 ^^y + 9 xY  24 a;?/* + 16 y\
Extract the cube roots of exs. 712.
8. m^ {^ m'^n + 12 m^Tv^ + 8 ?i^
9. 8 x8  84 cc^y + 294 xif  343 2/».
10. 8ic^ + 12a;5 + 18ic4 + 13a;3 + 9a;2 + 3^ + l.
11. 77^6 _ 3 ^^5 _ 3 ^4 _^ ;L;l_ ^8 ^ g ^2 _ 12 m  8.
12. x'  12ccs + 54 cc*  112 a;^ + 108 a;^ _ 48 a^ + 8.
Extract the fourth roots of exs. 13, 14.
13. 3^5^ ^' + I ^'2/ + I xy + ^x7f^ if y\
14. 16 a;4  96 xhj + 216 a^y  216 xf + 81 2/*.
■ Extract the fifth roots of exs. 15, 16.
15. 80a;^80a;^ + 32a;S40ic2_j_i0a;l.
16. x^'  5 ^8y _^ 5 ^Y  1 xY + A ^ V'  ^V 2/'
Extract the sixth roots of exs. 17, 18.
r. a« _ 12^5 + 60a*  160^8 + 240^^  192a + 64.
a6 _ 2 a^Z* + I a'b^  fa a%^ + ^j a^'' ^ab' + ^^^ b\
96 ELEMENTS OF ALGEBRA.
129. Square root by the formula f ^ + 2 f n + n^. The subject
is best understood by following the solution of a problem.
1. Required the square root of 4 a?* — 12 x^y + 9 y^.
Let / = the found part of the root at any stage of the
operation, and
n = the next term to be found.
Then (f+ny=f' + 2fn + n\ § 82
The work may be arranged as follows :
Root =±{2x''^y)
Power = 4 cc* — 12 £c V + 9 ^2 contains /^ + 2fn + n^
f= ^
2/=4£c2 Ux^y + dy"" " 2fn + n^
2f+n = 4:X^Sy Uxhy + dy"" = '^
Explanation. 1. If a root is arranged according to the powers of
some letter, the square obtained by ordinary multiplication will be so
arranged (§ 65).
2. .. the square is arranged according to the powers of x, so that
the square root of the first term shall be the first term of the root.
3. ••• 4 x* = the square of the first term, the first term is 2 x^.
4. Subtracting /2, the remainder, —12x^y + 9?/2, contains 2/n + n^.
5. Dividing 2fn{i.e.,  12x^y) by 2 /{i.e., 4a;2), w is found to be
3y.
6. •.• /2 = 4 a:*, and 2/n + n^ =  12 x'^y \ 9 y^, .. the sum of these
is the square of ±{2x^  Sy).
Check. Jjetx = y = l. Then ( 1)2 = 4  12 + 9 = L
We might, after a little practice, detach the coefficients.
In the above example it would be necessary to remember
that the powers of x decrease by two, while those of y
increase by one.
:EJ.g., 412 + 9 23
4
4 12 + 9 ±(2x''3y)
43 12 + 9
I
FACTORS. 97
2. Kequired the square root of
Root =±(a62 +2c)
Power = a2_2tt62 + &4^4o[c_4 52c__4c2 contains /2 + 2/w + n2
2/ =2 a 2a62 + 64+... u 2/n + ri2
2/+7i2a62 2a62f54 ^ ^^
2/ =2 a2 62 4 ac 4 62c+4 c2 contains 2/n + n2
2/+n=2a2&2j2c 4ac4b2c+4c2 = '^
Explanation. 1 . See p. 96 for explanation down to 2/= 2 a  2 62.
2. •.• /2 = a2, and
2/n + n2 =  2 a62 + 6*,
.. (/ + n)2 = a2  2 a62 + 6*, the square of a  62.
3. •.• a — 62 has now been found, it may be designated by/.
4. .. 4ac — 4 62c + 4 c2 contains 2/n + ^2, the square of a — 62
having been subtracted.
Check. Let a = 6 = c = 1. Then 22 = 12+1 + 44 + 4 =4.
Or let a = 1, 6 = 2, c = 3.
Then (1  4 + 6)2 = 32 = 9
and 1  8 + 16 + 12  48 + 36 = 9 ;
d so for any other arbitrary values.
y^ffU
130. Extension of the definition of root. If an algebraic
expression is not the product of r equal factors, it is still
said to have an rth root. In such a case the rth root to n
terms is defined to be that polynomial of n terms found by
proceeding as in the ordinary method of extracting the rth
root of a perfect rth power.
kl^L ■^•^•' ^^^^ square root of 1 — x to 5 terms is
IB ± (1  ix  1x2  ^^r,z _ _^x* ...).
IH In the same way we may speak of the square root of
ilTOmibers which are not perfect squares. Thus the square
root of 2 to two decimal places is 1.41 ; to three decimal
places, 1.414, and so on. We may also speak of the cube
98 ELEMENTS OF ALGEBRA.
EXERCISES. L.
Extract the square roots of exs. 116.
1. x^ + 2x^x + i.
2. l + 8a + 22a2 424a» + 9a^
3. 9(a^iy12(a^l)a^Aa\
4. x^6x^{4.x^ + 9x^12x+4:.
5. x^2 ax^ + aV _ 2 to^ + 2 abx^ + ^>l
6. 25a^ + 9b^ + c^ + 6bc10ca~S0ab.
7. 10:c* 10x8 12^5 + 5ic2 + 9cc«2cc + l.
8. 9 ic«  12 aa^^ + 4 a^x^ + 6 a^x^  4 a*cc* + a^x^.
9. 16  Sr/i  237?^2 + 227^1^ + 5m*  12m^ + 4m^
10. 9 a^^*  12 a'^b' + 4 a%' + 24 d^^^^^^^ _ iq ^3j4^8 __ ^g ^4^,2^6^
11. 9a^12 a%^ + 4 &« + 24 aV  16 b^c"^ + 16 c«  30 aH
+ 2()bH4.0cH\2^d\
12. 4 x^y"^  12 xHf + 9 i«y + ^.x'^ifz — Qxh/z + x'^y'^z'^ 
16 a;3z/2^8 + 24 xy^z^  8 cc^^^* + 16 i/z^ + 4 x^'yz  6 ic^^/^^ +
2 cc^T/^^ — 8 £c'^2/s* + x'^z'^.
13. 1 + a; to 4 terms.
14. 1 — 2 ic to 4 terms.
15. 4 + 2 a; to 4 terms.
16. 9 a^ + 12 ax to 2 terms.
17. Find x so that a* + 6a8 + 7a2 — 6<^4a; shall be a
perfect square.
18. Find m so that 4 cc* + 4 cc^ + mx^ 4 4 cc + 4 shall be
a perfect square.
19. Find m so that 9 a* + 12 a« + 10 a^ + w«^ + 1 shall
be a perfect square.
20. Show that the square root of 2\_{in + ny + (m^ + n*)']
is 2 (m^ + 71^ + m?z).
FACTORS. 99
131. The square roots of numbers are similarly found.
Kequiied the square root of 547.56.
Eoot =2 3. 4
Power = 5'47.56 contains p + 2fn + n^
f^ = 4 00.00
2/i =40 147.56 " 2/i% + V /i = 20
2/, + % = 43 1 29.00 = " %= 3
2/2 ==46 18.56 contains 2/2/^2 + ^2^^ /2 = 23
2/2 + ^2  46.4 18.56 = " n^= 0.4
Explanation. 1. •.• the highest order of the power is lOO's, the
highest order of the root is lO's, and it is unnecessary to look below
lOO's for the square of lO's.
2. Similarly, it is unnecessary to look below I's for the square of
I's, below lOOths for the square of lOths, etc.
3. The greatest square in the lOO's is 400, which is the square of
20, which may be called /i (read "/one "), the first found part.
4. Subtracting, 147.56 contains 2fn + w^, because /^ has been sub
tracted from/2 ^ 2fn \ n^, where / stands always for the found part
and n for the next order of the root.
5. 2fn + w2 is approximately the product of 2/ and n, and hence,
if divided by 2/, the quotient is approximately n. .. n = 3.
6. .. 2/ f n = 43, and this, multiplied by n, equals 2/n  n^.
7. ••• /2 has already been subtracted, after subtracting 2fn { n^
there has been subtracted p f 2/w + n^, or (/ + n)2, or 232.
8. Calling 23 the second found part, /2, and noticing that
f2=fi + m, it appears that 23^, or f^^, has been subtracted.
9. .. the remainder 18.66 contains 2/2^2 1 ^2^.
[10. Dividing by 2/2 for the reason already given, 712 = 0.4.
[11. .. 2/2 + ?i2 = 46.4, and 18.56 = 2/2^2 + n2^, as before.
^12. Similarly, the explanation repeats itself after each subtraction.
EXERCISES. LI.
Extract the square roots of exs. 16.
1. 958441. 2. 7779.24. 3. 32.6041.
4. 24.1081. 5. 0.900601. 6. 0.055696.
100 ELEMENTS OF ALGEBRA.
132. Cube root by the formula f ^ + 3 f ^n + 3 fn^ + n^.
Eequired the cube root oi S a^ — 12 a^b + 6 ab^ — b^.
Let / = the found part of the root at any stage of the
operation, and
n = the next term to be found.
Then (/ + ny =/' + Sfhi + 3fn^ + n\ § 82
The work may be arranged as follows :
Root = 2 a — 6
Power = 8 a^ 12 a^b\Q a¥¥ contains
P =
8a3 /3+ 3/2^ 4. 3/^2 +^3
3/2
3/n
3/2 + 3/n
12 a26+6 a6263 contains
+n2
+ n2
3/2n+3/n2+n3
12 a2
6ab
12a26a6
12a26+6a6263 =
+ 62
+ 62
Explanation. 1. The cube is arranged according to the powers of
a and 6 for a reason similar to that given in square root.
2. •.■ 8a^ = the cube of the first term, the first term is 2 a.
3. Subtracting f^, the remainder, — 12 a26 + 6 a62 — 6^, contains
3/2n + 3/n2 + n^
4. Dividing by 3/2 {i.e., 12 a^), n is found to be — 6.
5. •.•/= 2 a, and n =  6, .. 3/2 + 3/n + n2 = 12 cfi  6 a6 + 62.
6. Multiplying by w,  12 a26 + 6 a62  6^ must equal 3/% + 3/w2
+ n^. This together with/^ completes the cube of / + n.
Check. Let a = 6 = 1. Then is =: 8  12 + 6  1 = 1.
EXERCISES. LII.
Extract the cube roots of exs. 16.
1. 8a836a26+54a&2_27^'^
2. a^x''  12 a'bx'' + 48 ab^x''  64 b''x\
3. 1  6ic + 2l£c2  44£c8 +. 63£c*  Ux^ + 27 (r^
4. a^2 a^b + I a'^b^  f a^"" + /^ ^'^^  /t «^' + 7k ^'•
5. ^6  12 a% + 54 a'^^^^ _ 1^2 a^Z*^ + 108 a^J*  48 ab^ + 8 b\
6. iB8+3a;22^_e^2_^3^2/2_12cc?/+12ic+2/^6/+l27/8.
FACTORS.
101
133. The cube roots of numbers are found by the same
general method.
W Eequired the cube root of 139,798,359.
Root =519
Power = 139,798,359 cont's/3+3/2n+3/n2 + ?i8
P = 125,000,000
3/2
L
3/n
+ n2
3/2 + 3/n
+ n2
14,798,359 contains 3/2n + SM + n^
/i = 500
750,000
15,100
765,100
7,651,000 = 3/2n + 3/w2 + n^
780,300
13,851
794,151
7,147,359 contains 3/2n + 3/n2 + n^
h = 510
7,147,359 = 3/2n + 3/n2 + n^
na =9
Explanation. 1. •.• the highest order of the power is hundred
millions, the highest order of the root is lOO's (why ?), and it is unnec
essary to look below millions for the cube of 100 's. (Why ?)
2. Similarly, it is unnecessary to look below lOOO's for the cube of
lO's, below I's for the cube of I's, etc.
3. The greatest cube in the hundredmillions is 125,000,000, the
cube of 500. .. 500 may be called /.
4. Subtracting, 14,798,359 contains 3/2n + 3/n2  n^. (Why ?)
5. This is approximately the product of 3/2 and w, and hence if
ivided by 3/2 the quotient is approximately n. .. w = 10.
6. .. 3/n + n2 = 15,100, and 3/2 + Sfn + n2 = 765,100, and this,
multiplied by n, equals 3/2n  3/n2 + n^.
7. ••• P has already been subtracted, after subtracting 3/2w + 3/n2
I +n^ there has been subtracted (/ } n)% or 510^.
8. Calling 510 the second found part, /a, it appears that fz^ has
'A been subtracted. .. the remainder contains 3/2n + 3/n2 + n^.
■ft 9. The explanation now repeats itself as in square root.
II
102 ELEMENTS OF ALGEBRA.
EXERCISES. T.TTT
Extract the cube roots of exs. 14.
I. (a) 10,077,696. (b) 31,855,013. (c) 125.751501.
2. (a) 367,061.696. (b) 997.002999.
3. (a) 551. (b) 975. Each to 0.001.
4. (a) 2. (b) 5. Each to 0.0001.
BEVTEW EXERCISES. UV.
Extract the cube roots of exs. 13.
1. 1 — a; to 5 terms.
2. 64 — 48 X + 9 arHo 3 terms.
3. a» + 9 a^b + 36 a?b'' + 84 a«ft» + 126 a^h^ + 126 a***
+ 84 a*b^ + 36 a%'' + 9 a*« + b\
4. Factor x* + a;^ — 4ic* — 4.
5. Show that xyz (x^ + y*\z^ — {fz^ + s*a;« + x^f) =
(x^ — yz) {f — zx) (z^ — xy).
6. Divide the product ol x^ \x — 2 and x* + a: — 12 by
the sum oi 2 x^ \ 6x ^ 1 and 2 — x (10 + x).
7. Find the square root of
(x + 3) (X + 4) (x + 5) (X + 6) + 1 .
8. Solve the equation
7  2 J6  3[5  2(4  3 + 2x)]; = 1.
9. Find the square root of
(2 a by 2(2a^ 5ab + 2b^ + (a 2by.
10. Find the three roots of the equation x^ — x^ \ 1 = x.
II. Also of the equation a;» + 9 x^ + 8 a;  60 = 0.
12. If a = — 3, 5 = 0, c = 1, <^ = — 2, find the numerical
value of a 2\b + Sic 2a (a b)]\ 2 a (b\3c)\.
CHAPTER Vn.
fflGHEST COMMOX FACTOR AXD LOWEST
COMMON MULTIPLE.
L m^SST COMMON FACTOR.
134. The integral algebraic factor of highest degr^
cammon to two or more integral algebraic expressions is
called their IrighrBt i wiib i m ^Mrtor.
S.g., o^ is the higiwst common tictor ol c\riauid2an^,
a6 " " " {aHf " rt*6».
Consider, abo, 2(«»6»)aiid4(6«i^.
Here 2 («»  6») = 2 (a  6)(a2 + oft + ft*), or  2<&a)(tf>+afr+6^r
and 4<^ai*) = 4(6a)(6 + o), • 4(a6)(a+6).
6or6 — aisa commnn fatAar, and there hei^g no
edraic fMCtor, either is cdkd tike h^^iert eoHBon
uor. Tiicre La a commm nnmerieal fKtor, 2, hot audi fattaa hsie
hing to do with the a^efaiaic dir^iMli^ of tiie ezpraaaioDs, and
nee may be Defected.
In the last example, it is iK>t osnal to state both aBSWOS, a — b and
 a, because a —b= — 1(6 — a); that is, the two are the same
pnr tor a nnmeiical factor, and numencal £Mt(ffs are not eon
135. The arithmetical greatest common diTisor must not
confounded with the algebraic highest common factor,
ough these are often called by the same name. The
common factor has reference only to the degree of
erpression.
103
 conf
iKhest
IKexp
104 ELEMENTS OF ALGEBRA.
E.g. , consider the highest common factor otx^ — Sx + 2 and x^—x2.
Here a;2  3x + 2 = (x  2) (x  1), or (2  x) (1  x),
and x2x2 = (x2)(x + l) "  (2  x) (x + 1) ;
hence, the highest common factor is x — 2, or 2 — x. Now if x = 5,
the expressions become 12 and 18, and the highest common factor
becomes 3, or — 3, although 6 is the greatest common divisor of 12
and 18.
The highest common factor is occasionally used in reduc
ing fractions to their lowest terms.
136. Factoring method. The highest common factor of
expressions which are easily factored is usually found by
simple inspection.
E.g. , to find the highest common factor of x2 — 3 x + 2, x^ — x^  2 x,
and i x2 + I X — 3, we have :
1. x2  3x + 2 = (X  2) (X  1).
2. x3x22x = x(x2)(x + 1).
3. ix2 + x3 =i(x2)(x + 3).
4. .. the highest common factor is x — 2, or 2 — x.
EXERCISES. LV.
Pind the highest common factor of each of the following
sets of expressions :
2. 15 mnx^, 17 mx^yz, f abcx^^z.
3. l()xhjz, Ibaxhjz^, 20 amxz'\
4. x^ — 2/^, if — x^, x'^ — %xy + 1 y^.
5. x^ — y^, 2/^ — a^^ a^^ — i^y — iy^
6. x''4., x'^xQ>, 25xSx''.
7. 2x^ — xy — y^, 4: x^ \ 10 xy \ 4: y\
8. 6a^ + 19ab7b% 2 a^ \ ab  21 b\
9. 4a2(et868)^ ^ab''{3a''^ab{2b').
FACTORS AND MULTIPLES.
105
137. If the factors of one of several algebraic expressions
are known, but those of the others not, it is easy to ascer
tain, by division or by the Remainder Theorem, if the known
factors of the one are factors of the other.
E.g., to find the highest common factor of 1 — cc^ and llSx'^ — 4x'
+ 2X11L
Here 1  x2 = (1  x) (1 + x), or. (x  1) (x + 1).
But X — 1 is a factor of 113 x'^ — 4 x^ + 2 x — 111, by the Remainder
Theorem (§ 103), while x + 1 is not. .. x — 1 is the highest common
factor.
I
EXERCISES. LVI.
rind the highest common factor of each of the following
sets of expressions :
1.
x'  //^ x^  /.
2.
x^4., a;^ 4x2 16.
3.
^24, x^ + 7x2 + 100.
4.
cc^ + 1, x^ + ax^ + ax + 1.
5. x^  3 X 4 2, x2  9 X + 14.
6. x29x + 14, 2x^5x2441.
8. x2  4, 5 X* + 2 x^  23 x2  8 X + 12.
9. 2x2 5x7/ + 32/2, 6x3 23x^2/ + 25x7/2 6/.
(10. ^8 _ 53^ yi _ ^2^ 117 ^3 _ 117 ^2^ _ 231 ab + 231 h\
111. x^l, x2l, 293xS200x'^ + 7x850x225x25.
[12. 1x^ x^1, x^l^ZxZx", 247x2240x7.
[13. x^  32, 16  X*, x2  9x + 14, X*  4x2 + 6x  12,
[14. x« + 1, x2 + 2x + 1, x^ + 1, 324xs ^ 2^r^r^^ j^ 100x»
+ 204x227.
106 ELEMENTS OF ALGEBRA.
138. Euclidean method. In case the highest common fac
tor is not readily found by inspection of factors, a longer
method, analogous to one suggested by Euclid (b.c. 300)
for finding the greatest common divisor, may be employed.
139. This method depends upon two theorems :
1. A factor of an algebraic expression is a factor of any
multiple of that expression.
Proof. 1. Let a, 6, p, q be algebraic expressions, p and q being
the factors of b.
2. Then 6 = pq.
3. .. ab = apq. (Why ?)
4. I.e., if p is a factor of b, it is a factor of any multiple of &, as ab.
A similar proposition is readily seen to be true for num
bers. E.g., 5 is a factor of 35 ; and since multiplying 35
by any integral number does not take out this 5, therefore,
5 is a factor of any multiple of 35.
2. A factor of each of two algebraic expressions is a factor
of the sum and of the difference of any multiples of those
expressions.
Proof. 1. Let b = pq and 6' = pq'.
2. Then ab = apq " a'6' = a'pq'. (Why ?)
3. .. ab ± a'b' = apq ± a'pq' =p{aq ± a'q'). (Why ?)
4. I.e., if p is a factor of b and b\ as in step 1, then it is also a
factor of the sum and of the difference of any multiples of b and 6',
as ab and a'b'.
A similar proposition is true for numbers. E.g., 5 is a
factor of 60 and of 35, and also of the sum and of the dif
ference of any multiples of these numbers.
140. The Euclidean method will best be understood by
considering an example.
FACTORS AND MULTIPLES. 107
Required the highest coninion factor of
x^ — x^ {2x^ — x + 1 and x^ + x^ ^ 2x^ + x i1.
x4x3+2x2ic+la:44 x^+2x'^+ x + l[l
x4 x^+2x^ x + 1
2a;l 2x3 +2x
x2 +Ix4x3+2x2ic+llx2a;+l
x*
+
X2
X3 +
X3
X2
x+1
X
X2
X^
+ 1
+ 1
Explanation. 1. The h.c.f. of the two expressions is also a factor
of 2x3+2x, by th. 2 (§ 139).
2. It cannot contain 2 x, because that is not common to the two
expressions.
3. .. 2 X may be rejected, and the h.c.f. must be a factor of x2 + L
4. x2 f 1 is a factor of x* — x^ f 2 x2 — x + 1, by trial.
5. " " " 2x3 + 2x.
6. .. " " " x* + x3 + 2x2 + x + 1. (Why?)
7. .. " is the h.c.f. (Why?)
141. In order to avoid numerical fractions in the divi
sions, it is frequently necessary to introduce numerical
factors. These evidently do not affect the degree of the
highest common factor.
E.g., to find the highest common factor of 4x3 — 12x2 __ n^ — 3
and 6x3 13x2 + 9x 2.
6x313x2+ 9x2
2
4x312x2 + llx3l2x326x2 + 18x4[3
12x336x2 + 33x9
5 l0x215x + 5
~ 2x2 3x+l4x312x2+llx3 2x3
' " 4x3 6x2+ 2x
 6x2+ 9a;_3
 6x2+ 9a;3
Here the introduction of the factor 2 and the suppression of 5 evi
dently do not affect the degree of the highest common factor.
108 ELEMENTS OF ALGEBRA.
142. In practice, detached coefficients should be used
whenever the problem warrants.
E.g., to find the highest common factor of
3 xSy + 3 x^y + 2 x^y  x'^y  xy and 2x'^ + 9x^ + 9x'^ + 1 x.
Here x is evidently a factor of the highest common factor. It may
therefore be suppressed and introduced later, thus shortening the
work.
But ?/ is a factor of the first only, and hence may be rejected
entirely.
The problem then reduces to finding the highest common factor of
3x* + 3x3 + 2x2x 1 and2x3 + 9x2 + 9x + 7.
3+3 +
2
1 
1
2
23
2 + 9 + 9 + 76+ 6 +
4
2 
6 + 27 +
27 +
21
21 
23
23
2
2
42 +
46 +
46 +
4[21
42 + 189 + 189 + 147
 143l 
143
143
143
1 +
1 +
l2 + 9 + 9 + 72 + 7
2 + 2 + 2
7 + 7 + 7
.. x(x2 + x + 1) istheh.c.f.
7+7+7 '
143. The work can often be abridged by noticing the dif
ference between the two polynomials.
E.g., m the case of x*  2x3 + Sx^  8x + 6 and x*  4x3 + 3x2 _
6 X + 6. Here we have :
1
1
2 + 3
4 + 3
8 + 6
6 + 6
2)2
2
1  1
X2  1 = (X + 1) (X  1).
By the Remainder Theorem x  1 is a factor of each expression, and
X + 1 is not ; .. x — 1 is the highest common factor of the expressions.
FACTORS AND MULTIPLES.
109
144. The highest common factor of three expressions
cannot be of higher degree than that of any two ; hence,
the highest common factor of this highest common factor
and of the third expression is the highest common factor of
all three. Similarly, for any number of expressions.
EXERCISES. LVII.
Find the highest common factor of each of the following
sets of expressions :
1. ic» — 2 ic + 4, x^ \x^ + 4:X.
2. 2x^ + 2x4., x^3x{2.
3. x^ + 4., x^2x^'{x^ + 2x2.
4. x^4:0x + 63, x^7x^h63xSl.
5. x^ + y^
2/^ x' + xV + xy + y\
6. x«(6a: + l)£c, 4£c»2cc(3£c + 2) + 3.
7. £c415£c2 428x12, 2ic«15cc + 14.
Ks. Ix^ \^x'1x^\^,2x^x''2x^\.
IH 9. x^^x 117, a;*  13ic8  a;2 + I4aj  13.
I 10. 63 a*  17 a^ + 17 a  3, 98 a" + 34 a^ + 18.
11. a;2_^4a^21, cc2 + 20£c + 91, 2x'' ^ix^ 'l^ x.
12. '^x^l^x^^lx''2x, 6£c5llic* + 8iK»2ic2.
3. 9a24^>2 + 4^>cc2, 2^ ^ c" \Z ah 3hc 3ac.
4. {a  b) {a^  c2) (ac) {a^  b^), a^  b', ab  b^
— ac \ be.
5. cc310(a;2 + 3)+ 31a!, ^^^(x  11) + 2(19 :b  20),
a;3_9cr2 + 26£c24.
6. a*52 + 4 a^b^ + 3 a%'' 4.ab^4. b\ a^'b + 3 a%^  a'^h^
7. 3a^7 ab + 2b^ + bac6bc + 2c% 12 a^  19 ab
+ 5b'' + llacllbc + 2c^.
110 ELEMENTS OF ALGEBKA.
II. LOWEST COMMON MULTIPLE.
145. The integral algebraic multiple of lowest degree
common to two or more algebraic expressions is called
their lowest common multiple.
E.g.^ a^b^cd is the lowest common multiple of a^bc and ab^^d.
Similarly, ± (a + b)^ (a — b) is the lowest common multiple of
a2 b%b  a, and (a + 6)2. For
1. a2  62 = (a + 6) (a  6).
2. 6  a =  (a  6).
3. (a + 6)2 = (a + 6) (a + 6).
4. .. either (a + 6)2 (a — 6) or (a + 6)2 (6 — a) contains the given
expressions and is the common multiple of lowest degree.
The lowest common multiple of algebra must not be con
sidered the same as the least common multiple when
numerical values are assigned, ^.ff., the lowest common
multiple of a \ b and a — b is (a \ b) (a — b) ; but it a = 6
and b = 4:, the least common multiple of 6 + 4 and 6 — 4 is
simply 6 + 4.
146. So far as the algebraic multiple is concerned, numer
ical factors are not usually considered.
E.g., aWc is the lowest common multiple of 2 ab^c, icfib, and 15 a6.
The lowest common multiple is used in reducing fractions
to fractions having a lowest common denominator.
147. Factoring method. The lowest common multiple is
usually found by the inspection of factors.
E.g., to find the lowest common multiple of x2 _ 12 x + 27, x2 + x
 12, and 15  2 X  x2.
. 1. x2 _ 12 X + 27 = (X  3) (X  9).
2. x2 + X  12 = (X  3) (X + 4).
3. 152xx2= (x3)(x + 5).
4. .. t (x — 3)(x + 4)(x + 5)(x — 9) is the lowest common multiple.
In practice, the result should be left in the factored form.
FACTORS AND MULTIPLES. Ill
EXERCISES. LVIII.
Find the lowest common multiple of each of the follow
ing sets of expressions :
1. —lOa^xyz, hxSjz^j \a^xy^z.
2. x^ + 2/^ ^ + 2/j xy — x^ — y^.
3. a^ 4. ^2 _ 2 ab, 52 _ ^2^ a h.
4. 27  12a^ + a;2, ic2 + 2x  15.
5. ic* + 4, 2  cB^^ a;2 + 2, ic  V2.
6. a^2 ^ ic  12,  36 + 13 X  ic2, £c2  16.
7. ic^ + 2/^ + 3 £c?/ (cc + ?/), cc^ + 2/^ ^ + 2/
8. 2 ic?/ — ic^ — ?/^, 2xy \ x'^ \ y'^, x^ — y^, x ■\ y.
148. Highest common factor method. Since the highest
common factor contains all of the factors common to two
expressions, it may be suppressed from either of them and
the quotient multiplied by the other to obtain the l.c.m.
Proof. 1. Let x — af,
y = ¥,
in which / is the highest common factor of x and y.
2. Then the lowest common multiple is evidently abf\
i.e., it is y multiplied by a.
E.g., to find the lowest common multiple of 2^3 + Sx^ — 3x — 27
and 2 x3 + 12 x2 + X  45.
2x3 + 12x2+ x45
2x3+ 8x23x27
2 1 4 x2 + 4 X 
18
2x2 + 2x
 92x3 + 8x23x27x + 3
2 x3 + 2 x2  9 X
6x2 + 6x27
6x2 + 6x27
: (2 x3 + 12 x2 + X — 45) (x + 3) is the lowest common multiple.
112 ELEMENTS OF ALGEBRA.
EXERCISES. LIX.
Find the lowest common multiple of the sets of expres
sions in exs. 115.
1. a;i2 + a;^ x^'' + x\
2. 3 a«  11 ^2 + 4, 6 a2  a  2.
3. x' + Sx' + x + S, x^8x + 3.
4. 6a;2 + 13x + 6, 10cc23 + 13a;.
5. x^ + 2ax + a% x^ + ab ^ (a + b)x.
6. Gx^ + llx^'dx + l, 2a;2 + 3x2.
7. x^x^ + x^ — x4.,x^x^ + 2x8.
8. x^ + l + 3{x^ + x), x^ + l ^4:{x^ + x){6x^
9. 3x^15ax^ + a^x5a^, 6£c*  25aV  9 a*.
10. x^ + 20x + 91,352xx'',x^ + 6x^6x'' + 6x7.
11. 2a;^2x3a;24a;7, 2ic* + 6ic«17£cH8a;35.
12. x^ + x^ + x + 1, 2 x^  3 x^ \ 4:x^ + 2 x^  3 X + 4:.
13. a;^ + a;«a;*^6£c2_5^_7^ ic^  x«  ic^ + ic*  6 a;»
— X + 7.
14. cc"^ + 2a;«  3 £c^ + ic^ + 2 a;  3, x^ + 4 a;^  7 cc^ + a;^
+ 4(K7, x^ + 1,
15. 4^2(3 ^^2) (27 a + 18), 12a3 a(8a + 27) + 18,
6(3a2) + 27a«8.
16. Find all of the algebraic expressions whose lowest
common multiple is ic^ — 4 xy\
17. Prove that the product of the lowest common mul
tiple and the highest common factor of two expressions is
the same as the product of the two expressions.
18. Investigate ex. 17 for the case of three expressions.
19. Find the lowest common multiple of a^ — 1 and
a^ — 4 a + 3. Can the result be checked by letting a = 5,
7, or any odd number above 3 ? Explain.
FACTORS AND MULTIPLES. 113
REVIET77 EXERCISES. LX.
1. Factor x(x — 1) — a(a — 1).
12. Solve the equation 4 ic^ + 1 = 4 cc.
3. Solve the equation 6x^ + llx — 7 ~0.
4. Extract the square root of cc^ + 1 to 3 terms.
5. Give a complete description of this expression as a
function of x and y: x^ { 3 xSj + 4 xhj'^ + 3 xy^ + ?/*.
6. Show that the difference of the squares of any two
consecutive numbers is equal to the sum of the numbers.
7. Find the lowest common multiple of 2x^ \ x'^ \ ^x^
{4.x^ + 2x + S and 6x^  5x^ + 12x^  Sx^ + 5x  6.
8. Find the lowest common multiple otx^ — x^ — 2x — l,
2x''x'^2x'2xl, and Zx^ _ 4a;8 + 6a;2  7 a;  8.
9. Find the highest common factor of x* + 2 ic^ — 5 a;^ }
15a; + 12, a;4 + 5a;3 + 5cc2^8ic + 16, and ic^ + 6 a;8 + 10 a;^
+ 4a;16.
I^B10. In finding the highest common factor of two alge
■flfi'aic expressions, by what right may a factor be suppressed
I in one if it is not a factor of the other ?
I^B 11. The highest common factor of two expressions is 4 x"^
^^ a?, and their lowest common multiple is 4a;'^ — ha^x^ + a*.
j One of the expressions is 4 a;^ + 4 ax'^ — a^x — a^. Find the
other.
12. Assign such values to a and h that the arithmetical
least common multiple of a^ — b^ and a^ \ b^ \ 2 ah (a { b)
shall not be the value of the algebraic lowest common
igultiple.
I^rl3. Prove that the difference between the cubes of the
sum and difference of any two numbers is divisible by the
sum of the square of the smaller number, and three times
the square of the laiger.
■
CHAPTER VIII.
FRACTIONS.
149. The symbol ? in which b is not zero, is defined to
mean the division of a by b, and is called an algebraic
fraction.
Hence, the algebraic fraction  represents a quantity
which, when multiplied by ft, produces a.
The terms of the fraction  are a and b, a being called the
numerator and b the denominator, and either or both may be
fractional, negative, etc.
The case in which h equals zero is discussed later.
There are two definitions of a fraction usually given in
arithmetic : (1) The fraction y is a of the b equal parts of
unity ; (2) The fraction  is one bth. of a.
Neither of these arithmetical definitions includes, for example,
2 2 3
, , y=, etc., for "2 of the — 3 equal parts of unity " means
3 f v2
nothing, and "one V2th of 3" is equally meaningless. Hence the
broader algebraic definition.
In the first arithmetical definition above given, 6 names the part
and hence is called the denominator (Latin, namer), and a numbers
the parts and hence is the numerator (Latin, numberer). Hence the
origin of these terms.
The fraction  is, therefore, read "a divided by 6," although the
reading "a over 6" is generally used in various languages, and is
sanctioned by most teachers on the ground of brevity.
114
FRACTIONS. 115
I. REDUCTION OF FRACTIONS.
150. Theorem of reduction. The same factor may be intro
duced into or cancelled from both numerator and denomina
tor of a fraction without altering the value of the fraction.
Given the fraction ■> and m any factor.
To prove that  = — j that is, that the factor m may
mo a
be introduced into both terms of  or cancelled
from both terms of — r •
mo
Proof. 1. b = a. Def. of frac.
b
2. .*. mb = ma. Ax. 6
b
3..: ^ = ^. . Ax. 7
b mb
An algebraic fraction is said to be simplified when all
common algebraic factors, and hence the highest common
factor, of both numerator and denominator have been sup
pressed, and there is no fraction or common numerical
factor in either.
(j2 I 2 a6 4 &2
E.g., the fraction z — — is simplified when reduced to the
a^ + 0^
form — ; — by cancellme the factor a + b.
J a2  a6 + &2 ^ &
I^BBut the fractions and — are not simplified.
M " The student should notice that the theorem does not
allow the cancellation of any terms of the numerator and
denominator. No factor can be cancelled unless it is con
116 ELEMENTS OF ALGEBRA.
Usually the factors common to the two terms of the
fraction can be found by inspection and cancelled ; other
wise the highest common factor of both terms is found and
then cancelled.
— a^'^cd^
Examples. 1. Simplify the fraction — •
1. Cancelling a^^ 52^ ^^ ^nd d^, the fraction reduces to
h
2. And since there are no other common factors, and the terms are
integral, the fraction is simplified.
Check. Let a = 3, 6 = d = 2, c = 1. Then " 27 • 4 • 1 • 16  6
9.8.1.8 2
2. Simplify
1. This evidently equals
{a + hY
{a + 6) (a  h)
a\h
2. Cancelling a + 6, this reduces to
a —
3. And since there are no other common factors, and the terms are
integral, the fraction is simplified.
Check. Let a = 2, & = 1. Then f = f . (If a and h are given the
same values, the denominator becomes zero, a case excluded, for the
present, by the definition of fraction.)
3. Simplify 3^^ + 26. 77
3x^l{)x + l
1. A factor of each term of the fraction is a factor of their differ
ence, 36X84 (§ 139, 2).
2. Hence of 3 x — 7, because the terms of the fractions do not con
tain 12.
3. Hence, if there is a common factor, it is 3 x — 7, because this is
irreducible.
4. By substituting arbitrary values this is seen to be a probable
factor, and the fraction reduces by division to
x+ 11
x1 '
Check. Let x = 2. (Why not 1 ?) Then ^ = — .
FRACTIONS. 117
4. Simplify o <i , A 2 I rz nr '
1. Here the simple factors are not as easily determined as the
highest common factor, x^ \ x + 2.
2 X \ 7
2. Cancelling this, the fraction reduces to
3x + l
3. .. the fraction is, by definition, simplified.
Check. Let x = 1. Then ff = f .
If the student has not studied Appendix III, ex. 5 may
be omitted.
I
. Simplify a'{io)!'Hc~a)e^ai )
^ ^ (a — b)(b — c) (c — a)
. By the Remainder Theorem (§ 104) a — & is a factor of both terms
of the fraction. (We try a — 6 because if there is any common factor
it must be a — 6, 6 — c, or c — a.)
2. Hence, because both terms are cyclic, b — c and c — a are factors.
3. And since the numerator is of the 4th degree, the other factor is
a linear cyclic factor. Hence, it is n(« + 6 + c).
4. Hence, the numerator is n {a + b + c){a — b){h — c){c — a). But
by substituting the values a = 2, 6 = 1, c = 0, n is seen to be 1.
5. Hence, the fraction equals a + 6 + c.
Check. Let a — S, b = 2, c = l (values different from those used
— 12
for finding n). Then = 6.
151. General directions for simplifying fractions. The pre
ceding fractions were simplified in different ways. While
there is no general method of attack, and the student must
use his judgment as to the best plan to pursue, the follow
ing directions are of value :
1. Cancel monomial factors first, as in ex. 1.
2. Then see if common polynomial factors can he readily
discovered. Make free use of the RcTnainder Theorem. Com
pare ex. 2.
118 ELEMENTS OF ALGEBRA.
3. If common factors are not readily discovered, see if the
difference between the numerator and denominator can be
easily factored. If so, try these factors, using arbitrary
values or the Remainder Theorem, as in ex. 3.
4. Never perforvi a multiplication until compelled to.
Factor whenever possible. If the terms are cyclic and you
have studied Appendix III, apply your knowledge of sym
metry and homogeneity, as in ex. 5.
5. Let the Triethod by finding the highest common factor
be the final resort. For one who is skillful in factoring,
this tedious method ought rarely to be necessary. In ex. 5
students will probably use the Remainder Theorem instead
of the method suggested.
6. Always check the final result by substituting arbitrary
values or by some other simple device.
EXERCISES. LXI.
Simplify the following fractions and check each result:
, ab^c"^ gg  3 r^ + 2
_ 21aj109a;2
bc^J~a
a'^b^
a^'b'^
x» + 2/«
x' + y^
x" + x^y
£C*2/2
a'bcH''
 ab'^cd''
mx^y — mxy^
6.
£C* — y^
„ a^cH""
9 TVr. 10.
11. ^Z "Ta  12.
nx*y — nx^y^
3ic226x + 35
a« + g^ + 3 g  5
^2 _ 4 <^ + 3
Q>x'' + lxy^y''
Q> x^ + 11 xyir^y^
x^ x^lx + ^ ^
a;4 + 2ic« + 2ccl'
x^ + y''z'' + 2xy^
FRACTIONS.
119
13.
15.
17.
a^ + a'
2a
23.
a^ — a^ — 6a
3 x^y^ + 4 xy^
5 x^i/^ — 4 x^y
a^' + Sa^lO
Sa^ + 2a16'
x^ + x^y + ^,y ^
cc^ + x^y^ + icy*
m«  39 m + 70
7?i2 _ 3 ^^^ _ 70
x^ — xy — 12 7/^
ic^ + 5 ic?/ + 6 ?/^
14.
16.
18.
20.
22.
24.
x^ \x
212 a^
ic^ + 4a;2
+ 5 a; + 20
1
6^2
(1 + axy
'■(a^xy
x'5x'
'{7X8
2^3 5a
;2_^4iCl
a«
a^x*
a^ + a^a? 
_ a'^x^  a^x^
77^^ — 6 m
2 + 11m 6
2 7/^3
14w + 12
ic^ + (m
— 7i)x — inn
x (x \ m) — n (x { 7n)
25.
26.
27.
x^ { (a \ b) X \ ab
(x +
a)(x
+ h)(x
+ <^)
2a2_
10 a
■28
3a«
27.
7,2 + 21
a + 147
7?i2ic2
{m
, + ?/) m7?,a; + mn
V
;^ — (771 + 1) 7ia;2 f mn'^x
[Omit the following unless Appendix III has been studied.
28.
29.
30.
31.
(i\h c) + b^{ca) + c^{a  b)
abc (a — b)(b — c) (c — a)
(a — b) (b — c) (c — a)
a\b c) + b^{ca) + c\a  b) '
ab (a — b) + be (b — c) [ ca (c — a)
(a — b) (b — c) (c — a)
ab (a \ b){ be (b { c) \ ca (c + a)
(a {b)(b\ c) (g 4 a)
120 ELEMENTS OF ALGEBRA.
152. Reduction to integral or mixed expressions. Since the
fraction  indicates the division of a by h^ it may be
reduced to an integral form if the division is exact, and
to a mixed form if the degree of the numerator equals or
exceeds that of the denominator and the division is not
exact.
E.g., ^ =x — y, the division being exact.
x + y
x^ \ y^ 2 2/^
= X — y \ '■ — ; that is, the division of the remainder
X + y x\y
hy X { y is indicated.
Check. On the last result. Let x = y = \. Then f = 1 — 1 + .
EXERCISES. LXII.
Reduce the fractions in exs. 110 to integral or mixed
expressions, preferably by detaching the coefficients. Check
each result.
x'^ + y^ _ x^ \ if { z^ — ^ xyz
1. , ' 2. ■
x + y
a^ + 3 «2 _ 1
a^ + l
3a;2 + 2icl
3xl
^x'^xy^y^
x + y
x^ \^ x^ ^Vlx
+ 8
3. — Vri 4.
6.
7. . ^ ' ^ • 8.
9. '■—: ■ 10.
x^y ^z
4tx^\4.x^^2x\l
2x^ + x\l
x^ \ x^ + x^ — 6
x^ + x^ + x^ i X — 6
a' 2 ab + 52
x^ + 4.x^y 4 5xy^ + 2^
x + 2 ' x^ + 3xy + 2y^
11. Show that = 1 + a + a2 + ftS + a* + t^'
FE ACTIONS. 121
153. Reduction to equal fractions having a common denomi
nator.
Theorem. If Ti T) 7 ^^^ ^^2/ fractions whatever, and m
is any coTwrnon ^multiple whatever of b, d, f, it is possible to
reduce the given fractions to equal fractions having the
commo7i denominator m.
In arithmetic, for example, we can reduce the fractions
h h \h ^^ equal fractions having for their common denomi
nators 24, 48, 96 • . •.
Proof. 1.
'.' ??i is a multiple of b,
m = pb,
m = qd,
7n ^= rf
d,f, we may let
2.
a pa c qc
But  = =^j j= j'
b pb d qd
and ■^ = ^ §150
3.
a pa c _qc
' ' b m d 711
and — = —■) by sub
f m ^
stituting the values of step 1.
In particular, if m is the lowest common multiple of the
denominators, the fractions will be reduced to equal frac
tions having the lowest common denominator, a step of
great importance in working with fractions.
E.q.. to reduce the fractions and to equal fractions
^' X y x + 2/
having the lowest common denominator :
1. The l.c.m. of the denominators \s, {x + y){x — y).
2 a; + y ^ {x + y)'^
Kxy (X + 2/) (X  2/) '
xy {x  yf
122 ELEMENTS OF ALGEBRA.
EXERCISES. LXIII.
Reduce the following to equal fractions having the lowest
common denominator :
X y z_^ ^ oh_ — ^' a^
' yz zx xy ' cH c^d} de'^
3.
y z xy x'^y^
y + z z + X X \ y ' x^ \y^ x^ — y^ ^ — y
X + \ X — 1 X
' m^ + 6 ?/i + 8 2 m^ + 7 m + 6
2m — 2n 4 (w + n)
7. ' r~^ *
m^ — mn + 7i^ 5 (jn^ + ^^ + ^^)
a^P (a + by a^ + b^ + 2ab
^ a^b^' a^ + b^' {a^by
9x^ + 12xy5y^ 6x^ 11 xy ] 4^y\
^' 8x^xy10y^' 2x^5xy + 2y^'
X — y X \ y 2 x^y'^
10.
11,
y{xy) x^ + y{x + y) x^ — y^
2ic2 + 3cc4 x^2x^8x + 4.
ic3 + 2a;2 + 3ic + 4 x^2x^\8x + 4.
io ^ + 1 x + 2 x^8
x^ + 5x + 6 x'' + 4:X + 3 a;2 + 3£c + 2
a — 3 2ct 48 g + 5
* a29a + 18' a2 + a12' a^ + Sa + ls'
14. ^y 2^5 ^ 5^
(2/ + ^)('^ + x) x^ { zy \ zx i xy y'^ \ yz \ xy { xz
FRACTIONS. 123
II. ADDITION AND SUBTRACTION.
154. Theorem. Operations involving the addition and sub
traction of fractions can be performed upon the numerators
of equal fractions having a common denominator, the result
being divided by this common denominator.
Proof. 1. It has been proved in § 87 that
a b c _a \ b \ c
2. .'.if the given fractions be reduced to equal
fractions having the common denominator k,
the operations can be performed as stated in
the theorem.
For simplicity it is, of course, better to reduce to equal
fractions having the lowest common denominator.
Thus, with numerical fractions,
.2 _1_ 5 _ 4 1 5 _ 9 _ 3
Examples. 1. Kequired the sum of
b — c b \ c
1. The l.c.m. of the denominators is (6 + c) (6 — c).
2. ^= ^^ + ^>^ ■ §150
6  c (6 + c) (6  c)
o a _ {b — c)a
6 + c ~ (6 + c) (&  c)
4 . _^ + _^ = (& + c) g + (&  c) g ^^^
6  c 6 + c (6 + c) (6  c)
2 ah
{b + c){b c)
Check. If a = 1, 6 = 2, c = 1, then i + i = . It is not permis
sible to let b and c have the same values, because that would make the
common denominator zero, a case excluded for the present.
124 ELEMENTS OE ALGEBRA.
X X \ S X 2
2. Simplify the polynomial ^^— ^ + ^— j  ^^^ •
1. The l.c.m. of the denominators is x^ — 1.
X + 3 ^ (g + 1) (X + 3)
X  1 x2  1
g x2^ (xl)(x2)
X + 1 X2  1
X x + 3 X 2 _x + (x + l)(x + 3)(x  l)(x 2)
■ * X2  1 X1 ~X+ 1 ~ X2 1
_x + x2 + 4x + 8x2 + 3x2
~ X21
8x + 1
X21
C/iecA;. Let x = 2. Then 2 + s _ o _
155. In a case like ^ ^ 7 ^? it must be remem
X'' \ y ic^ + y
bered that the bar separating numerator and denominator
is a sign of aggregation.
In this case the result is ^^VJ^y) = ^^V^^V ^ _^JL_ .
X2 + 2/2 a;2 + y1 yp. + ^^2
EXERCISES. LXIV.
Simplify the following expressions, checking each result
by the substitution of such arbitrary values as do not make
the denominators zero :
1. ^+A4._^. 2^"^^ ^~^
he ca ah ' a — h a { h
„ 2x ^ bz ^ 2x , x2
32/^2 g^2^ • j^_^2 ' \^x2x^
1,1 x X 2xy
x + y x — y ' x\y x — y x'^ — if
FRACTIONS.
125
g + l Q^ + 2 a1
' a + 2 a{3 a{2'
' {xyf x + y xy
9.
10.
5x^1x^^x^ + 11 x1
2x^3x^ + 2x^1 x + s'
a — 4: a — 5 a — 3
a' 9a + 20 a" 11 a w"  1 a + 12
\. 1 _ 2(1 x) 1+^' _ 6x^(1 x)
^ ' 1 + x {i+xy'^{i + xf (i + xy
a^ + ab + b^ a^ab + b'^ 2b^ b'^ + a^
a + b ab "* a^  b^
L3.
+
+
{ab){ac) Q)c)(ba) (c  a) (c  b)
14.
+
+
(a — b)(a — c) (b — c)(b — a) (c — a){c — b)
xy yz . zx
+
+
(y  ^) (^  ^) (« x){x y) {x y){y z)
ax^ + byz ay'^ + bzx az^ + bxy
(x y)(x z) {ij z){y x) (z x)(z y)
X 2x:' + 7f 3 xy'' 3x^y^ 4:Xi/2 xhf  ?/
y xy xy xy^
L8.
19.
+
a{a — b)(a — c) b(b — a) (b — c) g(c — a) (c — b)
xy yz zx
{y + z){z + x) ' {z + x) (x + y) (x + y) {y + z)
2xyz
(^ + y){y + ^) (« + X)
126 ELEMENTS OF ALGEBRA.
III. MULTIPLICATION.
156. Theorem. The product of two fractions is a frac
tion whose numerator is the product of their numerators
and whose denominator is the product of their denomina
tors.
Given the two fractions z^ '
d
To prove that
Proof. 1. Let
2. Then bdx = b~d^ Ax. 6
a c
Td
_ ac
^bd
X
a c
^b'd'
bdx
= bd
b d
= ac, for b
a
~b
X
ac
~bd
a G
b'd
ac
~bd
3. = ac, for b  = a, by def . of
division
4. .'. ^ = F7* ^x. 7
bd
^ a G ac ^ ^
b d bd
Corollaries. 1. Similarly for the product of any num
ber of fractions.
c ao
2. The product a = — j as defined in § 52.
For iih = \, the identity  . ^ = ^ becomes «• = •
b d bd d d
Illustrative problem, z z^ — ^— — • — — —
^ x^12x\35 x^ 11x^12
_ {xh){xZ){xl){x^) _ x?>
~ (03  5) (ic  7) (£c  9) (ic  8) "" ic  9*
^, , 8 42 2
Check. =
24 56 8
FRACTIONS. . 127
EXERCISES. LXV.
Perform the multiplications indicated, simplifying the
results and checking as usual.
7 x^y'^ 18 xHf x^ — if x^ — xy \ y
^' 12 xy* 28icy' ' x^ ^if' x'^xy \ y
21 X x + y a^ + b'' + 2ab 1
8y + 8x 3 ab a'b^
x^y^ x^y"^ (aJ^b){x + y) a'^b'^
x^ — if (x + yY ' (a — b)(x — y) x + y
cc* — ?/* ^ — y ^
(x — yy x^ \ xy x^ \ y^
X* + x^y + xy^ + y^ ^ — y _
x^ { 2 xy \ y'^ x^ \ xy
g;^ + ^  12 g;^ + 2 a;  35
^* cc'^13ic + 40* cc2_^9x + 20
ic2 + 5 a; + 6 cc'^ + 9 ic + 20
10.
11.
a;2 + 7x412 ^2 + 11^^ + 30
2 ft^ + 5 ^ + 2 9a^ + 15a + 4
6a2^5a + l* 5a2 + 12a + 4
Reduction of integral or mixed expressions to fractional form.
157. Theorem. An integer can always be expressed as a
iction with any denominator.
For since 1 = 7'
128
ELEMENTS OF ALGEBRA.
158. Theorem. A mixed expression can always be written
in fractional form.
h ac b
For since « + = 1 —  § 157
b ac { b
a{ =
c c
§154
EXERCISES. LXVI.
Write the expressions in exs. 18 as fractions with the
denominators indicated, as in § 157.
1.
5, denominator 25 a.
2.
abc, '
' abc.
3.
^ + y, '
' a? 2/.
4.
x' + x'{x + l,
' x1.
5.
x^x^ + x^x + 1, '
' x + 1.
6.
a«  b%
' a^ + bK
7.
x^jxy^ y\
' x^xy + y\
8.
{ab){bc){ca), ^
i (a + b){b\i
Reduce the following to fractional forms, as in § 158,
checking each result :
9. 4. a
6ab2
Sb
11. CC^ + iC + 1 +
x1
13. 1 + a + a2
10. a + bi
a — b
12. ^8_3^_ 3a;(3a;)
x2
a^ +
14. x^ { 2 xy { 2/ 
a1
(x^  yy
x^ — 2xy \ y^
FRACTIONS. 129
159. Theorem. Any integral power of a fraction equals
that poiuer of the numerator divided by that power of the
denominator.
Given the fraction y? and the integer n.
To prove that ( ^ ) = i" *
/aV_ a a a
Proof. 1. I T 1 = T • T • T • • • to TO factors
Def. of power
_ aaa ■■ • to n factors „ ^^^ ^
2. = —7 ^— § 156, cor. 1
bob • • . to TO lactors
3. —J^' Def. of power
EXERCISES, LXVII.
Express the quantities in exs. 16 without using the
parentheses. Check each result.
\' \^ — yj ' \a — by ' \x — 3
(a + Z> + c V m'^ \m\l fp\q\f\
abc J ' (jn + iy ' \p—q—r)
Express the following quantities as powers of a fraction :
ft2 ^_ lOaJ + 25Z>2 a* + 9/>^ + ^a%''
8.
1 + 4 ic + 4 £c2
100£C4 + 20£C2 + 1
a;4 + 20ic2^100
4 a;2 4 9 ?/2 + 12 aj?/
81(a2 + ^;2) + 162a^>
a;6^X+3r«^(a;^ + l)
11  . .. , — ^ a;3 _ 3 a;2y 4 3 a;y^  y^
■ 4ic2f 92/'12a;y/ ' a;« + 3 ic^ + 3x + 1
130 ELEMENTS OF ALGEBRA.
Illustrative problems in multiplication. 1. To find the
, '. X — a X — 2a ^ x
product 01 > p—i and
■^ X [ a x\za x — a
1 o Pie^^~<*^"~2a X _ (x — a)(x — 2a)x
1. liy § loo, • • =: —
X + a x + 2a x — a (x \ a){x + 2a){x — a)
(x + a) (X + 2 a)
2 13 3
Check. Let x = 3, a = 1. Then =
4 5 2 45
2. To find the product of  +  and
a a
/a . 6\/a h\ /a\^ /b\'^
{ii)Q^i>Qr&
2. ^'^^ §169
C/iecA:. Let a = 1, 6 = 2. Then
\2/V2 / 4' 22 4
3. To find the product of ^ = zrx • —z z r '
x2 + 6x + 5 x2 + 8x + 15
x2 + 7x+12 x2 + 5x + 4
_ (x + l)(x + 5) (X + 3) (X + 5)
~ (X + 3) (X + 4) ■ (X + 1) (X + 4)
, by factoring
(X + 4)2
12 24 36
_ (x + l)(x + 5)(x + 3)(x + 5) ^gg
(x + 3)(x + 4)(x + l)(x + 4)
Check.
20 10 25
I
7. 1
FRACTIONS. 131
EXERCISES. LXVIII.
Perform the multiplications indicated, simplify each
result, and check.
■i(f)('^)
ri 2x "I rs _ 3x "I
a { h f a a — b a — b\
a — h \a \ h a a \ h J
\a bjG \a cj b \b cj a
9 A I 1 , ^ + M ^ ^^ + ^^ Y
10 ^ /^i^i^ 2 A , A_J:_
• (a + ^,)2 * \^«2 + ^2^ "^ (a + ^,)3 • y^ ^ b)~ a'b'''
I'+(i+9(M)(^;>
^* a;22aic + a' x' 2(b + c)x+(b + cy
f / a;^ a;«y xY 4a;?/^ 16 y^ \ / a; 2y\
• Vl6^' 12^«"^9«* 27^2"^ 81 y V2«' 3/
132 ELEMENTS OF ALGEBRA.
IV. DIVISION.
160. The fraction formed by interchanging the numer
ator and denominator of a fraction (of which neither term
is zero) is called the reciprocal of that fraction.
E.g., 2 Ls the reciprocal of ,  is the reciprocal of , and  is
^, . , . a 2 3 2 a
the reciprocal of  •
Evidently 1 and — 1 are the only numbers which are their own
reciprocals, respectively.
The term reciprocal is used only in relation to abstract numbers.
161. Theorem. To divide any number by a fraction is
equivalent to multiplying that number by the reciprocal
of the fraction.
Given the fraction  and the number a.
b ^
To prove that q ^7^  ■ q
a
Proof. 1. Let x = qT ■
2. .'. J ■ X = q, hj def. of division, or Ax. 6
^ b a b ^ , , ^
3. .'.•• X =' • q, by mult, by  • Ax. 6
a b a ^' ^ ^ a
4. .. x = 'q, since '7 = 1. §§156,150
a a b
O. . . q i =  ' q. Ax. 1
b a ^
162. Corollaries. 1. The reciprocal of a fraction
equals 1 divided by the fraction.
For 1 i. = _ . 1^ by the theorem.
a
a + b + c 1 , 1 , _L 1
^. = — aH bH c.
m m m m
For to divide by m is to multiply by its reciprocal.
FRACTIONS.
133
Illustrative problems. 1. Perform the following division :
27 3x
S(x'y^ ' xy
1.
27
8(x22/2)
Sx _x  y 27
xy 3 X 8 {X { y) {X  y)
§161
o
{xy)27
§156
SxS{xiy){xy)
3.
= , cancellinff 3 (x 
8x(x + 2/)' ^ ^
y)
§150
. .. . ^u„„ 27 6 9 ,_
9 ,
. 3
Check. Let x = 2,y = l. Then =  = , f or  ^ 6 =
83 1 16.3 8 16
). Perform the following division
X \ a x^ + a^
X a x^ { a^ x^ a^
x + a x3 + a3 X — a x + a
_ (x3 + a^) (x3  a^)
~ (x — a) (x + a)
= (x2 xa + a2) (x2 + xa + a^).
Check. Let x = 2, a = 1. Then
^^  ^^^ = (4  2 + 1) (4 + 2 + 1), for  f  = 21.
2 + 1 8 + 1 ^ nMTT^ g Q
§161
§156
3. Perform the following division :
b \ w' + h'
a — b a \b
a^ — ab
a2 + 62
a — h a + 6 (a — 6) (a + 6)
a2 + 62 a2 + 62 _ a2 ^ 62 a (a  6)
(a  6) (a + 6) a2  a6 (a  6) (a + 6) a2 + 62
a
161
a + 6
C/iecfc. Let a = 2, 6 = 1. Then f f f = .
iB4 ELEMENTS OF ALGfiBtlA.
EXERCISES. LXIX.
Perform the following divisions, simplifying each result,
and checking.
^' a^ + Sa 33 ' a' + 7 a  u'
x^ \ X (a ^ b) { ab _ /"x \ a\
'• x^\x{b + c')^ be ' \x\cJ
ai^^2_^2_^2ab {a + b  cf
^' (a + b \' cy ' abc
25x29
\54:X 2x) (54ic)(2a;)
fa a \ I a 1 — a\
' \l\a l — a)^\l + a «""/
(X x — l\ f x I ^ — l\
iC + l X J ' \x + 1 X )
f a" + b'' _ w"  b'\ ^ U^b _ ab \
• ^2^.2 a'^^b'') ''\ab a + bj
( la\3
\ a 3b
3. . ■13^ ^2^^5^_^,
b ^ 3ba ^J ' a3b
x'^Q>xy + 9i/ ^ f x^ — 9y^ ^ x^ { xy  Q y^ \
• x^^xy + 4.y^ ' \x'' + 4:y^^ x''xyQ»y^)'
10. f^^+_3^y 3.^^/_l i_\
\3ab 3abJ da'^ + b^ \3 a  b 3a^b)
' m + n ' \ W(a + b) ' _ 5x^y ' 7 (m^  72:') j
FRACTIONS. 135
V. COMPLEX FRACTIONS.
163. A fraction whose numerator, denominator, or both,
fractional is called a complex fraction.
a + 6 a + b
E.g., , , are complex fractions.
^ ' a^ ^ ab + b'^' b_+_c' ojb ^
be a
164. Complex fractions are simplified either by perform
[g the division indicated, or by multiplying both terms by
such a factor as shall render them integral.
a 4 6
c _ c^ a + b § 150
E.g.,
§§ 150, 156
c2
_ c
~ a — b
a + b
Or, = — , by multiplying both terms by c2,
a2 — 62 ^2 — 62
= , by cancelling a + b.
a — b
Check. Let a = 2, 6 = 1, c = 1. Then f = 1.
It is obvious that the latter plan is the better when the
multiplying factor is easily seen.
X22/2
E.g., to simplify —
y1
Multiplying both terms by xy2, this equals
y (g2  y^) _ y (x + y)
x(xy) ~ X
Check. Let X = 2, y = 1. Then f = .
136 ELEMENTS OF ALGEBRA.
EXERCISES. LXX.
a + b
1. Simplify .
a + b
2
^  y
2. Simplify
X
x + y
y
ab 1
3. Simplify 
5a + lb
a \l a — 1
4. Simplify ""] ^ + ^
a — 1 a + 1
5. Simplify ;+: 1+: .
1 + a^ ~" l + a*
6. Simplify ^(^~^)H<^ + b)
a + b a — b
7. Simplify the reciprocal of "' "^ ^^
a
a^ + b^
8. Simplify the reciprocal of — —
9 b 15 a
i
FRACTIONS.
137
Simplify the following expressions :
a
1 +
9.
10.
1 + « . (« + 1)^
a +
a^ + a + 1
1^a
(a + by — a^ — b^ a b
{a + bf a^b^' a^ _b2
b a
11.
12.
a^ + a%'' + Z** a^ b'
1 . ^. a — b
' (a + b)
a^' + h
1 +
7 ~r 9
*1
b^
13.
('d(';)'(';)('5)
+
('3('d
14. ^
x^ y^ x^ y^
x^ y^ x^ y^
f ^ + y
\xy
_^xy\fx^ y^
)
138 ELEMENTS OF ALGEBRA.
165. Continued fractions. Complex fractions of the form
b +
d +
are called continued fractions.
Such fractions are usually simplified to the best advantage
by first multiplying the terms of the last fraction of the
form by the last denominator, /, and so working up.
E.g., to simplify the fraction 
1
Multiplying the terms of by c, the original fraction reduces
1
to Multiplying the terms of this fraction by &c + 1, this
reduces to
6c + 1
6c + 1
a6c + a { c
1 2
Check. Let a = 6 = c = 1. Then = ■
1 +i 3
EXERCISES. LXXI.
a
1.
Simplify
1 +
1^:
a^
2
Simplify
1
X —
1
X — ■
1x
FRACTIONS. 139
3. Simplify ^
1
4. Simplify —
X
.+ 1
5. Simplify
6. Simplify
7. Simplify
x1
1
x^
x+ 1
X \l
X + y
x + y \
xy +
x + y
a
a
l + a+ ''
8. Simplify a' +
1 + a + ^2
2
a^+ '
a^ — ;
9. Simplify x { y {
a"
1
x \ y \
1
x + y +
10. Simplify (a + hf ^
(a  by +
x + y
1
(a + by +
1
(ab)
140 ELEMENTS OF ALGEBRA.
VL FRACTIONS OF THE FORM ^> ^> AND 
166. By the definition of fraction (§ 149) expressions
of division in which the divisor (denominator) is zero
were excluded. An interpretation of this exceptional case
will now be considered.
When the absolute value of a variable quantity can
exceed any given positive number, the quantity is said to
increase without limit, or indefinitely.
E.g., in the series , — , — r, , • • •, the values of the suc
cessive terms are 1, 10, 100, • • • . Hence, as the absolute values of the
denominators are getting smaller, the absolute values of the fractions
are getting larger and may be made to increase without limit.
The symbol for an infinitely great quantity is oo, read
'* infinity." This symbol must not, however, be understood
to have a definite numerical meaning. It is merely an
abbreviation for "a quantity whose absolute value has
increased beyond any assignable limit."
Hence, go + a = oo.
and — = 00.
a
In fact, the symbol oo is not subject to any of the common laws of
numbers.
167. If a is a constant finite quantity, the absolute value
of  can be made as small as we please by increasing x
sufficiently. That is,  can be brought as near as we
please. This is expressed by saying that the limit of > as
X increases indefinitely, is 0.
This is written,  == as x increases without limit, the symbol ==
being read " approaches as its limit,'
FRACTIONS. 141
x^ — a'^
168. The form  • The fraction has a meaninsr for
X — a ^
^2 qJI rj^ Q^
all values of x except x = a. But = (x \ a),
X — a X — a ^ ^
and as ic = ct it is evident that (x \ a) = 1 • (a { a)
X — a ^ ^ ^ ■^
— 2 a.
\
x^ — 1 X — 1
Similarly, = (x + 1), which = 2 as x = 1
X — 1 X — 1
X2— 4x + 4 X — 2. „. ,. u.r^ .o
— (x — 2), which =bO asx = 2 ;
x2 x2
X^  1 _ X  1
X — 1 ~ X — 1
(x2 + X + 1), which = 3 as X = 1.
But all these fractions approach the form ^ as cc ap
proaches the limit assigned, and in the several cases the
fractions approach different limits. And since the limits
are undetermined at first sight, § is said to stand for an
undetermined expressiori.
This is commonly expressed by saying that § is indeterminate.
le limit, however, can often be determined by simple inspection.
x^ — 1 .
[169. The fact that the limit of — is 2 as a? = 1 is
)ressed in symbols thus :
xlji
EXERCISES. LXXII.
Find the limit of each of the following expressions :
,8
1. ^
I
x —
i1 . 2. El^2
142 ELEMENTS OF ALGEBRA.
£c2 + 2a;8"
X
2a;8 "[ x^ ^x" hx \ ^ '^
2 J; ^* x'^x^?, J;
a;^ + 2a;8 "[ ^ x"" ^ 2 x'' \ 2 x ^ \ ^
170. The form • This form should be interpreted to
mean an expression whose absolute value is infinite.
For in the fraction  ? as x == the absolute value of the
X
fraction increases without limit.
Hence, the symbol j while without meaning by the com
mon notion of division, is interpreted to mean infinity.
171. The form — • This form should be interpreted to
mean an expression whose absolute value is zero.
For as x increases without limit,  = 0.
X
172. The form oo • 0. This form should be interpreted to
mean an undetermined (indeterminate) expression. (Why ?)
173. The relation of these forms to checks. The student
has been cautioned against substituting any arbitrary values
which make the denominator of a fraction zero. The reason
is now apparent.
12 1 1
E'.cr., = If X = 1, this reduces to oo — oo = ,
which checks nothing because oo has no exact numerical value.
Similarly in the case of
a  62 a2  6 ~ (a  62) {g?. _ 6)
If a = 6 = 1, this reduces to § + ^ = §, which checks nothing because
§ has no exact numerical value. But if a = 2 and 6 = 1, this reduces
to 3 + i = V.
FRACTIONS. 143
EXERCISES. LXXIII.
1. How should  be interpreted ? Why ?
2. Also .
3. Find the limit of : •
. n. a, b G
(a — b)(a — c)^ (b ~ c)(b — a) (c — a)(c — b)
What arbitrary values are excluded in the check ? Why ?
/T A fin (1
5. Similarly with — h — + — • Why is it
•^ b a^ — b^ a^ — b^ ^
no check to let a = ^ = any number ?
6. Similarly with
1 ^ _1_ ^ _i_ ^ {abY^{bcY^{caY
a — b b — G c — a 2(a — b) (b — g) (g — a)
7. Similarly with
r a b 2b^ \ a_
\_{a + bf a'b'''^ {a + bf{a  b) \' a
8. Show that
1 1 2a 2a'
+ b
1a 1 + a 1 + a'^ (1  «) (1 + a2^
2a^
(1 + a) (1 + a^) 16^4
9. Verify the following identity, (1) by actually adding,
(2) by the substitution of arbitrary values.
xYz^ (x^b^)(y^b'')(z^b^) (x^G^)(y^G^)(z^G^)
bH^ ^ b'^QP'C^ G^{fb^)
= X + y^ + z'^b^ c\
144 ELEMENTS OF ALGEBEA.
REVIEW EXERCISES. LXXIV.
1. What is the value of  + , when x =
a b~a a+b
b{b + a) '
2. Show why the arithmetic definition of fraction is not
sufficient for algebra.
3.
Simplify the expression  — 7 + y( + l)
4. Extract the square root of
(a) 9x743ic* + lli«V54/5 + 4/25iz;2.
(b) 1 + 4/cc + 20/£c8 + 25/^* + lO/ic^ + 24/^^ + 16/a^«.
(c) 178/720£c/72/ + 9?/V16£c2+4a;V497/215y/2(r.
5. Extract the cube root of
(a) 8xVa3 + 48ic7a2__96ic/rt + 64.
(b) \o.^l a%  ^^b^ + I a^c + 1 aZ»2 _^ 1 Z»2c + 1 c8 + 3 ac"
— \bc'^ — \ abc.
(c) l3cc/2 + 3icV25:rV4 + 3icV43icV8 +
5ic732  3a;V64 + 3 0^7256  ^V 5 12.
6. Prove that the sum of two quantities, divided by the
sum of their reciprocals, equals the product of the quanti
ties.
7. Show that by substituting 3 (cc + 1) /(cc — 3) for x in
either of the expressions (3 — 4 cc + x^) / (3 + x'^), 2 (3 + a^) /
(3 + x"^), it becomes identical with the other.
8. Eaise to the fourth power. Check.
9. Raise — [ ^^ to the sixth power. Check.
CHAPTER IX.
SIMPLE EQUATIONS INVOLVING ONE UNKNOWN
QUANTITY.
I. GENERAL LAWS GOVERNING THE SOLUTION.
174. An equation has already been defined (§ 16) as an
equality which exists only for particular values of certain
letters, called the unknown quantities.
E.g.^ x2 = 4 exists only for the two values x = + 2 and x = — 2.
175. An equation is said to be rational, irrational, integral,
or fractional, according as the two members, when like terms
are united, are composed of expressions which are rational,
irrational (or partly so), integral, or fractional (or partly
so), respectively, with respect to the unknown quantities.
E.g.^ X + Vs = is a rational integral equation ;
5 + i vx = is an irrational integral equation ;
 + 4 = X is a rational fractional equation ;
X
1
4 is an irrational fractional equation.
(X + 2)^
176. A rational integral equation which, when its like
terms are united, contains no term of degree higher than
the first with respect to the unknown quantities, is called
a simple or a linear equation.
E.g.^ X — 3 = 5, x2 + X — 1 = x2 + 2/, are simple or linear equations.
But Vx = 5,  = 2, are not as they now stand.
145
146 ELEMENTS OF ALGEBRA.
177. Equations which are not simple are, however, often
solved by the principles which govern the solution of simple
equations.
E.g., (x — l)(x — 2) = is an equation of the second degree. (Why?)
But it is satisfied only if
X  1 = 0,
or if X  2 = 0,
that is, if X = 1, or if X = 2. Hence, the solution of this equation of
the second degree reduces to the solution of two linear equations.
EXERCISES. LXXV.
1. What is meant by the roots of an equation ? (See § 16.)
What are the two roots of the equation x^ = 25?
2. What is meant by solving an equation ? Solve the
equations.
(a) 3ic + 5 = 0. (b) (x2)(x3)=0.
(c) (x + l)(x + 2)=0. (d) (x + 2)(x3)=0.
3. What is meant by an equation being satisfied ? What
values of x satisfy these equations ?
(a) (x + ^)(3x2)=0.
(b) (2£cl)(2ic + 3)=0.
(c) x(xl)(x 2) (ic  3) = 0.
4. What is meant by the members of an equation ? How
do they differ from the terms ?
5. Which of the following are simple equations with
respect to cc ?
(a) x^ { x^ ^ X — x^ — x^ = A.
(b) 3x^ + x + 7 = 2x^^x(x + 3).
(c) V^ + 4 = 7. (d) i^ = 3.
^ '^ ^ ^ X 2
(e) x^x + l=:0. (f) x(x + l)=^x^
¥
SIMPLE EQUATIONS. 147
178. Known and unknown quantities. It is the custom
to represent the unknow7i quantities in an equation by the
last letters of the alphabet, particularly by x, y, z.
This custom dates from Descartes, 1637.
179. Quantities whose values are supposed to be known
are generally represented by tlie^rs^ letters of the alphabet,
as by a, ^, c, • • • .
E.g., in the equation ax + 6 = 0, a and 6 are supposed to be
known. Dividing both members by a, x + h/a = 0, which is satisfied
if X = — b/a.
180. The solution of the simple equation has already
been explained (§ 17). The general case, not involving
fractional coefficients, will be understood from two illustra
tive problems and the series of questions in the following
exercises.
1. Given the equation 5ic — 2 = 3ic + 8, to find the
value of X.
1. 5x2 = 3x + 8.
Given.
2. 5x = 3x + 10.
Adding 2. Ax. 2
3. 2x = 10.
Subtracting 3 x. Ax. 3
4. x = 6.
Dividing by 2. Ax. 7
Check. Substitute 5 for x in the original equation, and
25  2 = ]
L5 + 8.
2. Given the equation 2 ax 
 a^ = ax \3 a^, to find the
value of x.
1. 2 ax  a2 == ax + 3 a^.
Given.
2. 2 ax = ax + 4 a^.
Adding a'^. Ax. 2
3. ax = 4 a\
Subtracting ax. Ax. 2
4. x = 4a.
Dividing by a. Ax. 7
Check. Substitute 4 a for x in the original equation, and
8 a2  a^ = 4
a2 + 3 a2.
148 ELEMENTS OF ALGEBRA.
181. From these illustrative problems it will be observed
that any term may be transferred from one member of an
equation to the other if its sign is changed. This opera
tion is called transposition.
E.g.., if ic + 2 = 7, transposing 2 we have
a = 7  2,
or x = 5.
It should be remembered, however, that the operation is really one
of subtraction, 2 being taken from each member by ax. 3.
In general, transposition is always an operation of subtraction or
addition.
EXERCISES. LXXVI.
The answers to the following questions will lead to the
understanding of the steps to be taken in the solution of
linear equations with one unknown quantity.
1. In which member do you seek to place the known
quantities, and in which the unknown ? Might this be
changed about ? What axioms are involved in this opera
tion ? (See ex. 1, steps 2 and 3, p. 147.)
2. Having done this, what is the next operation ? What
axiom is involved ? (See ex. 1, step 4, p. 147.)
3. State, then, the two general steps to be followed in
solving a linear equation with one unknown quantity.
4. How is the work checked ?
Solve the following equations, checking the results.
5. 12a;28 = 8f3ic. 6. 11 x = 2xl.
7. 27a; 127 = 11 19a;. 8. 2a; f 3 = 4ic f 5.
9. 4a; 34 = 22 3a;.
10. a;f2 + 3a;}4 = 5a; + 6.
11. 3a; + 4a; + 5a; = 6a; + 72.
SIMPLE EQUATIONS. 149
182. The axioms applied to the solution of equations. While
it is true that the solutions of equations depend upon cer
tain axioms (§ 22), it is necessary to consider the precise
limitations of these axioms before proceeding further.
183. Two equations are said to be equivalent when all of
the roots of either are roots of the other.
E.g., x + 4=2x5,
and x + 5 = 2 (a; — 2), are equivalent equations,
for x = 9 is a root and the only root of each.
But X = 2 and x^ = 4 are not equivalent equations, for — 2 is a root
of x2 = 4, but not of X = 2.
The necessity for a consideration of the limitations of
the use of the axioms is seen from the following :
Suppose 1. X = 2.
Then 2. x2 = 4, by ax. 8.
But a root of equation 2 is not necessarily a root of equation 1,
because while equation 2 is true when equation 1 is true., it is not
equivalent to equation 1.
184. Axioms 6 and 7. If equals are multiplied or divided
hy equals, the results are equal.
This is true, but it Tnust not he interpreted to mean that
if the two members of an equation are multiplied hy equals,
the resulting equation is equivalent to the given one.
E.g., if the two members of the equation
x1 = 6
are multiplied by x — 2, we have
(X  1) (X  2) = 5 (X  2),
or x28x + 12=0,
or (x  6) (x  2) = 0,
which has two roots, x = 2 and x = 6. Of these, only x = 6 satisfies
the original equation. Hence, the resulting equation is not equivalent
to the original one ; there has been a new root introduced.
150 ELEMENTS OF ALGEBRA.
185. A new root which appears in performing the same
operation upon both members of an equation is called an
extraneous root.
EXERCISES. LXXVII.
What, if any, extraneous roots are introduced by multi
plying both members of the following equations as indi
cated ?
1. cc + 2 = 5 hj x\3.
2. X — a = '' X \ a.
3. a;2  1 = " x^5x\6.
4. cc2 = 4ic + l " 3.
5. x5 = 5x 21 " X.
6. x^\x =(1 — xy " x.
7. 3x4. = 4:X3 " 21.
8. (x f ay = (x ay " fc ^ 1.
186. Just as an extraneous root may be introduced by
multiplying both members of an equation by equals, so a
root may also be lost by the same process.
E.g., it is not permissible to multiply the two members of the
equation
(X  6) (X  2) =
by , expecting thereby to obtain an equivalent equation, for we
X — Ji
should have x — 6 = 0,
which has only a single root, x = 6, whereas the original equation had
two roots, X = 6 and x = 2. Hence, the resulting equation is not
equivalent to the original one ; a root has been lost by multiplying
equals by equals.
In the same way, while if we multiply both members of the equation
x3  X =
by  , , , or , the results will be equal, it is not true
X x+1 X — 1 x2 — 1
that we shall obtain equivalent equations.
SIMPLE EQUATIONS. 151
187. Hence, it appears that multiplying the two members
of an equation f (x) = by a function of x does not, in general,
give an equivalent equation. The operation may introduce
an extraneous root, or it may suppress a root.
It should also be stated, in connection with extraneous
roots, that no value is considered a root unless it makes the
members identically equal. Hence, a value that makes
both members ififinite is not a root, for infinity is not iden
tically equal to infinity (§ 166).
2 X
E.g., 1 is not a root of = , for it makes each member
.^.^ xlx — 1
mfmite.
EXERCISES. LXXVIII.
Would each resulting equation be equivalent to the given
one, by multiplying both members as indicated below ?
1.
^4
X
by
X
i'
2.
'iel^'
x2.
3.
hl*h'
X.
4.
X^ 1
X1.
x1 x1
5.
x^_6x + 5
x^5x + 4.
5
x^5x + 4..
6.
x^5x + 6 =
1/(032).
7.
2x^5 = x^
1
1/(^ + 2).
8.
^2_3^_28.
=
1/(^7).
Q
x'^ + 9x + 14.
1
/^2 1 1J. ^ t
cc'^ + 14 ic + 49 X + 7 ^
jlO. Also by cc^ + 14 a; + 49 ; also by a; + 7.
162 ELEMENTS OF ALGEBRA.
188. Axioms 8 and 9. Like powers or like roots of equals
are equal.
This is true, but it must not be interpreted to mean that
the equation formed by taking like powers or like roots of the
members of a given equation is equivalent to that equation.
E.g.,\t x=l,
then x2 = 1 , or x2  1 = 0, or (x + 1) (x  1) = ;
but this equation has two roots, x = — 1, and x = + 1, and of these,
only X = + 1 satisfies the original equation.
Similarly, if x^ = 4,
it is true that x = 2 ;
but this equation is not equivalent to the original one. It should be
written x = + 2, and — 2.
Students are liable to make a mistake by omitting the
± sign in extracting a square root, thus losing a root.
E.g., in the equation
x2 + 2 X + 1 = 4,
extracting the square root, x + 1 = 2,
.. X = 1.
It should be x + 1 = + 2 and  2,
.'. X = 1 and — 3.
EXERCISES. LXXIX.
What extraneous roots are introduced by squaring both
members of the following equations ?
1. a; = 0. 2. ic + 3 = 3. 3. £c  2 = 2.
4. 2 a; = 9. 5. a; 5 = 0. 6. 4.x =28.
7. 205 + 1=3. 8. ^ = 1. 9. 1 + 1 = 2.
The discussion already given may be set forth in four
theorems. These theorems, with strict proofs, may be
found too abstract for most beginners, and hence they are
given in Appendix V.
SIMPLE EQUATIONS. ' 153
II. SIMPLE INTEGRAL EQUATIONS.
189. General directions for solution. From the sugges
tions in exs. 14, on p. 148, it appears that, to solve a
simple integral equation, we
1. Transpose the terms containing the unknown quan
tities to the first member, changing the signs (axs. 2, 3,
§§ 22, 181) ;
2. Transpose the terms containing only known quantities
to the second meiriber, changing the signs ;
3. Unite terms ;
4. Divide by the coefficient of the unknown quantity ;
5. Check the result by substituting in the original equation.
EXERCISES. LXXX.
Solve the following equations :
1. «,a7 + & = te + <^ 2. {x — V)i\ — X) =■ — x^
3. 8a; (7 a;) = 29. 4. 3cc  2(2  a^) = 21.
5. (2x){5x) = x\ 6. 3(a;l) = 4(a; + l).
7. 9x3(5a;6) = 30.
8. 2(aj + 3)3(x + 2) = 0.
9. x{x'^ + 1) =x (x^ — 1) + 9.
10. {x + 5)2 =: 21 a; + (4  xy\
11. 3a; + 14.5(a;3) = 4(a; + 3).
12. a;(a;l)a;(aj2)=2(a;3).
13. a;(l +a; + a;2) = a;3_^a:2 + 3a;17.5.
14. {x + 1) (a; + 2) = (a; + 3) (a; + 4)  50.
15. 2(a; + l) + 3(aj + 2) + 4(a; + 3)=101.
16. {x  2)2 {x 3)2 = (x 4)2 {x 6)2.
I
154 ELEMENTS OF ALGEBRA.
III. SIMPLE FRACTIONAL EQUATIONS.
190. If the equation contains fractions, these may be
removed by multiplying both members by the lowest com
mon multiple of the denominators. This is called clearing
the equation of fractions.
Unless the fractions are in their lowest terms before multiplying by
the lowest common denominator, an extraneous root is liable to be
introduced (§ 187).
It is not always advisable, however, to clear the equation of frac
tions at once, as is seen in the following illustrative problems.
Illustrative problems. 1. An equation which should be
^ 3 ic h 7
cleared of fractions at once : ^rr — I — 7=— = 8.
lo IT
1. The l.c.d. of the fractions is 15 • 17.
2. Multiplying both members by 1517, by ax. 6,
17 a; 61  15 a l 105 = 15 17 8.
3. Subtracting —511 105, and uniting terms,
32aj= 1986.
4. Dividing by 32,
0^ = 62,1,.
Check. ^ + ??^ = 8, or 3H + 4^^ = 8.
15 17
2. An equation which need not be cleared of fractions at
X 3 1
once: x = '
1. Adding f and uniting terms,
x = f.
2. Multiplying both members by f (or dividing both members by f),
« = ¥•
Check. 1^1 _ = .
7X12
8.
Transposing and uniting terms,
7x
4)
12
I
Dividing by 2 and multiplying by 7
x12
I
14 X
24 = 16x
30.
4.
f
Adding 24 15 X,
 X =  6.
h
Multiplying by  1,
x = Q.
Check. Hir% = l.
SIMPLE EQUATIONS. 155
3. An equation which should be cleared of fractions part
^ . 3a; + 7 2x4: x + 1
^.tatime: ^^ ___ = __.
1. Multiplying both members by 15,
Z. + 7'^^^^ = Bz + S. AX.0
Azs. 2, 3
Axs. 6, 7
Ax. 2
Ax. 6
4. An equation in which the fractions may be united to
, , , ; , . X x\l x — S x — 9
[vantage before clearing :  —    — j =  —    —  •
X — J x — 1 x — o x — 7
1. Adding the fractions in each member separately,
xaxx2+x + 2 x«15x + 56x«+ 15X54
I It will be noticed, in step 1, that the bar of a fraction is a symbol of
K'egation, and in adding fractions or in clearing of fractions this
t be tiiken into account.
■—
(x2)(xl) (x6)(x
7)
5 . 2 _ 2
(x2)(xl) (x6)(x7)
8. Dividing by 2 and cleai'ing of fractions,
xa_l3x + 42=x23x + 2.
4. .. 10x= 40.
(Why ?)
6. .. x=4.
(Why ?)
4 5 4 5
Check. =
2 323
156 ELEMENTS OF ALGEBRA.
5. An equation in which the fractions should be reduced
to mixed expressions before clearing :
5a;8 6xU _ xS 10.^8
1. Reducing to mixed expressions,
2 2 2 2
5 + ^ + 6 = 1  ^— + 10 +
X — 2 X — 7 X — 6 X — 1
2. Subtracting 11 and dividing by 2,
3. Adding the fractions in each member separately,
5 6
{X  2) (X  7) (X  6) (X  1)
4. Dividing by — 5, and clearing,
x27x + 6 =x2 9x + 14.
5. .. 2 X = 8. / (Why ?)
6. .. X = 4. (Why ?)
lO Of) A QO
Check. :L + __ = __ + _, or 6 + 6f = 2 + lOf.
EXERCISES. LXXXI.
First determine which seems the best method of solving
each of the following equations ; then solve and (except as
the teacher otherwise directs) check the solution by substi
tuting the roots in the original equation.
1.
ad
X — 1 X — b
ax + bx
 a; + l x3
3.
50 12 49
Ax^ X lO'
4. 2^1 = 2.415.
X — o
5.
2 + 3^^ 4
6. 0.5a:40.25rr = 1.5.
SIMPLE EQUATIOJSS.
X . 1
157
^ + ^ = 17^
5^8 ' 10
4ic 5x
9. ^r
10.
11.
12.
13.
14.
X — a _ (2 ic — ay
xb ~ {2x by '
1 + i^ i^ + 1
^»iC
a
iC
^«^
3bx
X
+a '
2b + x
6
x + 7
12
3x
4.x
4
3
+
X
2*
iC
2
X4:
X
6
15.
X 1 X
16. 1
17.
18.
19.
a { X {
X'
a — X
3 4
ic — 3
X — 9 cc — 6
b^ — aic
5 + a;Z' — ic_
&4(x b ~ a a^ — b^
5 a; 4 10.5 2x
x + 0.5 2ic + l
= 9.
158 ELEMENTS OF ALGEBRA.
20.
21.
22.
*
2
"3"
2
'3
l^' + l
%+x
f + ^
X ^x
2
+ 4
3
= 7
^x2
3
8
9
15
a; + 3
2x
+ 6
ra^ + 2
lx + 5
5
X —
6 85a;
23.
6 4 12
op; 2a;5 , 6x + 3 ^ 35
25. _^ + ___ = 5x.
2a; + 3 6a; + 22 _ 3a; + 17
5 15 ~5(la;)"
a — c> X a \h x
28 ^ + 46^ + ^* 4a7 + a + 2^ > ^ '
a; + a + ^ iz;[_(t — ^ ""*
29. 1 2 3 4
1 x2 x3 x4.
30 '^ + 3 a;6 _g; + 4 a; 5
a; + l a;4 a; + 2 a;  3
1 1
31.
(a; H 3) (a; + 5) (a; + 9) (a;  5)
32 1_+ ^^ 9  11 a; ^ 14(2 a;  3)^
5 + 7a; 5  7 a; "" 25  49 a;^ '
33. ^ 1 l + _^_ = o
2a; 1 a;3 a;2a;5
SIMPLE EQUATIONS.
159
34
35
36.
37.
38.
(9.
40.
1.
42.
43.
44.
45.
5 cc h 1 = 3 a; H ^ V 7.
5 4 ^1/ 3 8 \
Bx + S 2ic + 3~5Vic + 3 a; + 2/
, a;/, 3\ 6xf^ 6\ „,
^2(l4^j = Ti^7^h''^^
3a;'^2a; + l _ (7a;  2) (3a;  6) _9
35
+
10
a{a + b)x cu'lP' 2bx _ {5a + b)b
a^ — b'^ a + b b — a a — b
\xl^^xl = ^% + \^x\l^x.
13.r10 4a; + 9 7 (x  2) _ 13a;  28
36 18 12 ~ 17a; 66*
a 2(3a + 5) 8^4 15 _ 3(^ + 2) 1
a; + 1 a; — 1 a; — 2 ~ a; — 3 a; + 2*
TV(2^l)TV(3^2) = TV(^12),,\(a; + l).
(g + bfjx + 1) + (g + ^,)(a; + 1) + (^ + 1)
a + b + 1
= (« 4 by + (g + &) + 1.
a; H1 _lf a; + 1
3 ~2y~~Y~
x2
\{''^)
+
31
46.
5 3V^ 2 y ■ 60
3 a;5 + 12 a;* + 44 a;^ + 185 a;^ + 8 a; + 98
I
3 a;* + 18 a;« + 26 a;2 + 15 a; + 14
3a;^ + 44a; + 2
3a;2 + 6a; + 2
160
ELEMENTS OF ALGEBRA.
IV. IRRATIONAL EQUATIONS SOLVED LIKE SIMPLE
EQUATIONS.
191. It often happens that irrational equations can be
reduced to equivalent simple equations and thus solved.
E.g. , Vx = 2 can be reduced to the equivalent simple equation a = 4.
In applying ax. 8 it is possible, however, that extraneous roots may be
introduced (§ 185). That this is not the case in this instance is seen
by substituting the value of x in the original equation.
192. A question at once arises, however, in dealing with"
equations like
Vic'^ + 2 ic + 1 + Vic^ _ 2 ic + 1 = 4.
Shall this be reduced to
or shall only the positive roots be considered, as in
iC + l+£C — 1=4?
The former would give x = ±2, the latter only x = 2.
To answer this question, let ^f(x) and \/F{x), foi
brevity, represent the square roots of any two fimctionsi
of x, like those already mentioned.
Then it is evident that an irrational equation of the foi
s/f(x) + VF(x) =
involves four equations, viz.:
1.
2.
3.
4. ■\/f(xj^F(x)
where V/(£c) and ^F{x) represent, in these four equations,
only the positive square roots.
This is also seen in the case of Vi + Vo, which equals (± 2) + (± 3).
SIMPLE EQUATIONS. 161
m
"• 193. Hence, any root which satisfies any one of the four
equations is strictly a root of V/(a;) + ■y/F(x) = a.
By convention, however, only the roots which satisfy equa
tion 1 are usually considered.
For example, consider the equation ■\x — 2{ Va; — 5 = 1.
1. Vx  5 = 1  Vx  2. Ax. 3
2. .. x5 = l+x2 2 Vx  2. Ax. 8
3. .. 2Vx 2 = 4. Ax. 3
4. .. X  2 = 4. Axs. 7, 8
5. .. x = Q>.
Substituting 6 for x in the given equation,
Vi + Vl = 1, or (db 2) + (± 1) = 1.
While this is true in the form (+ 2) + (— 1) = 1, the root 6, by the
convention just given, is usually called extraneous.
194. Irrational equations can often be solved by isolating
the radical and then a pplying ax. 8. For example, consider
the equation Vic — 2 — Vcc — 5 = 1.
1. We first isolate the radical Vx — 2, by adding Vx — 5 to each
member.
2. .. Vx  2 = 1 + Vx  5.
3. Then, by squaring both members,
x2 = l+x5 + 2Va
l. Then, isolating the radical Vx — 5, by subtracting 1 + x — 5
dividing by 2,
1 = Vx6.
5. .•. \ z= X — 5, whence x = 6.
Check. V62 _ Ve  5 = 1.
L95. If the equation contains several irrational expres
is, there is no general rule for solution. The student
ist use his judgment as to which radical it is best to
llate first.
162 ELEMENTS OF ALGEBRA.
Illustrative problems. 1. Solve the equation
Va; + 1 — 4 Vcc — 4 + 5 Vic — 7 = 0.
1. Isolating the radical 4 vcc — 4 "by adding it to both members, we
have :
Vx + l + 5Vx7 = 4 Vx 4.
2. Squaring
X + 1 + 25x  175 + 10 Vx26x7 = 16x  64.
3. .. X  11 =  Vx2  6 X  7. (Why ?)
4... x222x+ 121 =x26x7. (Why?)
5. .. x = 8. (Why?)
Check. V94V4 + 5V1 =38 + 5 = 0.
2. Solve the equation Vcc = — 2.
Squaring, x = 4. But on substituting 4 for x, Vi = — 2. This
satisfies the equation because Vx is both + 2 and — 2. But since the
positive sign is usually taken with the radical (§ 193), 4 is usually
called an extraneous root and the equation is said to be impossible.
The equation — Vx = — 2 is not open to the same objection for it is
satisfied by x = 4.
EXERCISES. LXXXII.
Solve the following equations, designating such roots as
are usually called extraneous.
1. Vic + 2  Va; + 9 = 7.
2. y/x + Va \ X = a/ V^.
3. Vx + 19 + Vcc + S
4. 2 Vcc  1 + V4ic + 5 = 9.
5. V8^+52 V2xl = l.*
6. 4 ^x + 2  Vx + 7  5 Vic1.
SIMPLE EQUATIONS. 163
V. APPLICATION OF SIMPLE EQUATIONS.
A. Problems Eelating to Numbers.
Illustrative problems, 1. The sum of two numbers is 200,
and their difference is 50. Find the numbers.
1.
Let
X = the lesser number.
2.
Then
X + 50 = the greater number.
3.
And
X + X + 50 = the sum.
4.
But
200 = the sum.
5.
X + X + 50 = 200.  ,
6.
..
x = 75,
and
X + 50 = 125.
Check. The sum of 125 and 75 is 200, and their difference is 50.
Always clieck by substituting in the problem instead of
the equation, because there may have been an error in
forming the equation. The neglect to take this precaution
often leads to wrong results.
2. What number must be added to the two terms of
the fraction ^^ in order that the resulting fraction shall
equal f ?
1. Let X = the number to be added.
2. Then 1+x ^ 59_
23 + X 67
3. .. 67 (7 + X) == 59 (23 + x). (Why ?)
4. .. 469 + 67x= 1357 + 59 X.
5. .. 8x = 888. (Why?)
6. .. x = lll. (Why?)
Check. ^"^^^^ ^ H? ^ ^. That Is, if 111 is added to both
23 + 111 134 67
terms of the fraction 2^3 , the result equals ff . *
164 ELEMENTS OF ALGEBRA.
EXERCISES. LXXXIII.
1. What number is that which when subtracted from
28 gives the same result as when divided by 28 ?
2. Or, more generally, what number is that which when
subtracted from n gives the same result as when divided
by n ? Check by supposing that n = S, n = 28.
3. What number is that which when multiplied by 16
gives the same result as when added to 16 ?
4. Or, more generally, what number is that which when
multiplied by 7i gives the same result as when added to
n ? Check by supposing that n = 2, n = 16.
5. What number is that which when divided by 12
gives the same result as when added to 12 ?
6. Generalize ex. 5 and check. (See exs. 2, 4.)
7. What number is that which when subtracted from
25 gives the same result as when multiplied by 25 ?
8. Generalize ex. 7 and check. (See exs. 2, 4, 6.)
9. What number must be added to 3 and 7 so that the
first sum shall be f of the second ?
10. Or, more generally, what number must be added to
a and b so that the first sum shall be — of the second ?
n
Check by supposing that a = 3, b = 7 , 77i = 3, n = 4:.
11. Determine x, knowing that a^ — 5a^{4:a — x is
algebraically divisible by 2 a  1.
12. Divide the number 121 into two parts such that the
greater exceeds the less by 73.
13. Or, more generally, divide the number n into two
parts such that the greater exceeds the less by a.
SIMPLE EQUATIONS. 165
14. Divide the number 121 into three parts such that
the first exceeds the second by 85 and the second is four
times the third.
15. Divide the number n into three parts such that the
first exceeds the second by^ and the third by q. Check by
letting 7i = 10, i? = 1, q = 1.
16. What is the value of n if ; — —  = —  when
17. If each of the two indicated factors of the two
unequal products 5245 and 6637 is diminished by a cer
tain number, the two products are equal. What is the
number ?
\ 18. Divide the number 99 into four parts such that if
2 is added to the first, subtracted from the second, and
multiplied by the third, and if the fourth is divided by 2,
the results shall all be equal.
tt . 19. Or, more generally, divide the number n into four
parts such that if a is added to the first, subtracted from
the second, and multiplied by the third, and if the fourth
is divided by a, the results shall all be equal. Check by
letting n = 10, a = 1.
20. The square of a certain number is 1188 larger than
that of 6 less than the number. What is the number ?
21. The. square of 13 times a certain number, less the
square of 3 more than 12 times the number, equals the
square of 9 less than 5 times the number. What is the
number ?
22. What number must be added to each term of the
fraction y that it may equal the fraction ? Check by
letfmg a =z 3, h = 5, c = ^, d = 10.
166 ELEMENTS OF ALGEBRA.
B. Problems Relating to Common Life.
Illustrative problems. 1. What sum gaining 6^% of itself
in a year amounts to $157.50 in 2 yrs. ?
1. Let a; = the wwm&er of dollars.
2. Then 6J% x = the number of dollars of interest for 1 yr.
3. .. x + 26i%x = 157.50. (Why?)
4. .. 1.12^x^157.50.
5. .. X = 140. (Why ?)
Check. The interest on $140 for 2 yrs. at 6^% is §17.50, and hence
the amount is $157.50.
2. The cost of an article is $17.15, and this is 30% less
than the marked price. What is the marked price ?
1. Let X = the number of dollars of marked price.
2. Then 30% x = the number of dollars of discount.
3. .. x30%x = 17.15.
•4. .. 0.7x = 17.15. (Why?)
5. .. x = 24.50. (Why?)
Check. $24.50 less 30% of $24.50 is $17.15.
EXERCISES. LXXXIV.
23. What is that sum which diminished by 9% of itself
equals $1538.50 ?
24. How long will it take an investment of $6024 to
amount to $7658.01 at 3% simple interest?
25. A man invests f of his capital at 4% and the rest
at 3^%, and thus receives an annual income of $76. What
is his capital ?
26. A man invests onefourth of his capital at 5<^, one
fifth at 4%, and the rest at 3%, and thus secures an animal
income of $3700. What is his capital ?
SIMPLE EQUATIONS. 167
*' 27. A train traveling 30 mi. per hour takes 2 hrs. longer
to go from Detroit to Chicago than one which goes J faster.
What is the distance from Detroit to Chicago ?
28. A loaned to B a certain sum at 4^, and to C a suii^
$200 greater at 5% ; from the two together he received
$276 per annum interest. How much did he lend each ?
29. The interest for 8 yrs. upon a certain principal is
$1914, the rate being 3% for the first year, 3^% for the
second, 3f % for the third, and so on, increasing :^^ each
year. What is the principal ?
30. A bicyclist traveling a miles per hour is followed,
after a start of m mi., by a second bicyclist traveling h mi.
per hour, h > a. At these rates, in how many hours after
the second starts will he overtake the first ?
31. A capitalist has f of his money invested in mining
stocks which pay him 13%, ^ in manufacturing which pays
him 9%, and the balance in city bonds which pay him 3%.
What is his capital, if his total income is $26,640 ?
32. A man spends th of his income for food, th for
a
rent, th for clothing, th for furniture, and saves e dollars.
c (t
How much is his income ?
33. Two trains start at the same time from Buffalo and
New York, respectively, 450 mi. apart ; the one from New
York travels at the rate of 50 mi. per hour, and the other
0.8 as fast. How far from New York do they meet ?
34. Two trains start at the same time from Syracuse,
one going east at the rate of 35 nii. per hour and the
other going west at a rate \ greater. How long after start
ing will they, at these rates, be exactly 100 mi. apart ?
168 ELEMENTS OF ALGEBRA.
35. A train runs 75 mi. in a certain time. If it were to
run 2^ mi. an hour faster, it would run 5 mi. farther in the
same time. What is the rate of the train ?
36. A steamer can run 25 mi. an hour in still water. If
it can run 90 mi. with the current in the same time that it
can run 60 mi. against the current, what is the rate of the
current ?
37. The cost of publication of each copy of a certain
illustrated magazine is 6^ cts. ; it sells to dealers for 6 cts.,
and the amount received for advertisements is 10% of the
amount received for all magazines issued beyond 10,000.
Find the least number of magazines which can be issued
without loss.
38. A steamer and a sailboat go from M to N, the
former at the rate of 35 mi. in 3 hrs. and the latter at the
rate of 10 mi. in the same time. The sailboat has a start
of 3^ mi., but arrives at N 5 hrs. after the steamer. How
long did it take the steamer to go from M to N, and what
is the distance ?
39. Two engines are used for pumping water from dif
ferent shafts of a mine, their combined horse power being
represented by 108. The first engine pumps 22 gals, every
10 sees, from a depth of 310 yds. ; the second pumps 9 gals,
more in the same length of time from a depth of 176 yds.
Eequired the horse power of each.
40. There are two hoisting engines at a coalpit mouth,
the first capable of raising at the rate of 144 tons every 5
hrs. from a depth of 375 ft., and the second 80 tons every
3 hrs. from a depth of 540 ft. After the first had been
running If hrs. the Second began, and after 7 hrs. it had
raised from the bottom of the mine 11:J tons more than the
first. Eequired the depth of the mine.
SIMPLE EQUATIONS. 169
C. Problems Eelating to Science.
Illustrative problems. 1. Alcohol is received in the labo
ratory 0.95 pure. How much water must be added to a
gallon of this alcohol so that the mixture shall be 0.5 pure ?
1. Let X = the number of gallons of water to be added.
2. Then 0.5(1 + x) represents the alcohol in the mixture.
3. But 0.95 represents the alcohol in the original gallon.
4. .. 0.5(1 + x) = 0.95.
5. .. x = 0.9. (Why?)
Check. Adding 0.9 gal., there are 1.9 gals, of the mixture, 0.5 of
which is the 0.95 gal. of alcohol.
2. Air is composed of 21 volumes of oxygen and 79 >ol
mnes of nitrogen. If the oxygen is 1.1026 times as heavy
as air, the nitrogen is what part as heavy as air ?
1. 21 . 1.1026 + 79 X = 100. (Why ?)
2. .. x = 0.9727. (Why?)
EXERCISES. LXXXV.
41. How much water must be added to a 5% solution of
a certain medicine to reduce it to a 1 % solution ?
42. How much pure alcohol must be added to a mixture
of f alcohol so that ^^ of the mixture shall be pure alcohol ?
43. In midwinter in St. Petersburg the night is 13 hrs.
longer than the day. How many hours of day ? of night ?
At what time does the sun rise ? set ?
44. How many ounces of silver 700 fine (700 parts pure
silver in 1000 parts of metal) and how many ounces 900
fine must be melted together to make 78 oz. 750 fine ?
45. How many ounces of pure silver must be melted
with 500 ozT of silver 750 fine to make a bar 900 fine ?
170 ELEMENTS OF ALGEBRA.
46. How many pounds of pure water must be added to
32 lbs. of sea water containing 16% (by weight) of salt, in
order that the mixture shall contain only 2 % of salt ?
47. In a certain composition of metal weighing 37.5 lbs.,
18f % is pure silver. How many pounds of copper must
be melted in so that the composition shall be only 15.625%
pure silver ?
48. How many pounds of copper should be melted in
with 94.5 lbs. of an alloy consisting of 3 lbs. of silver to
4 lbs. of copper so that the new alloy shall consist of 7 lbs.
of copper to 2 lbs. of silver ?
49. What per cent of the water must be evaporated from
a 6% solution of salt (salt water which contains 6%, by
weight, of salt) so that the remain,ing portion of the mix
ture may be a 12% solution ?
50. The planet Venus passes about the sun 13 times to
the earth's 8. How many months from the time when
Venus is between the earth and the sun to the next time
when it is in the same relative position ?
51. Two bodies start at the same time from two points
243 in. apart, and move towards each other, one at the rate
of 5 in. per second, and the other 2 in. per second faster.
In how many seconds will they be 39 in. apart ?
52. Prom two points d units apart two bodies move
towards each other at the rate of a and h units a second,
respectively. After how many seconds will they be c
units apart for the first time (c < d)? together ? c units
apart for the second time ?
53. These bodies (of ex. 52) move, from the two starting
points, away from one another. How far are they apart
after t sees. ? When will they be e units apart (e>d)?
t
SIMPLE EQUATIONS. 171
54. If sound travels 5450 ft. in 5 sees, when the temper
ature is 32°, and if the velocity increases 1 ft. per second
for every degree that the temperature rises above 32°, how
far does sound travel in 8 sees, when the temperature is
70°?
55. Seen from the earth, the moon completes the circuit
of the heavens in 27 das. 7 hrs. 43 mins. 4.68 sees., and the
sun in 3(35 das. 5 hrs. 48 mins. 47.8 sees., in the same direc
tion. Required the time from one full moon to the next,
the motions being supposed to be uniform. Answer cor
rect to 0.0001 da.
56. In Spitzbergen (77° N. lat.) there is a certain part of
the year in which the sun does not rise, remaining con
stantly below the horizon ; there is also an equal length of
time during which it does not set. The period in which it
rises and sets is 1^ months longer than the period of con
tinued night. How many months in each of these three
divisions of the year ?
57. It is shown in physics that if ^ = the number of
seconds which it takes a pendulum to swing from one state
of rest to the next, through a small angle, then t = 7r \l/g,
where 7r=3^, <7 = 32.2, and Z=the number of feet of length
of the pendulum. Required the length of a 1second pen
dulum; of a 2seconds pendulum; of a pendulum which
oscillates 56 times in 55 sees.
58. It is proved in physics that if v = the velocity of a
body which started with an initial velocity of a ft. per
second and has gained in velocity / ft. per second for t
seconds, then v = a { ft. Suppose v = 15, a = 0, t = 5.
Find /. (This is one of many exceptions to the custom of
representing known quantities by the first and unknown
(quantities by the last letters of the alphabet.)
172 ELEMENTS OF ALGEBRA.
D. Problems Eelating to Mensuration.
The following formulas are proved true in geometry and
are probably already known to the student from his work
in arithmetic. They are inserted for reference.
Symbols.
TT = 3.14159 . . ., or nearly Sf
r = radius. a = area. b = base.
c = circumference. h = altitude (height).
Formulas.
Rectangle, a = bh. Triangle, a = ^bh.
The square on the hypotenuse of a rightangled triangle
equals the sum of the squares on the other two sides.
Circle, c = 2 irr. a = rrrr^.
Illustrative problem. What is the length of the radius of
the circle whose circumference is 62.8318 units ?
1. •.• = 27^*,
2. .. 62.8318 = 2 3. 14159 r.
3. .. 10 = r.
EXERCISES. LXXXVI.
59. What is the altitude of a triangle whose area is
7 sq. in. and whose base is 2 in. ?
60. What is the length of the base of a rectangle whose
area is 18 sq. in. and whose altitude is 2^ in. ?
61. From the top of a flagstaff a line just reaches the
ground ; if a line a yard long is tied to this (no allowance
being made for the knot), the whole line when tightly
stretched touches the ground 20 ft. from the staff. Ee
quired the height of the staff.
I
SIMPLE EQUATIONS. 173
62. What is the length of the radius of the circle whose
area contains 25 tt sq. in. ?
63. If the area of a triangle is 3 V3, and the base is
2 V3, required the altitude.
64. What is the length of the diameter of the circle
whose circumference is 157.0795 in.?
65. The perimeter of a rectangle is 14 in., and the base
is 33^% longer than the altitude. Eequired the length of
the diagonal.
66. Two rectangles of the same area have the following
dimensions ; the first, 15 ft. by 10 ft., and the second,
18 ft. by X ft. Eequired x.
67. What is the length of the radius of the circle the
number of square units of whose area equals the number
of linear units of circumference ?
68. The perimeter of a triangle is 75 in. ; the second
side is § of the first and the third is f of the first.
Required the length of each side.
69. The area of a triangle is 250 sq. ft., and the altitude
is 25% more than the base. Eequired the length of the
base. Is the resulting equation linear ?
70. The perimeter of a triangle is 24 in., the first side is
2 in. longer than the second, and the second is 2 in. longer
than the third. Eequired the length of each side.
71. A dock pile is  above water and ^ is driven into
the soil ; if the water at the dock is 7 ft. deep, what is the
entire length of the pile and how many feet are above
water ?
174 ELEMENTS OF ALGEBRA.
E. Historical Problems.
Many problems which were of considerable difficulty
prior to the introduction of our present algebraic symbols,
about the opening of the seventeenth century, are now com
paratively easy. They have considerable historical interest
as showing the state of the science at various periods, and
a few examples are here inserted.
EXERCISES. LXXXVII.
72. If 9 porters drink 12 casks of wine in 8 das., how
many will last 24 porters 30 das. ? (Tartaglia, a famous
Italian algebraist, about 1550 a.d.)
73. Demochares lived i of his life as a boy, ^ as a
young man, ^ as a man, and 13 years as an old man. How
old was he then ? (Metrodorus, 325 a.d.)
74. Of 4 pipes, the first fills a cistern in 1 da., the second
in 2 das., the third in 3 das., and the fourth in 4 das. How
long will it take all running together to fill it ?
75. In the center of a pond 10 ft. square grew a reed
1 ft. above the surface ; but when the top was pulled to
the bank it just reached the edge of the surface. How
deep was the water ? (From an old Chinese arithmetic,
Kiu chang, about 2600 b.c.)
76. A horse and a donkey, laden with corn, were walk
ing together. The horse said to the donkey : " If you gave
me one measure of corn, I should carry twice as much as
you; but if I gave you one we should carry equal buidens."
Tell me their burdens, most learned master of geometry.
(Attributed to Euclid, the great writer on geometry at
Alexandria, about 300 b.c.)
I
eq
i
SIMPLE EQUATIONS. 175
77. Heap, its whole, its seventh, it makes 19. (That is,
hat is the number which when increased by its seventh
equals 19? From the mathematical work copied by the
Igyptian Ahmes about 1700 b.c. from a papyrus written
ut a thousand years earlier.)
I
78. Find the number, ^ of which and 1, multiplied by ^
of which and 2, equals the number plus 13. (Mohammed
ben Musa AlKhowarazmi, the famous Persian mathemati
cian, 800 A.D. From the title of his book comes the word
Algebra, and from the latter part of his name — referring
his birthplace — comes our word Algorism.)
79. In a pond the top of a lotus bud reached ^ ft. above
e surface, but blown by the wind it just reached the
surface at a point 2 ft. from its upright position. How
deep was the water? (From a mathematical work by
Bhaskara, a Hindu writer of about 1150 a.d. The work
Kas named the Lilavati in honor of his daughter.)
80. Two anchorites lived at the top of a perpendicular
iff of height h, whose base was mh distant from a certain
•wn. One descended the cliff and walked to the town;
the other flew up a height, x, and then flew directly to
ibe town. The distance traversed by each was the same.
I^nd X. (Brahmagupta, a Hindu mathematician, about
^0 A.D.)
^ 81. An ancient problem relates that Titus and Caius sat
iOwn to eat, Caius furnishing 7 portions and Titus 8, all
f equal value. Before they began Sempronius entered
and they all ate equally and finished the food. Sempronius
then laid down 30 denarii (pence) and said : " Divide these
K[uitably between you in payment for my meal." How
176 ELEMENTS OF ALGEBRA.
F. Discussion of Problems.
196. Many problems can be suggested which admit of
mathematical solution, but whose practical solutions are
impossible by reason of the physical conditions imposed.
E.g.^ I can look out of the window 18 distinct times in 4 sees.
What is the rate per second ?
The answer, 4 times per second, while entirely correct from the
mathematical standpoint, is physically impossible ; for while I can look
out 4 times, I cannot look out half of a time.
The problem might easily be changed, however, so as to demand
the time required to look out once, the answer behig  of a second.
A similar absurdity appears in the result of the following
problem: A father is 53 yrs. old and his son 28. After
how many years will the father be twice as old as the son ?
We have the equation
53 + £c = 2(28 + ic),
whence cc = — 3.
We are now met by the necessity of
(1) interpreting the meaning of the answer — 3 yearn
after this time, or
(2) changing the statement of the problem so as to avoid
an answer which seems meaningless.
It is immaterial which course we take. We may say :
(1) — 3 years after this time shall be understood to mean
3 years before this time, which is entirely in harmony with
our interpretation of negative numbers (§ 29) ; or
(2) we may change the problem to read : " How many
years ago was the father twice as old as the son ? " For
this latter question the solution would be
53 a: = 2(28 a;).
.•.x = 3,
and the answer would be 3 years ago.
SIMPLE EQUATIONS. 177
The discussion of results of this nature is well illustrated
in an ancient problem known as the Problem of the Couriers.
A courier, A, travels at the uniform rate of a mi. per
hour from F ; after t hrs. a second one, B, starts in pursuit
from F and travels at the uniform rate of h mi. per hour.
After how many hours will B overtake A ?
Pr
hx
Solution. 1. Let x = the number of hours required.
2. Then •.• a (^ + x) = 6x, the distance B must travel,
at ,/. V '^
(^■' ba
Discussion. 1. If none of the quantities is zero, and 6 > a, the
denominator is positive and .. x is positive.
2. But lib = a, the denominator is zero and .. x is infinite (§ 170).
I.e., if they are traveling at the same rate B will never overtake A.
3. And if b <ja, tlie denominator is negative and .. x is negative.
I. e. , if B is traveling slower than A he will never overtake him. But
if the problem reads, " • • • after t hrs. B passes through P in pursuit,"
then the result would mean that they had been together hrs.
before reaching P. \ — a\
4. If either t = 0, or a = 0, the numerator is zero and .. x = 0,
except when 6 = a, in whicli case x is undetermined (§ 168). This is
evidently true, for ift = and they are traveling at tlie same rate they
will always be together. Or if a = and a = 6, then neither courier
is traveling at all, and hence they are always together at P.
EXERCISES. LXXXVIII.
Solve the following and discuss the results according to
the suggestions given above and in the problems.
1. A bicyclist starts out riding 10 mi. per hour, and is
followed after 30 mins. by a second riding 8 mi. per hour.
In what time will the second overtake the first ? 
178 ELEMENTS OF ALGEBRA.
2. A bicyclist starts from P, riding a mi. per hour ; after
t hrs. another follows and overtakes him in h hrs. At what
rate did the second one ride ? Discuss for ^ = A = 0.
3. A bicyclist starts from P, riding a mi. per hour ; he
is followed after t hrs. by a second rider traveling c times
as fast. After how many hours will the second overtake
the first ? Discuss for (1) c > 1, t^ 0, (2) c = 1, t^ 0,
(3) c < 1, t^ 0, (4) c = 1, t = 0.
4. Two trains going from San Francisco to Chicago, on
the same road, pass through Omaha, the first at 9.30 a.m.,
and the second at 10 a.m. The first train travels at the
rate of 50 mi. per hour, and the second 10% slower. At
what distance from Omaha are they together ?
REVIEW EXERCISES. LXXXIX.
1. Are x = 2 and x* = 16 equivalent equations ? Why ?
2. Show that if cc is a factor of every term of an equa
tion, is a root. E.g.^ x^ { 2x = 5x.
3. Solve the equation
3a2\a + 3[a2(aa2x)^\ = lla.
4. Show that if both members of an equation have a
common linear factor containing the unknown quantity, a
root can be found by equating this factor to zero.
5. What is the fallacy in this argument ?
1. Let X = a.
2. Then x^ = ax, multiplying by x.
3. Then x^ — a^ = ax — a^, subtracting a^.
4. Then (x \ a) (x — a) = a (x — a), factoring.
5. .'. 2 a (x — a) = a (x — a), hecsbuse x\ a = 2a.
6. .*. 2 = 1, dividing by a(x — a).
CHAPTER X.
SIMPLE EQUATIONS INVOLVINCx TWO OR MORE
UNKNOWN QUANTITIES.
197. A single linear equation containing two unknown
quantities does not furnish determinate values of these
quantities.
This means a single equation in which the similar terms have been
united. I.e., x + y = x { 3 is not included, because the x's have not
been united.
E.g., X — y = 1 is satisfied if x = 1 and y = 0, or if x = 2 and y = 1,
or if X = 3 and y = 2, etc.
198. But two linear equations containing two unknown
quantities furnish, in general, determinate values. Simi
larly, as will be seen, a system of three linear equations
(ontaining three unknown quantities, a system of four linear
^nations containing four unknown quantities, • • • a system
K n linear equations containing n unknown quantities,
urnish, in general, determinate values of all of these
uantities.
1
199. Equations all of which can be satisfied by the same
lues of the unknown quantities are said to be simultaneous.
E.g.,x{y = 7,x — y = S, are two equations which are satisfied if
X = 5 and y = 2. Hence they are simultaneous.
But x + y = 7 and x { y = S cannot be satisfied by the same values
of X and y, and hence they are not simultaneous.
The equations x + 2 ?/ = 6, Sx \6y = 9, are simultaneous ; but
each being derivable from the other they do not furnish determinate
values.
179
180 ELEMENTS OF ALGEBRA.
I. ELIMINATION BY ADDITION OR SUBTRACTION.
200. The solution of two simultaneous equations involv
ing two unknown quantities is made to depend upon the
solution of a single equation involving but one of the
unknown quantities. The usual process, by addition or
subtraction, is seen in the following solutions :
1. Solve the system of equations
L 4ic + 3y = 41.
2. Zx2y = l.
We first seek to give the ?/'s coefficients having the same absolute
values. This can be done by multiplying both members of the first by
2, and of the second by 3. Then
3. 8x + 6y = 82.
4. 9x62/ = 3.
Add equations 3 and 4, member by member, and
6. 17x = 85.
6. .. X = 5.
Substitute this value in equation 1, and
7. 4.5 + 3y = 41.
8. .. 32/ = 21.
9. .. 2/ = 7.
Check. Substitute these values in equation 2 (because 2/was obtained
by substituting in equation 1), and 3 • 5 — 2 • 7 = 1.
For brevity we shall hereafter use the expressions, in
solutions, "Multiply 2 hj 5," etc., meaning thereby, "Multi
ply both members of equation 2 by 5," etc.
201. When one of the unknown quantities has been made
to disappear (as in passing from steps 3 and 4 to step 5
above) it is said to be eliminated.
In the above solution y was eliminated by addition. The
quantity x may, however, be eliminated first, by subtraction,
as in the following solution.
SIMPLE EQUATIONS.
181
2. Solve the
system of equations
1. 4ic + 32/ = 41.
2. 3x2y = l.
3. •••
12x + Oy = 123, multiplying 1 by 3,
4. and
12x82/ = 4.
(Why ?)
5. .
11 y = 119, subtracting 4 from 3.
^....
2/ = 7.
(Why ?)
^7. ..
4a; + 21 =41.
(Why ?)
8. ..
4x = 20.
(Why?)
«. ..
x = b.
(Why ?)
Check. In which equation should these values now be substi
tuted ? (Why ?)
Other types are illustrated in the two problems fol
lowing,
3. Solve the system of equations
^ V r.
1. = 2.
3 2
X y „
2^4
It is not worth while here to clear of fractions. Simply multiply
)th members of the first by i, and
1 = 1
6 4
2x
». .*. — = 8, adding 2 and 3.
6. .. X = 12. (Why ?)
It is now apparent that y can easily be found and the results checked
in the usual way.
I
182 ELEMENTS OF ALGEBRA.
4. Solve the system of equations
X y 4t
X y
These are not linear equations because, when cleared of algebraic
fractions, they are of the second degree. But they can easily be
solved by the methods of linear equations as here suggested.
3. \^l = ^ ^^^y^)
4. .'.  = , subtracting 1 from 3.
X 4
5. .. 4 = », multiplying by 4 x.
Hence, y is easily found to be 2, and the results check.
EXERCISES. XC.
Solve the following, checking each result by proper sub
stitution :
9.
Ix
32/ =
:3.
5x
+ 7)/ =
:25.
x\
■17 2/ =
53.
Sx + y =
19.
6x5y:
12.
12 a; 11 2/
= 27.
X
7 2
X
2 +
F^^
3
X
X
4_
y~
5 31
y~ 3
15
2 '
5.
2.
3x + 5y = 5.
4,x3y = 26.,
4.
5x + 2y = l.
18x{8y = ll.
6.
l.Tx{l.ly = 13,
1.3cc0.l2/ = l.
8.
5 + 10^
^ 4.2/S
10 + 5"^'
0.
¥1=''
i+ir»
. ELIMINATION BY SUBSTITUTION AND COMPARISON.
202. After finding the value of one unknown quantity
addition or subtraction the other is usually, but not
necessarily, found by substitution. It is often more con
venient to find each by substitution, especially when one
of the coefficients is 1.
This method of elimination by substitution is illustrated
in the following solution :
1. Given x — ?/=— 5,
2. and 3 x + 2 ?/ = 45.
From equation 1 we have :
3. X = f ?/  5. (Why ?)
Substitute this value in equation 2, and
4. 2 y — 15 + 2 2/ = 45, from which
4 ?/ = 60.
6. .. y = 15.
From this x is found, by substitution, to be 5, and the results check.
It is not necessary that the coefficient oi x oi y should
1, although this is the case in which the method is most
requently employed. Consider, for example, the follow
ing solution :
1. Given 2x\by= 154,
2. and 30x2?/ = 0.
3. From equation 2, x = ^T^y.
4. Substituting, ^^y \ ^y — 154,
lence y = 20.
x = 2.
203. A special form of substitution occurs when the value
one of the unknown quantities is found in each equation,
nd these values are compared. This is called elimination
by comparison.
i
184 ELEMENTS OF ALGEBRA.
The method is illustrated in the following solution :
1. Given x — ^y = — 5,
2. and 3 x + 2 y = 45.
Solving equations 1 and 2 for x, we have :
3. x = f2/5,
4. and x= 15 — f y.
Substituting the value of x from step 3 in step 4, or, what is the
same thing, comparing the values of x (by ax. 1), we have :
5. f?/5 = 15^y.
6. .. f2/ = 20. (Why?)
7. .. 2/ = 15. (Why?)
8. .. X = 5, by substituting in step 3.
Check. Substituting in both of the original equations,
5 _ 1 . 15 = _ 6.
3 • 5 + 2 ■ 15 = 45.
EXERCISES. XCI.
Solve the following by substitution or comparison, check
ing the results as usual :
1.
x+y=:ll.
2.
X + y = s.
3 £c + 2 2/ = 44.
X — y =: d.
3.
x = y.
4.
X \ay — h.
3ic + 5?/ = 120.
ex \ y =■ d.
5.
xyl = 0.
6.
ax + hy = c.
2x + y29 = 0.
a'x + b'y = c'.
7.
x + y = 6912.
8.
x\2y = 30.
X = 4444 f y.
i^i2/ = 3.
9.
x\lly = 300.
10.
x\l^y = 26^^^.
llxy = 104..
42/^ = 44.
11. 1.543689 cc
y
= 1.543689.
aj 0.839286 7/
= 0.839286.
SIMPLE EQUATIONS. 185
III. GENERAL DIRECTIONS.
204. The following general directions will be found of
ome value, although the student must use his judgment
in each individual case.
1. If the equations contain symbols of aggregation, decide
whether it is better to remove them at once.
It is usually best to remove them, as in a case like ex. 18, p. 188.
But in a case like ex. 17, p. 188, it is evidently better to add at once.
2. If the equations are in fractional form, decide whether
it is better to eliminate without clearing of fractions.
See pp. 181, 182, illustrative problems 3, 4. Much time is often
wasted by clearing of fractions unnecessarily. This is also seen in
the example on p. 193.
3. If it seems advisable, clear of fractions and reduce
each to the form ax + by = c.
See illustrative problem 1, p.. 186. The same course will naturally
be followed with an example like ex. 8, p. 187.
4. If the coefficient of either unknown quantity is 1, it
is usually advisable to eliminate by substitution.
See illustrative problem 1, p. 186, steps 4, 6, 7. This is, however,
not often the case.
5. Otherwise it is generally best to eliminate by addition
or subtraction.
This is the plan usually employed.
I'
6. If the unknown quantity is in an exponent, follow the
Ian suggested in § 205.
It is here assumed that the root of the single equation
erived from the two given equations satisfies those equa
186 ELEMENTS OF ALGEBRA.
Illustrative problems. 1. Solve the system of equations
y
2. ? + 5 = ^ + 2.
X X
Here it is not best to attempt to eliminate without clear
ing of fractions. Multiplying both numbers of 1 by y,
3.
l+x + 3?/ = 5?/.
Ax. 6
4. ..
x = 2y \.
(Why ?)
5.
2 + 5x = ?/ + 2x, from 2.
(Why?)
6. ..
3 X — ?/=:— 2, from 5.
(Why ?)
7. ..
6^/ — 3 — ?/=— 2, substituting 4 in 6.
8. ..
'oy=\.
(Why ?)
9. ..
1 .> 3
(Why?)
Check.
d
Substituting in both given equations,
2 + 3 = 5, from 1,
i/ + 5 = i+2,
f = f , from 2.
and
205. Equations in which the unknown quantities appear
as exponents are called exponential equations.
Exponential equations of the following type are easily
solved by means of simple equations.
2. Solve the system of equations
60
1. •.•
a2a;.a3/ = a32,
2. ..
Qix + Zy — a^.
3. ..
2x + 3?/ = 32.
4. Similarly,
3x + 42/ = 44.
Solving,
X = 4, 2/ = 8.
SIMPLE EQUATIONS.
187
EXERCISES. XCII.
Solve the following, checking as usual :
= V^.
3.
X
m n
=p.
X y
5. p^"" jy"' =f'^.
9. ^"•^:65 2/^^20_
11. a^ + TV^/^^l.
 1 cc = 61.
2/
ri3. ^+^ = «.
a; — 2/Hl
y — X {1
x — y + 1
ab.
L5. ^' + ^
6.3.
3+56 ~'^*^''^
2.
4^ = 1
a b
X
m
3a;
~5~
6.
3 8~^'
4^ . 16^ = 2^°.
16^.22 2/ =41
4a; + 81
10^/ 17
12a; + 97
152/ 17
10. 7 a; 1 7/3=48.
5 ?/ + I a; = 26.
12. 17 a; 13?/ =.144.
23 a; + 19 2/ = 890.
14.
16.
+ 1 = 0.
X y
: — I = a— 0.
X V
xy
15
9x
32/ + 44
too.
188 ELEMENTS OF ALGEBllA.
17. a{x \ y) — h{x — y) = 2 a.
a{x — y) — b {x \ y) = 2h,
18. 10[ic + 9(3/8ic + 7)]=6.
5[a^ + 4(2/3^T2)]=l.
19.
a
xy =
a — c a
a +
6 a + 1
c^
c a
20.
5y
6
42/ig
3
1 03 20 
6"^ 3
22/.
X
6 +^
3
21.
2a;_
1/ "
29
14'
2/ + 4CC + 6 =
4^/^ + 13^7/
12ic2
42/3X
1
22.
7 + 3
^ =
= a; +
2/
5i.
11— £c Ax ^8y 
.2 ' 9 ^
— 2
= 8
(y
:r).
4:x^^2xyh288
6 2/^
= 2x
+ 3
2/
131,
2^ + 1322/
5a;
42/:
^22.
23.
24. 7y + 135. _ 3y + 2x16 .
4 ^ 3
,5y + 2x 3.T12 + 8y , 15 + 2?/ 4a;
SIMPLE EQUATIONS. 189
IV. APPLICATIONS OF SIMULTANEOUS LINEAR EQUA
TIONS INVOLVING TWO UNKNOWN QUANTITIES.
Illustrative problem. The sum of two numbers i$ 12
and 7 times the quotient of one divided by the other is 5.
Required the numbers.
1. Let
2. Then
X, ?/ = the numbers.
x + y = 12, and
3.
7 •  = 5, by the conditions of the problem.
4. ..
5. And
6. ..
7. ..
8. ..
y = 12 — X, from 2.
7 X = 5 2/, from 3*
7x = 605x.
12x = 60, andx = 5.
y = 7, from step 4.
Ax. 3
Ax. 6
(Why ?)
(Wliy?)
Check. The sum of 5 and 7 is 12, and 7 times f is 5.
EXERCISES. XCIII.
1. The sum of two numbers is 30 and their difference
is 17. Eequired the numbers.
2. What is that fraction which equals J when 1 is
added to the numerator, but equals ^ when 1 is added to
the denominator ?
3. A number of two figures is 5 times the sum of its
ligits. If 9 is added to the number, the order of its digits
reversed. Required the number.
4. A man invests $16,000 for 8 yrs. and $11,000 for
yrs., and receives from the two $8090 interest. Had the
[•st been invested at the same rate as the second and the
icond at the same rate as the first, he would have received
f310 more interest in the same times. Required the rate
tt which each was invested.
190 ELEMENTS OF ALGEBKA.
5. The sum of two numbers is s and their difference
is d. Required the numbers. From the result, deduce a
rule for finding two numbers, given their sum and their
difference.
6. The sum of two capitals, each invested at 5%, is
$12,000, and the sum of 5 yrs. simple interest on the
larger and 4 yrs. simple interest on the smaller is $2800.
Required the capitals.
7. Divide the two numbers 80 and 90 each into two
parts such that the sum of one part of the first and one
part of the second shall equal 100, and the difference of
the other two parts shall equal 30.
8. Two points move around a circle whose circumfer
ence is 100 ft. ; when they move in the same direction they
are together every 20 sees. ; when in opposite directions
they meet every 4 sees. Required their rates.
9. The boat A leaves the city C at 6 a.m. ; an hour
later the boat B leaves the city D, 80 mi. from C, and
meets A at 11 a.m. They would also meet at 11 a.m. if B
left at 6 A.M. and A 45 mins. later. Required their rates.
10. Of two bars of metal, the first contains 21.875%
pure silver and the second 14.0675%. How much of each
kind must be taken in order that when melted together
the new bar shall weigh 60 oz., and 18.75% shall be pure
silver ?
11. A marksman fires at a target 500 yds. distant and
hears the bullet strike 4J sees, after he fires ; an observer
standing 400 yds. from the target and 650 yds. from the
marksman hears the bullet strike 2^ sees, after he hears
the report. Required the velocity of sound and the
velocity of the bullet, each supposed to be uniform.
SIMPLE EQUATIONS. 191
12. Find two numbers the sum of whose reciprocals
is 5, and such that the sum of half of the first and one
third of the second equals twice the product of the two
numbers.
13. Two bodies are 96 yds. apart. If they move
)wards each other with uniform (but unequal) rates, they
rill meet in 8 sees. ; but if they move in the same direc
y^on the swifter overtakes the slower in 48 sees. Required
le rate of each.
14. The sum of two numbers, one of one figure and the
ither of five figures, is 15,390. Writing the fi^rst number
the first digit to the left of the second number gives a
lumber 4 times as large as that which is obtained by writ
ig it as the last digit to the right. Required the numbers.
15. A reservoir has two contributing canals. If the
first is open 10 mins. and the second 13 mins., 15 cu. yds.
of water flow in ; if the first is open 14 mins. and the
second 5 mins., 2.4 cu. yds. more flow in. How many
cubic yards of water per minute are admitted by each ?
16. A silversmith has two silver ingots of different
quality. He melts 13 oz. of the finer kind with 12 oz. of
the other, the resulting ingot being 852 fine (see p. 169,
ex, 44) ; but if he melts 1.5 oz. of the finer kind with
1 oz. of the other the resulting ingot is 860 fine. Required
le fineness of each original ingot.
17. It is shown in physics that if a body starts with a
relocity of u ft. per second, and if this velocity increases
ft. per second, then at the end of t sees, the body will
Lve passed over ut + ^ft^. Suppose / is uniform and
lat in the 11th and 15th sees, the body passes through
J4 ft. and 32 ft., respectively, find u and /.
192 ELEMENTS OF ALGEBRA.
V. SYSTEMS OF EQUATIONS WITH THREE OR MORE
UNKNOWN QUANTITIES.
206. In general, three linear equations involving three
unknown quantities admit of determinate values of these
quantities. For one of the quantities can be eliminated
from the first and second equations, and the same one from
the first and third, thus leaving two linear equations involv
ing only two unknown quantities. Similarly for a system
of four linear equations containing four unknown quanti
ties, and so on.
Illustrative problems. 1. Solve the following system of
equations :
1. 5x — Sy{4:Z==17.
2. 2x + 7y — 5z = 5.
3. 9x2yz = S.
We first proceed to elimiuate z from 1 and 2.
4. 25x  15 2/ + 20 z = 85, multiplying 1 by 5.
5. 8 a F 28 ?/  20 z = 20, multiplying 2 by 4.
6. .. 33x + 132/= 105. (Why?)
We now proceed to eliminate z from 1 and 3.
7. 36 X  8 y  4 2 = 32, multiplying 3 by 4.
8. .. ' 41 X  11 ?/ = 49, from 1 and 7.
We now proceed to eliminate y from 6 and 8.
9. 363 X + 143 2/ = 1155, multiplying 6 by 11.
10. 533 X  143 y = 637, multiplying 8 by 13.
11. .. 896x = 1792. (Why?)
12. .. x = 2. (Why?)
13. .♦. 2/ = 3, substituting in 6.
14. .. 2 = 4, substituting in 1.
Cfieck. Substitute in 2 and 3. (Why not in 1 ?)
4 + 21  20 = 5, and 18  6  4 = 8.
f
SIMPLE EQUATIONS.
2. Solve the following system of equations
ox 1 y 9z
2. ?+?
X y
67.
5 + £
y z
38.
193
We first proceed to eliminate  from 1 and 2.
4 4 4
\ 1 = 4, from 1.
5x ly 92
2,3 4 67 ,
— — , from 2.
9x 92/ 9z 9
4
li + ^ = 103.
5x 72/
(Why ?)
(Why ?)
(Why ?)
We then proceed to eliminate  from 2 and 3.
7.
u
X
i^ + " = _ 76, from 3.
2/ z
(Why ?)
8. ..
^  ^ =  9, from 2 and 7.
(Why ?)
We then proceed to eliminate  from 6 and 8.
8.
322
5x
+ ^ = 721, from 6.
72/
(Why ?)
L
456
7x
399 513 ^
— — = , from 8.
72/ 7
(Why ?)
11.
4534 4534 ^ ,, , ,_
= , from 9 and 10.
35x 7
(Why ?)
12. .'.
^ ^ A 1
= 5, and x = •
X 5
(Why ?)
13. ..
p7, and2/ = .
(Why ?)
14. ..
l=9,and.=.^.
(Why ?)
194 ELEMENTS OE ALGEBRA.
Check. Substitute in 1 and 2. (Why not in 3 ?)
1 + 11 = 1.
10 + 21+36 = 67.
The equations in ex. 2 are not linear in x, y, z (why ?),
and it is unwise to clear of fractions (why ?). The equa
tions are linear in ? > ? and it is better to solve as if
X y z
these were the unknown quantities.
3. Solve the following system of equations :
1. ic 4 2/ — s; = 6.
2. ic42/ + 2« = — 3.
3. x2y z = ^.
Frequently systems of equations offer some special solu
tion, as in this case.
Adding the equations, member by member,
4. 3x = 3.
5. .. x = l.
Subtracting 2 from 1, member by member,
6. 32 = 9.
7. .. 2 =  3.
8. .. 2/ = 2, substituting in 1.
Check. Substitute in 2 and 3.
1 + 26= 3.
14 + 3 = 0.
4. Solve the following system of equations :
1. w{2x\y — z = 4^.
2. 2w\x\y\z = l.
3. ^w — x\2y — z = l.
4. 4.W + 3xy\2z = l^.
SIMPLE EQUATIONS.
196
Eliminating z from 1 and 2,
5. 3w) + 3a; + 22/ = ll.
Also from 1 and 3,
6. 2«;3x + y=3.
Also from 1 and 4,
7. 6iw4 7a; + y = 21.
Eliminating y from 6 and 6,
8. to9x=17.
9. Also from 6 and 7, 2 w> + 5 cc = 12.
Eliminating w from 8 and 9,
10. jc = 2.
11. .. io = 1, substituting in 8.
12. .. 2/ = 1, substituting in 6.
13. .. 2 = 2, substituting in 1.
Check. Substitute in 2, 3, and 4. (Why not in 1 ?)
2+2 + 1+2 = 7.
32 + 22 = 1.
4 + 61+4= 13.
EXERCISES. XCIV.
Solve the following systems of equations
1. l+i=l.
X y
2.
 + f +  = 258.
^ + ^2.
X z
M+5^«^
y^ z 2
M + f = 296.
3. lx^y =
1.
4.
5x — 6y\4:Z = 15
llz7u =
1.
7x + 4:y3z = 19
4rz7y =
1.
x + y = l.
19x3u^
1.
a; + 6 « = 39
196 ELEMENTS OF ALGEBRA.
5. x + y = 16. 6. x\yz = 132.
z{x = 22. xy + z = 65.4.
y + z=2S. x + y + z =  1.2.
7. a* • a^ + 2 == a>\ 8. a^x + h^y + Cit; = ^i.
~ a^' ■ a^ + ^ = a^"^. a^x + h^y + Cg;^ = c^g.
9. a; = 21— 4?/. 10. ic + y + ;^ = 5.
z = 9^x, 3x5y + 7z = 75.
y = 64  7^5:. 9ic  11 ^ + 10 = 0.
11.
X y
L8 = i8.
a: ?/ ^
X y
^ = 25.
z
13.
3yl
4
5.^4;*;
4+3
3a; + l ^ 1
12. 7ic — 2;t; + 3ii = 17.
4?/ — 2« + V = 11.
5?/3x2i^ = 8.
4 7/  3 w + 2 V = 9.
3z + 8it = 33.
6 ^ .T 9
T~2"^5'
14 "^ 6 21 "^ 3
14. «4^+3?/ + 2. ^3x + y + 2 __ ^2« + lo_
15. 2 1^; + cc  10 ?/ + 0.5 ^ = 7.62.
3 ^  2 .X + 2 ?/ + 3 s; = 8.26.
w + 3x + 5y — z = 8.61.
6?^2ic + 32/+ 10^ = 25.51.
SIMPLE EQUATIONS.
197
VI. APPLICATIONS OF SIMULTANEOUS LINEAR EQUA
TIONS INVOLVING THREE UNKNOWN QUANTITIES.
Illustrative problem. A certain number of three figures
such that when 198 is added the order of the digits is
jversed ; the sum of the hundreds' digit and the tens' digit
the units' digit ; and the number represented by the two
jfthand digits is 4 times the units' digit. Required the
dumber.
1. Let a = the hundreds' digit, 6 the tens', c the units'.
2. Then 100 a + 10 6 + c = the number.
3. Then, by the first condition,
100a + 10& + c + 198 = 100c + 10 6 + a.
4. By the second condition, a \h = c.
5. By the third condition, 10 a + 6 = 4 c.
6. .. the equations are
a — c = — 2, from step 3.
a + & — c = 0, from step 4.
10a + & — 4c = 0, from step 5.
7. Solving, a = 1, & = 2, c = 3.
.. the number is 123.
Check by noting that 123 answers all of the conditions of the origi
nal statement.
EXERCISES. XCV.
1. What three numbers have the peculiarity that the
pm of the reciprocals of the first and second is ^, of the
irst and third ^, and of the second and third ^ ?
2. There is a certain number of six figures, the figure
units' place being 4; if this figure is carried over the
ther five to occupy the lefthand place, the resulting
lumber is four times the original one. Required the origi
lal number.
108 ELEMENTS OE ALGEBRA.
3. Divide the number 96 into three parts such that the
first divided by the second gives a quotient 2 and a
remainder 3; and the second divided by the third gives
a quotient 4 and a remainder 5.
4. The middle digit of a certain number of three figures
is half the sum of the other two ; the number is 48 times
the sum of the digits. Subtracting 198 from the number,
the order of the digits is reversed. Required the number.
5. Of 3 bars of metal, the first contains 750 oz. silver,
62^ oz. copper, 187 oz. tin; the second, 62 oz. silver,
750 oz. copper, 187^ oz. tin ; and the third no silver, 875 oz.
copper, 125 oz. tin. How many ounces from these bars
must be melted together to form a bar which shall contain
250 oz. silver, 562 oz. copper, and 187^ oz. tin ?
6. Of three bars of metal, the first contains 750 oz.
silver, 200 oz. copper, 50 oz. tin ; the second, 800 oz. silver,
125 oz. copper, 75 oz. tin ; and the third 700 oz. silver,
250 oz. copper, 50 oz. tin. How many ounces from these
bars must be melted together to form a bar which shall
contain 765 oz. silver, 175 oz. copper, and 60 oz. tin ?
7. Two bodies, A and B, start at the same time from the
points P and Q, respectively, and move at uniform rates
towards one another, B faster than A; at the end of
18 sees., and again at the end of 30 sees., they are 48 ft.
apart. Had they moved in the same direction, B follow
ing A, at the end of 40 sees, they would have been 48 ft.
apart. Determine their rates and the distance PQ.
Solutions by determinants. The treatment of simulta
neous linear equations by determinants is set forth in
Appendix VII, and should be taken at this point if time
allows.
SIMPLE EQUATIONS. 199
REVIEW EXERCISES. XCVI.
1. Solve the equation 2.25 x — 5 — 0.4a; + 2.6 = 2a; — 3.
2. By the Remainder Theorem ascertain whether 10 x^
— 13 ic^ — 5 a; + 3 is exactly divisible by 2 a; — 3.
3. Form an integral linear function of x which shall,
equal 37 when a; = 10, and 4 when x = — 1.
4. Form an integral linear function of x which shall
vanish when a; = 2, and which shall equal 4 when x = 3.
(If /(a;) = 7nx f n, then 2 m \ n = and 3 m { n = A.)
5. Show that the following set of equations are not
simultaneous and hence cannot be solved :
62x + 93tj = 31.
2x\3y = 4:.
ii + 1
6. Simplify ' + +^' 
a^lH aj + lH
X X
7. Write down by inspection the quotient of
3 + ^2 + — + 4 by 4 +  + 1. Check.
*Aj %K/ %Mj *Aj %aj
13 2 2 3
8. Multiply 1 + ^ ^7by — f4by detached
JLf JL JU *kj *Aj
)efficients. Check.
^. .^1 3?/ , 5^/' 52/« ,37/* . , 1 y
9. Divide — r + ^ ^ + ^2/ ^y ^
x^ x^ x^ x^ X ^ _ x^ X
y^ by detached coefficients. Check.
10. Solve the equation
2a;l 2a; + 5 2a; + l ^^ + 3 ^^
2a; + l 2a; + 7 2a; + 3 2a; + 5
200 ELEMENTS OF ALGEBRA.
11. Solve the system
0.2x{0.3yhOAz = 25.
0.3x + 0.7y{0.6z = 4.5.
0.4a; + 0.8v/ + 0.9^ = 58.
12. Solve the system
1 1 1_1 3 2_3 1_
xyzxyzxy
3
4
13. Solve the system
3x5y + 4.z = 0.5.
7x{2y3z = 0.2.
4:x \3y — z = 0.7.
14. Solve the system
x^y\z = 3.824.
1.25 a; + 23.8 y + 3.1z = 7.5276.
1.1^ +2y0.5z = l.S505.
15. The sum of three capitals is $111,000. The first is
invested at 4%, the second at 4:%, and the third at 5%,
and the total annual interest is $5120. If the first had
been invested at 2^(^q, the second at 3%, and the third at
4%, the total annual interest would have been $3710.
Eequired the capitals.
16. In each of three reservoirs is a certain quantity of
water. If 20 gals, are drawn from the first into the second,
the second will contain twice as much as the first ; but if
30 gals, are drawn from the first into the third, the third
will contain 20 gals, less than 4 times as much as the first ;
but if 25 gals, are drawn from the second into the third,
the third will contain 50 gals, less than 3 times the second.
How many gallons does each contain ?
CHAPTER XI.
INDETERMINATE EQUATIONS.
207. A linear equation involving two unknown quantities
can be satisfied by any number of values of those quantities.
E.g., in the equation x {■ y = b
can equal • • •  2,  1, 0, 1, 2, 3, 4, 5, 6, • • •
16 corresponding values of y being 7, 6, 5, 4, 3, 2, 1, 0, — 1, • • •.
But of course this applies only to equations after like terms are
: united, and not to an equation like x ^ y = x \ "1.
208. Equations like the above, which can be satisfied by
an unlimited number of values of the unknown quantities
are called indeterminate equations.
209. Since two equations containing three unknown
quantities give rise, by eliminating one of these quantities,
to a single equation containing only two, it follows that,
in general, Two equations, each containing three unknown
quantities, are indeterminate as to all of these quantities.
E.g., the two equations
2x + 3 2/ + 2 = 10,
3x + 22/ + 2 = 8,
give rise to the single equation
 X + 2/ = 2,
or to by \ z = 14,
or to 5 X + 2 = 4,
all three of which are indeterminate.
201
202 ELEMENTS OF ALGEBRA.
210. Similarly, it is evident that, in general, n linear
equations, each containing n + 1 or more unknown quanti
ties, are indetermifiate.
Koots of an indeterminate equation are often foimd by
simple inspection.
JE.g., to find the roots oi 2 x — 7 y = 6.
Let X = 0, 1, 2, 3, 4, • • ■
then the corresponding values of ?/ are , , , , , •••.
Similarly, find a set of roots of. x + 2 y { S z = 10.
Let z = 1 ;
then X {2y = 7 ;
and if x = 0, 1, 2, 3, • • •
the corresponding values of y are , 3, f , 2, • • • .
That is, the equation is satisfied if
z = l, x = 0, y = l,
or if z = 1, x = 1, y = 3, etc.
Similarly, we may start with 2 = 2.
211. Sometimes it is desirable to find the various positive
integral roots of an indeterminate equation. For practical
purposes these may be found by simple inspection.
E.g. , to find the positive integral roots of5x + 37/=19. Here x ";}> 3,
because if x > 3, and integral, y is negative.
If x = 3, 2, 1
then ?/ = a fraction, 3, a fraction.
.. x = 2, y = S are the only positive integral roots of the equation.
Graphs and discussion of equations. For those who have
the time, the study of the graphic representation of linear
equations, and the discussion of solutions (Appendix VIII)
are strongly recommended at this point.
I
INDETERMINATE EQUATIONS. 203
EXERCISES. XOVII.
1. Find three sets of roots of each of the following
equations :
(a) 10a; + 32/ = 4.
(b) 5x2y = 17.
(c) 5cc + 232/ = 100.
2. Find two entirely different sets of roots of each of
the following equations :
(a) xStj\4:Z = 20.
(b) 2x^10yz = 15.
(c) 8x7y^5z = 12.
3. Find all of the positive integral roots of each of the
following equations :
(a) x + y = 5.
(b) 2^ + 102/ = 30.
(c) Sx + 52/ = 20.
4. Find all of the positive integral roots of
x\2y + 3z = 14:.
5. Find three sets of roots of the following system of
equations :
X — 2y\4:Z = 5.
2x — y \z = 1.
6. Find a set of roots of the following system of
luations :
2w + 2x + 3y\z = 20.
Sw{3x\2y + 2z = 25.
4:W{5x — y — z = 6.
CHAPTER XII.
THE THEORY OF INDICES.
I. THE THREE FUNDAMENTAL LAWS OF EXPONENTS.
212. It has already been proved that, when m and n are
positive integers,
1. a"', a" = a'" + ". § 60
2. a'^.a''^ a"*". § 86
3. (a™)" = «""». § 75
It has also been stated (§ 125) that a^ means the square
root of a, a^ means the cube root of a, and, in general, an
means the nth root of a, but the reason for this symbolism
has not yet been given.
It is now proposed to investigate the meaning of the
negative and the fractional exponents ; that is, to find what
meaning should be attached to symbols like 3~^, 8% 16~*,
_ 1
a"", a «,••.
We shall then proceed to ascertain whether the three
fundamental laws given above are true if m and n are
fractional, negative, or both fractional and negative.
The necessity for this is apparent. We know that a*" • «"
= a"*+", if m and n are positive integers, because a is taken
first m times, and then n times, as a factor, and hence
1 ]^
m \ n times in all. But we do not yet know that an • am
+I 1 \ i_i
= a" »». Neither do we know that a"» : <x« = a'" ", nor that
a"* • a" = a"—", nor that a""* • a^ = a '"^n, etc.
204
t
THE THEOllY OF INDICES. 205
II. THE MEANING OF THE NEGATIVE INTEGRAL
EXPONENT.
213. The primitive idea of power (§ 8) was a product of
equal factors. The primitive idea of exponent was the
number which showed how many equal factors were taken.
According to this primitive idea the
3d power of a meant aaa, written a^,
2d " " " aa, " a^,
but there was no first power of a, because that is not the
product of any number of a's, nor any zero power, fractional
power, or negative power.
But since a^ means aaa, or a^ = a,
and a^ "
.'. it is reasonable to define a^ as
and " " " " ^0 "
I
d, in general, to define
being a positive integer.
214. For this reason we define
a^ to mean a,
a' " 1,
1
a " 
a"
n being a positive integer.
But it is evident that a ^ 0.
aa,
a
a"" Ta,
a,
((
a^^a,
1,
a
a 7 a,
1
— J
a
ii
1 Ta,
1
a
1
a
1
— J
206 ELEMENTS OF ALGEBRA.
Illustrative problems. 1. Express 2"^ as a decimal frac
tion.
1. 22 i3y definition, means — •
2.  =  = 0.125.
23 8
2. Express _ with positive exponents.
. ah ah « «„ ^
1. — —  means § 214
a3
2. This equals ^. § 161
3. Express g^ in the integral form.
X y
§150
x3y2 aj
2. =xi?/. §214
The expression cc 1?/ is as much a fraction as is  , but it is not in
the form of a common fraction.
4. Simplify (22)2.
1. 22 means — §214
<«"■ •■ ©■
§214
.22>
3. This equals (22)2 which equals 2*. §§ 161, 76
5. Simplify \_(2^)^y\
1. 2~i means J.
2. (i)i " 2.
3. 21 " i
4. .. the expression reduces to ^.
THE THEORY OF INDICES. 207
EXERCISES. XCVIII.
Express exs. 14 without exponents.
 ^ f p:' (?r — .■(r
Express exs. 59 with, positive exponents.
9. 2(t3, t^^ (cc)^.
Express exs. 1016 in the form of common fractions, with
positive exponents for the factors.
, . , .. , a^^^' 2^32
10. ^^;;r • 11.
12 ^ ^, . 13. £cV^HiC2/V.
14. [(£C«)«]«. 15. [(1  £C)2(1  £C2>)1_
16. a%^cH^, a'^b^'cP.
Simplify exs. 1720.
28 45 e7 68 ^6 ^a Jc
• 3
~^'l
4
76
• 77'
18.
ar'' c^
&«
19.

 a^
■ 1 + a
2 3 + «
3
3«^ + l
(
a)
20.

a2.
(2i)^
• 2
«.6t2.(^,
• 2) 2.
208  ELEMENTS OF ALGEBRA.
IIL THE MEANING OF THE FRACTIONAL EXPONENT.
215. We have now found the meaning of
1. The positive integral exponent greater than 1, the
primitive meaning of exponent;
2. The unit exponent ;
3. The zero exponent ;
4. The negative integral exponent.
216. It remains to find the meaning which should attach
to the fractional exponent.
The expression a'^ means aaaa,
and if the exponent is half as large,
a^ or aa is the square root of a^,
and if the exponent is half as large,
a^ or a is the square root of a^.
.'. if an exponent half as large indicates a square root,
a'^ should mean the square root of a.
Hence, a^ is defined to mean the square root of a, and,
1
in general, a'^ is defined to mean the 7ith. root of a.
217. The reason for this is also seen from the fact that
• . • «"* • «"* • • • to n factors = a'"" .
.'. (/^" • a» ' • • " " should equal a"^»'^ or «" or a.
1
.'. «" should be defined to mean the nth root of a.
p
218. And since a*"" = («"')«, so a't should be defined to be
r 1
identical with {a'^y.
p
Hence, we define a*i to mean the ^th power of the qth root
_H p
of a, and ^ ^ to mea?i the reciprocal of a*.
THE THEORY OF INDICES.
209
219. The following identities involving fractional expo
nents are also true and will now be proved.
1.
a"Z*«c« • • • = {(xbG •••)«.
Proved in § 220
2.
1 1 m
§ 221
3.
m pm
§ 222
4.
11 J_ 11^
§ 224
220. To prove that a»b»&^ ■ ■■ = (ahc ••■)«
1. Let
1 1
2. ..
1 1
X" = (a"6«)»
Ax. 8
3.
4.
1 1
= ab.
§76
§217
5. ..
x =
1 1 1
(a6)«, or a«&» =
1
(a6)«.
Axs. 9, 1
1 1 1
0. .. a«6»c»E
1 1
E (a6)«c« E
1
E (a6c)", and so or
I for any
number of factor^.
Similarly,
6"
221. To prove that (a"*)" = (a^y = a^.
1 112
1. {aaa • ■ • to m factors)" = a^a^a'^ • • • to m factors. ' § 220
1 1
2. I.e., (a"»)" = (««)"'.
1 ^
3. But (««)»*= a«. Def. § 218
Hence, a" may be considered either as the mth power of
the nth. root of a (as defined in § 218) or as the nth root
of the mth power of a.
210 ELEMENTS OF ALGEBRA.
But § 221 must be understood to apply only to the abso
lute values of the roots.
E.g.\
(42)* = 16* = ± 4,
but
(4*)2 = (±2)2=+4.
222. To
prove
m pm
that a« = op^.
1. Let
m,
X = a».
2. ..
x» = a"*.
Ax. 8 and § 221
3. ..
^pn ^ ^pm^
Ax. 8 and § 75
4. ..
pm
X = a^".
Ax. 9 and § 218
5. .. a'' = a^''. Ax. 1
Hence, both terms of a fractional exponent can be multi
plied or divided by the same number "without altering the
value of the expression.
223. The student should understand clearly that § 222 is
true not because the exponent is a fraction. The exponent
is merely an expression in the form of a fraction, and hence
a proof like that of § 150 has no application to this case.
The laws of fractions apply to fractional exponents only as
they are proved to do so.
11 JL i 1
224. To prove that («»»)» = a""» = («»)"«.
1. Let a; = (a'»)«.
}_
2. .. jc" = a"*. Ax. 8
3. .. x""^ = a. Ax. 8
j_
4. .. X = a"*". Ax. 9
11 j_
5. .. («'«)« = «"««, and similarly
1 j. j_
(a«)'» = a'»«.
THE THEORY OF INDICES. 211
EXERCISES. XCrX.
Find the absolute value of each of the expressions in
exs. 13.
1. 4^ 9^ 8'% 32^, 81^.
2. 25^ 125^, 32^, 64^, 625i
3. 16"^, 36"^, 343«, 1331"^, 14641?.
Write in integral form, with negative or fractional expo
nents, the expressions in exs. 49.
1
a a { b a 1
4. T +
V^ ^ab V^\ ^a *'
5. a^b {b\^ + VaT~b — ^a — b.
1 1 V^ + V^ 1 1
' a\U bVb V^ «^' ^'
7. Vl H a% v^l  a% ^a^ h b^, 1  V^,
8. ^2 _^ ( V^ 4 Vl f a)  (^3 + V'^)  V^.
9. ^ywF, vv^^, Vi + ^^Vc, ^1 5 (a + ^)'. •
Write the following without negative or fractional expo
nents, using the old form of radical sign (V) and the
common fraction :
m m + l
10. ah^, ah^, ic% £c 2 , x^y^, zK
11. a~^, a'h~^, x^y~i, ic"*"?/"*^".
I
12. ^^ a^a^^a^ +{ab'^)\
a"^ r a~i
212 ELEMENTS OF ALGEBRA.
IV. THE THREE FUNDAMENTAL LAWS FOR FRACTIONAL
AND NEGATIVE EXPONENTS.
225. Laws 1 and 2. To prove that
o m . on :=: o m — n
if m and n are fractional, negative, or both fractional and
negative.
a. Let them he fractional and positive. We have first to
p r_ Pj^r
prove that a'^a^ = a^ *.
1.
p r ps qr
a^'a' = a'^'a^'
§ 222
2.
1 1
= {apy • (a'^y
§ 221
3.
1
= (aP' ■ a^y
§ 220
4.
1
= (aP' + '^y
§ 60
5.
ps + qr p r
= a '^' , OT a'' ^
§ 221
3/
This shows that a case like V a^ • Va^ can be easily
handled by fractional exponents, thus :
2 4 2,4 _22
a^ a^ = a^ ^ = a^\
3/ — 5/—
To see that V«.^ • Va^ equals the 15th root of a'^^ is not so
easy by the help of the old symbols alone.
We have also to prove that a'> : a^^ a* *.
The proof is evidently identical with that just given,
except that the sign of division replaces that of multipli
cation in the first member, and the sign of subtraction that
of addition in the second member.
THE THEORY OF INDICES. 213
b. Let one exponent he negative and either integral or
fractional. We have then to prove that
1. a"* • a
2.
3.
We have also to prove that <x'" :«,"" = a"'~<^~'*> = a'"+".
The proof is evidently identical with that just given,
except that the sign of division replaces that of multipli
cation, and the sign of subtraction that of addition.
^m+(n)^
or
a"•""^
a""
§§ 214, 218
a""
§ 156, cor. 2
a— «.
§§ 86, 225, a
c.
fraci
Let both exponents be negative
ional. We have then to prove
a?ic? either integral or
that
a"'a" = ^_ „« + („)
= a^
[ — n
1.
§§ 214, 218
2.
_ 1
§ 156
3.
1
— ^m + «
§§60, 225, a
4.
= »"»".
§§ 214, 218
As an illustration of the value of these laws, consider
the case of — — : ^
Here we have
a~^'a~^ = a~ ^^"^^ = a~ ^,
or the 20th root of , a result not so easily reached by the
older notation.
214 ELEMENTS OF ALGEBRA.
226. Law 3. To prove that (a™)° = a™", if m and n are
fractional, negative, or both fractional and negative.
a. Let m be fractional or negative or both, n being a posi
tive integer.
1. From §§ 60, 225, it follows that
aPa^a" '• = aP + 9 + r+
if p, q, r, "■ are fractional, negative, or both fractional and
negative.
2. And ifp = q = r = ' = m, and there are n factors,
then , ,
(a'^y = a"""",
whether m is positive or negative, integral or fractional,
provided % is a positive integer.
b. Let m and n be positive fractions. We then have to
p r pr
prove that (ai)^ = ai^
1. Let
x = {a^)\
2. Then
p
Ax. 8 and § 221
3.
pr
§ 226, a
4. ..
a;9« = aP\
Ax. 8 and § 221
6. ..
pr
X = a^\
Ax. 9
6. ..
pr pr
c. Let n be negative and either integral or fractional, m
being positive. We have then to prove that (a™)"'' = a""""^.
1. (a"*)" = ^ § 214
2.  — § 75
3. = a"^. § 214
THE THEORY OF INDICES. 215
d. Let m be negative and either integral or fractional, n
being positive. We have then to prove that (a~™)° = a~™°.
I
a"/
1. («—)" = (^ j § 214
2. ^ §75
3. = a^, § 214
e. Let m and n &e negative and either integral or frac
tional. We have then to prove that (a"~™)~" = a™".
1. ^ar^^Q^~^ §214
2. ^(^l^i.y^(^.«)n §214
3. = a^^. § 75
The value of this law may be seen by the solution of a
few problems. Consider for example the case of
This expression, thus written in the older style, does not
strike the eye as simple ; but since 1 ? Va^ may be written
a~i, the expression reduces to {ai)^, which equals a.
Consider also the more complicated expression
have
_r 9 qr r^—q^ qr
X'(X ^ '')r^Q^ = XX «'" r^<l^ = XX^ = X^=1.
[To simplify this without the assistance of negative and
fractional exponents would be more difficult.
216 ELEMENTS OF ALGEBRA.
V. PROBLEMS INVOLVING FRACTIONAL AND NEGATIVE
EXPONENTS.
227. It has now been proved that we can operate with
expressions involving negative or fractional exponents just
as if these exponents were positive integers. Exercises
involving such exponents will now be given.
The student should see the distinct advantage in using
the fractional exponent instead of the old form of radical
sign, except in cases like the expression of a single root,
and in using the negative exponent, except in cases like the
expression of a simple fraction. This has been shown on
p. 215, but it is worth while to consider the matter further,
that the student may become entirely familiar with the use
of the modern symbols.
E.g., while it is easier to write Va than a% and  than ai,
because we are more accustomed to the forms Va and  , it is much
a
easier to see that
than to see that the equivalent expression
V (1  Va;2)3
Similarly, it is easier to recognize in
x's _l_ 2 x^"^ + 1 =
the quadratic form
x^ + 2 x^ + 1 = 0,
than to recognize it in
2 8/— 2 1/
Vx8 + 2 V^ +1 = 0.
It is doubtful if students would readily grasp the significance of
the form a^ + 2 a2 va + a Va ; but when written a^ + 2 a^ + a* it is
seen to be the square of a^ + a^.
THE THEORY OF INDICES.
217
I
^^K Illustrative problems. 1. Eemove the parentheses from
^^pB~^ H 2/"'^)""^, expressing the result with positive expo
^Rents.
B (Xl 2/l)2 = X2 ^y2. §226
W 2. Multiply x^ + x^ + 1 by x^ — x^ + 1.
Since we can multiply as if the exponents were positive, we have the
following :
Check.
x2 + xi + 1 3
g2 _ a;i + 1 1
X4 I X8 + X2
 X3  X2  X1
X2 + Xl + 1
x4 + x2 +1 3
Detached coefficients should be used in practice.
3. Divide x'^ + 3x'^ + Sx^ + 1 by x'^ + 1.
Since we can divide as if the exponents were positive, we have the
following :
Quotient = x2 + 2xi + 1
xi + lx3 + 3x2 4 3xi + 1
X3 + X2
Check. 8  2 = 4.
2x2 + 3xi
2 x2 + 2xi
+ 1
xi + 1
Detached coefficients should be used in practice.
4. Solve the equation x ^ — 3x o + 2=:0.
x2 3x + 2 = (x 2)(x  1),
x~*  3x~* + 2 = (x*  2)(x^ 
(x^ 2)(x3 _ 1) =0.
x~5 = 2, or x~^ = 1.
xi = 23 = 8, or xi = V^ = 1.
81
or X = 1, a::d these roots check.
218
ELEMENTS OF ALGEBRA.
EXERCISES, C.
Eemove the parentheses and simplify in exs. 18.
1. [(arf.
2. (ar' + yy.
1 _1_
4. [(«'» + '*)"'".(«»')«]'«*.
_i _i
6. \l(xY^T^~% («^^^^)*.
7. (2032 2/ 2)^ V64[(x2/)6]i
Express with positive integral or fractional exponents, in
simplest form, exs. 914.
9. y/a^b^p.
11. ■V^a2'"Z»3m2_
_^ 3/ —
13. a ib'^Wcd^
10. V^v^
m ( re /
12. v«y<
14. ^a"'b^'"c"'\
Perform the multiplications indicated in exs. 1519.
15. 3a~^4:a~^'2a^; ax—^y'^bx^'y'^.
16. a^a^; 3a^^c^4.a%^c^; a;^.(aj^).
17. 5y/x^2xhf', a'b^c^d''a'b^c'd\
18. (a;2 + 2xy + y^) (x'  2 xhf" + y').
19. (a;3 + 3 x'y + 3 icy + y^) ■ (x' + 2xhy + y^).
THE THEORY OF INDICES.
Perform the divisions indicated in exs. 2030.
21. 4:abh^:2b^ck
22. (ah^ — ah^ + 4 Jb^) : ah\
23. 4a;* + 11^"' 45 by 2ici3.
24. a" — a^ + 1 by a^ — a'^ + l.
25. (4 x^yl  9 xhjl) : (2 xJy^ + 3 aj^i).
26. x~^ + 2 a;2^i — 3 y^ by cc^ — y\
27. 3 a4 : 5 a't, x^ : J [xi^/'i (ccy)^]^^^.
28. 16cc8 + 6a32 + 5a;i6 by 2icil.
219
29. V^4 xh/z^ : [(1 : ^12a;V^«') • '^lOSxh/z^y
30. ic^ — 2ic*
4ic« + 19x2_3l£ci + 15 by a;"
3 1 . rind the remainder when a; ^ — 11 cc" ^ + 1 is divided
byxi1.
32. Also when x~^ + (a — 3) x'
divided by x~'^ — 3.
(b3a)x^3bis
33. Factor 2 x^ — 9 x^ — S x'^ \ 15, negative expo
nents being allowed in the factors.
34. Also 6a;« + x25a;i2.
35. Also 6a;^ + 17ic218a:i45.
36. Also 1 — 9a7^ — 486 x^, fractional exponents being
allowed in the factors.
220 ELEMENTS OF ALGEBRA.
VI. IRRATIONAL NUMBERS. SURDS.
228. Rational and irrational algebraic expressions have
already been defined (§ 98). But in algebra it is often
necessary to use numbers which are irrational.
229. A rational number is a number expressible as the
quotient of two integers.
E.g., 3 = f , 0.666 •• = , f
230. An irrational number is a number which is not
rational.
E.g., 2' or V2, (1 + 2^)^ or V^l + V2, V 1.
231. Irrational numbers which are not even roots of
negative numbers are often called surds, but in elementary
works the term is still further limited to irrational roots of j
rational numbers, or to such roots combined with rational
numbers.
E.g., V2 and 3 + Vs are the types here treated, but not v 2 + Vs j
and V— 5.
232. Surds are classified as follows :
1. According to the root index, as
quadratic, or of the second order, as Vs,
cubic, " " third "
" ^7,
quartic, or biquadratic,
" V^,
quintic.
« ^,
sextic
" V^,
and in general as
ntic, n being a positive integer.
" v^.
THE THEORY OF INDICES. 221
2. Similar or dissimilar (if they have a single term),
according as the surd factors are or are not the same.
E.g., 2V3, 4V3,  7 V3 are similar surds.
2 V3, 3 V2 are dissimilar surds.
V2 . V3, 5 V3 are similar as to V3 but dissimilar as to V2.
3. Pure or mixed (if they have a single term), according
as they do not or do contain either real factors or dissimilar
surd factors.
E.g., V3 is a pure surd, but 2 V3 and V5 • V3 are mixed surds.
4. According to the number of terms in the expression
when simplified, as
monomial surds, as 'v2, 3V2,
binomial " " V2 + Vs, 5 + V2,
trinomial " " 2 + Vs + y/j,
and, in general, polynomial surds.
5. According to simplicity. A surd is said to be in its
simplest form when all the factors that are perfect roots
are expressed without the root sign, when the index is as
small as possible, and there are no fractions under the
radical sign.
E.g., V9, Vi, V^, Va^x, are not in the simplest form. For
V9 = 3,
Vi = Vvl = V2,
VJ = V = VfT2 = ^ V2.
Va^ = aVx.
The fractional exponent is, in general, more convenient in
all operations involving surds. The two forms of the radical
symbol are used here in order that both may be familiar.
222
ELEMENTS OF ALGEBRA.
233. Convention as to signs. When we consider an ex
pression like V4 + V9 we see that it reduces to (±2)+(±3),
and hence to
+ 2 + 3 = 5,
+ 2  3 =  1,
2 + 3 = 1,
_ 2  3 =  5.
But for simplicity it is agreed among mathematicians
that in expressions of this kind only the absolute values oj
the roots shall be considered unless the contrary is stated.
Hence, Vi + Vg = 2 + 3 = 5 , but ± Vi ± ^9 = 5,  1, 1, or
 5. (Compare § 192.)
EXERCISES. CI.
1. Classify according to the index of the root :
(a) ■^. (b) ^. (c) a*. (d) x^.
2. Classify as similar or dissimilar :
(a) 2V2, 5V2, 8 21 (b) 2v^, V^, i^.
3. Select the surds from the following :
(a) V2. (b) 4*. (c) VV2 + V3.
4. Classify as pure or mixed :
(a) V47. (b) 3v/5. (c) ab^. (d) V2V^J
5. Classify according to the number of terms :
(a) ah^. (b) V2 + ^. (c) 2 + V3 + ^J
6. rind the value of each of these expressions :
(a) V4 + V9 + VI6. (b) V^ + V25 + ^ + ^/32.
(c) v^l728  Viii + Vi69  13.
THE THEORY OF INDICES. 223
234. Reduction of surds. It has been shown (§ 217) that
a = (a")". Hence, it follows that a number can be reduced
to the form of a surd of any order.
E.g., 2 = Vs, the form of a surd of the 3d order.
Similarly, V2 can be reduced to the form of a surd of the 5th order,
for 2' = (2^)^ or V^, or ^V^.
Similarly, Vi = Vis = Vie, a surd of the 10th order.
Hence, mixed surds can ahvays be reduced to pure surds.
E.g., :■ aVb=V^,
3V5 = y/S^6 = Vl35.
235. Since it is desirable to have the number under the
radical sign as small an integer as possible, it is often
necessary to reduce surds to their simplest forms (§ 232, 5).
Vl36= V33.5 =3 Vs.
\18 \32.2 \32.22 6
Hence, in the case of fractions under the radical sign we
multiply both terms by the smallest number which will make
pAe denominator the required power, then extract the indi
cated root of the denominator, and reduce the remaining
surd as much as possible.
E.g., ^/i^^ f±:^^lV39.
^' \13 \132 13
236. Since in multiplying surds it is desirable to have
them of the same order, it is often necessary to reduce
several surds to equivalent surds of the same order, the
order always being as low 'as possible.
E.g., \^ • V3 = 2^ • 3's = 2^ • 3^ = (23 . 32)* = Vs^ = V72.
224 ELEMENTS OF ALGEBRA.
EXERCISES. CII.
1. Reduce the following numbers to the forms of surds
of the orders indicated :
(a) 5, 3d order. (b) 2, 6th order.
(c) i, 4th « (d) 10, 5th «
(e) 11, 2d " (f) 12, 3d "
(g) 2, 2d " (h) 5,3d "
(i) 3, 5th ^' (j) 2, 6th «
2. Eeduce the following to pure surds :
(a) 2V3. (b) 3V2. (c) 2V^.
(d) 52*. (e) ah^c. (f) a^2^\
(g) 3 V2. V3 Vs. (h) ah^c.
3. Eeduce the following numbers to the forms of surds
of the orders indicated :
(a) ^abc% 9th order. (b) V'a^, 14th order.
(c) a/5, 30th " (d) 3* .15th «
(e) 5*, 20th " (f) 10* 15th "
(g) v'4, 8th " (h) V'S, 60th "
4. Eeduce the following to equivalent surds of the same
order, the order being as low as possible in each case :
(a) V^, Vb. (b) V3, ^, a/2.
(c) 2^, 3^ 4^. (d) v^, V3, A^.
(e) ah^, ah^. (f) ^, V^, V^.
(g) 7^, 9^, 111 (h) 2, V2, v^, v^, ^5.
THE THEORY OF INDICES.
226
237. Addition and subtraction of surds. Irrational expres
sions may evidently be added and subtracted the same as
rational expressions, by taking advantage of some con
venient miit.
Check.
E.g., aVx+ hVx — c\^ 1
3/ n/
 c^x + cVz
gVx 4 6Vx + cVz 3
2aVx + (26 c)Vx + cVz 4
Similarly, required the sum of V24, V54. and — V96. Here we
have, each surd being reduced to its simplest form,
V24= V4T6 = 2V6
V54= V9~^ = 3V6
 V96 =  V16.6 = 4V6
Hence, the sum is v6
Similarly, required the sum of Vs, V27, — 2 V2, and Vis. Here
we have 2V2 + sVs  2V2 + 4V3 = 7 Vs.
In general, however, the sums of surds can only be indi
cated as Vs + ^,  v^ + "V^.
I
EXERCISES, cm.
Simplify the following :
1. V72 + VIO8  V32  V243.
2. ■>^+ v^375 v^elS + lOv^.
3. V^ + ^^/a^^  ^a^^ ■ V&.
4. (a^b)^  a^^b + a%^c.
5. Vl47 + V243  V363 + V432  V507.
6. v^l715 4 V^3645 + ^6655 + V^640  39 ^5.
7. Vx^ { 5x^ + 6x^  4:x  S + Vx^  4.x^ {6x^ — ^x + 1.
226 ELEMENTS OF ALGEBRA.
238. Multiplication of surds. In general, products involv
ing irrational numbers must be indicated, as 3 V2, or
expressed approximately, as 3 V2 = 3 • 1.414 • • • = 4.24 • • •.
E.g., 3 V2.2 Vs = 3 V82 Vo §236
= 6V72. § 220
This result, while it still leaves a root to be extracted
and a multiplication to be performed, is more compact than
the indicated product 3 V2 • 2 Vs.
Similarly, to square 3 V2 + 2 Vs.
(3a^ + 2V3)2 = (3V2)2 + 2(3V2)(2V3) + (2V3)2 §69, 1
= 18 + I2V72+4 V9.
It' is understood that no results are to be expressed
approximately, in decimal form, unless so stated.
EXERCISES. CIV.
Perform the following multiplications :
1. 3V.2^. 2. V2^.v^.V3.
3. (35V3)2. 4. V^.^.^^.4^.
5. V7 . ^7 • ^. 6. {^/ab + Va + bf.
7. 2V2.3V3.5V6. 8. VT2I. V^.J/14641.
9. (2 + 8 V3)(45V3).
10. 3 V2.2^3.4V^.5V^.
11. V^/a + V^ . V^Va  V^.
12. (V2 + V3)(2 V25 V3).
13. 5 ■y/{a + 2 ^»)2 . 3 V(a + 2 bf.
THE THEORY OF INDICES. 227
239. Division of surds. To divide an irrational number
hy a rational number is equivalent to multiplying by the
reciprocal of the rational number, and hence it may be con
sidered as a case of multiplication.
E.g., IS merely  • (a + V6), or  +  V6.
240. Division by a surd usually reduces, without much
difficulty, to division by a rational number, as shown in the
following example :
To divide V2 + Vs by Vs, we have :
Vi + Vs _ V5(V2 + V3)
assuming that we can multiply both terms of the fraction by VE with
out changing the value, as we can in the case of rational multipliers
(§ 150). This equals
, or i(^10 + ^1^)
6
241. In the preceding example we have reduced the
•fraction to an equivalent fraction with a rational denomi
[nator. The process of rendering a quantity rational is
called rationalization.
The advantage of rationalizing the denominator is seen by consid
^ering the computation necessary to find the approximate value of
V2 + Vs
• Here there are three square roots to be extracted, fol
V5
lowed by one addition and by one division with a long divisor.
But in the case of ^ ( VlO + Vl5) there are only two square roots
to be extracted, followed by one addition and by one division with a
! short divisor.
242. The factor by which an expression is multiplied to
produce a rational expression is called a rationalizing factor.
E.g., Vs can be rationalized by multiplying it by V2.
228 ELEMENTS OF ALGEBRA.
243. Since the prohleyn of division by surds reduces to
that of the rationalization of the divisor, exercises in
rationalization will 'first be considered
Illustrative problems. 1. By what expression may ab^
be multiplied in order that the product shall be rational.
1. •.• X'* • X « = X,
2. .. a%^ ■ a}%^^ = ah.
3. .. a^~V~^ or a^h^^ is a rationalizing factor. There are evi
dently any number of rationalizing factors, since we may multiply this
one by any rational expression. This is, however, the simplest one.
2. By what expression may V a"* • V5^ be multiplied in
order that the product shall be rational ?
1. V#. W = o^lfi = ah^h.
2. Evidently a^h^h • a}'~%^~^ will equal a^^, a rational expression.
3. .. a^h^ is a rationalizing factor.
3. By what expression may a f V^ be multiplied in
order that the product shall be rational ?
1. ••• (X  2/) (X + y) = x2  y'^, § 69
2. .'. (a — \h) {a + Vo) = a^ — b, sl rational expression.
3. .. a — Vft is a rationalizing factor.
244. And, in general, the conjugate of a binomial quad
ratic surd (§ 69, 3) is a rationalizing factor of that surd.
4. Find a rationalizing factor for 'y/a ± VZ ± Vc.
\. :• {x + y + z) { X + y + z) {X  ij \ z) {x ^ y  z)
= 2 X^?/2 + 2 2/222 + 2 22x2 _ X*  2/4  2*,
2. .. any trinomial quadratic surd of the form v a ± \b ± vc
can be rationalized by multiplying it by the product of the other
three trinomials. E.g., the rationalizing factor for V2 — VS + Vs is
(^^ + Vs + V5) ( V2 f V3 + V5) ( V2 + V3  Vs).
THE THEORY OF INDICES.
229
EXERCISES. CV.
Find the simplest rationalizing factor for each of the
following expressions :
1. ahh^(h.
3. (^h^c^.
5. 2 + Vs.
7. 3V2.
9. V5I.
11. aJn'^c » .
13. Vt + Vs.
2. V7V5.
4. Va — V^.
6. Va + ^» + c.
8. V5V2V3.
10. V5 + V7 + Vil.
12. V2 + V7Vli.
14. Vi* + 6 + Va — 6.
Illustrative problems in division. 1. Divide Vl2 by V3.
§220
2. Divide V5 by ^2.
V5_ V53
■v/2 V22
=Ai
§220
6 h" ■ 53
\2^
= i V2000.
3. Divide V2 + V3 by V2  VS. That is, rationalize
V2 + V3
ithe denominator of the fraction — 7= 7=
V2 V3
1. The rationalizing factor for the denominator is evidently
V2 + Vs.
2_ (V2 + V3)(V2 + V 3)^ 2 + 2V6 + 3 ^ ^^ ^ ^ ^/^^^
(V2 + V3)(\^  V3)
23
230 ELEMENTS OF ALGEBRA.
EXERCISES. CVI.
Perform the divisions indicated in exs. 116.
1. 6:4v'24. 2. 24 : (2 Vt  6).
3. ^a%^'.2^^h. 4. 15V24:3V^.
5. 58:(8+V35). 6. 12 ^192 : 4 ■V^729.
7. 16V^^^':8V'^«^. 8. 90 : (5 V3  VSO).
9. 10Vl2:2 Vl8:4V8.
10. ^(a^ 2 6)^: V(a«  2 h).
11. (VT2 Vl8+ V6): V2.
12. (3V58 V2):(3 V34V5).
13. (18  16 V5) : (4  VS  2 V3).
14. (7 Vi2  4 V27) : (8 V3 + 2 V2).
15. (15 V8 + 10 Vt  8 V2 f 5) :  4 Vs.
16. (3 V3  2 V2) : (5 VS  3 V2  2 VS).
Rationalize the denominators of the fractions in exs.
1723.
17. i4±4 18. ^'
11^  5^ 23^ + 5^
7 + 3V7 3_5i_2^
19. p= 20. •
126 Vll 3 + 5i + 2^
21.
2 2m
22.
(^2 4 If + («;2 _ ^^t (^^ _^ ^)* + (a  ^) 
1
23.
a;(l a2)i_2/(l 4_a2^i
THE THEORY OF INDICES.
231
245. Roots of surds. The roots of perfect powers of
surd expressions can often be found by inspection or
extracted in the ordinary way.
1. To find the square root of a + 4 \fab + 4 J.
1. •••
2. ..
Check.
V/2 ^ 2fn + n^ = ± (/ + n) ,
V9 = ±3.
2. To find the fifth root of the perfect fifth power
a2  5 a%^ + 10 a/'h^  10 ab + ^ ahl^  h^.
This is readily seen to be a* — 6^. § 82
To check, let a = 6 = 1. Then 0^ = 0. If, however, we wish to
jheck the exponents, let a equal any square and 6 equal any cube.
?.gr., leta = 9, 6 = 8. Then
(3  2)5 z= 243  810 + 1080  720 + 240  32.
3. To find the square root of 7 + 4 Vs.
1. If this can be brought into the form/2 _j_ 2/n  71^, the root will
in the form ± (/ + w). § 69
2. We first make the coefficient of the second term 2, because of
le 2/n, and have 7 + 2 Vl2.
3. And ',• 12 is the product of 3 and 4, and 7 is the sum of 3 and
I, we have
V7 + 4V3 = V4 + 2 V3T4 + 3 =: ± ( Vi + V3) = ± (2 + V3).
Check. Square 2 + Vs.
4. To find the square root of 8 — 2 Vl5.
1. As in ex. 3 we attempt to bring tjhis into the form/2 ^ 2/n + v?.
2. •.• 15 is the product of 5 and 3, and 8 is their sum, we have
V82 Vl5 = Vs  2 Vl5 + 3 = ± ( V5  V3).
Of these results, only the positive one is usually considered in
)ractice.
Check. Square Vs — V3.
232
ELEMENTS OF ALGEBRA.
EXERCISES. CVII.
1. Extract the square roots of
(a) a2yf2ab^2h. (b) a2a'^oJ'.
(c) 3 a  8 VS^ + 16. (d) a*  2 a^&3 4. ^,1 .
(e) 26t V200a + 25. (f) x^^^x'^Hj
2. Extract the cube roots of
(a) 8  12 Va + 6 tt  a V^.
(b) a3V^^2^3Z>V^Z»l
(c) a;^  3ic2 v^ + 3£c Vy  V^.
(d) cc^ Vx — 3 x^ V?/ + 3 a; V^^y — y.
3. Extract the fifth roots of
(a) 1  52/^ + 10 y"  l^if + St/^  y\
(b) 3280 ^ + 80v^2/^407/v^^ + 10y/V^2/2
4. Extract the square roots of
(a) 82 Vt. (b) I + V2.
(c) 8 + VeO. (d) 9  4 V2.
(e) 10  V96. (f ) I + ^ V6.
(g) IOV7 + 32. (h) II2 + 40V3.
5. Extract the square roots of
(a) 2x42V^
(b) 2 a; + 2 Va;2 _ 1.
(c) ah — 2a ^ ah — a'^.
(d) x^ { X { y \ 2 X Vic + y.
^i ff
THE THEORY OF INDICES.
233
VII. THE BINOMIAL THEOREM.
246. It has been shown (§ 80, the proof being given in
Appendix I) that if n is a positive integer
(a + by = a" + na'^^ + ^—^ «"~
2^2
^ .(.l)(»2) ^,_3^3^
It was proved by Sir Isaac Newton that this is true even
if w. is negative or fractional. The proof is, however, too
: difficult for the student at this time.
Assuming that the binomial theorem is true whether ?i
positive or negative, integral or fractional, it offers a
raluable exercise in the use of negative and fractional
jxponents.
E.g., •.• (a + 6)«
,, n(n — l) „,, n(n — l)(n — 2)
2 2 • o
•. y/a+b={a\h)^
2 Z • o
= a^+iah ia^62
+tV«^6«
.. V5=(4 + l)^
=4^ + 1.4^1.4*
+ T^64^
= 2 +i J^
+ 5l^
_/i 1 ^\—a.
(l+X)3(''^^
= 1 +(3)x+^^i^
V,3(3l)(32)^, 1
= 1 3x +6x2
10x3 +
234 ELEMENTS OF ALGEBRA.
EXERCISES. CVIII.
1. Expand to four terms (1 + x)~^.
2. Also l/Vl£c.
3. Also Vil = V162 = 4 (1  1)^.
4. Find the 5tli term in the expansion of (1 — x)~^.
5. Also in the expansion of (1 + xy.
6. Also in the expansion of (1 — x)i.
7. Find VTo by expanding (9 + 1)^ to four terms,
reducing these to decimal fractions and adding.
8. Similarly for V82 = (81 + 1)^.
9. Similarly for ^ = (27 + 1)^
10. Similarly for V37 = (36 + 1)K
REVIEW EXERCISES. CIX.
1. Divide x^ — 4:X^a^ + Qx^a^ — 4:X^a^ + a'^ by x^ —
2 x^a^ + ak
2. Simplify 3 (a^ + x^y  4 (a"^ + x^) (a*  x^) + (a^ 
2x^y.
3. Simplify 3
4. By inspection find the square root of
(a) 4 a^ + 4 4 a^.
(b) at2af + 5al4ai + 4.
(c) cc + ?/ + ^ rf 2 x^y^ — 2 a;'^^^ — 2 ?/^^^.
(d) a^ + 4 aV + 10 aifi + 12 ahy + 9 ?/t.
THE THEORY OF INDICES.
5. Simplify (3* + 3^ + 3^ + 1) (3^  1).
235
6. Factor 36 x^ — 65 cc^ — 36, fractional exponents being
lUowed in the factors.
7. Also 4 a;^ — 4 x^y^ + 9 y^
8. Solve the equation x'^ \ ^ x'^ \ 2 = 0.
9. Also 4 x^  15 a;* + 14 = 0.
10. Also £c^ — 5 cc^ + 6 = 0.
11. Extract the square root of
12. Also of 25ic*30cc^3/ + 49icy24a;y + 162/*.
13. Extract the cube root of
aj6 _ 9 ^5 _^ 33 ^4 _ g3 ^3 ^ gg ^2 _ 3g ^i _^ 3
14. Also of
8ic2 4. 48a;t + 60cc"3  80a;  90 cc"^ + 108 x^  27.
15. Also of
8 a* + 48 a^b + 60 a^b^  80 a^Z^^ _ 90 a^b"^ + 108 ai^*^  27 ^»«.
16. Also of
■^^xi + ^\ici + a7~t + 7 a;T% + 3 a;i + xtV + 1.
17. If a^ = b% show that (t V = «^"^
\^ / V3
V3V2y VV3 + V2y
19. Simplify
, V3 + V2\2 , / V3  V2V
18. Simplify ( ,^ — 7= ) + ( := 7= ] •
CHAPTER XIII.
COMPLEX NUMBERS.
I. DEFINITIONS.
247. Certain steps in the growth of the number system
have already been set forth in § 24, but are here repeated
for reasons which will be obvious.
1. The positive integer suffices for the solution of the
equation a; — 3 = 0, since x = 3 satisfies the .
12 3
equation. We can represent such a number
by a line three units long, as in the annexed figure, the unit
being of any convenient length.
2. The positive fraction. If, however, we attempt to
solve the equation 3 x — 2 = 0, either we must say that
the solution is impossible or we must extend the idea of
number to include the positive fraction. Then ic = f sat
isfies the equation. We can represent such a number by
dividing a line one unit long into three parts and taking
two of them.
3. The surd. If we attempt to solve the equation
ic^ — 2 = 0, either we must say that the solu
tion is impossible or we must extend the idea
of number to include the surd. Then V2
satisfies the equation. We can represent
V2 by the diagonal of a square whose side
is one unit long. This is evident because the square on
the hypotenuse equals the sum of the squares on the two
sides of the rightangled triangle.
236
COMPLEX NUMBERS. 237
4. The negative number. If we attempt to solve the
equation a; + 2 = 0, either we must say that the solution
is impossible or we must extend the idea of niunber to
; include the negative number. Then x = — 2 satisfies the
equation. We can represent such a number by supposing
^tlie negative sign to denote direction, a direction opposite
that which we assume for positive numbers.
248. The numbers thus far described in this chapter are
jailed real numbers.
249. The imaginary number. If we attempt to solve the
jquation x''^ { 1 = 0, either we must say that the solution
is impossible or we must extend the idea of number still
further.
The equation
a;2 + 1 =
eads to
x^^ = l,
jvhich leads to
x = ±V
1,
'which cannot be a positive or a negative integer, fraction,
or surd (§ 126).
250. We call an even root of a negative number an
imaginary number.
The term " imaginary " is unfortunate, since these num
bers are no more imaginary than are fractions or negative
lumbers. We cannot imagine looking out of a window
[— 2 times or ^ of a time any more than V— 1 times. The
:" imaginary " is merely another step in the number system.
?he name is, however, so generally used that it should
jontinue to designate this new form of number.
To the ancients, negative numbers were as " imaginary " as V— 1
to us. It was only when some one drew a picture of V2 (see § 247, 3),
[of — 1, and later of V^l, that these were uuderstood.
238
ELEMENTS OF ALGEBRA.
251. As with fractions, surds, and negative numbers, it
is necessary to represent the imaginary graphically by a
line, or in some other
concrete way, in order
to make its nature clear
to the beginner. *
In this figure the
multiplication of + 1
by — 1 swings the line
OA^ through 180° to
the position OA^.
As a matter of custom
this line is supposed to
swing as indicated by the
arrows, opposite to the movement of clockhands, counterclockwise.
'
'
B.
■iaYn
Ba
A.
/
A^3
+iV=:i
V.
B,
—2
^;
f 2
B.
.2TP,
y
r
252. That is, since ( V— 1)^ means V— 1 • V— 1 or — 1,
the multiplication of + 1 by V— 1 • V— 1 swings f 1
th roug h 180° ; therefore the multiplication of + 1 by
V— 1 should be regarded as swinging it through half of
this angle, or 90°, to the position 0^2
Or we may say that since multiplication by V— 1 twice,
carries OA through 180°, therefore multiplication by V— 1
once should carry it through 90°.
Similarly, — 1 multiplied by V— 1 • V— 1, or — 1 mul
tiplied by — 1, swings OA^ the rest of the way around to
0^1 ; hence, — 1 multiplied by V— 1 should be looked
upon as swinging it to the position OA^.
253. Hence, we represent + 1 V— 1 (or f V^), +2 V^,
+3 V — 1, . • ., by integers on the perpendicular OY, upward
from 0, and iV^ {or V^), 2Vi, 3V^, •••,
hy integers on the negative side of this line, i.e., on OF',
downward from 0.
COMPLEX NUMBERS. 239
2 54. H ence, it appears that the symbols + V— 1 and
— V— 1 wre, like f and —, symbols of quality and may be
I looked upon as indicating direction.
E.g.^ +3 indicates 3 units to the right,
3 " " " left,
+ 3V^ " " up,
V — 3 V — 1 " " down.
li
255. Since Va6 = Va \/b, we say that V— 3 shall
[ual V31 = V3 . V 1. Hence,
Every imaginary number can be written in the form
a V  1, where a is real, though possibly a surd or a frac
tion, and v — 1 is the imaginary unit.
E.g., to represent 3 V— 1, we measure 3 un its upward from the
point on the li ne X 'X ; to represent — V— 2, we reduce this to the
form — V2 • V— 1, then construct a line equal to Vi, as in § 247, 3,
and lay this off on 0Y\
EXERCISES, ex.
Solve the following equations, expressing the results in
the form a V— 1.
1. x2 = 9. 2. 3cc2_^2 = 0.
3. 5a;2 = _5. 4. a;2V2 = 3.
5. ^2_^5 = 0. 6. 5x^ = 125.
7. a;2 + 4 = 0. 8. cc2 + 20 = 5.
Kepresent graphically the following imaginary numbers :
9. V^. 10. V^. 11. 5V^.
12. V32. 13. 3V^. 14. V2.V^.
15.  V 16. 16. 2V9. 17. ^V12.
or
{x2f
or {dc,
2 + V
^(^2
whence
or
240 ELEMENTS OF ALGEBRA.
256. The complex number. If we attempt to solve the
equation ic^ — 4cc + 5 = 0by factoring, we may write it in
theform ^^ _ 4^ + 4  ( 1) = 0,
(!)== 0,
V1)=0,
x = 2 V^Ti^
x = 2+ V^.
Hence, it appears that each root is the algebraic sum of
a real number and an imaginary.
Such a number is said to be complex.
257. As with positive and negative integers, fractions,
surds, and imaginaries, we proceed to make the nature of
the complex number more clear by resorting to a graphic
representation.
If we wish to represent the sum of 2 and — 3, we pass
" ';:::::"_::__;, from zero 2 units to the right
10 12 and then 3 units to the left,
and we say that the sum is the distance from to the point
where we stop.
The fact that the absolute value of the sum is less than the sum of
the absolute values of the addends is no longer strange to us, because
we have become accustomed to this in dealing with negative numbers.
258. Similarly, to represent the sum of 3 and 2 V— 1 we
pass from zero 3 units to the
right and then 2 units upward
(for 2 V— 1) and we say, as be
fore, that the sum is the distance
from to the point where we stop.
The fact that the absolute value of the sum is less than the sum of
the absolute values of the addends is no more strange than it is in the
case of 2 4 (— 3).
COMPLEX NUMBERS.
241
EXERCISES. CXI.
Represent graphically the following complex numbers :
1.4+ V^^. 2. 5  2 V^.
3. 5 + 2V^. 4. ^_V:r^.
5. 52 V^. .6.  3  3 V^^.
7. i + iV3Vl.
8.
iiV3.Vl.
259. Symbolism of complex numbers. Instead of writing
the symbol V— 1, the letter i is usually employed.
This letter, standing for imaginary, seems to have been first used in
this sense by Euler in 1777. ,
Then V^ = 2 V^ = 2 i,
V^ = i V3, etc.
Llso,
i' = l,
P =.!.{ =  i,
i'=(:ir={ir = i,
i^ =l.iz= i,
i^ = i.i = i^ = l,
{^ = — 1 . i = — i,
i^ = i.i = ({y = (l)=l;
, in general,
^^« = 1,
i' + ^ = i,
{' + ' = 1,
•4n + 3 ^ _ .•
I'" ' " =
EXERCISES. CXII.
Represent graphically the following complex numbers :
1. 2 + 3i. 2. 4 + 2i. 3. i^ + ^i.
4. i^ + i\ 5. i* + t^ 6. i^h2i\
242
ELEMENTS OF ALGEBRA.
II. OPERATIONS WITH COMPLEX NUMBERS.
260. Complex numbers are subject to all of the laws of
rational numbers and the operations do not materially
differ from those already
familiar to the student.
Illustrative problems. 1.
Represent graphically the
sum of 2 + 3 ^ and — 3 — t.
Starting from we lay off
+ 2 (to the right), then Si
(upward), OA being 2 + 3 i.
From A we then lay off — 3
(to the left), then — i (one
unit downward), reaching B.
Then the sum is OB, the distance from to the point where we stop.
2. Add 1, — Y + i * V3, and — ^ — ^i Vs ; then repre
sent the sum graphically.
1
 i ji Vs
Sum =
Graphically, we lay off 1 from
to A. From A we lay off — ^,
then, iiVi" (i.e., iM.73.,
or 0.87 i), reaching B. From B we lay off — ^, then — i i Vs, reach
ing O. Hence, the sum is zero.
Y
B
Av
A
3.
Multiply 2 + 3 ^
2 + 3i
3 2t
: by 3 
21.
6 + 9i
= 6 + 9i
=
6 + 9i
 4 i  6 i2
= 4ti
6(
1) =
64i
12 + 5i
Simply multiply by i as if it were any other letter, but in finally
simplifying remember that i^ = — 1.
COMPLEX NUMBERS. 243
4. Divide 12 + 5i by 32i.
Multiply both terms of the fraction
12 + 5i
32i
by the conjugate of the denominator. Then
(3 + 2 i) (12 + 50 _ 26 + 39 i _ 26 + 39 i _
(3 + 2i)(32i) ~94(l) ~ 13 ~ "^ *'
5. Cube — 1 + i * Vs.
••• (/ + n)8 =P + SPn + 3M + n\
.. (i + iiV3)3
= i + 3.iiiV3+3.(i).f.(l) + f(l).iiV3
= i + iV3 + ffiV3
= 1.
Hence, — i + i i V3 is a cube root of 1.
6. Extract the square root of — 16 + 30 *.
•.• a + 2 V^ + 6 = [±(Va + V6)]2, §245
id •.•  16 + 30 i can be written 9 + 2 V  9 • 25 + (  25),
I:
..  16 + 30i = 9 + 2 V 9 • 25 + ( 25)
= [±(3 + V325)]2
= [±(3 + 5i)P.
."• ± (3 + 5 i) is the required square root.
The solution is seen to consist simply of making the coefficient of
the square root 2, and then separating — 16 into two parts whose prod
uct is  225. (See § 245, 3.)
The addition (including subtraction) of complex numbers
has been represented graphically. It is also possible to
represent the other operations graphically, but the expla
I nation is too difficult for an elementary textbook.
I
7. Extract the square root of a^ + 2 abi — Z»^.
This is evidently the same as a^ + 2 ahi + (bi)^.
Hence, the square root is ± (a + hi).
244 ELEMENTS OE ALGEBRA.
EXERCISES. CXIII.
1. Find the following sums and represent each solution
graphically.
(a) 57 i and 5 + 7 i. (b)  2  3 ^ and 2 + 3 i.
(c) 1,  1, i, and  i. (d)  6 + 2i and 6 + 2 i
(e) 1,^ + ii^, i + iiV3, 1, i'iV3, and
ii^V3.
2. Multiply
(a) 3  4^ by 5 + 21. (b)  ^ + ^ ^ by i + ^ /.
(c) 2 + 9t by 9 + 2/. (d) 4 + 2^ by 42/.
(e) ^ + i/V3 by  i^V3.
3. Divide
(a) 10 by 3  i. (b) 4 + 22 i by 7 + i.
(c) 1 + 8 i by 2 + i. (d) 1 + 8 i by 2 + 3^.
(e) 7 + 61 i by 4 + 7 i. (f ) 3 + 6 i by 36 i.
4. Eaise the following to the powers indicated :
(a) p\ (b) (2 + 3 ly. (c) ( i  i ^^sy.
(d) (2 + .•)^. (e) (3  5 .)^ (f ) ( i + i V:=l)3.
(g) (« + My. (h) (2  7 i)^ (i) ( i  i V33)3.
5. Extract the square root of
(a) 3 + 4^'. (b) 5 + 121. (c) 5121
(d) _4528^. (e) 24  10 i. (f) 15  8 i.
(g) \^ + 21 i. (h)  i  ^^. (i)  f + i.
COMPLEX NUMBERS. 245
REVIEW EXERCISES. CXIV.
1. Simplify the expression
 i/[10 + 2 V5 4 ( VS + 1) i].
2. Also the expression (Vs { iy / i (— 1 \ V— 3)^
3. Also the expression
1 + V^ Y A + A ' / I v^ Y
2 A V2; V 2 ;•
4. By factoring, solve the equation 8a?^ — 35a^ + 12 = 0.
5. By the Remainder Theorem determine whether x — i
is a factor of x^ \ 5x^ \ 4.
6. Find the times between 4 and 5 o'clock at which the
hands of a watch are at right angles.
7. By factoring, find four different roots of the equation
1 = 0. (Two are imaginary.) Check.
I
I^P 9. Find to two decimal places the values of x and y in
the following :
" =7.935.
8. By substituting the three numbers
1, i + i^V3, ^^iVs,
)r X, show that they are the roots of the equation ic^ — 1 = 0.
^ , y _
3.579 ' 5.793
^ + I— = 5.397.
1
9.753 7.539
10. The sum of two numbers is 16, and the sum of their
reciprocals is double the difference of their reciprocals,
hat are the numbers ?
CHAPTER XIV.
QUADRATIC EQUATIONS INVOLVING ONE
UNKNOWN QUANTITY.
I. METHODS OF SOLVING.
261. A quadratic equation (or equation of the second
degree) involving one unknown quantity is an equation
which can be reduced to the form ax^ ^hx \ c = ^, a, b, c
being known quantities and a not being zero.
E.g., 3ic2 + 2ic + 3 = 0,
a;2 + 1 = 0,
■ia^2_^ic V2 = 0,
are quadratic equations involving one unknown quantity.
So is the equation
2x^ + Zx^5x + l ={2x^ + l){x 1),
because it can be reduced to the form ax^ \ bx + c = 0.
Similarly for
although, in general, multiplication by any f{x) is liable to
introduce an extraneous root (§ 185).
But • a;2 + 4 (c  5 =
is not a quadratic equation ; neither is
2ic8 + ic + l =x^ + x'' + ^x,
nor x'' + x + l={x + l){x 1).
246
QUADBATIC EQUATIONS.
247
The equation x^ \ x^ \ 4 =
is not a quadratic equation in x, but it is one in x^, for it
the same as
So
(xy + (x^) + 4 = 0.
++2=0
X' x
I, without reduction, a quadratic equation in > or x~^, and
(^a + xy \2(a + x^ + 3 =
a quadratic equation in a. + x% and
x^ \x + 3 Vic2__^ ^ 4
a quadratic equation in Vic^ + cc.
262. The quadratic equation ax"^ ^ bx ^ c = is said to
be complete when neither b nor c is zero ; otherwise to be
incomplete.
The coefficient a cannot be zero, because the equation is to be a quad
ratic (§ 261).
jB.flr., x2 + 2 X — 3 = is a complete quadratic equation,
It x2 _ 3 ^
x2 + 2 X = are incomplete.
Older English works speak of an equation of the form
ax^ + c = as a pure quadratic,
id ax^ + So:; + c = as an affected quadratic.
The following are further examples of complete (aifected)
[uadratic equations :
(x  1)^ + (^ _ l)i + 5 = 0, in (x  1)*;
11 ir
— + —= + 7 = 0, in V^;
£c* + 2 ic^o + 1 = 0, in x^^.
248 ELEMENTS OF ALGEBRA.
263. Solution by factoring, (a) The type
(ax + b) (ex + d) = 0.
One of the best methods of solving the ordinary quadratic
equation is by factoring, as already shown in § 123.
Illustrative problems. 1. Solve the equation
a^2 + 16 X + 63 = 0.
1. This reduces to (x + 9) (x + 7) = 0. § 119
2. This is satisfied if either factor is zero, the other remaining finite
(§ 123). Hence, either
X + 9 = 0, or X + 7 = 0.
3. .. X = — 9, or X = — 7.
Check. Substituting these values in the original equation (§ 189),
81  144 + 63 = 0,
49  112 + 63 = 0.
2. Solve the equation 2x^ = 1.
1. This reduces to x^ = . Ax. 6
2. .. X = ± V = ± i Vii. Ax. 9, § 235
That is, it is not worth while to factor as in ex. 1. But the problem
can be so solved ; for
x2  I = 0.
••• («  ^) (« + ^) = 0
, ... X =: ± V = ± i Vli.
Check. Substituting in the original equation,
2.^.14 = 7.
3. Solve the equation Qx^ — lx + 2 = 0.
1. This reduces to (2 x  1) (3 x  2) = 0. § 120
2... 2xl =0, or 3x2 = 0. §123
3. .. 2x= 1, or 3x = 2,
and X = i, or x = f .
Check. I _ I + 2 = 0, f  Y + 2 = 0.
QUADRATIC EQUATIONS.
249
EXERCISES. CXV.
Solve the equations :
1. x^ = X.
2. X
7 6cc.
3. ^i = 6.
X^ X
5. 9a;2_l = 0.
7. ic2_j_i7^^0.
9. x''2xlb = ().
11. cc2 + 5x14 = 0.
13. x^ + 19x + lS = 0.
15. a;212a;85 = 0.
17. x''22x\121 =0.
19. ic2 _ 24 x + 143=^0.
1 1
6. x^ = 2{12~^x).
8. 8ica;2_12 = 0.
10. X (10 + £c) =  21.
12. 6a;247ic + 2 = 0.
14. ic2 + 26 X :  120.
16. £c (4 — ic) + 77 == 0.
18. 3cc210x + 3 = 0.
20. 10:z;2 + 29cc = 10.
264. (b) The type (x + a) (x  a) = 0.
It frequently happens that it is easier to arrange the first
lember as the difference of two squares than to factor in
the form suggested on p. 248, especially when the numbers
re such that the linear factors involve surds.
E.g. , to solve the equation x^ + 4 x + 1 = 0. Here x^ + 4 x are the
rst two terms of a square, x^ + 4 x + 4. The equation may be written
x2 + 4x443=0,
(X + 2)2 3 = 0,
(X + 2 + V3) (X + 2  V3) = 0,
ice we are not confined to the domain of rationality (§ 107) in our
)lutions.
.. X + 2 + V3 = 0, or X + 2  V3 = 0,
id X =  2  V3, or X =  2 + V3.
Check. 4±4V3 + 38t4V3 + 1=0.
250
ELEMENTS OF ALGEBRA.
265. The addition of an absolute term to two terms so
that the trinomial shall be a square is called completing the
square.
E.g.^ to complete the square of x^ 4. 2 x we must add 1 ; to complete
the square oix"^ {■ x we must add ^.
266. Since {x { ay = x'^ \ 2 ax { a'^, it is seen that the
quantity which must be added to x^ + 2 ax to complete the
square is the square of half the coeffi
cient of X.
E.g.^ to complete the square for cc2_j_8x,
add 16, £c2 + 8 X + 16 being (x + 4)2. To com
plete the square f or x + 6 Vx with respect to
Vx, add 9, X + 6 Vx + 9 being ( Vx + 3)2.
From the annexed figure it is readily seen
that if we have x2 + cix + ax, or x2 + 2 ax, the
square on x + a will be completed by adding a2 in the corner.
a ax
a' i
X x==
X
ax
a
EXERCISES. CXVI.
Complete the squares in exs. 116.
1 2
1 i + 
£C^ X
3. cc — V^.
5. x'^ + ^x.
7. x'^ — lx.
9. 4.x'' {^x.
11. x^lOOx.
13. 9ic2 4.36x.
15. 100x2 + 20cc.
2. ^+2^.
a^ a
4. x'^Qx.
6. x^ + 30x.
8. x''Mx.
10. cc2 + 10 X.
12. x'' — 2x', x'^ + ^x.
14. (xiy + 4.{xl),
16. {x + ay + 2(x + a).
17. In general, to complete the square for x'^ +px what
must be added ?
QUADRATIC EQUATIONS.
251
Illustrative problems. 1. Solve the equation
1. Completing the square for x^ + 3 x, the equation may be written
x2 + 3x + 1^ = 0.
2. .. (x + 1)2^ = 0.
3. .. (aj + +i)(x + fi) = 0,
(X + 2) (X + 1) = 0.
4. .. X = — 2, or — 1.
Check. 46 + 2 = 0, 13 + 2=0.
2. Solve the equation x — Va; + 1 = 0.
1. x=V^
+ 1.
Ax. 3
2. .. x2 = X + 1, or x2  X  1 = 0.
Axs. 8, 3
3. .. x2x + ^f = 0.
4. ...(a;i+iVg)(xiiV5) = 0.
5. .. » = i ±
iVs.
Check. i±V5V3±iV5
= i±^V5iV6±2V5
'=^±V5iVl±2V5 + 5
§245
= ±iV5(l± V5) = 0.
EXERCISES. CXVII.
Solve the equations :
X 25
3.
3. a;22x = 2.
5. cc2 _ 9 ^ _ 1 ^ Q^
T,. x''7x + 5 = 0.
(9
2. 1/
4. x'' + 6x + 2 = 0.
6. x26ic + 2 = 0.
8. a^2_^io^__5^o.
9. a;2_pi0ic + 25 = 0.
252 ELEMENTS OF ALGEBRA.
267. Solution by making the first member a square. The
method of § 264 may be modified by making the first
member the square of a binomial of the form x + a.
E.g., to solve the equation cc^ + 4 cc + 1 = 0.
The first member would be a square if the 1 were 4, i.e.,
if 3 were added. Hence, adding 3 to both members,
1. ic2 + 4 ic + 4 = 3. Ax. 2
2. .. (x + 2y = 3,
3. .. 03 + 2 = ±V3, Ax. 9
4. .. x = 2±sfZ.
Check. ( 2 ± Vsf + 4 ( 2 ± Vs) + 1
= 4T4V3 + 38i4V3 + l=0.
268. It therefore appears that the equation x^ + px + q =
can be solved by
1. Subtracting q from each member ; then
2. Completing the square, by adding the square of half
the coefficient of x (§ 266) to each member ; and then
3. Extracting the square root of each member and solv
ing the simple equations which are thus obtained.
The ± sign in step 3 of the above solution is placed only
in the second member, because no new values of x would
result if it were placed in both members.
Suppose it were placed in both members. Then
± (x I 2) = ± Vs ; that is
(1) + (x + 2) = } V3, whence x =  2 + V3,
(2) +(x + 2) = ~V3, " x = 2V3,
(3)  (X + 2) = + V3, "  X = 2 + v'S and .. X =  2  V3,
(4) (x + 2)=V3, "x= 2V3 " .•.x = 2 + V3.
That is, X = — 2 ± V3, as in step 4 of the solution.
QUADKATIC EQUATIONS.
263
Illustrative problems. 1. Solve the equation
x^ { X + 1 = 0.
1. x^ + x= 1.
2. X2 + X + i =  1 + i =  f .
3. x + i=±iiV3.
4. .. x= iiiiVa.
Check. ( i T ii V3) + ( i ± ii V3) =  1.
Ax. 3
Ax. 3
Ax. 9
Ax. 3
2. Solve the equation x^ + 3x+ Vx^ + 3a; + 723 = 0.
1. This may be written in quadratic form, thus,
x2 + 3 X + 7 + Vx2 4 3 X + 7  30 = 0,
quadratic in Vx^ + 3 x + 7. This quantity may now be represented
jy y, for simpUcity, and
2. 2/2 + 2/  30 == 0.
3. .. y''hy^l = H^
4. .. y + h=± ¥•
6. .. y=l± u = 5, or  6.
6. .. Vx2 + 3 X + 7 = 5, or  6.
This evidently gives rise to two quadratic equations in x. First
consider the case ot y = 6.
7. Then x2 + 3x + 7 = 25.
8. .. x2 + 3x18 = 0.
9. .. (X + 6) (X  3) = 0, and X =  6, or S,
results which easily check. ^
If y = — 6, we have
10. x2 + 3 X + 7 = 36,
11. whence x^ + 3 x + f = ^f ^.
12. .. x+ f = ±f V5, andx=  I ±1 Vs.
This pair of results checks, provided we remember that
Vx2 + 3 X + 7 = 5 or  G.
For, substituting 5 and  6 for Vx^ + 3x + 7, we have
18 + 523 = 0,
29  6  23 = 0.
254 ELEMENTS OF ALGEBRA.
3. Solve the equation 2 a;'^ — 2 cc = 5.
1. x^x = ^. Ax. 7
2. a;2  X + I = V Ax. 2
3. X  J = ± i VlT. Ax. 9
4. x = i(l'±VlT). Ax. 2
Check. (6 ± VlT)  (1 ± vTT) = 5.
It is often possible, in cases of this kind, to avoid fractions by the
exercise of a little forethought. This equation may be written
r. 4x24x = 10.
2'. .. (2 x)2  2 (2 x) + 1 = 11, a quadratic in 2 x.
3'. .. . 2 X  1 = ± Vll.
4'. .. 2 X = 1 ± VlT.
6'. .. x = i(l ± Vll).
EXERCISES. CXVIII.
Solve the equations :
1. x^^x = l. 2. 6X + 4.0 — x^ = 0.
3. a;2 + 8a; = 65. 4. x^ + ^x %» = 0.
5. x^ + 0.9 x = 8.5. 6. 2.5£c24fa; = 304.
7. 3ix^4.x = 96. 8. ic2_,_i32a; = _l33l.
9. x'' + 6x + 25 = 0. 10. 7ic25x 150 = 0.
11. 4x2_5^_pg2 = 0. 12. 4..05x^7.2x = U76.
13. (a; + a)2f2(£c + a)+l = 0.
14. ( a; + y_3 ^ +  1 + 2 = 0.
(. + ^y3(x + l) + 2
15. (ic2 + 2a;)2_3(ic2 + 2ic) + 2 = 0.
16. (x^^xiy + 4:(x^ + xl) + 4. = 0.
17. (cc + 4) (12 ic  5) + 4^ = (7ic2  10)8  12.75a;,
QUADRATIC EQUATIONS.
255
269. Solution by formula. Every quadratic equation can
ll)e reduced to the form ax^ \hx \ c = ^ (§ 261).
This equation can be solved by any of the methods already
suggested and it will be found that
h^h^
4taG.
Hence, the roots of any quadratic equation which has
)en reduced to the form ax^ j bx + c = can be written
'down at sight.
E.g., the roots of
6x2 _ I3x + 6 = are — ± — V( 13)2
26 26 ^ '
= {i±^ Vl69  144
Similarly, the roots of
2 3
4.66
+ 1 = are  =
X X
22 22 ^ '
42.1
= f ± i V9  8
= f ± i = 1 or i.
X = 1 or 2.
270. In particular, the roots of
x^ +px \ q = are x = — ^ ±^ ■\Jp^ — 4:q.
E.g., the roots of x2 + x + 1 = are  i ± ^ Vl  4
= i±iiV3.
271. The formulas
X = 
^±JVb^
2a 2a
4ac,
X =
__P
±iVp^4q,
are so important that they should be Tnemorized and freely
used in the solution of such quadratic equations as are not
:readily solved by factoring.
256 ELEMENTS OF ALGEBRA.
EXERCISES. CXIX.
Write out, at sight, the roots of equations 130, and then
simplify the results.
1. a;23a; + l = 0. 2. ic^ + 6ic + 2 = 0.
3. cc^ + dcc — 4 = 0. 4:. x^ — 5x\l =0.
5. x^ + 2x + 2 = 0. 6. x2 + 2cc24 = 0. .
7. ic22ic + 3 = 0. 8. x''5x36 = 0.
9. ic^ + 2 ic  3 = 0. 10. 02 + 7 cc  44 = 0.
11. x^5xS6 = 0. 12. ^2 + 10 a^ + 5 = 0.
13. x^ + 7x\l() = 0. 14. x^ — 4.x 12 = 0.
15. 12ic2 + aj6 = 0. 16. x''^ 4.x 45 = 0.
17. cc27ic + 12 = 0. 18. x23cc28 = 0.
19. 3x^2x\l = 0. 20. ;:c2 16a; + 60 = 0.
21. 4:X^\5x + 6 = 0. 22. ic2 _^ 10 a: + 21 = 0.
23. 2cc2 + 3x + l = 0. 24. 6cc237cc + 6 = 0.
25. ic22.l£cl = 0. 26. 6x^\5x56 = 0.
27. i»2llic60 = 0. 28. a;2 _^ 0.6 ic + 0.3 = 0.
29. a;2  10 a; + 16 = 0. 30. r^^ + 0.7 a + 0.1 = 0.
31. What are the roots of the equation ax^ jbx \c = 0,
iiP = 4ac?
32. Show that if b'^ — 4 ae is negative the two roots are
complex.
33. Show that if S^ — 4 ac is positive the two roots are
real.
34. Show that if Z'^ — 4 ac is a perfect square the two
roots are rational.
QUADRATIC EQUATIONS.
257
272. Summary of methods of solving a quadratic equa
ftion. From the preceding discussion it appears that a
luadratic equation is solved by forming from it two simple
ruations whose roots are those of the quadratic.
E.g., to solve the quadratic equation
re may write it in the form
(x + 3) (^ + 4) = 0,
rhence cc + 3 = 0, or x + 4 = 0, two simple equations
fhose roots, — 3, — 4, are those of the quadratic.
Or we may write it in the form
rhence [(x + ) + i] [(x + J)  ^] = 0,
md therefore x +  + ^ = 0,
0,
fwo simple equations whose roots, — 3,
[quadratic.
Or we may write it in the form
x' + ix\{iy = {\f,
f whence cc + 7 — i
4, are those of the
4, are those of the
two simple equations ivhose roots, — .
\quadratic.
Or we may simply write out the results from a formula
[obtained by one of the above methods.
For expressions easily factored the first method is the
llDest; otherwise it is usually better to use the formula at
(once.
258 ELEMENTS OF ALGEBRA.
Illustrative problems. 1. Solve the equation
a; + 3 X + 1 _ Sx — 5 3a; — 3
x\5~x + 3~3x — 7~3x — 5
The denominators are such as to suggest adding the fractions in
each member separately before clearing of fractions. Then
* = i
(a; + 3)(x + 5) (3x5)(3x7)
2. Multiplying by i (x + 3) (x + 5) (3 x  5) (3 x  7),
(3x5)(3x7)=(x + 3)(x+_5). Ax. 6
3. .. 8 x2  44 x + 20 = 0, (Why ?)
or 2x2llx + 5 = 0.
4. This is easily factored (§ 263), and
(X  5) (2 X  1) = 0.
6. .. X = 5 or i.
Check. For x
2. Solve the equation \ \ = 0.
X — 1 X — 2 X — 3
Multiplying by (x — 1) (x — 2) (x — 3) we have
1. 3x2 12x + 11=0.
2. This is not so easily factored as in the first problem; hence,
applying the formula (§ 271), we have
X =  ^^ ± — V( 12)2 _ 4 . 3 • 11
23 23 ^ '
= 2±iV3.
Check. ^ H ^ +
l±iV3 ±iV3 _l±^V3
li_ 1i
= f TiV3± V3f TiV3 = 0.
3. Solve the equation x^ \ 2 x = 0.
This factors into x (x + 2) = 0, whence x = or  2.
And, in general, if x is a factor of every term of an equation, x =
is one root.
QUADRATIC EQUATIONS.
259
EXERCISES. CXX.
Solve the following :
1.
1 2 13
x + l 1x 4.x 1
2.
3 2 1
3x 2x l3ic
3.
2cc4l xj1 x6
x + 1 x + 2~ x1
4.
4:X X { 1 X \ 5
2icl X £c + 4
5.
111
a — X a — 2x a — 5x
6.
^ ^ + 1 2 = 0.
x^1 a;lx + l
7.
2 2 ^ 16 34 2 76 5
3^ 2 15~69^ 115^ "^6
8.
x2a x3b x''6ab_
2a 3b 6ab
9. V2a; + V3 + a?  Vll + ic = 0.
10. (1 + 2 x)^  (3 + x)* + (2  a:)^ = 0.
6\5x __ 3x4 57 X _ :§! ^ ^
^^' 4(5x) 5(5+x) ~^25a;2 105
4 (2 — \/x) Vx — 35 ,
12. ^= ^ =  +
3x'
13.
Va; + a3 2 + Vx 4( V^ + ic) (2 + Vx)
4(2 + V^) _ ■\/x[x 3x^
+
V^  cc 2  Vic 4 ( Vic  ir) (2  V^)
260 ELEMENTS OF ALGEBRA.
II. DISCUSSION OF THE ROOTS.
273. The number of roots. The roots of the equation
ax^ + 6ic + c = have been shown to be
2a 2a
This shows that every quadratic equation has two roots.
It is also true that no quadratic equation has more than
two different roots.
For, suppose the equation x^ \ px \ q = has three dif
ferent roots, Ti, rg, rg. Then by substituting these for x
we have
1. ri^ + pi\ { q = 0,
2. r^^+pr^ + q = (),
3. rg^ + pr^ + 2 = 0, whence
4. ri^ — rg^ + i? (ri — r^) = 0.
Dividing by r^ — r^, which by hypothesis 9^ 0,
5. ri + ^2 + jp = 0.
Similarly, taking equations 2 and 3,
6. r2 4^3+i? = 0,
7. .. ri — Tg = 0, by subtracting. Ax. 3
But this is impossible because, by hypothesis, 7\ ^ r^.
Hence, it is impossible that the equation shall have three
different roots, and so for any greater number.
It must be observed, however, that a quadratic equation
need not have two different roots. For example, the equa
*^°^ x^4.x + 4. =
reduces to (x — 2)(x~2)= 0,
and the roots are 2 and 2; that is, the equation has two
roots, but they are equal.
QUADRATIC EQUATIONS.
261
4ac is
274. The nature of the roots. The expression V^
jailed the discriminant of the quadratic equation
ax^ + ^a; + c = 0.
In this discussion a, h, c are supposed to be real.
If the discriminant is positive, the two roots are real and
mequal.
For then — 77— ±  — V^^ — 4 ac can involve no imaginary.
2a 2a ^ J
In particular, if the discriminant is a perfect square, the
two roots are rational.
For then V^^ — 4 ac is rational.
If the discriminant is zero, the two roots are equal.
For then
Ya^2a^'
Aac =
2a
±0.
In this case, — 7 is called a double root.
2a
If the discriminant is negative, the two roots are complex.
For then
2a 2a
4c ac contains the imaginary
I V^>2 _ 4 ac.
Since the two complex roots enter together the instant
that y^ becomes less than 4 ac, we see that complex roots
iter in pairs.
For example, in the equation
the roots are real, since 3^ — 4 (— 7) is positive.
In 2a;2 + a;3 =
the roots are rational, since 1 — (— 24) is a perfect square.
In 3ic2^2a; + l =
bhe roots are complex, since 4 — 12 is negative.
262 ELEMENTS OF ALGEBRA.
275. Since the equation ax^ ^ hx \ c =. ^ has for its
roots — TT + ^r V^>^ — 4 ac and — p— V^*^ — 4 ac, it
2a 2a 2 a 2a '
follows that
Hence, any quadratic function of x can he factored
1. In the domain of rationality^
if the discriminant is square ;
2. In the domain of reality,
if the discriminant is positive ;
3. In the domain of comjdex numbers,
if the discriminant is negative ;
4. Into two equal factors,
if the discriminant is zero.
Illustrative problems. 1. What is the nature of the roots
of the equation ic^ + cc + l = 0?
•.• 6^ _ 4 c^c = 1 — 4 = — 3, the two roots are complex.
2. What is the nature of the roots of the equation
•.• 62  4 ac = 36  36 = 0, the two roots are equal.
3. What is the nature of the roots of the equation
4ic2 + 8aj + 3 = 0?
••• 62 — 4 ac = 64 — 48 = 16, the roots are real, unequal, and rational.
4. Can fix) = hx^ ^Zx 1 \y^ factored ?
••• ft'^ — 4 ac = 9 + 140 = 149, which is not a square, /(x) cannot be
factored in the domain of rationality.
QUADRATIC EQUATIONS.
263
EXERCISES. CXXI.
What is the nature of the roots of equations 110 ?
1. 5 a;2 + 1 = 0.
3. x""  X + 1 = 0.
5. Sx^\x + 7 = 0.
7. 7a;2a;3 = 0.
9. i x"^ \ X + 1 = 0.
2. a^x^ \ I — ax = 0.
4. 2 a;2  a;  20 = 0.
6. 3x''h4:X + 5 = 0.
8. a^2__50a:4.625 = 0.
10. 12a^2_i2a: + 3 = 0.
Of the following functions of x select those which can be
itored in the domain of rationality and factor them.
11. 3a;2_7.
13. 6x^^x — l.
15. 7x'' + 2x6.
17. 6 ^2 _^ 7 a: 3.
19. 2x^^3x4:,
21. 40x2 + 34^ + 6.
23. 80x2 + 70^ + 60.
25. 65x2263x42.
12. 2x2 + 7x + 3.
14. 2ic25x + 3.
16. 55x^27x\2.
18. llx223ic + 2.
20. 132 tt^ + 51 « _ 21.
22. 121.r2 + llx + 12.
24. 56x2 + 113^ + 56.
26. 105x2 246x + 33.
Reduce the following to the form ax^ + &x + c = 0, and
}tate the nature of the roots :
27.
x +
Vx
X
 V
4.
X
^1
0.
28 (^ + «)^ ^ + ^ = 3
(« — ^•)2 a — b
2x + b 4:X — a
30. . = 0.
31. 2
1 (x  1)^
32.
a
Vx
2xb
20 V^
Vx — 5 Vx
= 3.
264 ELEMENTS OF ALGEBRA.
276. Relation between roots and coefficients. The roots of
the equation x^ +2)x + q = are
P
iK2 =  ^  i Vp2 4,q,
Their sum is x^ + x^ = — p,
and their product XiX^ = ( ~ ^ ) ~ (i V^^ — 4 g')^
_ ^^ pp — ^q
~T 4
That is, m an equation of the type x^ + px + q = 0,
1. The sum of the roots is the coefficient of x with the
sign changed;
2. The product of the roots is the absolute term.
These relations evidently give a valuable check upon our
solutions. Any solution which contradicts these laws is
incorrect.
E.g., if the student finds the roots of the equation x^ — x — 30 =
to be — 6 and 5, there is an error somewhere in the solution, because
their sum is not the coefficient of x with its sign changed.
EXERCISES. OXXII.
Solve the following, checking by the above laws.
1. a;2 + 1 zz. 0. 2. ic2  1 = 0.
3. £C2 + iC = 0. 4. x2  ic  1 = 0.
5. cc26a; + 8 = 0. Q. x^x2 = 0.
1. x''6x + 4. = 0. 8. ic2  17 a; + 16 = 0.
9. x^12x + 21 = 0. 10. a;2 + 24a; + 144 = 0.
QUADRATIC EQUATIONS.
265
277. Formation of equations with given roots. Since if
X = Ti and X = r^,
len X — ri = 0, " x — r2 = 0,
[and hence (x — r^(x — T^ = 0, a quadratic equation ; there
of ore it is easy to form a quadratic equation with any giyen
roots.
E.g.^ to form the quadratic equation whose roots are 2 and — 3.
1. •.• X = 2, .. X  2 = 0.
2. ••• X =  3, .. X + 3 = 0.
3. .. (x  2) (x + 3) = 0, or x2 + X  6 = 0.
Similarly, to form the equation whose roots are i ± i i.
1. •.• x = i + ii, .. xiii = 0.
2. V x=:iii, .. X J + ii = 0.
3. .. (x — i  i i) (x — i + i i) = 0, and this may, if desired, be
dtten in the form
X2 _ X + t\ = 0,
16x216x45 = 0.
EXERCISES. CXXIII.
Form the equations whose roots are given below.
2. V2, Va
4. V2,  3.
6.  7,  8.
8. ^Vh,\Va.
a b
10. 3 + 2*, 3 2i.
12. 5 + 3 i, 5 — 3 i.
1. 14. a + 2 V^, a~2
1.
§>!•
3.
i, — i.
5.
3,  11.
7.
a a
2'2'
9.
a ±2 hi.
11.
i±i'iV3.
13.
a / —  a
1.
266 ELEMENTS OF ALGEBRA.
IIL EQUATIONS REDUCIBLE TO QUADRATICS.
278. Thus far the student has leained how to solve
any equation of the first or second degree involving one
unknown quantity, and simultaneous equations of the first
degree involving several unknown quantities.
It is not within the limits of this work to consider gen
eral equations of degree higher than the second. It often
happens, however, that special equations of higher degree
can be solved by factoring, as already explained, or by
reducing to quadratic form.
A few of the more common cases will now be considered,
some having already been suggested in the exercises.
279. The type ax^" + bx" + c = 0. This is a quadratic
in aj", and (§ 269)
whence
X" = — 7— ± — V^/"^ — 4 ac,
Za la
^ Za la
Illustrative problems. 1. Solve the equation
a;6 + 10a.3 + 160.
This is a quadratic in x^ and is easily solved by factoring.
■ ... (x3 + 8) (x3 + 2) = 0,
.. x3 =  8, or  2.
..x =  2, or  V2.
CJieck for x =  \/2. 4  20 + 16 = 0.
We might also solve by the above formula, thus :
3/ ■
x= V 5 ± i VlOO  64
= V3^, or V^s = 2.
QUADRATIC EQUATIONS.
267
2. Solve the equation x^ \ x ^1 \ 1 — ^ = 0.
X x^
This may be arranged
(x + j +(» + ] — 1=0, a quadratic in x + 
Solving (§ 270),
X +
^ = i±iV5.
x'^{i±i^b)x + l=0,
id (§ 270)
X = i{V5 1 ±i VlO + 2 VS),
i( V5I ±i VlO2 V5).
3. Solve the equation x~^ \ x~^ — 2 = 0.
Tliis is a quadratic in x~^. Solving by factoring,
x~'^= 1, or  2.
1
1, or
(2)4
Check for x
(2)2 + (2) 2=0.
(2)4
If (— 2)4 had been written 16, there would appear to be an extra
leous root, but by writhig it (— 2)4 we know that the 4th root is — 2.
4. Solve the equation cc'^ = 21 + Va;^ — 9.
This may be arranged
(x2  9)  (x2  9)^  12 = 0.
The solution often seems easier if y is put for the unknown expres
ion in the quadratic. Here, let y = (x^ — 9)*. Then
y^y 12 = 0,
(2/4)(2/ + 3) = 0,
rhence ?/ = 4, or — 3.
x29= 10, or (3)2,
id x2 ^ 25, or 9 + (3)2,
id X = ± 5, or ± V9 + (  3)2.
Check for X = ± V9 + (3)2. 9 + ( 3)2 = 21 + V9 + ( 3)2  9,
)r 18 = 21  3, because V( 3)2 =  3. If the (3)2 were written 9,
fthere would appear to be an extraneous root.
268 ELEMENTS OF ALGEBRA.
6. Solve the equation (x^ + x { 3) (x^ ^ x \ 5) = 35.
In equations of this kind there is often an advantage in letting y
equal some function of x. Here, let 2/ = x^ f x + 3. Then
y(2/ + 2) = 36,
or 2/2 + 2 2/  35 = 0,
or {y + 7) (y  5) = 0,
whence 2/ = — 7, or 5.
Hence, x^ + x + 3 = — 7, or 5,
and each of these equations can be solved for x.
It would answer just as well tolet y = x^ + x \ 5, in which case we
should have {y — 2)y = 35.
EXERCISES. CXXIV.
Solve the following :
1. Vic  1 = cc — 1. 2. a;ic*20 = 0.
3. 7x4.x^ 20 = 0. 4. x^ + x^20 = 0.
5. a;«28a;« + 27 = 0. 6. 7 x^ \ x^  350 = 0.
7. x^ + x — 6x^ = 0.
8. x^ + 5xl =  ^ •
x^ + 5x + l
9. (ic2 + 3)2 + (x^ I 3) _ 42 = 0.
10. x(a + b)x^2a(ab) = 0.
11. (x^\2x\3)(x^ + 2xh6) = 2.
12. (x2 + 3a;4)(ic2 + 3aj + 2) + 8 = 0.
13. V^/(21 V^) + (21 V^)/V^ = 2.5.
^4 1 ■ 1 6 ^
QUADRATIC EQUATIONS.
269
280. Radical equations have already been discussed (§ 191)
the special case in which they lead to simple equations,
id several problems have been given in connection with
^the study of quadratics.
Whenever they lead to quadratic equations their solution
IS possible, and a few cases somewhat more elaborate than
[those already given will now be considered.
Illustrative problems. 1. Solve the equation
2a;2_.3^_3V2ic2_^3:c42 =
1. This may be arranged
0.
2x2 + 3x43V2x2 + 3x4 + 2 = 0.
2. Let y = V2x2 + 3x4.
3. Then y^ Sy + 2 = 0.
4. .. (y2){yl) = 0.
5. .. y = 2, or 1.
6. .. 2x2 + 3x4 = 2, or 1,
two quadratic equations in x, which give
X =  I ± i V57, 1, or  .
Check for X =  . 2.5 _ j^ _ 3 _ 2 = 0.
2. Solve the equation x — 1 = 2 { 2x~^.
1. This may be written
(V^i)(V^ + i)'^^ + ^^^o.
Vx
2. Or (V^+ l)('Vx 1 4=') = 0.
^ Vx/
3. .. Vx+1 = 0, and Vx = 1, andx
land X
2
Vx
0,
(1)2, orVx
Vx — 2 = 0, a quadratic in Vx.
4. .. (Vx2)(V^+1) = 0.
5. .. Vx = 2, and x = 4, or Vx = — 1, and x = {— 1)2.
.. there are three roots, two being alike, 4, (— 1)2, (— 1)2. All
three are easily seen to check. The reason for writing ( — 1)2 instead
of + 1 is explained on p. 267, exs. 3 and 4.
270 ELEMENTS OF ALGEBRA.
3. Solve the equation Va; + 3 — Vic + 8 = 5 V^.
1. 2a;+ 11 2 V(x + 3)(x + 8) =25x. Ax. 8
2. ..  2 Vx2 + 11 X + 24 = 23x  11. Ax. 3
3. .. 21 x'^ — 22 X + 1 = 0, squaring, etc.
4. .. (21 X  1) (x  1) = 0, and x = ^\, or I.
In checking, each root is found to be extraneous. This might have
been anticipated because in squaring the first member of step 2 the
( — 2)2 was called 4, and hence, when the result was placed under the
radical sign for checking, and the root taken as positive, a failure to
check was natural.
Had the original equation been Vx + 3 + V x + 8 = 5 Vx, tlie root
1 would have checked ; had it been — Vx + 3 + Vx + 8 = 5 Vx, the
root 2X would have checked.
EXERCISES. CXXV.
Solve the following :
1.
Va; + 3  V;r  4  1 = 0.
2.
x^J^x = 4.i~ VlO  X''  X.
3.
Vl +4^; Vl 4x = 4 Vaj.
4.
Vx2 8x{31+(x4.y = 5.
5.
x'^\5xW = Wx^\5x + 2.
6.
7.
Va;  2 + V3 + ic  VlO + ^ = 0.
xi(x^ _ l)i _ 2 x (x^  1)^  ^ == 0.
8.
Vl + 2 a^  V4 + a; + V3  a, = 0.
9.
Vcc + 8 + Vx  6  V8 a^  10 = 0.
10.
V4a;2 + 2V2ic Vl4 4.x = 0.
11.
3 Vic^  7 X + 12 = V7 • Vx  7 ic + 12.
12. V(ic  1) (x  2) + V(a'  3) (.X  4) = V2.
QUADRATIC EQUATIONS.
271
281. Reciprocal and binomial equations. A reciprocal equa
jtion is an equation in which the coef&cients of the terms
jquidistant from those of highest and lowest degree, respec
Lvely, have the same absolute value and have the same
jigns throughout or opposite signs throughout.
E.g., the following :
ic2  1 = 0,
ax^ + hx^ — bx — a = 0,
ax^ — bx^ \ cx^ — bx ^ a = 0,
x^ { x^ h x^ i x'^ + X + 1 = 0.
They are called reciprocal, because they are unaltered
rhen for the unknown quantity is written its reciprocal.
Kg., when  is written for x in the equation
becomes
ax^ \ bx \ a = 0,
X"^ X
rhich, by multiplying both members by x^, reduces to
a { bx \ ax^ = 0,
le original equation.
282. Since x can be replaced by  > the roots of reciprocal
mations enter in pairs, each root being the reciprocal of
the other root of that pair, excepting the two roots + 1 and
1, each of which is its own reciprocal.
E.g., x"^ \ X \ 1 = has for its roots
^1 =  i + i * Va,
x^ = ^^i Vs,
id each is the reciprocal of the other, because their prod
ict is 1 (§ 162).
272 ELEMENTS OF ALGEBRA.
So ic^ + 1 = has for its roots the reciprocals i and — i.
Similarly in the case of x^ — 2 x^ — 2 x \ 1 = 0. Here
x^ + l2x(x\l) = 0,
whence (x { 1) (x^ — x } 1 — 2 x) = 0,
and therefore a? + 1 = 0, and x = — 1,
or cc2  3 ic + 1 == 0, and a; = I ± 1 Vs.
In this case, f + ^ V5 and  — ^ Vs are reciprocals,
because their product is 1 (§ 162), and the other root,
— 1, is its own reciprocal. And in general, in the case of
reciprocal equations of odd degree, one root is always its
own reciprocal.
This is seen in the case of x^ — 1 = 0.
283. Reciprocal equations can often be reduced to equa
tions of lower degree by the factoring method set forth in
the preceding example, or by dividing by some power of the
unknown quantity, as in the following case :
Solve x^ j x^ + x^ { X + 1 = 0.
Divide by x^, and ^ 2
X2 + X + 1++ — = 0,
X x2
an equation already considered (§ 279).
It reduces to ( x +  j + ( x +  j — 1 = 0,
a quadratic in x j Solving for x +  , we have
X X
x + ^=i±iV5. §270
two quadratics in x.
These equations may now be solved for x, each giving two values.
The final roots are four in number, as would be expected. They are
given on p. 267, and in more complete form on p. 273.
QUADRATIC EQUATIONS.
273
284. Equations of the form x"" \ p = are called bino
[mial equations. In this case, no restriction is placed on p ;
jit may be positive, negative, integral, fractional, real, imag
finary, etc.
The solution of binomial equations in which p = ±l evi
dently depends upon the solution of a reciprocal equation.
E.g., x5l=0
luces to (x  1) (x4 + x3 + x2 + X + 1) = 0,
whence x — 1 = 0, and x = 1,
3r x* + x3 + x2 + X + 1 = 0,
a reciprocal equation, the one just considered in § 283, with four roots.
Since if cc^ — 1 = 0, or ic^ = 1, a: is the fifth root of 1, and
3mce cc^  1 = has 5 roots (§§ 279, 283), viz.:
Xi = l,
CC2 =: 1 ( V5  1 + ^ VlO + 2 V5),
xs = i(V5  1  ^VlO + 2 VS),
x, = l(^5l+ ^ VlO  2 VS),
x, = l(~^5li VlO  2 V5),
therefore, there are 5 fifth roots of 1.
Similarly, there are 2 square roots of any number, 3 cube
roots, • • ■ n nth. roots.
Thus the two square roots of 1 are evidently +1, — 1,
jwhich may be obtained by extracting the square root
[directly or by solving the equation o;^ — 1 == 0.
The three cube roots are readily found by solving the
[equation cc^ — 1 = 0.
Here a:;^ — 1 =
[leads to (x — 1) (x"^ \ x { 1) = 0,
whence a; — 1 = 0, and x = 1,'
)X x^\x + l = 0, solved in § 282.
274 ELEMENTS OF ALGEBRA.
EXERCISES. CXXVI.
Solve the following :
1. 2i»2 + 5cc + 2 = 0.
2. £C» + cc^ + ic 4 1 = 0.
3. 10a;229cc + 10=:0.
4. 2x^3x^Sx\2 = 0.
5. x^ + x^ — 4.x^ } X + 1 = 0.
6. x^  x^ — 4.x^ — X \ 1 = 0.
7. x^ + 4:X^ + 2x^ + 4.x \l = 0.
8. ic*  5 ic^ + f I a:^  5 cc + 1 = 0.
9. X* — I ic^ — y cc^ — I X + 1 = 0.
10. 2x^9x^ + Ux''9x + 2 = 0.
11. 12 a;* + 4x3 41x^ + 40; + 12 = 0.
12. x^ + 1 = 0. 13. x^ + 1 = 0. 14. x^  1 = 0.
15. What are the 2 square roots of 1 ? the 3 cube roots ?
the 4 fourth roots ?
16. What are the 3 cube roots of 8 ?
17. What are the 6 sixth roots of 1 ?
18. Show that the product of any two of the fourth roots
of 1 equals one of the four roots^ and that the cube of either
imaginary root equals the other.
19. Show that the product of any two of the cube roots
of 1 equals one of the three roots, and that the square of
either complex cube root equals the other.
20. Show that the sum of the 2 square roots of 1, the
sum of the 3 cube roots, the sum of the 4 fourth roots, the
sum of the 5 fifth roots, are all equal to zero.
QUADRATIC EQUATIONS.
275
285. Exponential equations have already been considered
§ 205. Only in certain cases can tliey be solved by
lear or quadratic methods.
E.g., 2^: 8 = 16:1.
This may be written
2x^ — ^x __ 24
rhence x^ — ^ x =^ 4c,
jiving a: = 4, or — 1. Each result checks.
The equation 2^ + ^ + 4^ = 8 may be written
2 . 2^ + 22^ = 8,
)r (2^)2 + 2 (2^)  8 = 0, a quadratic in 2^.
Hence, solving, 2^ = 2 or, — 4.
If 2^ = 2, X = 1, a result which checks.
If 2^ = — 4, we cannot find x.
EXERCISES. CXXVII.
Solve the following :
1. 64^: 2 = 4.
3. 2.2^' + i = 2.
5. 2^.16^ = ^1^.
7. 2 • 4^^^ = 2^^^
9. (4V8)" = 23 + ».
11. '^ = 1,
13.
m^^ = 1.
15. 25
9 . rfix . 9K
2. 3"':8P = (3ii)'.
4. 32^ 9^ = 27^' 3.
6. a'^:{a'=f = {ay.
8. (3f3 = 27i^
10. 26^* + '^ = 3^^* 2^+
12. 9^9^ = ^.
14, a^'^^' . (ay' = \'
2/25^^
276 ELEMENTS OF ALGEBRA.
IV. PROBLEMS INVOLVING QUADRATICS.
Illustrative problems. 1. What number is 0.45 less than
its reciprocal ?
L Let
X = the number.
2. Then
x =  0.45.
X
3. ..
aj2 + o.45x
1 = 0.
4. ..
x=  0.225 X ± 0.5 Vo.2025 + 4
= 0.8, or 1.25.
Check. 0.8
= L250.45.
1.25= 0.80.45.
Hence, either result satisfies the condition. But if the
problem should impose the restriction "in the domain of
positive numbers," — 1.25 would be excluded ; if " in the
domain of negative numbers," 0.8 would be excluded ; if
" in the domain of integers," both results would be excluded
and no solution would be possible.
2. A reservoir is supplied with water by two pipes. A, B.
If both pipes are open, i of the reservoir will be filled in
2 mins. ; the pipe A alone can fill it in 5 mins. less time
than B requires. Find the number of minutes in which
the reservoir can be filled by A alone.
1. Let X = the number of minutes required by A.
2. Then ■ a; + 5 = " " " B.
3. Then  = part filled by A in 1 miu. ,
and = " " B " 1 min.
x + 5
4... ?+ '
X ic + 5
5. .. 11 x2 + 7 X  120 = 0, or (x  3) (11 x + 40) = 0.
6. .. X = 3, or  f ft.
QUADRA. :C EQUATIONS.
277
Here each root satisfies the equation ; but the conditions
)i the problem are such as to limit the result to the domain
positive real numbers. Hence, — f ^, being meaningless
this connection, is rejected.
3. The number of students in this class is such as to
Satisfy the equation 2 a;^ — 33 a? = 140. How many are
there ?
1.
2 x2  33 a;  140 = 0.
(X  20) (2 X + 7) = 0.
X = 20, or  .
Here, too, the conditions of the problem are such as to
it the result; this time to the domain of positive inte
gers. Hence, " —  of a student," being meaningless, is
rejected.
4. A line, AB, 3 in. long, is produced to P so that the
rectangle constructed with the base AF and the altitude
»P has an area 14.56 sq. in.
rind the length of BF.
P'
1. Let X
= the number of inches in BP.
2. Then the area R
= (3 + x)x = 14.56.
3. .•.x2 + 3x 14.56 = 0.
4. .. X = 2.6, or  6.6.
Here we are evidently not
lited as in probs. 2 and 3.
?he negative root may be
iterpreted to mean that AB is produced to the left.
?P' is — 5.6 in., i.e., 5.6 in. to the left, and the rectangle
jcomes E', which is, however, identically equal to B.
278 ELEMENTS OF ALGEBRA.
EXERCISES. CXXVIII.
In each exercise discuss the admissibility of both roots.
A. Eelating to Numbers.
1. What number is y\ of its reciprocal ?
2. What number is ^^ greater than its reciprocal ?
3. What is the number which multiplied by f of itself
equals 1215 ?
4. Separate the number 480 into two factors, of which
the first is  of the second.
5. The sum of a certain number and its square root is
42. Required the number.
6. Find a number of which the fourth and the seventh
multiplied together give for a product 112.
7. Onefourth of the product of f of a certain number
and I of the same number is 630. Find the number.
8. The square of 5 more than a certain number is
511,250 more than 10 times the number. Required the
number.
9. The product of the numbers 2ic3 and 4cc6, written
in the decimal system, is 115,368. What figure does x
represent ?
10. Separate the number 3696 into two factors such that
if the smaller is diminished by 4 and the larger increased
by 7 their product will be the same as before.
11. Of three certain numbers, the second is f of the first,
and the third is  of the second ; the simi of the squares
of the numbers is 469. What are the numbers ?
QUADRATIC EQUATIONS. 279
B. Relating to Mensuration.
For formulas see p. 172.
12. How many sides has a polygon which has 54
[iagonals ?
13. The area of a rectangle is 120 sq. in., and its diagonal
17 in. Required its length and breadth.
14. The base of a triangle of area 16.45 sq. in. is 2.3 in.
iore than the altitude. Required the base.
15. The length of a rectangle of area 70 sq. in. is 3 in.
)re than the breadth. Required the dimensions.
16. Divide a line 16 in. long into two parts which shall
)rm the base and altitude of a rectangle of 63.96 sq. in.
17. The hypotenuse of a rightangled triangle is 10
and one of the sides is 2 in. longer than the other.
jquired the lengths of the sides.
18. In a rightangled triangle one of the sides forming
le right angle is 6 in., and the hypotenuse is double the
ler side. Find the length of the other side.
19. A square and a rectangle have together the area 220
sq. in. The breadth of the rectangle is 9 in., and the
length of the rectangle equals the side of the square.
Required the area of the square.
k 20. From the vertex of a right angle two bodies move
on the arms of the angle, one at the rate of 1.5 ft., and the
other 2 ft., per second. After how many seconds are they
50 ft. apart ?
I
21. What is the result if, in the preceding example, 1.5,
, and 50 are replaced by m., n, d'i
280 ELEMENTS OF ALGEBKA.
22. A square is 78 sq. in. greater than a rectangle. The
breadth of the rectangle is 7 in., and the length is equal to
the side of the square. Required the side of the square.
23. If the sides of a certain equilateral triangle are
shortened by 8 in., 7 in., and 6 in., respectively, a right
angled triangle is formed. Required the length of the side
of the equilateral triangle.
24. If two sides of a certain equilateral triangle are
shortened by 22 in. and 5 in., respectively, and the third
is lengthened by 3 in., a rightangled triangle is formed.
Required the length of a side of the equilateral triangle.
25. On an indefinite straight line given two points, A
and B, d units apart, to find on this line a point, F, such
that AP^ = BP ' AB. Draw the figure showing the posi
tions of the two points. (This is the celebrated geometric
problem of " The Golden Section.")
26. Four places. A, B, C, D, are represented by the
corners of a quadrilateral whose perimeter is 85 mi. The
distance BC is 24 mi., and CD is 14 mi. The distance
from ^ to D by the way of B and C is y\ as great as the
square of the distance from A direct to D. How far is it
from Ato B? also from ^ to D ?
27. About the point of intersection of the diagonals of a
square as a center, a circle is described ; the circumference
passes through the midpoints of the semidiagonals; the
area between the circumference and the sides of the square
is 971.68 sq. in. Required the length of the side of the
square. (Take tt = 3.1416.)
28. A mirror 56 in. high by 60 in. wide has a frame of
uniform width and such that its area equals that of the
mirror. What is the width of the frame?
QUADRATIC EQUATIONS. 281
C. Eelating to Physics.
29. If a bullet is fired upward with a velocity of 640 ft.
[per sec., the number of seconds elapsing before it strikes
the earth is represented by t in the equation = 320 t — ^gf'^
^in which ^ = 32 ft. Find t.
30. Two points, A and i?, start at the same time from a
fixed point and move about the circumference of a circle in
)pposite directions, each at a uniform rate, and meet after
sees. The point A passes over the entire circumference
b 9 sees, less time than B. Required the time taken by
fA, and also by B, in passing over the whole circumference.
31. It is shown in physics that if two forces are pull
ing from a point, P, and are represented in direction and
tensity by the lines PA, PB, the resultant force is repre
ented by PC, the diagonal of their parallelogram. Two
brces, of which the first is 23 lbs. greater than the second,
t at right angles from a point. Their resultant is 37 lbs.
equired the intensity of each force.
32. Two forces, of which the first is 47 lbs. less than
the second, act at right angles from a point. Their result
ant is 65 lbs. Required the intensity of each force.
33. It is proved in physics that if a body starts with a
velocity ("initial velocity") of u ft. per sec, and if this
increases a ft. per sec. (the " acceleration "), then in t sees,
the space s described i^ s = ut { ^ at^. Suppose the initial
velocity is 40 ft. per sec, and the body moves with an ac
celeration of — 2 ft. per sec, find when it will be 400 ft.
from the starting point.
^^B' 34. Suppose a body starts from a state of rest, and the
^H acceleration is 18 ft. per sec, find the time required to pass
^H over the first foot ; the second ; the third. (See ex. 33.)
I
282 ELEMENTS OF ALGEBRA.
35. Two points, A and B, start at the same time from a
fixed point and move about the circumference of a circle
in the same direction, each at a uniform rate, and are next
together after 8 sees. The point A passes over the entire
circumference in 18 sees, less time than B. Required the
time taken by A iji passing over the whole circumference.
36. It is shown in physics that if h = the number of feet
to which a body rises in ^.secs. when projected upward
with a velocity of it ft. per sec, then h = ut — \ gf, where
g = 32. Find the time that elapses before a body which
starts with a velocity of 64 ft. per sec. is at a height of
28 ft.
37. A body is projected vertically upward with a velocity
of 80 ft. per sec. When will it be at a height of 64 ft. ?
(See ex. 36.)
D. Miscellaneous.
38. A reservoir can be filled by two pipes, A and B, in
9 mins. when both are open, and the pipe A alone can fill
it in 24 mins. less time than B can. Required the number
of minutes that it will take A alone to fill it.
39. A reservoir has a supply pipe, A, and an exhaust pipe,
B. A can fill the reservoir in 8 mins. less time than B can
empty it. If both pipes are open, the reservoir is filled in
6 mins. Required the number of minutes which it will
take to fill it if A is open and B is closed.
40. Two travelers, A and B, set out at the same time
from two places, P and Q, respectively, and travel so as to
meet. When they meet it is found that A has traveled
30 mi. more than B, and that A will reach () in 4 das., and
B will reach P in 9 das. after they meet. Find the dis
tance between P and Q.
QUADRATIC EQUATIONS.
283
1. Solve
REVIEW EXERCISES. CXXIX.
X^ + 111^ _ X
2. Solve  +  =  + ■
2 a; 3 ic
3. Factor (x"" + x  ISy  49.
4. Factor x^6x^37x + 210.
2x ^ 4tx — 3 _ _
5. Solve 7 H ^i 9 = 0.
6. Solve
7. Solve
.r — 4 X \ 1
xi y fx
x + l) \x +
2\2
2
a.
X — h
= 0.
8. Simplify
^, , /x2  11
9. Solve —
\ X'^ { X
X — b X — a cr — ax
2x'x{2 4 .r^  1
4ic3 + 3x + 2 2xl
^ + 19 Y_ 3(2^
11 ; 2 + x
10. Solve 18 (x + iy{x + 2)2 = 8 (x  3)2(ic + l)^.
11. Solve {x 3y3{x 2)^ + 3{x  If  x^ = ^  x.
12. Find the square root of
13. li x"^ \xy+ z = 0, and iv"^ \ luy + « = 0, where x^w,
prove that w + x + ?/ = 0. (Subtract and factor.)
14. Find the lowest common multiple of
(^,2 ^ c''  a'' + 2hc) {c + a  h)
.aad
{a^
^ ^2hG)(a\h + c).
CHAPTER XV.
SIMULTANEOUS QUADRATIC EQUATIONS.
I. TWO EQUATIONS WITH TWO UNKNOWN QUANTITIES.
286. 1. When one equation is linear. While this is not
a case in simultaneous quadratics, since one equation is
linear, it forms a good introduction to the general subject.
In this case, one of the unknown quantities can be found
in terms of the other in the linear equation, and the value
substituted in the quadratic. The problem then becomes
that of solving a quadratic equation.
E.g.^ to solve the system x — 2 2/ = 3.
0:2 I 2/2 = 26.
(Why ?)
Here^
tve have
1.
X ■■
= 3 + 2 2/, from the first equation.
2. ..
(3 + 2^)2 + 2/2 =
:26.
3. ..
52/2 + 12 2/ 17 =
:0.
4. ..
(52/+ 17) (2/ 1) =
:0.
6. ..
y=  V, or 1.
6. ..
x = 3 + 22/= 
¥, or 5.
Check,
, for X = 
 ¥, y =  V
 ¥ + ¥ = ¥ =
:3.
¥/ + ¥/ = %¥ =
:26.
In checking, the roots must be properly arranged in pairs.
E.g.^ in the preceding example
X = — J/ when and only when y = — ¥,
and x = 2 " " " " 2/ = 5.
284
SIMULTANEOUS QUADRATIC EQUATIONS. 285
EXERCISES. CXXX.
Solve the following systems of equations :
1. xly = 2.
xy = ^.
3. X + 2/ = 9.
xy = 45.
5. X — y = 24:.
xy = 4212.
7. x + y = 1.25.
xy = 0.375.
9. x^ + y^ = 1274:.
X = 5y.
11. 5(x + y) = xy.
xy = 180.
13. x\y = 6.
xy = 2592.
15. ^x' + iy' = ll.
^x\^y = 5.
17. x^\xy + y^ = 63.
xy = 3.
19. (7 + ^) (6 + 2/) =80.
x + y = 5.
21. a;2 + 7/2 = 500.
3.
xy
2. x + y = 100.
x?/ = 2400.
4. X — y = 11.
6/ic = 2//10.
6. tV^' + 2/' = 122.
^a:;3/ = 13.
8. 2x^{y^ 100 = 0.
jxy50 = 0.
10. 3a; + 4?/8 = 10.
x^ — y^ = — 5.
12. ia;2+ 1^/260 = 0.
i ^ + i 2/  5 = 0.
14. 14a^2_i22 7/2 = 100.
x = Sy.
16. 27 ic + 33 2/ 60 = 0.
Sx^\10y^18 = 0.
18. 0.01ic2 + 0.52/2 = 0.
0.1;z;0.25y3 = 0.
20. 0.01 0^2 + 400?/ 25 = 0.
0.5x + y10 = 0.
22. a? + ?/ — 4 = 0.
« 2/
286 ELEMENTS OF ALGE13RA.
287. 2. When both equations are quadratic. In this case,
X can be found in terms of ?/ in either equation, but, in
general, the value will involve y'^. In this case, the value
of X substituted in the other equation will inyolye 2/*? and
hence the result will he an equation of the fourth degree.
E.g., given the system x^ — y^ = — S.
From the first equation
x= ± Vy2 _ 3.
Substituting in the second,
2(?/23)±3V2/23 + 7/ = 7.
Isolating the radical, squaring, and reducing, we have
2 ?/ + 2 7/3  30 ?/2  13 ?/ f 98 = 0,
an equation of the fourth degree.
288. Hence, in general, two simultaneous quadratic equa
tions involving two unknown quantities cannot be solved hy
means of quadratics.
It is only in special cases that such systems admit of solu
tion by quadratics, and four pairs of roots should aliv ays be
expected.
A few of the more common of these special cases will
now be considered.
EXERCISES. CXXXI.
To what single equations of the fourth degree do the
following systems reduce ?
2. y'^ \ 2 X — xy = 5.
x"^ \ x \ y = 4:.
4. x'^ \ xy \ y^ \ X — 5 = 0.
2x^ + y^x + y3 = 0.
1.
x + y^ = 11.
3.
2x^i3xt/ =
:0.
x'32/ + y =
:0.
SIMULTANEOUS QUADRATIC EQUATIONS. 287
289. When one equation is homogeneous. In this case a
)lution is always possible. For if ax^ + hxy + cy^ = is
le homogeneous equation we can divide by y^ and have
Qu X Of "V
—^ + h \ c = 0, a quaxlratic in  • Hence,  can be
f y y y
mnd and x will then be known as a multiple of y, and
this value can then be substituted in the other equation.
E.g.^ to solve the system
1. x2 _ 5 a;^/ + 2/2 = 0.
2. x2 + 3x4y + 4=0.
2 5/x
(^o(^^)=o■ .
X \ ^ , y
5. ..  = , or 2, and x = , or 2y.
» 2/ 2 2
V
Substituting x =  in eqiuation 2, we have
6. ^ + ^_4j, + 4 = 0.
7. .. ^2 _ 10^ + 16 = 0.
8. .. ?/ = 2, or 8, and .. x = ^ = 1, or 4.
Substituting x = 2y in equation 2 and reducing, we have
10. .. y= _;,t jiVl5.
11. .. x = 22/= ^±^iVTb.
12... jc = l, 4, i + iiVIs, iiiVl5,
and y = 2, 8, i + izVl5, iiiVi5,
these roots being taken in pairs in the order indicated.
Check. All of these roots check. While the substitution of the
complex roots takes time and patience, it is the only method of deter
mining the correctness of the solution.
288
ELEMENTS OF ALGEBRA.
EXERCISES. CXXXII.
Solve the following systems of equations :
1. CC^ 4 2/2 _ Irj^y _ Q^
x\y = a.
2. Zx'^^xyy'' = h.
x^ — 2xy + y^ = 0.
3. Bx^ \4txy — y^ = 0.
x^ + X + y = 5.
4. x^ \ xy + X — y = — 2.
2x^ — xyy^ = 0.
5. x^ + 3xy + 3x~y = 2.
x^ + 2xy3y^ = 0.
6. x'y^ + x + y=is,
86 (x'' + 7/^= 97 xy.
7. 2x^ + 3xy + 4.y = 18.
x^ + 4:xy = 12 if.
8. 3x^^^xy^3xy = 3.
x^ \ xy = 0.
9. x^ + 4:x + 3y + y''= 2.
x(x}2y)15y^ = 0.
10. x(x + y) + y(y^x) = 4:X7/.
^{^ + y) + y + x = 24..
11. x'^3x\4.y + 2xy = 24..
x^\3xy = 4. tf.
12. 147iz;2 + 196a:?/ + 57?/2 = 0.
x'' + 2xy + 33 = 0.
SIMULTANEOUS QUADRATIC EQUATIONS. 289
290. When both equations are homogeneous except for the
ibsolute terms. In this case a solution is always possible
quadratics. For if
a^x^ + bixy + c^y^ = d^,
id a^x^ + b^XT/ + Czi/ = d^,
re can multiply both members of the first by d^, and of
le second by di, and subtract, and
(a^dz — a^di) x^ + (b^d^ — b^di) xij + (c^d^ — CgC^i) 3/^ = 0.
lis may now be treated as in § 289.
JBgr., to solve the system
1. x2 + 3x?/2?/2 = 2.
2. 2 x2 _ 5 x?/ + 6 2/2 = 3.
Multiplying both members of equation 1 by .3, and of equation 2 by
^, and subtracting, we have :
3. x2 _ 19 JC2/+ 18 7/2 = 0.
This equation is easily reducible. If it were not, we should divide
by 2/2 and proceed as in § 289.
4. .. (a; 18?/)(x?/) = 0.
5. .. X = 18y, or y.
Substituting 18?/ for x in 1, we have
6. 324 2/2 + 54 2/2  2 ^2 = 2.
7. .. 2/ = ± i V^  ± 9V ^^»
and X = 18 2/ = ± /t ^^•
Substituting ?/ for x in 1, we have
8. 2/2 + 3 2/2  2 2/ = 2.
9. .. y = ± 1, whence x = ± 1.
Check. All of these results check.
E.g., try X = ± 57 ^^7, 2/ = ± 9V ^^7.
Substituting these values in equation 1,
H + fTl8=2.
Substituting in equation 2,
290 ELEMENTS OF ALGEBRA.
29J.. Since § § 289 and 290 depend upon finding the value
of ; or of ? we can also solve by letting  = v, ot7/ = vx,
y X X
then finding v.
E.g., ill the preceding example we had the system
1. x^ + 3 x^/  2 ?/2 = 2.
2. 2 x2  5 X2/ + 6 ?/2 z= 3.
Let  = u, OT y = vx. Then, from 1, we have
X
3
x2 + 3 ux2  2 ^2x2 = 2.
4. /.
'
1 + 3 u  2 u2
Similarly,
from 2
, we have
5.
2 x2  5 vx2 + 6 U2x2 = 3.
6. ..
^
2  5 u + 6 u2
Equating
the values of x^,
7.
2 3
l+3u2u2 25w + 6d2
Reducing,
8.
18 v2 _ 19 u 4 1 = 0,
or
{18ul)(t; 1) = 0.
9. ..
v = j\, or 1.
10. ..
7/ = vx = tVx, or X.
This is substantially the same as step 5 of the preceding solution
(p. 289), and the rest of the work is as given there.
In the same way we may let  = v, or x = vy. We should then
have, from equation 1,
1)22/2 I 3 vy"^  2 ^2 _ 2.
v2 + 3 u  2
Q
Similarly, from 2, y^ =
2 v2 _ 5 u + 6
Equating these values of y^, v can be found as above.
SIMULTANEOUS QUADRATIC EQUATIONS. 291
EXERCISES. CXXXIII.
Solve the following systems of equations :
1.
x'^ + 2xy = 39.
xy + 2y^ = &b.
2.
x^ + 3xy = 2.
3y^ + xy = l.
3.
x'^ + Zxy = 54.
xy{.4.y^ = im
4.
2x^ + 3xy = 21.
xy {y"^ = 4.
5.
7}i^x^ \ n^y^ = q^.
x'^ y'
6.
7a;25xy = 18.
2/' 2/
7.
3xy + y^lS =
0.
8.
x'^xy + y^ = 21.
4.x''{xy7 =
0.
y''2xy = 15
p
9. x!^ + xy \7/ = 139. 10. aic2 4 ^, (a;^ + t/) = m.
5 y^ — 4:xy = — 75. mf" f c? (a^^ + ?/^) = /?.
11. x^2xy ^y'' = 51. 12. 3 cc^  5 x^/ + 2 t/^ = 14.
169 x^ + 2 2/' = 177. 2x''hxy ^Zy'' = Q>.
13. 2ic2 + 2a;2/ + 2/2 = 73. 14. 32?/2_2£c?/ll = 0.
ic2 + ic?/ + 2/ = 74. x2 + 42/^ = 10.
15. 3x2 + 13a;y + 8?/2 = 162.
x^ — xy + y'^ = 7.
16. (3a; + ?/)(3 2/ + a;)=384.
(x  2/) (x + 7/) = 40.
17. 3x2 + 4x2/457/248 = 0.
4 x^ 4 5 x?/ — 36 = 0.
18. 2x2 + 3x7/ 37/2 + 124 = 0.
7 x2  XT/  2/^ + 49 = 0.
292 ELEMENTS OF ALGEBRA.
292. When the equations are symmetric with respect to the
two unknown quantities. In this case a solution is always
possible by quadratics. The solution is accomplished by
letting X = u \ V, and y = u — v, and first solving for u
and V.
E.g., given the system
1. x^ + Zxy \y^ = Al.
.2. x^ + y^ + X + y = 32.
Let x = u + V and y = u — v. Then, by substituting in 1, we have
3. 5 w2 _ u2 _ 41^ or v^ = 5u'^ 41.
Substituting in 2,
4. u^\v^ + u = 16.
Substituting here the value of v^ from 3,
5. Qu^ + u67 = 0,
or (6w + 19)(tt3) = 0.
6. .. u=  V, or 3.
Substituting this value of m in 3,
7. v= ±1 V329, or ± 2.
8. .. X = w + V = , 6, or 1, four values as we should
expect (§ 287). ^
9. Since the equations are symmetric with respect to x and y, y
must have the same values, always arranged so that x { y shall equal
2 u. (Why ?)
„  19 + V329  19  V329 ^ ,
10. ..for x = , ,5,1,
6 ' 6 ' ' ' .
 19  V329  19 + V329 , ,
we have y = , ,1,5
D O
All of the results check.
It should be noticed that a set of equations like
xy = l, a;2 + 2/2 z= 25,
is symmetric with respect to x and — y. Hence, if
x = 3, or — 2, y = 2, or — 3.
SIMULTANEOUS QUADRATIC EQUATIONS. 293
EXERCISES. CXXXIV.
Solve the following systems of equations :
1. jc2 _^ 2/^ = 41. 2. x^ ^xy ^y^ = 19.
X — y — \. X \ y = 5.
3. x^xy + y'^ = 3. 4. x^ \ y^ \ 3 (x { y) = 4:.
x^{xy + y^ = T. Sx^ + 4,xy + Sy^ = 3.
5. x + Vxy 4 2/ = 14. 6. x^ — 2.5 xy \ y'^ = 0.
x' + xy{y' = 84. 2(x\ yf = 3.6 (x' + y').
1 , 1 55 o
 +  = 7. y^x 5
X y
9. ic (ic + 2/) — 40 = 0.
y(y + x)~60 = 0.
10. 2ic2__^2/ + 2 2/'79.58.
a;2 2 a:?/ 4 2/' = 21.29.
p. 293. 3. When equations above the second degree are involved.
In general, such systems cannot he solved by quadratics,
although they can be solved in special cases.
E.g., x3 + x2?/ + 2/3 = 11.
xy = \.
Here x = y — \; hence,
(yl)3 + (yl)2 2/ + 2/'rz:ll,
or 3 2/^ — 5 ?/2 + 4 ^ — 12 = 0, a cubic equation.
Now a cubic equation may sometimes be solved by factoring, as
here, for this reduces (§ 104) to
,^ (?/2)(3 2/2 + 2/ + 6) = 0,
^^fcwhence y = 2, or \{l ±i ^71),
294 ELEMENTS OF ALGEfiRA.
294. If the equations are symmetric with respect to the
unknown quantities, they often yield to the method given
in § 292.
E.g., to solve the system
1. x^\y^ = 91.
2. x + y = 7.
Let x=:w + v, y = u — V. Then
3. 2 M^ + G uv^ = 91, from 1.
4. w = , " 2.
5. .. i^ + 21 «2 ^ 91, and V = ± i
6. .: X = u + V = 4, or 3, and .. y = S, or 4, by symmetry.
This system is easily solved in other ways, as by dividing the mem
bers of 1 by the members of 2, etc.
EXERCISES. CXXXV.
Solve the following systems of equations :
1. x^\if = 72. 2. x^^y^ = 97.
X \ y = 6. X \ y = 1.
3. ic* + 7/4 = 337. 4. x^y^ = 219.
X — y :=!. X — y = 3.
5. x^ \ y^ = 4:14,9. 6. x"^ + y^ ^ xy (x + y) = 15i.
x + y = 9. ' cc« + 2/'3(a;' + 2/') = 50.
x^ y^ ^y ^x 2 .
11 = 1 1 1_10
X y ' X y xy
I 4
9. Vic + 2/ + —j= — 4 = 0.
x" + y^ 34
x«/ 15
V
SIMULTANEOUS QUADRATIC EQUATIONS. 295
295. Special devices will frequently suggest themselves,
but it is not worth while to attempt to classify them. A
few are given in the following illustrative problems.
1. Solve the system
1. x^'if' \ xy — ^^^.
2. a;2 + ?/2 = 5.
From 1 we have
3. (xy — 2) {^y + 3) = 0, whence xy = 2, or — 3.
4. Addmg 2 xy = 4 or — 6 to, and subtracting it from, the respec
tive members of 2, we have
5. x2 4. 2 xy + 2/2 = 9, or  1.
x22xy + 2/2 = 1, " 11.
6. .. X + y = ± 3, or ± i,
X  ?/ = ± 1, " ± ViT.
Adding, and dividing by 2,
2 2
i+ VIT i VTT i + vn ^VTT
2,1,1,2, ^ , 2 ' 2 ' 2 '
On account of symmetry, y must have the same values, arranged so
as to satisfy step 6.
I vTT i + Vii  i ViT i+ Vn
.•.y_l,2, 2, 1, ^ , ^ , 2 ' 2 ■
All of the results check.
E.g.^ consider the last ones,
i  Vn  i + vTi
x = ^ , y = ^
Substituting in equation 1,
/ i _ viT  i + vn y t  Vn  i + vn ^
V 2 ' 2 ^ 2 " 2
::3 (_ 3)2 + (_ 3) _ 6 = 9  3  6 = 0.
Substituting in equation 2,
/_i_ViTV /t + ViiV io + 2iVn io2iViT
296 ELEMENTS OF ALGEBRA.
2. Solve the system
1. X = a V£C + y.
2. y = b^x + y.
Adding,
3.
x + y = {a^b)Vx + y,OT
X + 2/  (a + 6) Vx + 2/ = 0, or
4.
Vx + 2/ ( Vx + y  a + 6) = 0.
5. ..
Vx + 2/ = 0, or a + 6.
Substituting in
1 and 2,
X = 0, or a (a + b).
y = 0, " 6 (a + 6).
The results check.
3. Solve the system
1. ic* + icy + 2/* = 481.
2. £c^ + iC2/ + 2/2 = 37. 
Factoring 1, by § 114,
3. (x2 + xy + y2) (a;2  xy + 2/2) _ 431.
4. .. 37 (x2 xy + 2/2) = 481, or
x^xy + y^ = 13.
Subtracting from 2,
5. 2xy = 24, whence xy = 12.
Adding to 2, and subtracting from 4,
6. x^ + 2xy + y'^ = 49.
x22x2/ + ?/2 = 1.
7. .. X + 2/ = ± 7.
X  2/ = ± 1.
8. .. x = 4, 4, 3, 3, y = 3, 3, 4, 4.
Graphs. For the graphic representation of quadratic
equations, and for the discussion of the number of roots
of simultaneous quadratic equations with two unknown
quantities, see Appendix IX. If Appendix VIII has been
studied, this may be taken at this point.
SIMULTANEOUS QUADRATIC EQUATIONS. 297
MISCELLANEOUS EXERCISES. CXXXVI.
Solve the following systems of equations :
1. x^ *+ y^ = h. 2. a;2 — £C2/ + / = 124.
X { y = a. x^ — y^ = 4:4c.
3. x^ + y^ = 3x. 4:. yVy = 17 Vy + Ax.
X2 { y^ = X. x^ = 4 V^/ + 17
X.
5. x'' + y^ = 25xY 6 ^' + 2/'a^2/ = l
12 cci/ = 1 xy = 1.
7. V^ + V^ = 12. 8. 3(x2 + 2/2) = io(a; + 2/).
x^}2f = 3026. 9 (ic^ + 2/0 = 34 (a;* + y^).
9. (ic2 + a;^/ + 2/^ ^^^ 4 2/^ = 185.
(x2  cc?/ + y^) V^M^ ^ 65_
2/ dVx 81 _ Vic
X 2/ ^2/
"^ + 3^^ =
X y
xWy
^ y
^x.
1. Vx^ + 144 + Vy^ + 144 .
= 35.
= 144.
2. Vic2  2/^  Vcc^
+ f + ^
= 0.
Vx + y — Vcc — ?/ = 1.5.
13. X + y — 2Vxy — ■\/x{^=2.
Vx + V^ = 7.
14. V^ + V^ = x — y = x — Vxy + ?/.
15. ic2_ g^^_^9^2_4^_j_12?/=i4.
ic^  2 ic?/ + 3 ^2  4 X + 5 2/ = 53.
298 ELEMENTS OF ALGEBRA.
II. THREE OR MORE UNKNOWN QUANTITIES.
In general, three simultaneous quadratic equations involv
ing three unknown quantities cannot he solved by quadratics.
Many special cases, however, admit of such solution.
The same is true if one equation is linear and the other
two are quadratic, or if one is of a degree higher than 2.
If, however, two are linear and the other quadratic, a
solution is possible by quadratics, as in illustrative prob
lem 2 on p. 299.
Illustrative problems. 1. Solve the system
1. ^xy = 2x + 2y.
2. 2yz = 3y + 2z,
3. 4.ZX = ^z — 3x.
Dividing both members of 1,2, 3, by xy, yz, zx^ respectively, we
have
4.
^hl
5.
 \4
6.
4 = 5 _?•
X z
Adding 5 and 6,
7.
6 = ' + '
X y
Eliminating ?/, with 4 and 7,
8.
3 =  , whence x
X
.■.y=:2, z = 3.
Check.
6 = 2 + 4,
12 = 6 + 6,
12 = 15  3.
SIMULTANEOUS QUADRATIC EQUATIONS. 299
2. Solve the system
1. x + y — 2z = — 9.
2. Sx + 2y\z = 9.
3. a;2 + 2/' + ^' = 30.
Eliminating z from 1 and 2,
9  7 X
4. y = —  — •
o
Eliminating y from 1 and 2,
27 X
5. 2 = •
5
Substituting
4 and 5 in 3, and reducing,
6.
6x2 12a; + 4 = 0, or
(x2)(5x2) = 0.
7. ..
X = 2, or f.
2/ =  1, or H.
z = 5, or 6^^.
Check for the second set of values.
l + H10=9.
1 + f f + 5.V = 9.
A + iM + Hm = Hir~ = 30
EXERCISES. CXXXVII.
Solve the following systems of equations
4:y^ = 9 xz.
2.
ic^ + y + 052/ = 19
x^ = S6yz.
7/2 + ^2 _^ 7/^ = 37
9z^ = 4: xy.
^2 _^ ic2 + ^x  28
x^ 4 y' 5
xyz 6
4.
cc?/^ 9
x^y 2
^2 + a;2 5
xijz 3
^2/^ =2
2/ + ^
2/2 + ^2^13
xyz 18
iC2/^ 6
^ + ic 7
300 ELEMENTS OF ALGEBRA.
III. PROBLEMS INVOLVING QUADRATICS.
EXERCISES. CXXXVIII.
1. The difference of two numbers is 11, the sum of their
squares 901. What are the numbers ?
2. The sum of two numbers is 30, the sum of their
squares 458. What are the numbers ?
3. Pind two numbers whose sum, whose product, and
the difference of whose squares are all equal.
4. The sum of the squares of two numbers is 421, the
difference of the squares 29. What are the numbers ?
5. A certain fraction equals 0.625, and the product of
the numerator and denominator is 14,440. Required the
fraction.
6. The sum of the areas of two circles is 24,640 sq. in.,
and the sum of their radii is 112 in. Required the lengths
of their radii.
7. The product of the numbers 2cc3 and 4=7/6, in which
X and y stand for the tens' digit, x being twice y, is 103,518.
What are the tens' digits ?
8. If a certain twofigure number, the sum of whose
digits is 11, is multiplied by the units' digit, the product is
296. Required the number.
9. Three successive integers are so related that the
square of the greatest equals the sum of the squares of the
other two. Required the numbers.
10. Separate the number 102 into three parts such that
the product of the first and third shall be 102 times the
second, and the third shall be  of the first.
i
SIMULTANEOUS QUADRATIC EQUATIONS. 301
11. Two cubes have together the volume 407 cu. in., and
e sum of one edge of the one and one of the other is
1 in. Required the volume of each.
12. If the product of two numbers is increased by their
the result is 89 ; if the product is diminished by their
m, the result is 51. Required the numbers.
13. One of the sides forming the right angle of a right
gled triangle is f the other, and the area of the triangle
5082 sq. in. Required the lengths of the sides.
14. There are two numbers such that the product of the
st and 1 more than the second is 660, and the product of
e second and 1 less than the first is 609. What are the
numbers ?
15. A sum of money at interest for 5 yrs. amounts to
$4600. Had the rate been increased 1% it would have
amounted to $40 more than this in 4 yrs. Required the
capital and the rate.
16. The product of the numbers £cl7 and 2?/2, in which
X stands for the hundreds' digit of the first and ij for the
tens' of the second, and in which y = x \ 3, is 83,054.
Required the values of x and y.
17. Find a twofigure number such that the product of
the two digits is half the number, and such that the dif
ference between the number and the number with the digits
interchanged is  of the product of the two digits.
18. In going 1732.5 yds. the front wheel of a wagon
makes 165 revolutions more than the rear wheel; but if
the circumference of each wheel were 27 in. more, the front
wheel would, in going the same distance, make only 112
revolutions more than the rear one. Required the circum
ference of each wheel.
302 ELEMENTS OF ALGEBRA.
19. The floor of a certain room has 210 sq. ft., each of
the two side walls 135 sq. ft., and each of the two end walls
126 sq. ft. Eequired the dimensions of the room.
20. A certain cloth loses ^ in length and ^i^ in width by
shrinking. Eequired the length and width of a piece which
loses 3.68 sq. yds., and which has its perimeter decreased
3.4 yds. by shrinking.
21. A rectangular field is 119 yds. long and 19 yds. wide.
How much must the width be decreased and the length
increased in order that the area shall remain the same
while the perimeter is decreased 24 yds. ?
22. Two points move, each at a uniform rate, on the arms
of a right angle toward the vertex, from two points 50 in.
and 136.5 in., respectively, from the vertex. After 7 sees.
the points are 85 in. apart, and after 9 sees, they are 68 in.
apart. Eequired the rate of each.
23. There are two lines such that if they are made the
sides of a rightangled triangle the hypotenuse is 17 in. ;
but if one be made the hypotenuse and the other a side, the
remaining side is such that the square constructed upon it
contains 161 sq. in. How long are the two lines ?
24. There is a fraction whose numerator being increased
by 2 and denominator diminished by 2, the result is the
reciprocal of the fraction ; but if the denominator is in
creased by 2 and the numerator diminished by 2, the result
is ly^ less than the reciprocal. Eequired the fraction.
25. If the numerator of a fraction is decreased by 2, and
the new fraction added to the original one, the sum is If ;
if the denominator is decreased by 2, and the new fraction
added to the original one, the sum is 2^l. Eequired the
fraction.
SIMULTANEOUS QUADRATIC EQUATIONS. 303
REVIEW EXERCISES. CXXXIX.
2. Form the equation whose roots are 0, i, i.
3. Solve w" /a^ = a% b^b^ = (b^y, d" I d" = l/c».
4. Solve X \ y =. a ^b, x /a — y /b = a /b — b / a.
5. Solve ^y + ^z = 11, 3^ + 6ic = 9, ^x^y = 4..
6. Construct an integral quadratic function of x such
that /(2) = and /(3) = 0.
7. Simplify
\ {x""^ ■ x^'Y • (£c« i x'^Yl/ \ (cc«x^)« = (x^' + 'Yl.
Solve the following :
8. x^ { x^y^ + y^ = 61.
42
x^ — xi/ 4 y^ =
" "^ xy
9. X { y = 2xy = x^ — y"^.
10. X \ y \ (x { y)^  12 = 0.
x^ + 2/'  45 = 0.
3
I
11. ^—4^ — .
x\y X — y 4
2x3 + 6V = ^(^'2/T
12. (3 cc + 4 ?/) (7 a;  2 2/) + 3 a; + 4 ?/ = 44.
(3 X + 4 2/) (7 a;  2 2/)  7 X + 2 2/ = 30.
13. 17 (cc + 7/)"^  7 (x + yfx^ = l^x{x.\ y)"^.
(x  y)^ = yl
CHAPTER XVI.
INEQUALITIES.
MAXIMA AND MINIMA.
296. Having given two real and unequal numbers, a and
h, a — b cannot be zero, li a — h i^ positive, a is said to
be greater than b ; if negative, a is said to be less than b.
E.g., 3 > 2 because 3 — 2 is positive,
2>3" 2(3)is positive,
• 8<2" 3(2)is negative.
If a > 0, then a is positive, and if
a<0, " " " negative.
297. The inequalities a> b, c> d are called inequalities
in the same sense, and similarly for a<b, c< d. But a>b,
G<d are called inequalities ih the opposite sense, and similarly
for a Kb, G>d.
298. In this chapter the letters used to represent numbers
will be understood to represent positive and real finite
numbers, except as the minus sign indicates a negative
number.
299. Just as we distinguish two classes of equalities,
(1) equations and (2) identities, so in inequalities we have
two classes, (1) those which are true only for particular
values of a quantity called the unknown quantity, and (2)
those which are true for all values of the letters.
JB.gr. , X + 2 > 3 is true only when a > 1, but a + & > & is always true.
304
INEQUALITIES.
305
300. If a variable quantity, x, cannot be greater than a
constant, m, but can equal it or approach indefinitely near
it in value, then m is called the maximum value of x.
Similarly, if ic < m but can equal it or approach indefi
nitely near it in value, then m is called the minimum value
of X.
E.g., (x — 1)^<^0, because it is the square of a real quan
tity and hence cannot be negative. But {x — iy can equal
by letting a; = 1. Hence, is the minimum value of
(X  1)^.
Since we shall need the subject of inequalities in only a
few cases in our subsequent work, we shall present but a
few of the fundamental theorems. It is evident, however,
that the subject is an extensive one, covering simple inequali
ties, quadratic inequalities, etc., together with simultaneous
inequalities corresponding to simultaneous equations.
301. The axioms of inequalities. The following axioms
have already been assumed and used :
Ax. 4. If equals are added to unequals, the sums are
unequal in the same sense.
Ax. 5. If equals are subtracted froin unequals, the re
mainders are unequal in the same sense.
These are easily demonstrated, thus :
1. If a > 6, then a — his, positive.
2. Then a 'b = a \k k h
= {a + k){k^h)
and this expression is positive.
3. .. a + k>h\k.
Similarly for ax. 5.
§296
§296
Theorems. Three important theorems of inequalities will
now be proved, the first two corresponding to axs. 6 and 8.
306 ELEMENTS OF ALGEBRA.
302. Theorem. If unequals are multiplied hy equals, the
products are unequal in the same or in the opposite sense,
according as the multiplier is positive or negative.
Proof. 1. If a> h, then a — h is positive. § 296
2. Then k(a — b) is positive,
and —k(a — b) is negative. § 296
3. .*. ka — kh is positive,
and — A;a —(— >t&) is negative.
4. .'. ka^kb,
and ~ka< kb. § 296
In this discussion the multiplier is supposed to be neither zero nor
infinite.
303. Theorem. i/'a>b, then a'">b'".
Proof. 1. a — b'\s positive. § 296
2. .. {a^^ + a'^^ H \ ab'''^ + b'^'^) {a — b)
is positive, because the multiplier is evidently
a positive quantity.
3. .".a"* — ^'" is positive, because this is the prod
uct of the expressions.
4. .. a"'>^"'. § 296
304. Theorem. // a ^^ b, a^ + b^ > 2 ab.
Proof. 1. (<x — ^)^ > 0, because (a — by is positive, being
the square of a real number. It is not 0, for
a4^b.
2. .. a^2ab + b''>0.
3. .. a'' + b^>2ab.
Evidently a'^ + 6'^ = 2 a6, if a = 6.
INEQUALITIES. 307
Illustrative problems. 1. Prove that ic^ > 2 x — 1, if aj^tl.
We have x2 + 1 > 2 x, by § 304.
2. x^'^*^ + yP^'^> x^Y ^ x'^yp, if x4^y.
1. This is true if xP + 'i — xPyi + yp + 'i — x^yP is positive.
2. Or if xP (x9 — yi) — yP {xi — yi) is positive.
3. Or if {xP — yP) {xi — yi) is positive.
4. But both factors are positive if x > ?/, and both factors are nega
ive if X < ?/, and in either case their product is positive.
3. Which is greater, 2 + V3, or 2.5 + V2 ?
1.
2 + Vs = 2.5 + ^2, according as
2.
7 + 4V3I8I + 5V2, squaring.
§303
3.
Or as
 1^ + 4 V3 1 5 V2.
Ax. 5
4.
Or as
49/g  10 Vs 1 50.
§303
5.
Or as
10V3A
Ax. 6
6. But a negative number is less than a positive one.
..2 + V3<2.5 + V2.
CC 1 1 iC
4. Solve the inequality 2ic —  + >3aj —  + •
1. 12x2x + 3>18x2 + x. §302
2. .. — 9x>5. Ax. 6
3. .. 3; < f , and f is the maximum value. § 300
Check. If X = f , the inequality becomes an equation. If x > f , the
sense of the inequality is reversed.
5. Solve the inequality a^ — 5 a? + 6 < 0.
1. (x — 2) (x — 3) < 0, and hence is negative.
2. The smaller factor, x — 3, is negative, and the other positive.
3. .. x>2 and x<3, or 2<x<3.
308 ELEMENTS OF ALGEBRA.
6. Show that the minimum value of x^ — 8 cc + 22 is 6.
1. Let x^ — 8x }22 = y, in which we have to find the minimum
value of y.
2. Then x^  8 x + 22  ?/ = 0. .
3. .. X = 4 ± Vy — 6, and y cannot be less than 6 without making
X complex.
7. Divide the number 6 into two parts such that their
product shall be the maximum.
1. Let X and 6 — x be the parts.
2. Then x (S — x) = ?/, in which we have to find the maximum value
oiy.
3. Solving for x, x = 3 ± v 9 — y, and y cannot be greater than 9
without making x complex.
4. When y = 9, x = 3 ; .. the parts are 3 and 3.
Check. 33 = 9; but 2 (6  2) = 2 • 4 = 8, a smaller number.
EXERCISES. CXL.
1. What is the nature of the inequality resulting from
subtracting unequals from equals ? Prove it.
2. Investigate the addition of unequals to unequals.
3. Also the subtraction of unequals from unequals.
4. Show that the maximum value of 4 a; — a:;^ is 4, and
that 2 is the value of x which makes this f(x) a maximum.
5. If f(x) = x^ { X \ 1, show that x = — 0.5 renders
f(x) a minimum, and find the minimum.
6. Prove that, in general, x^ \ l>x^ \ x. What is
the exception ?
7. Also that (x + yy>4: xy.
8. Solve the inequality x'^ \hx'> — ^.
INEQUALITIES. 309
9. Prove that {a + h){b \ c) (c + a)>% abc.
10. Prove that the mininmrn value of a;^ — 10 cc + 35 is 10.
11. Solve the inequality 5xH2>3cc +  — 7. Check
[the result.
X — 3
12. Solve the inequality > 0. Check.
13. Required the length of the sides of the maximum
rectangle of perimeter 16.
14. Prove that if the sum of two factors is k, a constant,
^he maximum value of their product is k^/A.
15. Show that if a square is inscribed in a square whose
area is 16, its corners lying on the sides of the larger square,
its area <: 8.
16. If a, h, G are three numbers such that any two are
together greater than the third, then
a'' + h^ + c^ <2 ah + 2hc + 2 ca.
17. Solve the inequality x^ — 3x< 10.
18. Solve the inequality x(x, — 10) < 11.
19. Find the maximum value of 8 cc — x^, and also the
value of X that renders this f{x) a maximum.
20. Find the minimum value of x(x \ 10), and also the
value of X that renders this f(x) a minimum.
21. Required the area of the largest rectangle having the
perimeter 20 inches. How do the sides compare in length ?
22. Required the area of the largest rectangle having the
perimeter p inches. How do the sides compare in length ?
CHAPTER XVII.
RATIO, VARIATION, PROPORTION.
I. RATIO.
305. The ratio of one number, a, to another number, h, of
the same kind, is the quotient  •
S2 2
Thus, the ratio of $ 2 to $5 is — , or , or 0.4, but there is no
$5 5
ratio of $2 to 5 ft., or $10 to 2. Here, as elsewhere in algebra, how
ever, the letters are understood to represent pure (abstract) number.
A ratio may be expressed by any symbol of division, e.g.,
by the fractional form, by ^, by /, or by : ; but the symbols
generally used are the fraction and the colon, as y? or a:b.
306. In the ratio a:h, a \b called the antecedent and h the
consequent.
307. The ratio h:a \& called the inverse of the ratio a : b.
308. If two variable quantities, x, y, have a constant
ratio, r, one is said to vary as the other.
E.g., the ratio of any circumference to its diameter is ;r = 3.14159;
hence, a circumference is said to vary as its diameter.
X
If  = r, then x = ry. The expression " x varies as y "
is sometimes written xccy, meaning that x — ry.
li X = r ' ■> ic is said to vary inversely as y.
310
If RATIO, VARIATION, PROPORTION. 311
309. If two variable quantities, x, ij, have the same ratio
as two other variable quantities, x', y\ then x and y are said
to vary as x' and y'. And if any two values of one variable
quantity have the same ratio as the corresponding values
of another variable quantity which depends on the first,
then one of these quantities is said to vary as the other.
E.g., the circumference c and diameter d of one circle have the
same ratio as the circumference c' and diameter d' of any other circle ;
hence, c and d are said to vary as c' and d\
If two rectangles have the same altitude, their areas depend on
tlieir bases; and since any two values of their bases have the same
ratio as tlie corresponding values of their areas, their areas are said to
vary as their bases.
310. Applications in geometry. Similar figures may be
described as figures having the same shape, such as lines,
squares, triangles whose angles are respectively equal,
circles, cubes, or spheres. It is proved in geometry that
in two similar figures
1. Any two corresponding lines vary as any other two
correspo7iding lines.
2. Corres20O7iding areas vary as the squares of any two
corresponding lines.
3. Corresponding volumes vary as the cubes of any two
corresponding lines.
E.g., in the case of two spheres, the circumferences vary as the
radii, the surfaces vary as the squares of the radii, the volumes vary
as the cubes of the radii.
These facts are easily proved. Let s, s' stand for the surfaces of
two spheres of radii r, r', respectively. Then we know from mensu
ration that
8 = 4 7irr"2, and s' = 4 7rr'2,
s _ 4 Ttr"^ _ r^
Hence, the surfaces vary as the squares of the radii. In like manner
the volumes might be considered.
312 ELEMENTS OF ALGEBRA.
Illustrative problems, 1. If the ratio of x^ to 3 is 27, find
the value of x.
•.• — = 27, .. x2 = 3 • 27 = 81, .. X = ± 9, and each value checks.
o
2. If a sphere of iron weighs 20 lbs., find the weight of
a sphere of iron of twice the surface.
1, Let ri, ri be the respective radii.
2. Then 4 itr^ = i • 4 Ttr<^^ because the surface of a sphere = 4 itr'^.
(p. 311.)
3. ... *:?=V2.
4. And ••• the volumes (and hence the weights) vary as the cubes of
the radii (§ 310), and •.• *^ = (V2)3 = 2 V2.
5. .'. the second sphere weighs 2 v2 times as much as the first.
2 V^ • 20 lbs. = 56.57 lbs., nearly.
EXERCISES. CXLI.
1. The ratio of 625 to x^ is 5. Find x.
2. Find X in the following ratios :
(a) 4.:x^=9. (b) ^c^ : 27 = 300. (c) cc = y^^ : x.
, t^ X _ . . 0\)
3. Find X in the following ratios :
(a) ==2.4. (b)s = y («) 13^ = 49 •
(d) 7:x = 4.9. (e) (r^ : 5 = V
4. One cube is 1.2 times as high as another. Find the
ratio of (1) their surfaces, (2) their volumes.
5. The surfaces of a certain sphere and a certain cube
have the same area. Find, to 0.01, the ratio of their vol
umes.
I .„...__ ™
^H^ 311. Applications in business. Of the numerous applica
^^ftions of ratio in business, only a few can be mentioned, and
f^^ot all of these commonly make use of the word " ratio."
In computing interest, the simple interest varies as the
time, if the rate is constant ; as the rate, if the time is con
stant ; as the product of the rate and the . number repre
senting the time in years (if the rate is by the year), if
neither is constant.
I.e., for twice the rate, the interest is twice as much, if the time is
constant ; for twice the time, the interest is twice as much, if the rate
is constant ; but for twice the time and 1.5 times the rate, the interest
is 2 • 1.5 times as much.
The common expressions " 2 out of 3," " 2 to 5," " 6 per
cent " (merely 6 out of 100) are only other methods of stat
ing the following ratios of a part to a whole, f , f , yf ^, or
the following ratios of the two parts, ^, f , ^\.
E.g., to divide $100 between A and B so that A shall receive 2 out
of every $3, is to divide it into two parts
(1) having the ratio 2 : 1, or
(2) so that A's share shall have to the whole the ratio 2 : 3, or
(3) so that B's share shall have to the whole the ratio 1 : 3.
EXERCISES. CXLII.
1. Divide $1000 so that A shall have $7 out of every 8.
2. Divide $500 between A aHd B so that A shall have
$0.25 as often as B has $1.25.
3. The area of the United States is 3,501,000 sq. mi.,
and the area of Russia is 8,644,100 sq. mi. Express the
ratio of the former to the latter, correct to 0.01.
4. The white population of the United States in 1780
was 2,383,000 ; in 1790, 3,177,257 ; in 1880, 43,402,970 ;
in 1890, 54,983,890. What is the ratio of the population
in 1790 to that in 1780 ? in 1890 to that in 1880 ?
314 ELEMENTS OF ALGEBRA.
312. Applications in physics, (a) Specific gravity. The
specific gravity of any substance is the ratio of the weight
of that substance to the weight of an equal volume of some
other substance taken as a standard.
In the case of solids and Uquids, distilled water is usually taken as
the standard. Thus, the specific gravity of mercury, of which 1 1
weighs 13.596 kg, is 13.596, because a liter of water weighs 1 kg, and
13.596 kg :1 kg = 13.596.
In the case of gases either hydrogen or air is usually the standard.
The following table will be needed for reference :
Specific Gravities.
Mercury, 13.596. Silver, 10.5.
Nickel, 8.9. Gold, 19.3.
Weights of Certain Substances.
1 1 of water, 1 kg. 1 cm' of water, 1 g.
1 cu. ft. of water, about 62. 5 lbs.
Example. What is the weight of 1 cu. in. of copper ?
1. 1 cu. ft. of water weighs 62.5 lbs.
2. .. 1 cu. in. of water weighs 62.5 lbs = 1728.
3. .. 1 cu. in. of copper weighs 8.9 • 62.5 lbs. f 1728, or 5.15 oz.
EXERCISES. CXLIII.
1. What is the weiglit of a cubic foot of gold ?
2. What is the weight of 1 cm^ of nickel ? of silver ?
3. The specific gravity of ice is 0.92, of seawater 1.025.
To what depth will a cubic foot of ice sink in seawater ?
4. From ex. 3, how much of an iceberg 500 ft. high would
show above water, the crosssection being supposed to liave
a constant area ?
v"
1
A
F
A 1
P
f
^
P
A'
P'
B~
w'
^
1
w
P
RATIO, VARIATION, PROPORTION. 315
313. (b) Law of levers. If a bar, ^^, rests on a fulcrum,
^F, and has a weight, w, at .
f.^, then by exerting enough
[pressure, p, at A the weight
,can be raised. In the first
igure the pressure is down
ward (positive pressure) ; in
^he second it is upward (neg
itive pressure).
There is a law in physics that, if p', tv' represent the
lumber of units of distance AF, FB, respectively, and p,
the number of imits of pressure and weight, respectively,
shen Pp[_.
ww'
In the first figure p, to, p% w' are all considered as positive ; in the
jcond figure p is considered as negative because the pressure is up
i^ard, and \o' is considered as negative because it extends the other
my from F. Hence, the ratio pp' : low' = 1 in both cases.
Example. Suppose AF = 25 in., FB = 14 in., in the
first figure. What pressure must be applied at A to raise
a weight of 30 lbs. at 5 ?
25 w
1. By the law of levers =— = 1.
^ 1430
2. .. p = ^ = 10.8, and .. the pressure must be 16.8 lbs.
25
EXERCISES. CXLIV.
1. Two bodies weighing 20 lbs. and 4 lbs. balance at the
ends of a lever 2 ft. long. Find the position of the fulcrum.
2. The radii of a wheel and axle are respectively 4 ft.
[and 6 in. What force will just raise a mass of 56 lbs., fric
ition not considered ?
B16 ELEMENTS OE ALGEBRA.
REVIEW EXERCISES. CXLV.
1. In each figure on p. 315, what must be the distance
AF in order that a pressure of 1 kg may raise a weight of .
100 kg 3 dm from i^'? I
2. If a sphere of lead weighs 4 lbs., find the weight of a
sphere of lead of (1) twice the volume, (2) twice the sur
face, (3) twice the radius. '
3. A nugget of gold mixed with quartz weighs 0.5 kg ;
the specific gravity of the nugget is ^.h^ and of quartz 2.15.
How many grams of gold in the nugget ?
4. A vessel containing 1 1 and weighing 0.5 kg is filled
with mercury and water ; it then weighs, with its contents,
3 kg. How many cm^ of each in the vessel ?
5. What pressure must be exerted at the edge of a door
to counteract an opposite pressure of 100 lbs. halfway from
the hinge to the edge ? onethird of the way from the hinge
to the edge ?
6. Explain Newton's definition of number : Number is
the abstract ratio of one quantity to another of the same
kind. What kinds of numbers are represented in the fol
lowing cases : 5 ft. : 1 ft., 1 ft. : 5 ft., the diagonal to the
side of a square, the circumference to the diameter of a
circle ?
7. The depths of three artesian wells are as follows :
A 220 m, B 395 m, C 543 m; the temperatures of the
water from these depths are: A 19.75° C, B 25.33° C,
C 30.50° C. Rom these observations, is it correct to say
that the increase of temperature is proportional to the
increase of depth ? If not, what should be the tempera
ture at C to have this law hold ?
RATIO, VARIATION, PROPORTION. 317
The Theory of Ratio.
314. A ratio is called a ratio of greater inequality, of
equality, or of less inequality, according as the antecedent
is greater than, equal to, or less than the consequent.
315. Theorem. A ratio of greater inequality is dimin
ished, a ratio of equality is unchanged in value, and a
ratio of less inequality is increased by adding any positive
quantity to both terms.
Given the ratio a : b, and p any positive quantity.
J.KI yi
uvc
b+p > b
clUOUi Uillg
«/» u ,
< ^'
Proof
1.
a \p < a
b+p> b
according
as
ab + pb ^ ab \ ap.
§ 302
ax. 6
2.
Or as 2)h = ap, or as ^ =
a.
§301
3.
I.e., as a^b.
316.
q n
Theorem. // r = 
•^ b d
e
~f '
then
each of
these
, a + c He + ••
ratios equals ^_^^_^^_^^^
Proof
1.
Let T = k Then k = ^ =
b a
e
f~
•••.
2.
•*•
a = kb,
c = kd,
e = kf
3.
.'. a\ c \ e + ■
• ^ = k{b +
d+f +■■■).
Ax. 2
4.
a \ c { e \ •
' ' b + d\f\
c
e
Ax. 7
318 ELEMENTS OF ALGEBRA.
EXERCISES. CXLVI.
1. Prove that the product of two ratios of greater in
equality is greater than either.
2. Consider ex. 1 for two ratios of equality ; of less
inequality. Then state the general theorem and prove it.
3. Find the value of x, knowing that if x is subtracted
from both terms of the ratio ^ the ratio is squared.
4. Is the value of a ratio changed by raising both
terms to the same power ? State the general theorem
and prove it.
5. Prove (or show that it has been proved) that the
value of a ratio is not changed by multiplying both terms
by the same number.
6. As in § 315, consider the effect of subtracting from
both terms of a ratio any positive number not greater than
the less term. State the theorem and prove it.
a ^ 5 b a + 6b ^
7. Which is the greater ratio,
a{ 6b a + 7b
8. Which is the greater ratio, —^j or ~ ?
y — 2x 3y — 2x
« TTTi, •■uo.t i J. a \ b \ G a — b \ c ^
9. Which IS the greater ratio, ; — — ? or ; ?
a — b — c a ib — c
.r.T£^ ^ ^ ^y ^ a^ \ b"^ + c^ ab \bc + cd
10. It 7 =  = ) prove that — ; ; — — •
b G d ^ ab\bG\cd b'^ + c^ \ d^
.. ^f ^ ^ ^ 1 ' 4.u^7 3a45c — 6e
11. It J =  =  = k, prove that k =
b d f "'^^ ^"^ 3b + 5d6f
a
b
12. If  > ) the letters standing for positive numbers,
prove that >V^i:p^>^
RATIO, VARIATION, PROPORTION. 319
II. VARIATION.
317. It has already been stated (§ 308) that the expres
lons ^^x varies as y," ^' x varies inversely as y," simply
lean that the ratios x:y, x\) are respectively equal to
)ine constant. These are merely special cases of x =/(?/);
)Y x:y — k reduces to a:; = ky, whence ic is a function of y\
1 k
milarly x\ — k reduces to a; = > whence cc is a function
■ y y
Although there is nothing in the theory of variation which
not substantially included in the theory of ratio, the
)hraseology and notation of the subject are so often used
■ in physics as to require some further attention.
Two illustrations from physics will be given in this con
nection, the one relating to the pressure of gases and the
other to electricity. While neither requires much algebra
for its consideration, each offers an excellent illustration of
the use of variation in physics. No preliminary knowledge
of physics is necessary, however, to the work here given.
318. Boyle's law for the pressure of gases. It is proved in
physics that if j^ is the number of units of pressure of a
given quantity of gas, and v is the number of units of
volume, then p varies inversely as v when the temperature
remains constant.
This law was discovered in the seventeenth century by Robert Boyle.
E.g.^ if the volume of a gas is 10 dm^ under the ordinary pressure
of the atmosphere (" under a pressure of one atmosphere "), it is
 as much when the pressure is 2 times as great,
i u u u u u u ^ u
n ^
n times " " ' " " " "  "
n
the temperature always being considered constant.
320 ELEMENTS OF ALGEBRA.
Example. A toy balloon contains 3 1 of gas when exposed
to a pressure of 1 atmosphere. What is its volume when
the pressure is increased to 4 atmospheres ? decreased to ^
of an atmosphere ?
1. •.• the volume varies inversely as the pressure, it is ^ as much
when the pressure is 4 times as great.
2. Similarly, it is 8 times as much when the pressure is i as great.
3. .. the volumes are 0.75 1 and 24 1.
EXERCISES. CXLVII.
1. If a cylinder of gas under a certain pressure has its
volume increased from 20 1 to 25 1, what is the ratio of the
pressures ?
2. A certain gas has a volume of 1200 cm^ under a pres
sure of 1033 g to 1 cm^. rind the volume when the pres
sure is 1250 g.
3. A cubic foot of air weighs 570 gr. at a pressure of 15
lbs. to the square inch. What will a cubic foot weigh at a
pressure of 10 lbs. ?
4. Equal quantities of air are on opposite sides of a
piston in a cylinder that is 12 in. long ; if the piston moves
3 in. from the center, find the ratio of the pressures. Draw
the figure.
5. A liter of air under ordinary pressure weighs 1.293 g
when the barometer stands at 76 cm. Find the weight when
the barometer stands at 82 cm, the weight varying as the
height of the barometer.
6. If the volume of a gas varies inversely as the height
of the mercury in a barometer, and if a certain mass occu
pies 23 cu. in. when the barometer indicates 29,3 in., what
will it occupy when the barometer indicates 30.7 in. ?
RATIO, VAKIATION, PROPORTION.
321
1319. Problems in electricity. The great advance in elec
■icity in recent years renders necessary a knowledge of
! such technical terms as are in everyday use.
When water flows through a
pipe some resistance is offered
due to friction or other impedi
ment to the flow of the water.
A certain quantity of water
flows through the pipe in a second,
and tliis may be stated in gallons
or cubic inches, etc.
A certain pressure is necessary
to force the water through tlie
l)ipe. This pressure may be meas
ured in pounds per sq. in., kilo
grams per cm2, etc.
Hence, in considering the water
necessary to do a certain amount
of work (as to turn a waterwheel)
it is necessary to consider not
merely the pressure, for a little
water may come from a great
height, nor merely the volume,
nor merely the resistance of the
pipe ; all three must be consid
ered.
When electricity flows through
a wire some resistance is offered.
This resistance is measured in
ohms. An ohm is the resistance
offered by a column of mercury
1 mm2 in crosssection, 106 cm
long, at 0°C.
A certain quantity of electric
ity flows through the wire. This
quantity is measured in amperes.
An ampere is the current neces
sary to deposit 0.001118 g of silver
a second in passing through a cer
tain solution of nitrate of silver.
A certain ^^t'essure is necessary
to force the electricity through the
wire. This pressure is measured
in volts. A volt is the pressure
necessary to force 1 ampere
through 1 ohm of resistance.
Hence, in considering the elec
tricity necessary to do certain
work it is necessary to consider
not merely the voltage, for a little
electricity may come with a high
pressure, nor merely the amper
age, nor merely the number of
ohms of resistance ; all three must
be considered.
The names of the electrical units mentioned come from the names
of three eminent electricians, Ohm, Ampere, and Volta.
322 ELEMENTS OF ALGEBKA.
320. It is proved in physics that the resistance of a wire
varies directly as its length and inversely as the area of its
crosssection.
That is, if a mile of a certain wire has a resistance of 3.58 ohms,
2 mi. of that wire will have a resistance of 2 • .3.58 ohms, or 7.16 ohms.
Also, 1 mi. of wire of the same material but of twice the sectional
area will have a resistance of i of 3.58 ohms, or 1.79 ohms.
From these laws and definitions, the most common prob
lems and statements concerning electrical measurements
will be understood.
EXERCISES. CXLVIII.
1. If the resistance of 700 yds. of a certain cable is
0.91 ohm, what is the resistance of 1 mi. of that cable ?
2. The resistance of a certain electric lamp is 3.8 ohms
when a current of 10 amperes is flowing through it. What
is the voltage ?
3. If ibhe resistance of 130 yds. of copper wire ^^ in. in
diameter is 1 ohm, what is the resistance of 100 yds. of
^1^ in. copper wire ?
4. The resistance of a certain wire is 9.1 ohms, and the
resistance of 1 mi. of this wire is known to be 1.3 ohms.
Required the length.
5. Three arc lamps on a circuit have a resistance of
3.12 ohms each; the resistance of the wires is 1.1 ohms,
and that of the dynamo is 2.8 ohms. Find the voltage for
a current of 14.8 amperes.
6. The resistance of a dynamo being 1.6 ohms, and the
resistance of the rest of the circuit being 25.4 ohms, and
the electromotive force being 206 volts, find how many
amperes flow through the circuit.
RATIO, VARIATION, PROPORTION. 323
Theory op Variation.
If xccj and y QC z, then xcc z.
xccy, then x = hy. § 308
2/ cc 2;, then y = k'z. § 308
X = ky = kk'z. Substn.
xc^z. § 308
Note that in step 2 we cannot use the same constant as in step 1.
E.g., if the edge of a cube varies as the diagonal of a face, and the
diagonal of a face varies as the diagonal of the cube, then the edge
must vary as the diagonal of the cube.
321.
Theorem
Proof
1.
If
2.
If
3.
.'.
4.
322. Theorem.
//x
ocyz, the?i y
ocx/z.
Proof. 1.
X = kyz.
(Why ?)
2. ..
y = \x/z.
Ax. 7
3. ..
y oc X / z.
§308
E.g.., if the area of a rectangle varies as the product of the (numbers
representing the) base and altitude, then the base varies as the quotient
of the (number representing the) area divided by the (number repre
senting the) altitude.
323. Theorem. If w ocx and y oc z, then wy oc xz.
Proof. 1. w = kx and y = k'z. (Why ?)
2. .. wy = kk'xz. (Why?)
3. .". wyccxz. (Why?)
E.g., it the surface of a sphere varies as the square of the diameter,
and ^ of the radius varies as the radius, then the product of the surface
and ^ of the radius varies as the product of the radius and the square
of the diameter.
324 ELEMENTS OF ALGEBRA.
324. Theorem. If xccj when z is constant, and if x ccz
when J is cofistant, then x oc yz when both j and z vary.
To understand this statement consider a simple illustration : The
area of a triangle (p. 172) varies as the altitude when the base is con
stant, and as the base when the altitude is constant ; but it varies as
the product of their numerical values when both base and altitude vary.
Proof. 1. Let the variations of y and z take place sepa
rately.
2. Let X change to x' when y changes to y\ z
remaining unchanged. Then
■ xccy,
x^^y_
x' y'
3. Let x' change to x" when y' remains unchanged
and z changes to z'. Then
x' z
x" z'
. X x' X ^ y z yz
4. .. — • — > or —J equals —, ? or ^^
x' x" x" y' z' y'z'
5. I.e., X changes to x" as yz changes to y'z', or
x oc yz.
Illustrative problems. 1. If xccy, and if x = 2 when
y = 5, find x when y = 11.
•.• xccy means that x — ky, .. 2 = fc • 5, and k = f. .. x = ly.
When ?/ = 11, X = I • 11 = 4.4.
2. The volumes of spheres vary as the cubes of their
radii. Two spheres of metal are melted into a single
sphere. Kequired its radius.
1. vi = kr^ and v' = kr'^. § 308
2. .. the volume of the single sphere is k (r^ + r'^).
3. Call v" this volume, and r" the radius ; then
v" = k{r^ + r'^) = kr"^.
4. .. r"^ = r3 if r'3, and .. r" = {r^ + r'^)K
RATIO, VARIATION, PROPORTION. 325
EXERCISES. CXLIX.
1. li xocz and y ^z, prove that xy oc z"^.
2. li xccz and y ccz, prove that x + y ccz.
3. li x { y ccx — y, prove that x^ \ y^ cc xy.
4. li w <x.x and y <^z, prove that w /y otz x/z.
5. If 10 0? + 3 ?/ = 7 cc — 4 2/, show' that xocy.
6. If a^ Gc h^, and if cc = 3 when y = 5, prove that
7. If a? oc ?/, and if cc = a when y = b, find the value of
X when y = c.
8. If X GC ?/, and if cc = 7 when ?/ = 11, find the value
of X when y = 7.
9. If xccy, prove that pxcc2)y, p being either a con
stant or a variable.
10. What is the radius of the circle which is equal to
the sum of two circles whose radii are 3 and 4, respectively ?
11. Prove that the volume of the sphere whose radius is
6 is equal to the sum of the volumes of three spheres whose
radii are 3, 4, and 5, respectively.
12. The illumination from a given source of light varies
inversely as the square of the distance. How much farther
from an electric light 20 ft. away must a sheet of paper be
removed in order to receive half as much light ?
13. Kepler showed that the squares of the numbers rep
resenting the times of revolution of the planets about the
sun vary as the cubes of the numbers representing their
distances from the sun. Mars being 1.52369 as far as the
earth from the sun, and the time of revolution of the earth
being 365.256 das., find the time of revolution of Mars.
826 ELEMENTS OF ALGEBRA.
III. PROPORTION.
325. The equality of two ratios forms a proportion.
Thus, f = f, a : b = c : d, x/y = m/n, are examples of proportion.
The symbol : : was formerly much used for = .
326. There may be an equality of several ratios, as
1:2 = 4:8 = 9:1 8, the term continued proportion being
applied to such an expression.
Three quantities, a, b, c, are said to be in continued proportion
when a : b = b : c.
327. There may also be an equality between the products
of ratios, as § • f =  • y, such an expression being called a
compound proportion.
328. In the proportion a : b = c : d, a, b, c, d are called
the terms, a and d being called the extremes and b and c the
means. The term d is called the fourth proportional to a, b, c.
329. In the proportion a:b = b:c, b is called the mean
proportional between a and c, and c is called the third pro
portional to a and b.
330. If one quantity varies directly as another, the two
are said to be directly proportional, or simply proportional.
E.g., at retail the cost of a given quality of sugar varies directly as
the weight; the cost is then proportional to the weight. Thus, at
4 cts. a pound 12 lbs. cost 48 cts. , and 4 cts. : 48 cts. = 1 lb. : 12 lbs.
331. If one quantity varies inversely as another, the two
are said to be inversely proportional.
E.g., in general, the temperature being constant, the volume of a
gas varies inversely as the pressure, and the volume is therefore said
to be inversely proportional to the pressure.
RATIO, VARIATION, PROPORTION.
327
Illustrative problems. 1. What are the mean proportionals
between 5 and 125 ?
1. . •5 = ^.
X 125
2. .. 625 = x2.
3. .. ± 25 = X, and both results check.
2. What is the fourth proportional to 1, 5, 9 ?
1. 1=1
5 X
2. .. a; =3 5 • 9 = 45, and the result checks.
3. What number must be added to the numbers 1, 6, 7,
18 so that the sums shall form a proportion ?
1 +x 7 +x
1.
Check. y5^
6 + X 18 + X
18 + 19 X + x2 = 42 + 13 X + x2.
x = 4.
EXERCISES. CL.
1. State which of the following, other things being equal,
are directly and which are inversely proportional :
(a) Volume of gas, pressure.
(b) Price of bread, price of wheat.
(c) Distance from fulcrum, weight.
(d) Amount of work done, number of workers.
2. Given 1.43 : x = 4.01 : 2, find the value of x.
3. Also m27 :x = x:4:S.
4. What are the mean proportionals between 1 and 1 ?
5. Also between 1 + ^ and 2(1 — i), where i = v — 1 ?
6. What is the third proportional to 1 + i and — 2 ?
328 ELEMENTS OF ALGEBRA.
332. The applications of proportion are found chiefly in
geometry and physics. Other methods are now generally
employed for business problems.
In the two illustrative examples below, the first three
steps are explanatory of the statement of the proportion
and may be omitted in practice. In the first problem the
ratios are written in the fractional form in order that the
reasons involved may appear more readily.
Illustrative problems. 1. The time of oscillation of a
pendulum is proportional to the square root of the number
representing its length ; the length of a 1sec. pendulum
being 39.2 in., what is the length of a 2sec. pendulum ?
1. Let X = the number of inches of length.
2. Then = the ratio of the lengths.
39.2 ^
3. And I = the ratio of the corresponding times of oscillations.
4. '.• the time is proportional to the square root of the number
representing the length,
. ^ _2
V39y2 ~ 1 '
5. .. ^ = , whence x = 39.2 • 4 = 156.8. Axs. 8, 6
39.2 r
6. •.' X = the number of inches, .. the pendulum is 156.8 in. long.
2. A mass of air fills 10 dm^ under a pressure of 3 kg
to 1 cm^. What is the space occupied under a pressure of
5 kg to 1 cm^, the temperature remaining constant ?
1. Let X =the number of dm^ under a pressure of 5 kg to 1 cm^.
2. Then x : 10== the ratio of the volumes.
3. And 5 : 3 = the ratio of the corresponding pressures.
4. •.• the volume is inversely proportional to the pressure,
.. X : 10 = 3 : 5.
5. .. X = 10 • 3 : 5 = 6. Ax. 6
6. ••• X = the number of dm^, .. the space is 6 dm^.
RATIO, VARIATION, PROPORTION. 329
EXERCISES. CLI.
1. How long is a pendulum which oscillates 56 times a
linute ?
2. A cube of water 1.8 dm on an edge weighs how
lany kg?
3. If a pipe 1.5 cm in diameter fills a reservoir in 3.25
tins., how long will it take a pipe 3 cm in diameter to
ill it?
4. If a projectile 8.1 in. in length weighs 108 lbs., what
the weight of a similar projectile 9.37 in. long ?
5. If a metal sphere 10 in. in diameter weighs 327.5 lbs.,
rhat is the weight of a sphere of the same substance 14 in.
Rn diameter ?
6. Of two bottles of similar shape one is twice as high
as the other. The smaller holds 0.5 pt. How much does
the larger hold?
7. If a sphere whose surface is 16 tt cm^ weighs 5 kg,
what is the weight of a sphere of the same substance whose
surface is 32 tt cm^ ?
8. If the length of a 1sec. pendulum be considered as
1 m, what is the time of oscillation of a pendulum 6.4 m
long? 62.5m long?
9. A body weighs 25 lbs. 5000 mi. from the earth's
center. How much will it weigh 4000 mi. from the center ?
(Weight varies inversely as the square of the distance from
the earth's center.)
10. The distance through which a body falls from a state
of rest is proportional to the square of the, number repre
senting the time of fall. If a body falls 176.5 m in 6 sees.,
how far does it fall in 3.25 sees. ? in 1 sec. ? in 2 sees. ?
330 ELEMENTS OF ALGEBRA.
Theory of Proportion.
333. Theorem. In any proportion in which the 7iumbers
are all abstract, the product of the means equals the product
of the extremes.
Proof. 1. If J = > then, by multiplying by bd,
2. ad = be. Ax. 6
334. Theorem. If the product of two abstract numbers
equals the product of two others, either two may be made
the means and the other two the extremes of a proportion.
Proof. 1. If ad = be, then, by dividing by bd.
Similarly,  =  , etc.
a c
335. Theorem. If Siih = c:d, then a : c = b : d.
The proof is left for the student.
The old mathematical term for the interchange of the means is
"alternation." The first proportion is "taken by alternation" to
get the second. The term, while of little value, is still used.
336. Theorem, i/* a : b = c : d, then b : a = d : c.
The proof is left for the student.
The old mathematical term for this change is "inversion."
337. Theorem. If a,:h = c.d, then a + b:b = c + d:d.
The proof is left for the student.
The old mathematical term for this change is " composition."
338. Theorem. 7/" a : b = c : d, then a — b:b = c — d:d.
The proof is left for the student.
The old mathematical term for this change is "division."
RATIO, VARIATION, PROPORTION.
331
339. Theorem. If
a : b = c : d, then a.+ b:a — b = cHd:c — d.
Proof. 1. ^ = ^L_. §337
d
2. ^ = ^ §338
d
_ a\b a — h c\d c — d
4... ^ = '^ §161
a — c — d
The old mathematical term for this change is "composition and
ivision."
There is sometimes an advantage in applying this principle in solving
Jtional equations. E.g., given the equation
x23x + 1 ~ic2 + 4x2'
2 x2 2 x2
6x 2  8x + 4
.. X = 0, or f .
340. Theorem. The mean proportionals between two num
bers are the two square roots of their product,
a _x
X b
x" = ab. § 333, or ax. 6
Proof. 1.
2. ..
3. ..
X
±Va6.
Ax. 9
Illustrative problems. 1. li a:b = c:d, prove that
a\b\c{d: b\d = c\d: d.
1. This is true if ad { bd + cd { d^ =^ be + bd + cd \ d^. § 334
2. Or if ad = be. Ax. 3
3. But ad = 6c. § 333
4. .. reverse the process, deriving step 1 from step 3, and the origi
nal proportion from step 1.
332 ELEMENTS OF ALGEBRA.
o a 1 ^1, ^ Va; + 2 + Va;  3 . .
2. Solve the equation  , , = l*.
V^+2  Vic  3
We may clear of fractions at once, isolate the two radicals, and
square ; but in this and similar cases § 339 can be used to advantage.
Writing the second member  and applying § 339, we have
2 Va + 2 _ 5^
2 Vx  3 ~ 1
2.... ^ = 25.
X — 3
3. .'. x + 2 =25x75.
4 . 7. _ 'TT
*• • • ^ — 2¥
Check. Substitute  for x in the original equation, and reduce ;
then ^ ^ _ 1
— 7= — l^
2Vf
3. Find a mean proportional between 1 + ^ and — 2 — 14 i.
1. By § 340 this equals ± V(l + i) (  2  14 i)
2. =±Vl216i
3. = ± 2 V34i
4. = ± 2 V4  2 V 4  1
5. = ± 2 (2  i). § 245
EXERCISES. CLII.
1. Find the value of ic in 2 : 3 + i = cc : 5.
2. Find the third proportional to 1 — V2 and 1 — 3 V2.
^4 ^4
3. If a:b = c:d, prove that a^ \b^\c^{d^ = — —: ;•
a\o c\d
4. Also that f^ = /
b +d d
5. Also that be { cd:c — a = *ihcd + cd^ — da^ : c<^ — &c.
6. Also that Va — b : ^c — d — V7i — Vi : Vc — V5.
II
RATIO, VARIATION, PROPORTION. 333
7. li a:h = h:c, prove that a \ c>2b.
8. li a:b = b:c^ prove that (a \ G)b is a mean propor
tional between a^ + b'^ and Z»^ + c^.
9. Find the two mean proportionals between
(a) 2 and 98. (b) 50 and  2.
(c) 3 and 432. (d)  7 and  847.
10. Given 1Q> — ^ x :3 = 2 + x : x, to find x.
\
\
11. Given Vic + 7 + Vcc — 7 : Vic + 7 — Vx — 7 = 6:1,
to find cc.
12. Given 1 + x .13  x = x — 2 \ x^ 21 = x + 4.:S1  x^,
to find ic.
13. Given a — 5: ^_ / l=a;;6tH^H 7' to find ic.
2ab a — b
14. Given 3a2 + 2a68^>2:5a2 + 4a^'12^»2 = x:5a6^>,
to find x.
r^ ^ ab ^ ab ^ „
15. Given X'.y^aArb — : a — b\ > and x\y:a^
= 2 : 1, to find x and y.
16. Given Va; — 5 : V7 + x = 1 : 2, to find x.
17. Find the value of x in
^ + 4tx  x"" :^  4. X + x"^ = 2 + X :2  X.
ax + cy ay \ cz az + ex
18. It ; 7^ = ~ — — = — ) prove that each of
by \ dz bz \ ax bx {ay
these ratios equals ;•
^ b + d
^^ a — b b — c c — a a {b + c
19. If T~ — ~^ = = ■, > prove
ay \ bx bz \ ex ey \ az ax \ byj ez
that each of these ratios equals
x + y + z
CHAPTER XVIII.
SERIES.
341. A series is a succession of terms formed according
to some common law.
E.g., in the following, each term is formed from the preceding as
indicated :
1, 3, 5, 7, ••■, by adding 2;
7, 3, — 1, — 5, • • • , by subtracting 4, or by adding — 4 ;
3, 9, 27, 81, • • •, by multiplying by 3, or by dividing by ^;
2, 2, 2, 2, ■ • . , by adding 0, or by multiplying by 1 .
In the series 0, 1, 1, 2, 3, 5, 8, 13, • • •, each term after the first
two is found by adding the two preceding terms.
342. An arithmetic series (also called an aritlimetic pro
gression) is a series in which each term after the first is
found by adding a constant to the preceding term.
E.g.., — 7, — 1, 5, 11, • • •, the constant being 6,
2, 2, 2, 2, ..., " " " 0,
98, 66, 34, 2, •• •, " " " 32.
343. A geometric series (also called a geometric progres
sion) is a series in which each term after the first is formed
by multiplying the preceding term by a constant.
E.g., 3,  6, +12,  24, • • •, the constant being  2,
10, 5, 2i, U, ..., " " " i,
2, 2, 2, 2, ..., " " " 1.
344. The terms between the first and last are called the
means of the series.
334
SERIES. 335
I. ARITHMETIC SERIES.
!345. Symbols. The following are in common use :
n, the number of terms of the series.
•' sum " " " "
fij hi hi • ' • tni the terms of the series.
tn particular, a, or <i, the '1st term, and Z, or ^„, the jith or last term,
d, the constant which added to any term gives the next ; d is usually
called the difference.
346. Formulas. There are two formulas in arithmetic
series of such importance as to be designated as fundamental.
1. t^, or I z= a + (ji — 1) d.
Proof. 1. ^2 = a + (i, by definition.
tz = t2\d = a\2d.
t^ = tz\d = a + Zd.
2. .. tn = tni\d = a + (n  1) d.
3. Or l = a + {n\)d.
E.g., the 50th term in the series 2, 7, 12, 17, • • • is 2 + 49.5 = 247.
Z Li
Proof. 1. s = a + (a + (Z) + (a + 2 d) + • • • (?  d) + I.
2. Hence, s^l \ {I  d) + {I  2 d) + ■ • • (a + d) + a,
by reversing the order.
3. .. 2 s = (a + Z) + (a + + • • • (a + 0 Ax. 2
4. .. 2 s = n (a + Z), ••• there is an (a + I) in step 3 for each of the
n terms in step 1.
E.g., the sum of the first 50 terms of the series 2, 7, 12, 17, •••,
of which I has just been found, is
60(2 + 247)^g^^g
2
ELEMENTS OF ALGEBRA.
347. It is evident that from formulas I and II various
others can be deduced.
E.g., given d, Z, s, to find n. The problem merely reduces to that
of eliminating a from I and II, and solving for n.
1. From I, a = 1 — {n — \)d.
2. Sabstitutlng in 11. s = "P''^" '>"].
3. ... „._^l±i'.„ + L^ = 0.
d d
2d
Illustrative problems. 1. Which term of the series 25, 22,
19, . • . is  125 ?
1. Given a = 25, d= S, 1=  125, to find n.
2. •.• l = a + {nl)d,  125 = 25 + (n  1) ( 3).
3. Solving, n = 51!
2. Insert arithmetic means between 5 and 41 so that the
4th of these means shall have to the next to the last, less 1,
the ratio 1:2.
1. The means are 5 + d, 5 + 2 (?, • • • 41  2 d, 41  d.
2 . 5 + 4(Z ^1
41 2d  1 2'
3. .. d = S, and the means are 8, 11, 14, 17, • • • 35, 38.
3. The sum of three numbers of an arithmetic series is
12 and the sum of their squares is 56. Find the numbers.
In this and similar cases it is advisable to take x — y, x, x \y, y
being the common difference. In the case of four numbers it is advis
able to take X — oy, X — y, X + y, X ]^y,2y being the difference.
1. {xy)^x+{x + y) = 12, .. X = 4.
2. (x  vY + x2 + (x + 2/)2 = 56, .. 3 x2 + 2 2/2 = 56.
3. .. y=±2.
4. .. the numbers are 4 ^ 2, 4, 4 ± 2 ; that is, 2, 4, 6, or 6, 4, 2.
SERIES.
337
348. The following table gives the various formulas of
'arithmetic series, and these should be worked out from
formulas I and II by the student.
Given.
To FIND.
Result.
1
2
3
4
a d n
ads
an s
dns
I
l = a\{nl)d.
l^id±V{aid)^ + 2ds.
I = 2s/n a.
l = s/n + {n l)d/2.
5
6
7
8
9
10
11
12
a d n
adl
a n I
dnl
S
s = in[2a + {nl)d].
s = i{l\a) + {l2a^)/2d.
s = in{a^l).
s = in[2l~{nl)dl
dnl
dns
dls
n I s
a
a = l{nl)d.
a = s/n — ^{n — l)d.
a = id±V{l + idy^2ds.
a = 2 s/n I.
1.3
14
15
16
a n I
a n s
als
n I s
d
d = {la)/{nl).
d = 2(san)/(w2  n).
d = {l^ a^)/{2sla).
d = 2{nl s)/(n2 n).
17
18
19
20
adl
ads
a I s
dls
n
n = {l a\d)/d.
n ^ [d  2 a ±V{2 a  df + 8 ds] /2 d.
n = 2s/{a + l).
n = [d ^ 21 ±V{21 + d)^  Sds]/2d.
ELEMENTS OF ALGEBRA.
Illustrative problem. Find the number of terms in the
arithmetic series whose first term is 25, difference — 5, and
sum 45.
We may substitute in formula 18, but it is quite as easy to use the
two fundamental formulas which the student will carry in his mind.
1. From I, fi=25 + (nl)(5) = 306w.
2. " II, 45 = n.
2
3. .. w2  11 n + 18 = 0.
4. .. {n  2) (n  9) = 0, and n = 2, or 9.
The explanation of the two results appears by writing
out the series.
25, 20, (15, 10, 5, 0,  5,  10,  15).
The part enclosed in parentheses has for its sum.
Hence, the sum of 2 terms is the same as the sum of 9 terms.
EXERCISES. CLIII.
1. Find ^200 ill the series 1, 3, 5, • • • .
2. Find s, given a = 4:0, n = 101, d = 5.
3. Find s, given a = 1, I = 200, n = 200.
4. Given t^ = — li and t^^ = 59^, find d.
5. Find t^o in the Series 540, 480, 420, • • • .
• 6. Find n, given s = 29,000, a = 4.0, I = 540.
7. Insert 7 arithmetic means between — 5 and 11.
8. Insert 12 arithmetic means between — 18 and 125.
9. Find s, given a = 14:, n = S, d = — 4:. Write out
the series.
10. How many multiples of 17 are there between 350
and 1210 ?
SERIES. 339
11. What is the sum of the first 200 numbers divisible
by 5 ? by 7 ?
12. Show that the sum of any 2n\l consecutive integers
is divisible by 2 ?i + 1.
13. What is the sum of the first 50 odd numbers ? the
first 100 ? the first n ?
14. What is the sum of the first 50 even numbers ? the
first 100 ? the first n ?
15. Given 1 = 11, d = 2, s = 32, to find 7i. Check the
result by writing out the series.
16. How long has a body been falling when it passes
through 53.9 m during the last second ?
17. Suppose every term of an arithmetic series to be
multiplied by A; ; is the result an arithmetic series ?
18. The sum of four numbers of an arithmetic series is
and the sum of their squares is 20. Find the numbers.
19. The sum of four numbers of an arithmetic series is
12 and the sum of their squares is 116. Find the numbers.
20. The sum of three numbers of an arithmetic series
is 21 and the sum of their squares is 179. Find the
numbers.
21. Find five numbers of an arithmetic series such that
the sum of the first and fifth is 46, and that the ratio of the
fourth to the second is 1.3.
22. $100 is placed at interest annually on the first of
each January for 10 yrs., at 6%. Find the total amount of
principals and interest at the end of 10 yrs.
23. Find the 7ith. term and the sum of the first n terms :
(a) 1 + 34.^ + .... (b) 11 + 9 + 7+ ....
340" ELEMENTS OF ALGEBRA.
II. GEOMETRIC SERIES.
349. Symbols. The following are in common use :
w, s, a, I and ti, t^, ■ • • tn, as in arithmetic series ;
r, the constant by which any term may be multiplied to produce
the next ; r is usually called the rate or ratio.
350. Formulas. There are two formulas in geometric
series of such importance as to be designated as fundar
mental.
I. t^, 01 I = ar^'K
Proof. 1. ^2 = ctr, by definition.
^3 = t2r = ar^.
ti = t^r = ar^.
2. .. tn = tnir — ar'''^.
3. Or 1 = ar»i.
S.g.i the 7th term of the series 16, 8, 4, • • • is
_ ^^" — a _ Ir — a
Proof. 1. s = a ^ ar \ ar^ + • • • + ar^—^ + ar"^~^.
2. .. rs — ar \ ar^ + • • • + ar"^—^ + ar'^ — ^ + ar^^
by multiplying by.r.
3. .. rs — s = ar^ — a, by subtracting, (2) — (1).
dyn Q^
4. .. {r — 1) s = ar^ — a, and s = , by dividing by (r — 1).
p * J 1 7 Ir — a
5. And ••• ar^ = ar'^^ • r = Ir, .. s =
r1
E.g., the sum of the first 7 terms of the series 16, 8, 4, • • • , of which
I has just been found, is
SERIES.
341
351. It is evident that from formulas I and II various
)thers can be deduced.
1
E.g.^ giveu Z, a, n, to find r. •.• I = ar^—\ ., r = {l/a)»—\
Given n, Z, s, to find a. The problem reduces to that of eliminating
from I and II and solving, if possible, for a.
1. From II, r =
2. Substitute this in
I, and z = a(iy \
a{s — a)"!
0.
Here it is impossible to isolate a. When the numerical values of
s, w are given, a can frequently be determined by inspection.
Eor example, given w = 4, Z = 8, s = 15, to find a. Here
.873 = a(15a)3,
and a evidently equals 8, or 1. Either value checks, for the series
may be 8, 4, 2, 1, or 1, 2, 4, 8.
Illustrative problems. 1. Find the sum of five consecutive
powers of 3, beginning with the first.
1. Here a = S, r = S, n = 6.
2. s = (a>«  a) / (r  1) = (3 • 35  3) /2 = 363.
2. Of three numbers of a geometric series, the sum of
•the first and second exceeds the third by 3, and the sum
of the first and third exceeds the second by 21. Find the
numbers.
1. Let X, xy, xy^ be the numbers.
2. Then x + xy = xy^ + 3, or x + xy — S = xy^.
3. And X + xy^ = xy + 21, or — x {■ xy + 21 = xy^.
4. .. x + xy — S = — X {■ xy + 21, or x = 12.
5. .. 4 ?/2 _ 4 y _ 3 = 0, by substituting in 2.
6. .. {2y + l){2yS) = 0, and y= 1, or f.
7. .. the numbers are 12, — 6, 3, or 12, 18, 27. Each set checks.
342
ELEMENTS OF ALGEBRA.
352. The following table gives the various formulas of
geometric series. They should be worked out from formulas
I and II by the student, excepting those for n. The for
mulas for n require logarithms and may be taken after
Chap. XIX.
GiVKN.
To FIND.
Results.
1
ar n
I = ar''\
2
ar s
I
l=[a + {rl)s]/r.
3
a 71 s
Z(s0«' a(sa)"i=0.
4
r n s
l = {rl)srn^/{rnl).
5
ar n
s = a(r«l)/(rl).
6
7
an
a n I
s
s = (Wa)/(rl).
71 « 1 1
8
r n I
s = Z(r»l)/(r«r»i).
9
r n I
a = l/r»K
10
r n s
a = 8(rl)/(rl).
11
rls
a
a = rl{r  l)s.
12
n I s
Z (s _/)«!_ a (sa)«i = 0.
13
a n I
r= {l/a)^>^^.
14
an s
r^sr/a + {sa)/a = 0.
15
a I s
r = {sa)/{sl).
16
n I s
rn _ sr^'^/{s l) + l/{sl) = 0.
17
arl
n = (log I  log a) /log r + 1.
18
ar s
n = {log[a + (r l)s] loga}/logr.
19
a I s
n
n = (log I  loga)/[log (sa) log (s  1)] + 1.
20
rls
n = {log Z  log [ir  (r  1) s] } /log r + 1 .
SERIES. 343
EXERCISES. CLIV.
1. The sum of how many terms of the series 4, 12,
S6, is 118,096?
2. Find the sum of the first ten terms of the series
3^, 2^.3V
3. Find the geometric mean between
(a) 1 and 4. (b)  2 and  8.
4. Find the sum of five numbers of a geometric series,
the second term being 5 and the fifth 625.
5. What is the fourth term of the geometric series
I whose first term is 1 and third term ^V ?
6. The arithmetic mean between two numbers is 39 and
the geometric mean 15. Find the numbers.
7. Prove that the geometric mean between two numbers
is the square root of their product (§ 343).
8. Prove that the arithmetic mean between two unequal
positive numbers is greater than the geometric mean.
9. To what sum will $1 amount at 4^ compound
interest in 5 yrs. ? (Here a = $1, r = 1.04, n = 6.)
10. In ex. 9, suppose the rate were 4% a year, but the
interest compounded semiannually ?
11. The sum of the first eight terms of a certain geo
metric series is 17 times the sum of the first four terms.
What is the rate ?
12. Find the 10th term and the sum of the first ten
terms of the series:
(a) l,hh (b) 1,2,4,8,....
(c) 1,2,4,.... (d) 32,16,8,4,....
344 ELEMENTS OF ALGEBRA.
353. Infinite geometric series. If the number of terms is
infinite and r<l, then s approaches as its limit
(§ 167). ^ ~ ''
This is indicated by the symbols s = , n being infinite.
1 — r
The symbol == is read " approaches as its limit " (p. 140).
Proof. 1. ■.* 7' < 1, the terms are becoming smaller, each
being multiplied by a fraction to obtain the next.
2. .'.1 = 0, and .*. Ir = 0, although they never reach that
limit.
3. .'. s == } by formula II.
r — 1 ^
4. ..8 =  J by multiplying each term of the frac
tion by — 1.
E.g., consider the series 1, i, i, •••, where n is infinite. Here
a 1
5 =: , or  — — , or 2. That is, the greater the number of terms,
1 — r 1 — i
the nearer the sum approaches 2, although it never reaches it for
finite values of n.
EXERCISES. CLV.
1. Given s = 8, a = 4:. Find r.
2. Given s = 10^, r = ^. Find a.
3. Given s = l,r = §§99. Find a.
4. Given s = 155, r = 2, n = 5. Find a.
5. Given s = 124.4, r = S, n = 4^. Find a.
6. Find the limits of the following sums, n being infinite :
(a) 20 + 10 + 5 + 2^ + .... (b)  + i + /,,4T^^ + .
(d) 10 + 1 + 0.1 +0.01 + ....
SERIES. 346
354. Circulating decimals. If the fraction y\ is reduced
the decimal form, the result is 0.272727 • • , and similarly
le fraction ^ = 0.152777 • •. The former constantly
jpeats 27, and the latter constantly repeats 7 after 0.152.
When, beginning with a certain order of a decimal frac
jtion, the figures constantly repeat in the same order, the
lumber is called a circulating decimal, and the part which
3peats is called a circulate.
A circulate is represented by a dot over its first and last figures.
0.272727 •  • is represented by 0.27 ;
0.152777 " " " 0.1527.
A circulating decimal may be reduced to a common frac
ion by means of the formula s =  > as follows :
•^ 1 — r
1. To what common fraction is 0.27 equal ?
1. 0.27 = 0.27 + 0.0027 + 0.000027 +  • • •
2. This is a geometric series with a = 0.27, r = 0.01, n infinite.
0.27 27 3
3.
1  0.01 99 11
2. To what common fraction is 0.1527 equal
1. 0.1527 = 0.152 + 0.0007 + 0.00007 +   • = 0.152 + a geometric
series with a = 0.0007, r = 0.1, n infinite.
^ 0.0007 7
' "^"1 0.1~9000'
3. To this must be added 0.152, giving 0.152, or if^§, or ^.
EXERCISES. CLVI.
Express as common fractions :
1. 0.3. 2. 0.045. 3. O.OOOl.
4. 0.147. 5. 1.2375. 6. 5.0504.
7. 0.045. 8. 2.003471. 9. 0.23456.
346 ELEMENTS OF ALGEBRA.
III. MISCELLANEOUS TYPES.
355. Of the other types of series, some can be treated by
the methods which have just been considered.
Illustrative problems. 1. Defining a harmonic series as one
the reciprocals of whose terms form an arithmetic series,
insert three harmonic means between 2 and 4.
This reduces to the insertion of three arithmetic means between 
and ^.
1. ••• a = i, n = 6, and I = i,
2. .. , = ^ + 4d, and d=j\.
3. .. the arithmetic series is ^, ^^, f , y\, J,
and " harmonic " 2, 2f, 2f, 3^, 4.
2. Sum to 20 terms the series 1, — 3, 5, — 7, 9, — 11, ... .
Here the odd numbers of the terms form an arithmetic series with
d = 4, and the even ones form an arithmetic series with d = — 4.
There are ten terms in each set. Summing separately, we have
190  210 =  20.
3. What is the harmonic mean between a and b ?
1. If yt is the harmonic mean, , ,  must form an arithmetic
. , 1, ' a' h' b
series (ex. 1).
2... 1_1 = 1_1.
h a b h
3..: h^^""
a + b
E.g., the harmonic mean between 3 and 4 is Y For, taking the
reciprocals of 3, y, and 4, we have i, ^j, , or gj, i^, and 2?, which
form an arithmetic series.
4. rind the sum of 71 terms of the series 1, 2x, 3x^,
Ax\ ....
SERIES. 347
Here the coefficients form an arithmetic series and the x's a geo
letric. Such a series is oalled arithmeticogeometric.
Let s = 1 + 2 X + 3 x2 + . . . + (71  1) x"2 + nx«i ;
[then xs= x + 2x2 H \ {n  2)x»2 + (n  l)x"i + nx«.
Subtracting,
(1  x) s = 1 + X + x2 + • ■ • + x«2 + x«i — nx".
_ 1 — X" x»
•■■'"(1X)2~''(1X)"
EXERCISES. CLVII.
1. Sum the series 3, 6, • • • 3 (?^ — 1), 3 ti.
2. Sum to 2 7i, terms the series 1, — 2, + 3, — 4, • • • .
3. Sum the series 1, 4 ic, 7 x^, 10 x^, • • jio n terms.
4. Sum the series 1, — 3, +5, — 7, H to 2n terms.
5. Insert a harmonic mean between 2 and 2 ; between
2 and 2.
6. Prove that no two unequal numbers can have their
arithmetic, geometric, and harmonic means equal, or any
two of these equal.
7. Show that the sum of the first n terms of the series
1, — 2, +4, — 8, H 16, • • • is ^ (1 ± 2"), the sign depending
on whether n is odd or even.
8. Find the sum of 1 + 2 a; + 3 ic^ + 4 cc^ H to n terms
by writing the series (1 + a; + £c^ H ) \ (x \ x^ \ x^ ] )
■} (x^ \ x^ ] ) + (x^ \ ), etc., summing each group sepa
rately, and adding the sums.
9. The number of balls in a triangular pile is evidently
lH(l + 2) + (l+2 + 3)H , depending on the number
of layers. How many balls in such a pile of 10 layers ?
CHAPTER XIX.
LOGARITHMS.
356. About the year 1614 a Scotchman, John Napier,
invented a scheme by which multiplication can be per
formed by addition, division by subtraction, involution by
a single multiplication, and evolution by a single division.
357. In considering the annexed series of numbers it is
apparent that
1. V 23.25 = 28,
8 . 32 = 28 = 256.
.. the product can be found by adding the
exponents (3 + 5 = 8) and then finding what
28 equals.
2. •.• 29 : 23 = 26,
512 : 8 = 64.
.. this quotient can be found from the table by a single subtraction
of exponents.
3. •.• (25)2 ^ 25 . 25 = 21^
322 ^ 1024.
4. •.• V2io = V25 . 25 = 25,
Vi024 = 32.
5. The exponents of 2 form an arithmetic series, while the powers
form a geometric series.
In like manner a table of the powers of any number may
be made and the four operations, multiplication, division,
involution, evolution, reduced to the operations of addition,
subtraction, multiplication, and division of exponents.
348
20 = 1
26
= 64
21 = 2
2"
= 128
22 = 4
28
= 256
23 = 8
29
= 512
24 = 16
210
= 1024
25 = 32
211
= 2048
I
LOGARITHMS. 349
I
358. For practical purposes^ the exponents of the powers
to which 10, the base of our system of counting, must be
raised to produce various numbers are put in a table, and
these exponents are called the logarithms of those numbers.
In this connection the word power is used in its
broadest sense, 10'* being considered as a power, whether n
is positive, negative, integral, or fractional. The logarithm
of 100 is written " log 100."
E.g., 103 :.^1000, /. log 1000=3. 102 ^iqo, .. log 100 =2.
10'> =1, .. log 1 =0. 101 = 10, ..log 10=1.
10i=i, ..logO.l =1. 102= — , ..log 0.01= 2.
10' ^ 102' ^
\{y?s^6^ that is, the thousandth root of lO^'^i, is nearly 2,
.. log 2 = 0.301, nearly.
Although log 2 cannot be expressed exactly as a decimal
fraction, it can be found to any required degree of accuracy.
EXERCISES. CLVIII.
1. What is the logarithm of 10" « ? of 1000^ ? of 10^ ?
2. Whatis the logarithm of 10^ 10«? of lOMO^ ?
3. What is the logarithm of ^lO"* • 10^ ■ 10« ? of vio ?
4. What is the logarithm of lO^lO^lO^? of 0.001 of
1010*? of 10^ 10^. 109?
5. Between what two consecutive integers does log 800
lie, and why ? also log 3578 ? log 27 ?
6. Between what two consecutive negative integers does
log 0.02 lie, and why ? also log 0.009 ? log 0.0008 ?
7. If the logarithm of 2 is 0.301, what is the logarithm
of 21000 9 (2 = 10^^^% .•.2"o« = ? .. the logarithm of
21000 ^ 9\
350 ELEMENTS OF ALGEBRA.
359. Since 2473 lies between 1000 and 10,000, its loga
rithm lies between 3 and 4. It has been computed to be
3.3932. The integral part 3 is called the characteristic of
the logarithm, and the fractional part 0.3932 the mantissa.
That is, lo^BB^^, or 1033932 =2473, ..log 2473 = 3.3932.
... 103.3991!. 101=102.8932^ .. 1028932 =247.3, .. log 247.3 = 2.3932.
Similarly, ioi.3982 =24.73, ..log 24.73 = 1.3932.
*" '"^ 1003932 =2.473, ..log 2.473=0.3932.
10039321=0.2473, ..log 0.2473=0.39321.
360. It is thus seen that
1. The characteristic can always he found hy inspection.
Thus, because 438 lies between 100 and 1000, hence log 438 lies
between 2 and 3, and log 438 = 2 + some mantissa.
Similarly, 0.0073 lies between 0.001 and 0.01, hence log 0.0073 lies
between — 3 and — 2, and log 0.0073 = — 3 + some mantissa.
Since 5 lies between 1 and 10, log 5 lies between and 1, and equals
+ some mantissa.
2. The mantissa is the same for any given succession of
digits, wherever the decimal point may he.
Thus, log 2473 = 3.3932, and log 0.2473 = 0.3932  1.
3. Therefore, only the mantissas need he put in a table.
Instead of writing the negative characteristic after the mantissa,
it is often written before it, but with a minus sign above ; thus, log
0.2473 = 0.3932  1 = 1.3932, this meaning that only the character
istic is negative, the mantissa remaining positive.
Negative numbers are not considered as having loga
rithms, but operations involving negative numbers are
easily performed, ^^.g, the multiplication expressed by
1.478 • (— 0.007283) is performed as if the numbers were
positive, and the proper sign is prefixed.
LOGARITHMS. 351
EXERCISES. CLIJC.
1. What is the characteristic of the logarithm of a
number of three integral places ? of 6 ? of 20 ? of 7i?
2. What is the characteristic of the logarithm of 0.3 ?
of any decimal fraction whose first significant figure is in
the first decimal place ? the second decimal place ? the
20th ? the nth ?
3. From exs. 1, 2 formulate a rule for determining the
characteristic of the logarithm of any positive number.
4. If log 39,703 =4.5988, what are the logarithms of
"(a) 39,703,000? (b) 397.03? (c) 3.9703?
(d) 0.00039703? (e) 0.39703? (f) 3970.3?
361. The fundamental theorems of logarithms.
I. The logarithm of the product of two numbers equals
the sum of their logarithms.
1. Let a — 10"*, then log a — m.
2. Let b = 10% " log b = n.
3. .*. ab = 10"' + ", and log ab = m { n = log a { log b.
Thus, log (5x6)= log 5 + log 6.
II. The logarithm of the quotient of two numbers equals
the logarithm of the dividend minus the logarithm of the
divisor.
1. Let a = 10"*, then log a =m.
2. Let b = 10«, " log b = n.
a 10"' ^^ ^ . a
3. ..  = — — = lO"*" and log = m — n.
b 10"
Thus, log (40 ^ 5) = log 40  log 5.
352 ELEMENTS OF ALGEBRA.
III. The logarithin of the nth power of a number equals
n times the logarithm of the number.
1. Let a = lO"*, then log a = m.
2. .'. a" = lO""*, and log a" = nm = n log a.
IV. The logarithm of the nth root of a number equals
— th of the logarithm of the number.
1. Let a = 10"*, then log a = m.
  ml
2. .'. a" = 10", and log a" = — =  • log a.
n n
Th, III might have been stated more generally, so as to include
X
Th. IV, thus : log a^ =  • log a. The proof would be substantially
the same as in ths. Ill and IV.
EXERCISES. CLX.
Given log 2 = 0.3010, log 3 = 0.4771, log 5 = 0.6990,
log 7 = 0.8451, and log 514 = 2.7110, find the following :
1.
log 60.
2. log 24.
3.
log 7«.
4.
log^.
5. log 625.
6.
log 7*.
7.
log ^3^.
8. log^.
9.
log 35.
10.
log 5141
11. log 1.05.
12.
log 257.
13.
log 1050.
14. log 154,200.
15.
log V'514.
16.
log 10.28.
17. log 154.2.
18.
log 3.598.
19.
log 0.3084.
20. log 30.84.
21.
log 15.421
22. log
1799 [= log ft. 514
m
23. Show how to find log 5,
given log 2.
LOGARITHMS.
353
362. Explanation of table. Given a number to find its
logarithm. In the table on pp. 354 and 355 only the man
tissas are given. For example, in the row opposite 71, and
under 0, 1, 2, • • • will be found :
N
1
2
3
4
5
6
7
8
9
71
8513
8619
8525
8531
8537
8543
8549
8555
8561
8567
This means that the mantissa of log 710 is 0.8513, of
log 711 it is 0.8519, and so on to log 719. Hence,
log 715 = 2.8543, log 7.18 = 0.8561,
log 71,600 = 4.8549, log 0.0719 = 2.8567.
And ••• 7154 is ^^ of the way from 7150 to 7160, ..log
7154 is about ^^ of the way from log 7150 to log 7160.
.'. log 7154 = log 7150 + y\ of the difference between
log 7150 and log 7160
= 3.8543 + T% of 0.0006
= 3.8543 + 0.0002 = 3.8545.
Similarly, log 7.154 = 0.8545,
and log 0.07154 = 2.8545.
The above process of finding the logarithm of a number of
four significant figures is called interpolation. It is merely
an approximation available within small limits, since num
bers do not vary as their logarithms, the numbers forming
a geometric series while the logarithms form an arith
metic series. It should be mentioned again that the man
tissas given in the table are only approximate, being cor
rect to 0.0001. This is far enough to give a result which
is correct to three figures in general, and usually to four,
an approximation sufficiently exact for many practical com
putations.
354
ELEMENTS OF ALGEBRA.
N
1
2
3
4
5
6
7
8
9
0000
0000
3010
4771
6021
6990
7782
8451
9031
9542
1
0000
0414
0792
1139
1461
1761
2041
2304
2553
2788
2
3010
3222
3424
3617
3802
3979
4150
4314
4472
4624
3
4771
4914
5051
5185
5315
5441
5563
5682
5798
5911
4
6021
6128
6232
6335
6435
6532
6628
6721
6812
6902
5
6990
7076
7160
7243
7324
7401
7482
7559
7634
7709
6
7782
7853
7924
7993
8062
8129
8195
8261
8325
8388
7
8451
8513
8573
8633
8692
8751
8808
8865
8921
8976
8
9031
9085
9138
9191
9243
9294
9345
9395
9445
9494
9
9542
9590
9638
9685
9731
9777
9823
9868
9912
9956
10
0000
0043
0086
0128
0170
0212
0253
0294
0334
0374
11
0414
0153
0492
0531
0569
0607
0645
0682
0719
0755
12
0792'
10828
0864
0899
0934
0969
1004
1038
1072
1106
13
1139
1173
1206
1239
1271
1303
1335
1367
1399
1430
14
1461
1492
1523
1553
1584
1614
1614
1673
1703^
1732
15
1761
1790
1818
1847
1875
1903
1931
1959
1987
2014
IG
2041
2068
2095
2122
2148
2175
2201
2227
2253
2279
17
2304
2330
2355
2380
2405
2430
2455
2480
2504
2529
18
2553
2577
2601
2625
2648
2672
2695
2718
2742
2765
19
2788
2810
2833
2856
2878
2900
2923
2945
2967
2989
20
3010
3032
3054
3075
3096
3118
3139
3160
3181
3201
21
3222
3243
3263
3284
3301
3324
3345
3365
3385
3404
22
3424
3444
3464
3483
3502
3522
3541
;i5fi0
3579
3598
23
3617
3636
3655
3674
3692
3711
3729
3747
3766
3784
24
3802
3820
3838
3856
3874
3892
3909
3927
3945
3962
25
3979
3997
4014
4031
4048
4065
4082
4099
4116
4133
26
4150
4166
4183
4200
4216
4232
4249
4265
4281
4298
27
4314
4330
4346
4362
4378
4393
4409
4425
4440
4456
28
4472
4487
4502
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
4942
4955
4969
4983
4997
5011
5024
5038
32
5051
5065
5079
5092
5105
5119
5132
5145
5159
5172
33
5185
5198
5211
5224
5237
5250
5263
5276
5289
5302
34
5315
5328
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
5&58
5670
37
5682
5694
5705
5717
5729
5740
5752
5763
5775
5786
38
5798
5809
5821
5832
5843
5855
5866
5877
5888
5899
39
5911
5922
5933
5944
5955
5966
5977
5988
5999
6010
40
6021
6031
6042
6053
6064
6075
6085
6096
6107
6117
41
6128
6138
6149
6160
6170
6180
6191
6201
6212
6222
42
6232
6243
6253
6263
6274
6284
6294
6304
6314
6325
43
6335
6345
6355
6365
6375
6385
6395
6405
6415
0425
44
6435
6444
6454
6464
6474
6484
6493
6503
6513
6522
45
6532
6542
6551
6561
6571
6580
&590
6599
6609
6618
46
6628
6637
6646
6656
6665
6675
6684
6693
6702
6712
47
6721
6730
6739
6749
6758
6767
6776
6785
6794
6803
48
6812
6821
6830
6839
6848
6857
6866
6875
6884
6893
49
6902
6911
6920
6928
6937
6946
6955
6964
6972
6981
N
1
2
3
4
5
6
7
8
9
LOGARITHMS.
355
N
1
2
3
4
5
6
7
8
9
50
6990
6998
7007
7016
7024
7033
7042
7050
7059
7067
51
7076
7084
7093
7101
7110
7118
7126
7135
7143
7152
52
7160
71 G8
7177
7185
7193
7202
7210
7218
7226
7235
53
7243
7251
7259
7267
7275
7284
7292
7300
7308
7316
54
7324
7332
7340
7348
7356
7364
7372
7380
7388
7396
55
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
58
7634
7642
7019
7657
7664
7672
7679
7686
7694
7701
59
7709
7716
7723
7731
7738
7745
7752
7760
7767
7774
60
7782
7789
7796
7803
7810
7818
7825
7832
7839
7846
61
7853
7860
7868
7875
7882
7889
7896
7903
7910
7917
62
7924
7931
7938
7945
7952
7959
7966
7973
7980
7987
63
7993
8000
8007
8014
8021
8028
8035
8041
8048
8055
64
8062
8069
8075
8082
8089
8096
8102
8109
8116
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
8555
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
8722
8727
8733
8739
8745
75
8751
8756
8762
8768
8774
8779
8785
8791
8797
8802
76
8808
8814
8820
8825
8831
8837
8842
8818
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
9025
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
922"^
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
9590
9595
9600
9605
9609
9614
9619
9624
9628
9633
92
9638
9643
9647
9652
9657
9661
9666
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
9956
9961
9965
9969
9974
9978
9983
9987
9991
9996
N
1
2
3
4
5
G
7
8
9
356
ELEMENTS OF ALGEBRA.
In all work with logarithms the characteristic should he
writte?i before the table is consulted, even if it is 0. Other
wise it is liable to be forgotten, in which case the computa
tion will be valueless.
Illustrative problems. 1. Find from the table log 4260.
The characteristic is 3.
The mantissa is found to the right of 42 and under 6 ; it is 0.6294.
.. log 4260 = 3.6294.
2. Find from the table log 42.67.
The characteristic is 1.
log 42.7 = 1.6304
log 42.6 = 1.6294
difference = 0.0010
tV of 0.0010 = 0.0007
.. log 42.67 = 1.6294 + 0.0007
= 1.6301.
EXERCISES. CLXI.
From the table find the following :
1. log 28.
4. log 2.34.
7. log 8940.
10. log 3855.
13. log 1003.
16. log 23.42.
19. log 75.551
22. log 0.2969.
2. log 443.
5. log 6.81.
8. log 43.41.
11. log 2.005.
14. log 3.142.
17. log Vl28.
20. log 0.0007.
23. logO.01293.
3. log 9.823.
6. log 700.3.
9. log V^125.
12. log 9.8211
15. log 24,000.
18. log 0.2346.
21. log 0.00323.
24. log 0.000082.
LOGARITHMS.
357
363. Given a logarithm, to find the corresponding number.
The number to which a logarithm corresponds is called its
antilogarithm.
E.g., :• log 2 = 0.3010, .. antilog 0.3010 = 2.
The method of finding antilogarithms will be seen from
a few illustrations. Referring again to the row after 71
on p. 355, we have :
N
1
2
3
4
5
6
7
8
9
71
8613
8519
8525
8531
8537
8543
8549
8555
8561
8567
Hence, we see that
antilog 0.8513 = 7.1, antilog 5.8531 = 713,000,
antilog 2.8567 = 0.0719, antilog 1.8555 = 0.717.
Furthermore, '.• 8540 is halfway from 8537 to 8543,
.*. antilog 2.8540 is about halfway from antilog 2.8537 to
antilog 2.8543.
.'. antilog 2.8540 is about halfway from 714 to 715.
.. antilog 2.8540 = 714.5.
Similarly, to find antilog 1.8563.
antilog 1.8567 = 0.719 1.8563
antilog 1.8561 = 0.718 1.8561
6 2
.. antilog 1.8563 = 0.718f = 0.7183.
The interpolation here explained is, as stated on p. 353,
merely a close approximation ; it cannot be depended upon
to give a result beyond four significant figures except when
larger tables are employed.
This is sufficient in many numerical computations. £^g,
we speak of the distance to the sun as 93,000,000 mi., using
only two significant figures.
358
ELEMENTS OF ALGEBRA.
EXERCISES. CLXII.
rrom the table find the following :
1. antilog 0.3234.
3. antilog 2.9193.
5. antilog 3.9286.
7. antilog 0.8996.
9. antilog 3.9320.
11. antilog 1.9850.
13. antilog 10.5445.
15. antilog 0.9485  4.
17. antilog 0.6120  2.
2. antilog 2.4271.
4. antilog 5.2183.
6. antilog 1.7929.
8. antilog 4.7834.
10. antilog 2.0000.
12. antilog 0.7076.
14. antilog 3.6987.
16. antilog 0.6585 .6.
18. antilog 0.9290  3.
364. Cologarithms. In cases of division by a number n
it is often more convenient to add the logarithm of  than
to subtract the logarithm of n. The logarithm of  is called
the cologarithm of n.
•.• log  = log 1 — log n = — log n,
.'. colog w = — log n.
Also, colog w — 10 — log n — 10, often a more convenient form
to use.
E.g., .: log 6 = 0.7782.
colog 6 =  0.7782.
This may also be written 10  0.7782  10, or 9.2218  10.
The object of this is seen when we consider the addition of several
logarithms and cologarithms ; it is easier to add if all the mantissas
are positive, subtracting the lO's afterwards.
In general, colog w = lOp— logw — 10_p; that is, we may use 10,
20, or any multiple of 10, as may be most convenient.
LOGARITHMS. 359
The cologarithm can evidently be found by mentally sub
tracting each digit from 9, excepting the righthand signifi
cant one (which must be subtracted from 10) and the zeros
following, and then subtracting 10.
E.g., to find colog G178.
9. 9 9 9 10
log 6178 = 3. 7 9 9
colog 6178 = 6.
2 9 110,
To find colog 41.5.
9.
9 9 10
log 41. 5= 1.
6 18
colog 41. 5 = 8.
3 8 2 010
To find colog 0.013.
9.
9 9 9 10
log 0.013=:^
113 9
colog 0.013 = 11. 8 8 6 1  10 = 1.8861.
In case the characteristic exceeds 10 but is less than 20,
colog n may be written 20 — log n — 20, and so for other
cases ; but these cases are so rare that they may be neglected
at this time.
The advantage of using cologarithms will be apparent
from a single example :
317 • 92
To find the value of n io '
d17o • U.lo
Using Cologakithms. Not using Cologarithms.
log 317= 2.5011 log 317 = 2.5011
log 92 = 1.9638 log 92 = 1.9638
colog 6178 = 6.2091  10 log(31792) = 4.4649
colog 0. 13 = 10.8861  10 log 6178 = 3. 7909
log 36. 32= 1.5601 log 0.13 = 1.11.39
log(6178. 0.13) = 2.9048
log (317 92) = 4.4649
31792 _„,, log(6178. 0.13) = 2.9048
OD.OZ.
61780.13 ' ■ log 36.32= 1.5601
360 ELEMENTS OF ALGEBRA.
365. Various bases. Thus far we have considered loga
rithms as exponents of powers of 10. Bait it is evident
that any other base might be taken. Logarithms to the
base 10, such as we have thus far considered, are sometimes
called common or Briggs logarithms, the latter designation
being in honor of Henry Briggs, who is said to have
suggested this base to Napier.
If 2 were the base, log 8 would be 3, because 2^ = 8.
Similarly, log 16 would be 4, and so on.
Where a different base than 10 is used (which is not the
case in practical calculations), or where more than one base
is used in the same discussion, the base is indicated by a
subscript ; thus, logg 32 = 5, because 2^ = 32.
366. Computations by logarithms. A few illustrative
problems will now be given covering the types which the
student will most frequently meet. It is urged that all
work be neatly arranged, since as many errors arise from
failure in this respect as from any other single cause.
Since tt enters so frequently into computations, the follow
ing logarithms will be found useful :
log TT = 0.4971, log  = 1.5029.
TT
0.007^
1. Find the value of
0.03625
log 0.007 = 0.8451  3
3. log 0.007 = 2.5353  9
colog 0.03625 = 11.4407  10
13.9760  19
= 0.9760 6 = log 0.000009462.
.. 9.462 . 106 = Ans.
It will be noticed that the negative characteristic is less confusing
if written by itself at the right.
LOGARITHMS. 361
2. Find the value of 0.09515*.
* log 0.09515 = 0.9784 2.
••• the characteristic (— 2) is not divisible by 3, this may be written
log0.09515 = 1.9784 3.
Then i log 0.09515 = 0.6595  1 = log 0.4566.
.. 0.4566 = Ans.
3. Given a, r, I, in a geometric series, to find n. Compute
the value if Z = 256, a = 1, r = 2.
1. From § 350, I = ar^K
2. .. log Z = log a + (n  1) log r. § 361
logr
log 256 = 2.4082
logl =0, log 2 = 0.3010;
2.4082 4 0.3010 = 8.
4. .. n = 8 + 1 = 9.
, ^. ^ , • , ^ 2.706 • 0.3 • 0.001279
4. Find the value oi ^TrxF^
2 706 • 3 • 1 279
This may at once be written — ' ' 10 8, thus simplifying
the characteristics. Then
log2.706 = 0.4324
log 3 = 0.4771
log 1.279 = 0.1069
colog 8.609 = 9.0650  10
log 1.206 = 0.0814
.. 1.206 . 108 = Ans.
5. Given 2^ = 7, find x, the result to be correct to 0.01.
X log 2 = log 7.
.... = '"81 = 2:^ = 2.81.
log 2 0.3010
This division might be performed by finding the antilogarithm of
(log 0.8451 — log 0.3010), a plan not expeditious in this case.
362 ELEMENTS OF ALGEBRA.
6. The weight of an iron sphere, specific gravity 7.8, is
14.3 kg. Find the radius.
■y = 1 7tr^ • 1 cm^ = volume in cm^.
.. weight = f ;rr8 • 7.8 . 1 g = 14,300 g.
r = ( ) , the number of centimeters of radius.
\ 4 7r. 7.8 /
log 3 = 0.4771
log 14,300= 4.1553
colog4= 9.397910
colog 7t = 9.5029  10
colog7.8= 9.1079  10
3  2.6411
log 7.593= 0.8804
.'. radius = 7.593 cm.
EXERCISES. CLXIII.
In the following exercises give the result to four signifi
cant figures.
1. Find the value of 37 ^V.
2. Given x^ = x' :15. Find x.
3. Find the value of (32/29)1
4. Find the value of Vtt • 5.927.
5. Find the value of (5.376 /7r)i
6. Find the value of (37/2939)^*.
7. Given 227,600 = 7'*^ Find n.
8. Find the value of ^2 ^2 : VTO.
9. Find the value of (3.64/ 7.985) «.
10. Find the value of v 4.257» V^OS.
11. Find the value of (1402/3999)^
LOGARITHMS.
363
12. Find the value of VlOO.
13. Find the value of (22.8 h 0.09235)^.
14. Find the value of (24.73^  31.97*)^.
15. Find the value of (44 • 8.37)^  0.227^.
16. Find the value of 4 irr'^, when r = 2.06.
17. Also of I TTV^.
18. Given x : 5.127 = 0.325 : 2936. Find x.
19. Find the value of * a^bir, when a = 19.63, b = 19.57.
20. Given a, r, s, in a geometric series, show that
_ log [ct + (r — 1) 6] — log a
log r
and compute the value of n when a = 1, r = 2, s = 511.
21. Also, given r, Z, s, show that
_ log I — log [Ir — (r — 1) 5]
logr
+ 1.
Compute the value of n when r = 3, I = 729, s == 1092.
22. Also, given a, I, s, show that
log I — log a
log (s a) log (5 Z)
Compute the value of n when a = 3, I = 729, s = 1092.
23. Find the values of V2, ^"v^, ^"v^, v^, each to 3
decimal places. Which of these is greatest ? From this
it may be inferred that the value of n that makes 'Vn
greatest is about what?
24. Solve the equation 5^ — 6. (First take the loga
rithm of each member.)
25. Also the equation Vs = 10.
CHAPTER XX.
PERMUTATIONS AND COMBINATIONS.
367. The different groups of 2 things that can be selected
from a collection of 3 different things, without reference to
their arrangement, are called the combinations of 3 things
taken 2 at a time.
E.g.^ representing the .3 things by the letters a, 6, c, we can select
2 things in 3 ways, a&, ac^ he.
In general, the different groups of r things which can be
selected from a collection of n different things, without
reference to their arrangement, are called the combinations
of n things taken r at a time.
So the combinations of the 4 letters a, h, c, d, taken 3 at
a time, are abc, abd, acd, bed ; taken 2 at a time, ab, ac, ad,
be, bd, cd.
EXERCISES. CLXIV.
1. What is the number of combinations of 5 things
taken 2 at a time ? Represent them by letters.
2. What is the number of combinations of 5 things
taken 3 at a time ? Represent them by letters.
3. Write out the combinations of the letters w, x, y, z,
taken 4 at a time ; 3 at a time ; 2 at a time ; 1 at a time.
4. How does the number of combinations of 6 things
taken 2 at a time compare with the number taken 4 at a
time?
364
I
PERMUTATIONS AND COMBINATIONS. 365
368. The different groups of 2 things which can be
selected from 3 things, varying the arrangements in every
possible manner, are called the permutations of 3 things
taken 2 at a time.
E.g., the permutations of the letters a, &, c, taken 2 at a time, are
a6, 6a, ac, ca, be, cb.
In general, the different groups of r things which can be
selected from n different things, varying the arrangement
in every possible manner, are called the permutations of n
things taken r at a time.
In all this work the things are supposed to be different, and not to
be repeated, unless the contrary is stated.
369. The number of combinations of n things taken r at
a time is indicated by the symbol C". The number of per
mutations of n things taken ?• at a time is indicated by the
symbol P".
EXERCISES. CLXV.
1. Show that Ft = 12.
2. Show that P^ = 2 PI.
3. ShowthatP = 2.(7^
4. Find the value of P^ ; of PI
5. Show that Cl = n, and C^ = 1.
6. Show that Pf = 3, and in general that P '{ = n.
7. Using the letters a, b, c, show that CI = 3.
8. Write out the permutations of the letters of the word
time, taken all together.
9. Write out the permutations of the letters a, h, c, d
taken 2 at a time; 3 at a time.
366 ELEMENTS OF ALGEBRA.
370. Theorem. The number of permutations of ii different
things taken t at a time ^s n (n — 1) (n — 2) • • • (n — r + 1).
Proof. 1. Since we are to take r things we may suppose
there are r places to be filled.
The first place may be filled in any one of n ways.
Thus, with a, b, c, d, we may fill the first place with a, 6, c, or d.
2. For every way of filling the first place there are
n — 1 ways of filling the first and second.
Thus, if the first place be filled with a, we may fill the first and
second with ab, ac, ad.
3. ,'.forn ways of filling the first place there are
n(n — 1) ways of filling the first two.
E.g., ab, ac, ad,
ba, be, bd,
ca, cb, cd,
da, db, dc,
giving 4 • 3 =3 12 ways in all.
4. For every way of filling the first two places
there are n — 2 ways of filling the first, second,
and third.
Thus, if the first 2 places be filled with ab, the first 3 can be filled
with abc, abd, i.e., in 4 — 2 ways.
5. .'. for n(n — 1) ways of filling the first two
places there are n{n — l)(n — 2) ways of fill
ing the first three.
E.g.f abc, abd, adc, adb,
acb, acd, bca, bed,
bda, bdc, cda, cdb,
and the same with the first two letters interchanged in each.
PERMUTATIONS AND COMBINATIONS. 367
6. Similarly, the number taken 4 at a time is
n(n — 1) (n — 2) (n — 3), and the same reason
ing evidently shows that the number of permu
tations of n things r at a time is
n (71 l)(7i2)(nr 1)
or 7i(7i — l)(7i — 2)'(n — r { 1).
Corollary, i/n = r, P^ = n(n — !)•• 3 2 • 1. HeTice,
the 7iumher of permutations of n things taken all together is
n(nl)(n2)...3.2.1.
EXERCISES. CLXVI.
1. Find the value of P^.
2. Find the value of P^.
3. Prove that Plz\ =  Pi.
n
4. Prove that Pl = P'l.' P^z^.
5. Find the value of Pf ; of Pg. Prove this by writing
out the permutations of the letters a,b,c,.
6. Show from the theorem (§ 370) that P^. is greater as
^r is greater.
7. Show from the corollary that P^J is the product of all
I integers from 1 to 7i inclusive.
8. Find the number of permutations of the letters of
L the word 7iu7nher taken all together.
f 9. Find the number of permutations of the letters of
the word courage taken 3 at a time ; taken all together.
10. By writing out the permutations and the combina
tions of the letters a, b, c, d, e, taken 2 at a time, ascertain
how P compares with C^.
368 ELEMENTS OF ALGEBRA.
371. Factorials. The product
n{n  1) (n  2) (n  3) S '2 '1,
that is of all integers from 1 to n inclusive, is called fac
torial n.
Thus, factorial 3 = 1.23 = 6,
4 = 1 . 2 . 3 . 4 = 24, etc.
Factorial n is represented by several symbols. In writing
it is customary to use \n, this being a symbol easily made.
In print, on account of the difficulty of setting the \n, it is
customary to use the symbol n ! or (especially in Germany)
Un.
n is a Greek letter corresponding to P, and may be thought of as
standing for product.
We shall use in print only the symbol nl
372. It therefore appears that
(1) F^ = n\
(2) P" ^K^^l)(^ 2)32l _ nl
(nr){nrl)...321 (n  7)l
EXERCISES. CLXVII.
10'
1. Showthat P\o = — '. 2. Show that 5! = 120.
3. Find the value of ^. 4. Also of ^ • ^ • ^
o ! 10 ! D ! 21
5. Prove that nl = n(n  1) (n  2)  (n  3)1
6. Prove that (niy = n\n  iy(n  2y ■ ■3'' 2^ 1.
7. In how many ways can 10 persons be placed in a
row?
PERMUTATIONS AND COMBINATIONS. 369
373. Theorem. The number of permutations of n differ
ent thi7ujs taken v at a time, when each of the n things may
be repeated, is n^.
Proof. After the first place has been filled, the second
can be filled in n ways, since repetition is allowed.
So for the subsequent places.
Hence, instead of having
P = n{n  1) {n  2) ■ • {n  r + 1),
we have nnn • n = n^.
EXERCISES. CLXVIII.
1. Find the value of P\, repetitions being allowed.
2. Find the value of PI, repetitions being allowed.
3. How many numbers are there containing 4 digits ?
4. How many ways are there of selecting 3 numbers
from 50 on a combination lock, repetitions being allowed ?
5. How many ways are there of selecting 3 numbers
from 10 on a combination lock, repetitions being allowed ?
6. Show that P^, repetitions being allowed, is n"". From
this tell how many 9figure numbers are possible, all zeros
being excluded.
7. From ex. 6, how many 10figure numbers are possible,
zeros being admitted except in the highest order.
8. How many possible integral numbers can be formed
from the digits 1, 2, 3, 4, or any of them, repetitions of the
digits being allowed ?
9. The chance of guessing correctly, the first time, the
three numbers on which a combination lock of 100 numbers
is set, is 1 out of how many ?
370 ELEMENTS OF ALGEBRA.
374. Theorem. The number of combinations of n different
things taken t: at a time is
n(nl)(n2).(n r + 1)
r!
Proof. 1. For each combination of r tilings there are r\
permutations.
2. .'. for C" combinations there are C^ x r\ per
mutations.
3. But it has been shown that this number of
permutations is
n{n  l){n  1) • {n  r + 1). § 370
4. .. 6':^ X r! = ?^(7^l)(/i2).••(7^r + l),
and C« = ^K^l)(^2)(r^r + l)
Corollaries. 1. C" = P^'^/rl
n\
2. Cl
r\ (n — r)\
For we may multiply both terms of the fraction
n{n — 1) {n — 2) • ■ {n — r + 1)
r\
hj {nr)\, giving
n{nl){n2){nr + l){nr){nrl).^.2l
r\(n — 7')l
which equals — '■
r\{n — r)\
This is a more convenient formula to write and to carry in mind.
Practically, of course, it gives the same result as the other. E.g.^
By the theorem, 0% = 5jAl5 •
by the corollary, C = ^'^'^"^•^ .
3.212.1
PERMUTATIONS AND COMBINATIONS. 371
EXERCISES. CLXIX.
1. If Pi = 3,628,800, find n.
2. Find the values of P^; of P'^; of CI
3. If P% = bQ>, find n, and explain why there should be
two results.
4. In how many ways can 3 persons be selected from a
class of 20 ?
5. In how many ways can the letters of the word cat
be arranged ?
6. Prove that C;? = C„!!.^, by substituting in the formula
of § 374, cor. 2.
7. What is the number of combinations of 20 things
taken 5 at a time ?
8. In how many ways can the letters of the word
number be arranged ?
9. How many numbers can be formed by taking 4 out
of the 5 digits 1, 2, 3, 4, 5 ?
10. How many triangles are formed from 4 lines, each of
which intersects the other 3 ?
11. How many changes can be rung with a peal of 7
bells, a particular one always being last ?
12. In how many ways may the letters of the word
united be arranged, taken all at a time?
13. How many changes can be rung with a peal of 5
bells, using each bell once in each change ?
14. In how many ways can a consonant and a vowel be
chosen out of the letters of the word numbers ?
372 ELEMENTS OF ALGEBRA.
15. How many numbers between 2000 and 5000 have
the hundreds figure 7 and are divisible by 2 ?
16. In how many ways may the letters of the word
rate be arranged, taken any number at a time ?
17. In how many ways can 5 persons be seated about
a circular table, one of them always occupying the same
place ?
18. How many different arrangements (permutations)
can be made by taking 5 of the letters of the word tri
angle ?
19. On an examination 15 questions are given, of which
the student has a choice of 10. In how many w^ays may he
make his selection ?
20. How many different arrangements can be made of
the letters of the word algebra, it being noted that two of
the letters are alike ?
21. There are four points in a plane, no three being in
the same straight line. How many straight lines can be
drawn connecting two points ?
22. How many different signals can be made with 5
different flags, displayed on a staff 3 at a time ? 4 at a
time ? 2 at a time ? altogether ? any number at a time ?
23. Suppose a telegraphic system consists of two signs,
a dot and a dash ; how many letters can be represented by
these signs taken 1 at a time ? 2 at a time ? 3 at a time ?
4 at a time?
24. Prove that the number of permutations of 7i different
things taken r at a time is n — r + 1 times the number of
permutations of the n things taken ?' — 1 at a time.
CHAPTER XXI.
THE BINOMIAL TPIEOREM.
375. The binomial theorem is stated in § 80, and a proof,
which may be used in connection with that section, is given
in Appendix I.
It is now proposed to consider this theorem in the light
of Chapter XX.
376. Theorem. If the binoviial a + b ^s raised to the nth
power, n integral and jjositive, the result is expressed by the
formula
(x + a)" = X" + C? x"ia + C^ x"2a2
+ Cjx"'V H C^^Lixa"^ + a".
Proof. 1. By multiplication we know that
{x + a){x\ h)
^ x"^ \ {a { h) X \ ab,
{x + a){x{ b) {x + c)
^x^\(a + b\c)x^+ (ab \bc\ ca) x + abc,
{x { a){x + b) {x + c){x + d)
= x^\{a + b + G + d)x^
4 (ab + ac { ad \ bo \ bd \ cd) x^
+ (abc + abd \ acd + bed) x + abed.
There is evidently a law running through all
these expansions, relating to the exponents and
the coefficients of x.
373
374 ELEMENTS OF ALGEBRA.
2. We might infer from step 1 that if there were
n factors, the product would have for the coef
ficient
of ic", 1 ;
of cc"~"^, a \ b { c'n;
of ic""^, the combinations of the letters a,b,n,
taken 2 at a time ;
of x""^, the combinations of these letters taken
3 at a time ;
of X, the combinations of these letters taken
7i — 1 at a time.
3. This inference is correct ; for the term con
taining a;" can be formed only by taking the
product of the x's in all the factors, and hence
its coefficient is 1.
The terms containing ic"^ can be formed only
by multiplying the ic's in all but one factor by
the other letter in that factor; hence the x'''^
term will have for its coefficient (a + b \ 7i).
The terms containing a;"^ q^^^ i^q formed only
by multiplying the x's in all but 2 factors by
the other letters in those factors, i.e., by a and
b, a and c, a and d, etc. ; hence the cc"~^ term will
have for its coefficient (ab { ac \ ad \ ).
The reasoning is evidently general for the rest
of the coefficients.
4. If, now, we let a = b — G =  = n, we have
(x f ay = x''+ C\x''\i + C^x«2a2
+ C^ic" V \
+ C^l^xa''^ + ft".
THE BINOMIAL THEOREM. 375
As stated in § 246, the binomial theorem is true whether
n is positive or negative, integral or fractional. While the
proof of this fact cannot satisfactorily be presented without
the differential calculus, the fact itself should be recog
nized.
The following exercises will serve to recall the applica
tion of the theorem, although they do not differ materially
from those already met by the student in the exercises
following §§ 80, 246.
Illustrative problems. 1. Required the square root of
1 \ X to 3 terms.
1. •.• (a + 6)» = a» + na«  16 + ^'^^^?— ^ a" 262 4 ...,
2. .. (1 + x)^ = 1^ + i • 1~ ^ • X + ^ ^^ ~ ^^ • 1~ ^ • x^ + • • •
= l+iXlx2 + ....
2. Expand to 4 terms {a — 2 b)^.
1. V (x + ?/)« = x'* + nx'^^y + ^^^~ ' x'^^y^
n{nl){n2) ^ „ „
23 . ^ ^
2. .. (a2 6)3 = a'^ + (3)a4(2 6)
= a3 + 6 a46 + 24 a''^b^ + 80 a%^ +
3. Expand to 3 terms (1 + x) *.
As above, (1 + x)i = 1 + { i)x + (~ ^) (" ^ " ^) 3.2 4.
= 1 ~ixh^%x^.
376 ELEMENTS OF ALGEBRA.
EXERCISES. CLXX,
Expand the following binomials :
1. (cc + 5)^ 2. (x^2af.
(^ a;\i« (a h\
5. (40 + 1)^ 6. {?>a\hy.
7. (1 + xy, to 4 terms.
8. {a + V)^, to 4 terms.
9. \ d^ — £c^, to 3 terms.
10. (1 + x)~^, to 4 terms.
11. (1 — 2 a)*, to 4 terms.
12. (3 a:  2?/)^, to 4 terms.
13. v'.Si = (32  1)^, to 3 terms.
14 —== = (1 + ic)~ , to 4 terms.
' Vl + a;
15. (1 — x)~'^, to 5 terms, checking by performing the
division z
16. (1 — x)~^, to 5 terms, checking by performing the
division t: o'
1  2 a: + ic^
17. (1 { x)~^, to 5 terms, checking by performing the
division •
1 + 2 a: + a;2
APPENDIX.
I. PROOF OF THE BINOMIAL THEOREM FOR POSITIVE
INTEGRAL EXPONENTS (p. 57).
If n is a positive integer
{a + Z»)» = a" + na''^h +
+
n{nV) {n  2)
23
3^3 _^
Proof. 1. The law is evidently true for the 2d power, for {a + h)^
= a2 + 2 a6 + h'^, or, as the theorem says,
= a25o + 2 aifti + a^62. § 69
2. It is also true for the 3d power, for (a + 6)3 = a^ + 3 a'^h + 3 a&2
+ 63, or, as the theorem says,
= aPW + 3 a26i + 3 ai62 + ^0^3. § 69
3. Now if it were true for the fcth power we should have (a + 6)*
and if this were multiplied by a + 6 we should have
4. (a + 6)^ + i
= #+1 + A:
+ 1
#6
fc(fcl)
afc162 +
A; (fc  1) (A;  2)
2.3
k{k\) •
a*263 +
a^+i + (fc + i)a% 4 ^A±il^a^i62
(fc + l)fc(fcl) ^,_
23
■253
5. But here we see that i/ the theorem were true for any power ^ as
the fcth, it would be true for the next higher power, as the {k + l)th.
0. But the theorem is true for the 3d power (step 2), and
.. it is true for the (3 + l)th or 4th power, by step 5 ;
.. " " " (4 + l)th " 5th " " "
and so on for all integral powers.
377
378 ELEMENTS OF ALGEBRA.
IL SYNTHETIC DIVISION (p. 67).
If the divisor is a binomial of the first degree, there is
often a considerable gain by resorting to a form of division
known as synthetic.
The process is best understood by following the solution
of a problem.
Required the quotient of £c^ — 3 cc^ + 3 ic + 4 by x — 1.
The ordinary long form would be as follows, the heavy
numerals being the ones reserved in the synthetic form
given below :
x^
2x
+ 1
x^
x^
3x^
Ix^
\3x
+ 4
2x^
H2a'
a; + 4
X 1
o rem.
This may be abridged by writing the quotient below, as
follows :
3x^ +3x +4
lx%\2x, 1
2x \l\ 5 rem.
Here the first term of the quotient, x"^, is multiplied by
— 1, this product subtracted from — 3x'^ and the remainder
immediately divided by x to get the next term, — 2x, and
so on.
Since it is easier to add than to subtract, it is usual to
change the sign of the second term of the divisor and add.
Doing this, and detaching the coefficients, we have the
common form for synthetic division, as follows :
1
+ 1
APPENDIX. 379
13+3+4
12 1
1 — 2 + 1 ; 5 rem.
Check. Let x = 2. Then (8 12 + 6 + 4 5) ^1=44 + 1.
In case any powers of a letter are wanting in arranging
according to descending powers of that letter, zero coeffi
cients should be introduced as usual.
EXERCISES. CLXXI.
Perform the following divisions by the synthetic process,
detaching the coefficients, and checking in the usual way.
1. a^ + b^ hj a + h.
2. x^ — y^ by x — y.
3. a^ — 4 <x + 3 by a — 1.
4. £c^ — 2 x^ + 1 by a; + 1.
5. 1 + ic + x^ + ic^ by 1 + a;.
6. x^  29 X + 190 by x  10.
7. ic^ + 3 x'^a — 4 xa^ by x — a,
8. x^ + 2 x^ — 4 X + 1 by x — 1.
9. x^  3 x^ + 2 X + 6 by X + 1.
10. x« + 3x2 + 3x + 28 by x + 4.
11. 5 x'^ + 4 x^ + 3 ic^ + 2 £c + 1 by X + 1.
12. a*  4 a^^» + 6 a%^  4 a6» + ^»* by a h.
13. x^ — 10 ic^ + 9 by X + 3 ; also by x — 3.
14. 3 x^  2 x^  7 X  2 by X + 1 ; also by x  2.
15. 2x2 + 3x?/ — 2?/2 by X + 2?/; ^Iso by y — 2x.
380 ELEMENTS OP ALGEBRA.
III. THE APPLICATIONS OF HOMOGENEITY, SYMMETRY,
AND CYCLOSYMMETRY (p. 73).
The applications of homogeneity, symmetry, and cyclo
symmetry are very extensive and they materially simplify
the study of algebra. The principle which lies at the foun
dation of these applications is as follows :
If two algebraic expressions are homogeneous, sijmmetric,
or cyclic, their sum, difference, product, or quotient is also
homogeneous, symmetric, or cyclic, respectively.
The truth of this principle follows from the definitions
and from previous proofs. Eg, by the law of the forma
tion of the product of two polynomials it appears that each
term of one factor is multiplied by each term of the other ;
hence, if one factor is homogeneous and of the third degree
and the other is homogeneous and of the second degree, then
the product must be homogeneous and of the fifth degree.
The converse is not necessarily true. E.g., the sum of
two nonsymmetric expressions may be symmetric, as the
sum a^ + h"^ \ G and c(c — 1).
These considerations suggest some valuable checks on the
four fundamental operations. Since algebraic expressions
are often homogeneous, symmetric, or cyclic, these checks
will be of service throughout the study of the subject.
E.g., the product of x'^ \ y^ and x \ y \s x^ {■ xy^ + x'^y + y^. This
may be checked by arbitrary values, or by noticing that the product
must be homogeneous, of the third degree, and symmetric as to x and y.
In the same way the square root of a^ + 6^ + c^ + 2 a& + 2 6c + 2 ca
must be symmetric, the product of (a — 6) (6 — c){c — a) must be cyclic
and the quotient of 27 a^h^  c^ by 3 a6  c must be symmetric as to a
and 6 ; otherwise there must be an error in the operation.
It so happens that many of the expressions dealt with in higher
algebra are, or can be made, symmetric or homogeneous, or both, and
hence the value of these checks becomes the more apparent as the
student progresses in mathematics.
APPENDIX. 381
EXERCISES. CLXXII.
Perform the operations here indicated, checking each by
substituting arbitrary values and also by (1) homogeneity,
(2) symmetry, or (3) cyclosymmetry, as seems best.
1. (x^J^i/){x^xY + y^)
2. (2x + yz){2xij + z).
3. (81 a'^h'^  256 c'')  (3 a^> + 4c).
4. (a \ h \ c) (he { ca \ ah) — abc.
5. (a;* + x^i/ + if) ^ (x^ + 2/^ + ^y)
6. — (a — b) (b — c)(c — a) (a + b { c).
7. a\bc)irb'(ca)\c\ab).
8. a^ (b — c){ b^ (c — a) + c\a — b).
9. —(a — b) (b — c) (c — a).
10. a''(b  c)\b''{c  a){ c'ici h).
11. {a^J^b^ + 1^ ah) (« + & + 1).
12. (x  y) (x^ + xhj + xY + xy' + y')
13. (£C^  18 xY + 2/') ^ (^^  .y' + 4 ^2/)
14. (a;2 _^ ^2 ^ ^2 _ ^^ _ ^^ _ ^^>) ^^ _l_ ^ _^ ^^^
15. (a; + 2/  2^)2 + (y + ^  2a^)2 + (^ + X 2yy.
16. (7^  2 Z  37/z)2 + (Z  2 m  3 A;)2 + (m  2 Z;  3 Z)^.
17. (^^2 _ ^2 _ ^2 _^ ^2 _^ 2 ^c  2 acZ) f (a + ^»  c  d).
18. (i? + 2 + r)' + (i>'  2'  r)« + (2'  r  py
19. (a; + ?/ + ^)^  (?/ + ^  cr)^  (^ + ir  ?/)«
(x + y zy.
382 ELEMENTS OF ALGEBRA.
Symbolism of symmetric expressions. Since the terms of
a symmetric expression are so closely related in form, it is
often necessary to write only the types of these terms.
E.g., if a trinomial is symmetric as to a, b, and e, and if
one term is ab, the others are at once known to be be and ca.
The term ab is therefore called a type term.
The Greek letter IS (sigma, our S) is used to mean " the
sum of all expressions of the type • • • ."
E.g., in f(x, y, z), %xhj means "the sum of all expres
sions of the form xhj " which can be made from the three
given letters.
That is, ^xhj = x'^y + x^z + y^x + y^z + z'^x + z'^y. This
polynomial is called the expansio7i of ^x^y.
If these same three letters are under discussion,
^x^ = x^ + y^ + z% but (2ic)2 = (x + y + zf.
In case of any doubt, the letters under discussion are
written below the %, thus :
2 (a + 6) = (a + 6) + (6 + c) \ {c \ a)
ahc
S (x2 + y) = (x2 + y) + (2/2 + x).
xy
If an expression is known to be cyclic, % has a slightly
different meaning. It then stands for " the sum of expres
sions of this type, ivhich can be formed by a cyclic hder
change of the letters.''^
E.g., if only cyclic expressions involving three letters are
under discussion,
%(ab) = {ab) + {bc) + {Ga),
instead of
{ab) + {ba) + {bc) + {cb) + (ca) + {a  c) ;
and
%a{b + c2 af = a{b + c  2 af +b{c + a  2bY
■\c{a + b2cy.
APPENDIX. 383
EXERCISES. CLXXIII.
Expand the expressions in exs. 18.
1. ^xy, where only x, y, z are involved.
I
2. %a%,
u
a, b
a
3. %{a^hY
ii
a, b, c
((
4. ^aJ)"
u
u
((
5. %xhfz.
xyz
6. %a}
ahc
'  3 abc.
7. %a^\2^ab.
8. ^a^
+ 3 %a% + 6 abc.
In cyclic functions involving only a, b, e, what is the
expansion of the expressions in exs. 914 ?
9. ^a(b + c). 10. ^a'^(bc).
11. :Sa'(^>' c«). 12. %a%\ab).
13. 2a'(ab + c). 14. [^C^^  c)^(^' + c  2 a).
Show that the following identities are true, by expanding
both members. Those involving negative signs are cyclic.
Except as stated to the contrary, only a, b, c are involved.
15. ^a{bc)^0. 16. (:^ay^a^ = 2:^db.
17. (^ay=^a^{:$2ab. 18. ^[(^ay  ^a^2 = %ab.
a •■■ d a •■■e
19. ■%(ab)(a + b c)^0.
20. '$(aby=3(ab)(bc)(ca).
21. %a\b c) = {ab){bc){c a).
22. %ab (a — b) = —(a — b)(b — c) (c — a).
23. (Sa) (^ab)  aZc = (a + ^) (Z» + c) (c + a).
24. (2a) {^a^ + 2abc={a + b) (b + c) (c + a) + 2a^
384 ELEMENTS OF ALGEBRA.
Illustrative problems. The preceding principles render it
easy to simplify certain expansions which would otherwise
require considerable labor. The process will be understood
from a few problems. ^
1. Expand {a + b \cy.
1. The expression is symmetric and homogeneous.
2. .. the expanded form contains only the types a^, a&, with numer
ical coefficients.
3. ,. it is of the form ni'Za'^ + nSa6, where we have to determine
m and n.
4. Considering the expression as a binomial, (a + 6 + c)2, we shall
evidently have cC^ \ 2 ah \ Jfi {■ some terms which do not contain cC^
or ah.
5. .. the coefficients of the type a^ are all 1, and those of the type
ah are all 2. .. m = 1, w = 2.
6. .. the result is Sa^ ^ 2 Sa6, or a^ + h^ \ c^ ^ 2 (ah \ ca \ he).
Check. Let a = 6 = c = 1. Then
32 = 12 + 12 + 12 + 2 (1 + 1 + 1) = 9.
2. Simplify
{a + b + cf ^{a + h  cf + {h + c  ay + {c \ a  bf.
1. As in problem 1, the types are a2, a6, and the expanded form is
m2a2 + nliah, where we have to determine m and n.
2. In the four. trinomials we have a2, a2, (— a)2, (— a)2, or 4a2, as
shown in problem 1. .. m = 4.
3. Also 2 a&, 2 ah,  2 ah,  2 a6, or • ah. .•.n = 0.
4. .. the result is 4 Sa2, or 4 {a^ + h^ + c2).
Check. Let a = 6 = c = 1. Then
32 + 12 + 12 ^. 12 ^ 4 (12 + 12 + 12) = 12.
This particular problem is so simple that there is no great gain by
using the S symbolism.
APPENDIX. 385
3. Expand (iSa)^ where ^a = a + h + c + d\e^ .
1. What can be said of (Sa)^ as to symmetry ? homogeneity ?
2. .. the expanded form contains only what types ?
3. .. it is of what form, and what coefficients are to be determined ?
(See problem 1.)
4. What are these coefficients in the expansion of (a + h)^ ?
5. Will the addition of other letters, as c + cZ + e + • • • , affect these
coefficients of cfi and ah ?
6. .. what values have the coefficients m and w, and what is the
result ?
4. Expand (Sa)^, where %a^a{h[c\d\e\.
1. The types are evidently of the third degree, and therefore must
be a3, cfih, abc. (Why ?)
2. In expanding (a + b + c)^, we have (§ 69)
'a+V^ + 8 a + b^ ■ c + 3a + 6 • c^ + c^,
in which the coefficient of a^ is evidently 1, of a^b is 3, and of abc
(found only in 3 a + b^ • c) is 6.
3. The addition of other letters, d + e + ■ • •, will not affect the
coefficients of a^, a^b, or abc.
4. .. (Sa)3 = 2a3 + 3 Sa26 + 6 Sa6c.
5. Expand (x { y { zy —(y \ z — xy — (z + x — yY
{x + y zy.
1. What are the types ?
2. •.• we have x^, — ( — x)^, — x'\ — x^, what is the coefficient of Sx^ ?
3. ■.• we have 3x2?/, — 3x2?/, _ ^_ Sx^y), — 3x2?/, what is the coef
ficient of 2x2?/ ?
4. •.• we have 6xyz (as in problem 4), — (— 6xyz), — (— Gxyz),
— (— 6x2/2), what is the coefficient of Xxyz ?
5. .. the result is 24x?/z.
Check. Let x = ?/ = z = 1. Then, etc.
386 ELEMENTS OF ALGEBRA.
EXERCISES. CLXXIV.
]§ is limited to three letters in each of the following
exercises, except as otherwise indicated.
1. Expand (%ay. 2. Expand (^ay.
a ■■• d
3. Show that, if 5a = 0, (^ay = 4 (^aby.
4. Show that %a ■ (^a^  ^ab) = %a^ — S abc.
5. Show that, if 2a = 0, :S(a + by + ^a^ = 0.
6. Show that (a + b) (b + c) {g + a) = %a%^2 abc.
7. Simplify {a  b  cy + (b  a  cy + {c  a  by.
8. Show that %x • (Sic  2 cc) • {%x  2ij)  {%x  2z)
= 2 %xY  ^x\
9. Simplify (a  2b  3 cy \ (b  2c  Say
^(^c2aSby.
10. Show that (— a + b + c){a — b + c)(a \b — c)
= '%a^(b + c)^a^2abc.
11. Show that %{ab) = 0.
12. Show that {a[b [c){— a + b \c)(a — b \c)
{a + bc) = ^2 a'b''  ^a".
13. Show that {a + &)(& + c) (c + a) = %ab'^ + 2 a^»c.
14. Show that 2a • ^a^ ^ a& (a + 6) + be (b + c)
+ ca (c + a) + 2a^.
15. Show that 2a • ^ab = a'^{b + c) + b'^ (c + a)
+ c^ (a + ^) + 3 a^>c.
16. Show that (2a  2 a) (2a  2 5) (2a 2 c)
= a2(5 + c) + ^'(c + a) + c2(a + ^)  ^a^  2 a6c.
APPENDIX. 387
IV. APPLICATION OF THE LAWS OF SYMMETRY AND
HOMOGENEITY TO FACTORING (p. 88).
Since many of the expressions in mathematics are sym
metric or homogeneous or both, the application of the laws
of symmetry and homogeneity is of great importance.
E.g., to factor ac^ + ha? + c62 _ ah"^ — bc^ — ca^, it should be noticed
that
1. It is homogeneous, of the third degree, and cyclic.
2. .. either it has 3 linear factors, a — b being one (why ?), or else
it has 1 linear factor, a + b + c (why ?) and 1 quadratic factor. (Why ?)
3. And •.• it vanishes for a = 6, .. a — 6 is a factor, and ..b — c and
ca. (Why ?)
4. There are no more literal factors (why ?), but there may be a
numerical factor n.
6. Then ac^ + ba^ + cb'^  a¥  be? ca? = n{a  b){b  c){c r a),
and if a = 2, 6 = 1, c = 0, this reduces to
2 =  2 • n,
whence w = — 1.
6. .. the expression equals — {a  b) {b — c) {c — a).
Check by letting a = 3, 6 = 2, c = l, or other values.
EXERCISES. CLXXV.
Factor the following :
1. ^x\yz). 2. 2a'(6'c2).
3. ^x\7/z). 4. ^a\b^c^).
5. (^af  :^a^ 6. ^a"^  2 %a%\
7. {%a) {%ab) — abc. 8. %ah {a \ b) { 2 abc.
9. S« (J)'' + c2) + 2 abc. 10. %a (b  c)'^ + 8 abc.
11. %a{b\cY^abc. 12. a^  b^ j^ c^ + ^ abc.
13. ^{ab){a + b  cf. 14. %(a  b){a \b 2 o)^.
388 ELEMENTS OF ALGEBRA.
15. %a^(b + c) + abG'S,a. 16. 4: a^""  (a"" + b""  c^.
17. (^xy + ^x^  ^(x + yy.
18. (S,xy + %x'  ^(x + yy.
19. a^ \ b^ \ c^ — 3 abc. One factor must be a ± 5 or ^a.
(Why ?)
20. %a^ ^ 3(a \ b){b \ c){c + a).
21. %(a — by, "Z referring to a, b, c.
22. ^(a^  by, 2 referring to a, b, c.
23. {%a){^ab){a + b){b + c){c + a).
24. X {f  z^) + 7/ (z^ x^)jz (x^  if).
25. {s  ay +{s by  (.s  cy, where s = a + b •} c.
26. 2a\b  c), i.e., a'ib  c) + ^'^(c _ a) + c'^{a  b).
27. {xay{b c) + {xby{ca) + {xcy{ab).
28. {a + b){a by ^ (b + c) {b  cy + ((; + a) (c  ay.
29. 2 (a  Z*) (a2 + Z*^)^ ^.g.^ (^^ _ ^^ (^^2 _^ ^2^
+ {bc) (b^ + c2) + (c  «) (c^ + 6^2).
30. (x42/ + ^)'(^ + 2/^)'(^ + ^^)'
(z + x tjy.
31. (ic  a) (x b)(a b) + (x  b) {x  c) (b  c)
+ (x — c)(x — a) (c — a).
32. a(b + c) (b^ + c''a^)\b(c^ a) (c^ + a^  b^)
+ c(a + b)(a^^b^c'').
33. Find three factors, only, of (x — yy" + '' f (y — zy'' + ^
+ {Z — CC)2» + 1.
34. Also of (5ic)2" + i  :§a;2" + i, 2 referring to x, y, z.
APPENDIX. 389
The type Sx^ + S 2 xy, the square of a polynomial.
Since C^xf = %x'' + '^2xy (p. 385), it follows that ex
pressions in the form of "^x^ \^2xy can be factored.
E.g., x2 4 2/2 + 22 + 2 x?/ + 2 ?/2 + 2 zx = (X + ?/ 4 z^.
Check. 9 = 82.
Similarly, 4 a2 + 9 62 + c2  12 a6  6 6c + 4 ca = {2 a  3 6 + c)2. ,
Check. Let a = 6 = 1, c = 2. Then 1 = 12.
EXERCISES. CLXXVI.
Factor the following :
1. 4 x^ + 9 2/2 + 1 + 12 ic?/ + 6 // + 4 ic.
2. 1 + 4 «.2 + 9 6* + 4 «; + 6 ^.2 + 12 ah\
3. 4 + 16 a^ + 25 Z>* + 16 a + 20 U' + 40 ah\
4. 5a;2 + 2/4 + 9 + 2.T?/2V5 + 6ic V5 + 62/'.
5. a^ + // + 9 c« + ^' + 2 a^ + 6 ac^ + 2 a^^ _^ g ^,2^3
+ 2 ^2^ + 6 c^f^.
The following miscellaneous exercises review some of the
elementary cases of factoring.
6. 4 x'^ + 8 iy + 9 ?/^ 7. 4 x^  4 a^2^2 _^ 9 y4_
8. 5 0^2 _^ 5 + 3 ic + 3 ic^ 9. xhf  4 .t'^ + 4  y.
10. a^j^h^ 1^ + 2 a^lP. 11. 1 + 4 a;y  4 /  xl
12. x'^a + a?/2 + hy''" + ^ic^. 13. x^ + 144 — 16 x^ — 9 x\
14. % 4 2 Zx + 2 a?/ + 4 «x. ■
15. x2 4 wxy — 4 i^?/2 — 4 x?/.
16. x^ 4 10 X + 2/' + 10 ?/ 4 25 4 2 xy.
17. x2  12 X 4 4 y/2 4 36  24 y + 4 x?/.
390
ELEMENTS OF ALGEBRA.
V. GENERAL LAWS GOVERNING THE SOLUTION OF
EQUATIONS (p. 152).
Theorem. If the same quantity is added to or subtracted
from the two members of an equation, the result is a7i equiva
lent equation.
Given A = B, an equation, and C any quantity.
To prove that A± C = B ± C i?, 2a\ equivalent equation.
Proof. 1. If for certain values of the unknown quantities,
A and B take numerically equal values, it is
evident that A± C and B ± C must also take
equal values.
2. .*. any root oi A = B is also a root of ^ ± C
= B±a
3. If for certain values of the unknown quantities
A±C and B±C take numerically equal values,
it is evident that A and B must also take equal
values, because we obtain their values by sub
tracting from the equal values oi A± C and
B dt C the same number.
4. .•. any root of A ± C = B ± C is also a root of
A = B.
5. Since a,ny root oi A = B is also a root of ^ ± C
= B±C, and any root of A±C = B±C is
also a root of ^ = ^, it follows that the two
equations are equivalent.
Corollary. Every equation can be put into the form
A = 0.
For iu subtracting from the two members of an equation a quantity
equal to its second member, an equivalent equation is obtained of which
the second member is 0.
APPENDIX. 391
Theorem. If the tivo members of an equation are multi
plied or divided by the same quantity, which is neither zero
nor capable of becoming zero or infinitely great, the result is
an equivalent equation.
Given the equation A = B, and the factor C, which by
the conditions cannot be or infinitely great.
To prove that AC = BC is an equation equivalent to
A^B.
Proof. l.\A=B, .'.AB = 0.
2. .. C(AB) = 0. Ax. 7
3. Every root of 1, making A — B = 0, must also
make C{A — B) = 0, because C is not infinitely
large.
If C = 00, then C {A — B) would be undetermined, by § 172.
4. .'. every root of 1 is a root of 2.
5. Conversely, every root of 2, making C(A — B)
= 0, must also make A — B = 0, because C is
not zero.
If C = 0, then A — B would equal §, an undetermined quantity
by § 168.
6. .". every root of 2 is a root of 1.
7. From 4 and 6, the equations are equivalent.
The necessity for the limitations on the value of the multiplier is
evident from a simple example. In the equation
X2 + X =
we cannot expect to get an equivalent equation by dividing by x
or multiplying by , for x = and  = co, and the simple equation
x + 1 =0
is evidently not equivalent to the quadratic equation x^ + x = 0.
392 ELEMENTS OF ALGEBRA.
Theorem. If the two members of a rational fractional
equation are multiplied by the lowest common denominator
of the fractions, the result is, in general, an equivalent
equation.
Proof. 1. The equation can be transformed so that the
second member is 0.
2. The first member, being a rational fractional
expression, can then be reduced to the form ?
in which B is the lowest common denominator
of the fractions, after they are added and
reduced, and hence is prime to A.
A
3. .'. the equation can be reduced to — = 0, the
B
members of which it is proposed to multiply
hy B.
4. There can be no values of th^ unknown quan
tity which make A and B zero at the same
time, since B is prime to A.
A
5. .'.in order that — = it is necessary and suffi
cient that A = 0. .'. the equation ^ = is
A
equivalent to the equation — = 0.
To illustrate the theorem, consider the following cases :
4
In the equation  = x, it is legitimate to multiply by x, giving 4 = x,
whence x =  2, or — 2, either root satisfying the original equation.
But in the equation — = 1 , it is not legitimate to multiply by x, for
then x2 = X, and x^ — x = 0, whence x (x  1) = 0, x = 0, or 1. But
X = does not satisfy the original equation, because g does not neces
sarily equal 1.
x"^ } 5x j 6
Similarly we cannot solve = 3 by multiplying by x  2.
X  2
APPENDIX. 393
Theorem. If both members of an equation are raised to any
integral power, the resulting equation contains all of the
roots of the given equation, but in general is not equivalent
to it.
Given the equation A = B.
To prove that the equation A"^ = B"" contains all of the
roots of the equation A = B, but in general is
not equivalent to it.
Proof. 1. From ^ = ^ it follows that A — B = 0.
2. From ^'" = ^" it follows that ^'"  B"" = 0.
3. But whether m is odd or even, A'" — B"" con
tains the factor A — B.
4. .". equation 2 becomes
{A  B) (^'"1 + A"'''B + ...)= 0,
and is satisfied hy A = B.
5. .". equation 2 contains the roots of equation 1.
6. But from equation 4,
and hence A"^ = B'" contains other roots than
A = B, and hence is not equivalent to it.
To illustrate let x = 2 ;
squaring, x'^ = 4,
an equation containing the root
x = 2,
but also the extraneous root
x= 2.
If we cube, x^ = 8,
and this again contains the root
x = 2,
but it also contains the extraneous roots
X =  1 ± V ~3.
394 ELEMENTS OF ALGEBRA.
VI. EQUIVALENT SYSTEMS OF EQUATIONS (p. 185).
It has been shown that the solution of a system of equa
tions is made to depend upon the solution of a second sys
tem derived from the first. But it has not yet been shown
that extraneous roots are not introduced by this operation.
Two systems of equations, each having the same roots as
the other, are called equivalent systems.
Theorem. Given a system of two equations
(1) f(x,y)=0, F(x,y) = 0,
and a, b two numbers (b ^ 0), then
(2) a.f(x,y) + br(x,y) = 0, f(x,y) =
is an equivalent system.
Proof. 1. •.' a solution of system (1) makes both f(x, y)
and F(x, y) equal zero, it makes both a /(x, y)
and b ■ F(x, y) equal zero, and hence satisfies
system (2).
2. •.•a solution of system (2) makes
and a.f{x,y)^bF{x,y)^^,
it must therefore make a /(x, y) = 0,
and hence, b F(x, y) = 0,
and hence, F(x, y)=0, ■,• b^O.
Hence, it is a solution of system (1).
This theorem justifies the solution of two simultaneous
linear equations by addition, subtraction, and substitution.
For it shows that we may multiply the members by any
numbers (a, b), add or subtract (since b may be negative)
the equations member for member, and combine this result
with the equation f(x, y) = 0.
APPENDIX.
395
VII. DETERMINANTS (p. 198).
The practical solution of simultaneous linear equations,
while possible by the methods already given, is frequently
tedious. For this reason mathematicians often resort to a
simpler method, that of determinants.
The theory of determinants is comparatively modern, and
although it is not practicable to enter into the subject at
any length at this time, the elementary notions are so
simple and so helpful, and the applications so common,
that a brief presentation of the subject will be of value.
The symbol
a^bo
is merely another way of writing
aib^ — a^bi. The symbol is called a .determinant, and the
letters a^, a^, b^, b^ are called its elements.
This is a determinant of the second order ; i.e., there are two ele
ments on each side of the square. It will be noticed that the expanded
form is simply the difference of the diagonals.
In a determinant, the horizontal lines of elements are
called roivs, the vertical ones columns.
In the above determinant the rows are ai, h\ and a^, 62 ; the columns
are and , .
&2
When the determinant
a^b^
is written in the form
a^^ — a^bi it is said to be expanded.
It is understood that the expanded form is to be simplified in all
12 31
cases. E.g., while =27 — 53, the result should be stated
as — 1. '
EXERCISES. CLXXVII.
1. Expand the following determinants :
a b
y X
X y
b a
ay
b X
X b\
y a
396
ELEMENTS OF ALGEBRA.
2. Also the following :
13
42
)
1 4
32
>
23
41
)
24
31
3.
Also the following :
10
50
19
Oa
00
20
' 7
>
18 30
)
27
J
h
)
a h
4. From exs. 1 and 2, state what changes can be made
in a determinant of the second order without changing its
value.
5. From ex. 3, what is the value of a determinant if
either a row or a column is made up of zeros ?
6. Expand the following determinants :
12 6
57
8 1
23
56
78
7. Expand the following determinants
«i ^1
a^ a^
h h
8. From ex. 7, state the effect on the value of a deter
minant of the second order of changing the rows into
columns and the columns into rows.
9. Expand tlie following determinants :
ai b^
«! + h
h.
«! b^
h
^2 f>2
>
a^ + b^
h.
5
a^ — ^2
h.
10. From ex. 9, state the effect on the value of a deter
minant of the second order, of increasing the elements of
one column by the corresponding elements of another, or of
diminishing the elements of one column by the correspond
ing elements of another.
I
APPENDIX.
397
11. Expand the following determinants
aiZ>i
2a^b^
mai bi
as h
?
2a^h^
}
ma^ b^
12. From ex. 11, state the effect on the value of a deter
minant of the second order, of multiplying the elements of
a column by any number.
A determinant of the second order is no more easily
written than is its expanded form. But one of the third
order (one with three elements on the side of the square)
is materially more condensed than is its expanded form.
<Xi Z*! Ci
is the general form of a deter
The symbol
^2 ^2 ^2
(^z bg C3
minant of the third order, and it stands for
<^J^2^3 + fh^'zCi + ^z^xC^ — a^b^fix — a^b^c^ — afi^c^.
The expansion of a third order determinant is easily
written by following the ^
arrows in this arrangement. ^""^ ^^x
This method of expansion
holds only for determinants
of the second and third orders,
all that we shall treat in this
work.
The fact that the student is
rarely called upon, in elementary
algebra, to solve a system of more
than three simultaneous linear
equations makes it undesirable to enter, at this time, upon the theory
of determinants of an order hiiher than the third.
398
ELEMENTS OF ALGEBRA.
EXERCISES. CLXXVIII.
1. Expand the following detefrminants, the rows of the
first being the columns of the second :
123
147
456
J
2 58
789
3 69
2. Also the following :
«! hy Ci
ai (^ ag
a^ h^ Cg
J
h b, b.
as bs Cg
Ci C2 Cg
3. From exs. 1 and 2, state the effect on the value of a
determinant of the third order, of changing the rows into
columns and the columns into rows.
4. Expand the following determinants :
246
135
7 89
J
2 + 446
1+335
7 + 889
'
646
435
15 8 9
5. Also the following :
^1 &1 Ci
^2 b^ C2
J
as ^3 C3
(Xi + b^ bi
«2 + ^2 ^2
as + ^3 ^3
I
6. From exs. 4, 5, state the effect on the value of a de
terminant of the third order, of increasing the elements of
one column by the corresponding elements of another.
7. Expand the following determinants :
ab
000
cd
>
a b c
ef
def
APPENDIX.
399
8. From ex. 7, what is the value of a determinant of the
third order if either a row or a column is made up entirely
of zeros ?
9. Expand the following determinants :
«! bi Cx
mai bx Ci
^2 h C2
)
ma.i ^2 <^2
dz h Cg
mag b^ Cg
10, From ex. 9, state the effect on th^ value of a deter
minant of the third order of multiplying the elements of a
column by any number.
In the preceding exercises certain general theorems have
been proved by the student for determinants of the second
and third orders. These will now be presented formally,
the proof, however, referring only to determinants of these
orders.
Theorem. The value of a determinant is unchanged if the
rows are changed to columns and vice versa.
Given the determinant
Proof. Each expands into
dib^Cz + a^b^Cx + «3^i<?2 — «^3^2Ci — a^biC^ — axb^c^.
The proof for the determinant of the second order is left
for the student. Take the determinants
ax bx Cx
a^ ^2 C2
.
«3 h Cz
ax a^ a^
determinant
bx b^ bs
Cx C2 Cg
ax bx
^2 ^2
and
ax a^
bx 62
and expand.
400
ELEMENTS OF ALGEBRA.
Theorem. If each element of a column {or row) of a
determinant is multiplied by any factor, the determinant
is multiplied by that factor.
Proof. Consider the determinants
ai bj_ Ci
ai^i
^2 h C2
and
«3 ^8 C3
a^b^
By the law of expansion every term of the ex
panded form contains one a (i.e., a^, a^, or a^)
and only one ; hence, if every a is multiplied by
m,, the 7n will appear once and only once as a
factor of every term of the expanded form.
Similarly, for any column or row.
13 2
For example, consider the determinant L „
6 /
If we multiply either
This
column or either row by 2, the determinant is multiplied by 2
is seen by expanding
62
3 4
64
3 2
10 7 '
5 14 '
57 '
10 14
the results being 42 — 20 = 22 in each case, while the original deter
minant equals 21 — 10 = 11.
Theorem. If a column (or row) is m,ade up entirely of
zeros, the determinant equals zero.
Proof. As in the precefding theorem, every term of the
expanded form contains an a ; hence, if every a
is zero, the expanded form vanishes. Similarly,
for any other row or column.
1 20
This is seen, for a special case, in the determinant 4 7 which
360
expands into l70t4.6.0f3.02l60420370 = 0.
The same may be seen in the case of determinants like
1241 103
00
02 '
etc.
APPENDIX.
401
Theorem. If each element of a column is multiplied by
any number, and a^ded to the corresponding element of any
other column, the value of the determinant is not changed.
«! bi Ci
a<i c>2 C2
^3 b^ Cs
«1
+ mbi
h
c.
^2
+ mbz
h.
C2
as
4 mbs
h
C3
Given the determinant
To prove that it equals the determinant
Proof. Expanding the second determinant, it equals
(^i + mb^ b^c^ + (02 + ^^^2) ^3^1 + (^3 + mh^ b^c^
— (a^ 4 mb^ b^Px — {a^ + mh^ b^c^ — {ax + mb^ b^c^
which equals
ai^2^3 + «2^3'?l + ^3^lC2 — ^3^1 — «2^lC3 " «1^3C2
the other terms all cancelling out.
That is, the two determinants are equal.
The proof is the same whatever columns (or
rows) are taken, and for the second order as well
as for the third.
Corollaries. 1. The elements of any coluinn (or row)
may be added to or subtracted from the corresponding ele
ments of any other column (or r^w) without changing the
value of the determinant.
For m may equal 1 or — 1.
2. If ttvo columns (or rows) are identical, the determinant
equals zero.
For, if the elements of one are subtracted from the corresponding
elements of the other, a column (or row) will be composed of zeros.
3. If the elements of one column are the same inultiples
of the corresponding elements of another, the deternninant
equals zero. ( Why ?)
402
ELEMENTS OF ALGEBRA.
Dlustrative problems. 1. Expand the determinant
27 25
42 41
Subtracting the second column from the first, the determinant
1 2 25 1
equals L . J = 82  25 = 57.
1 1 41 1
This is much easier than finding the value of 27 • 41 — 42 • 25.
2. Expand
8 21
6 15
Factoring the second column by 3, and then subtracting it from
the first, we have
65
'' = 3! ^1 = 3(57)
1 5
3. Expand
10 17 3
20 16 4
30 15 5
Subtracting the first row from the second and that from the third,
10 17 3
10 1 1
10 1 1
= 0.
(Why ?)
General directions for exj)anding determinants.
1. Remove factors from columns or rows.
2. Endeavor to make the absolute values of the elements
as small as possible by subtracting corresponding elements of
rows or columns, or multiples of those elements.
3. Endeavor to bring in as inany zeros as possible.
4. Endeavor to make the elements of two columns (or
rows) identical, so that the determinant may be seen to be
zero (if that is its value) without expanding.
5. After thus siinplifying as much as possible, expand.
APPENDIX.
403
EXERCISES. CLXXIX.
Expand the determinants or prove the identities as
indicated.
1.
121
112
21 1
2.
97 96
63 62
.
3.
3 9
13 39
•
Oil
14 1
aba
4.
125
137
5.
4 16 4
97 5 17
6.
a2 b^ ab
a^ b' ab^
•
1 a a^
7.
1 b b^
1 c c'
= {ab)ib c) (c  a).
a c
a \ b c G
8.
a b
b c
= 2 abc. 9.
a b \ e a
b b c \ a
= 4 abc.
p + c2 ab ca
10.
ab c^ + a^ be
ca be a^ + b'^
= Aa%^c^
Application of determinants to the solution of a system of
two linear equations.
On solving the system
a^x + b^y = Ci,
a2X + b2y = C2,
the roots are found to be
11 =
«]
Gi 
2_
a^
^.
aib2 — ^2^1
404 ELEMENTS OF ALGEBRA.
It is at once seen that
, made up
aibi
a2 02
64
11
t
11
16
7
— 7
3
11
14
11
17
 7
10
7
1. Each denominator is the determinant
of the coefficients ofx and j.
2. The numerator for x is the same determinant ivith c
put for a (the coefficient of x).
3. The numerator for j is also the same determinant,
ivith c put for b (the coefficient of y).
Illustrative problem. Solve the system
3a; + lly/ = 64,
nxly = U.
Herex^'^^^U '^  ^ L 16( 28  11) ^ 8,^^ ^
2(4955) 104
.. y = 5, by substitution.
EXERCISES. CLXXX.
Solve by determinants, checking in the usual way.
. 24^ 93 . „ 49 56 ^
1. \ = 41 2. = 7.
X y ^ X y
1?31 = 1. ^^ = 21i.
X y X y ^
3. 23£c302/ = 2. 4. 23a^ + 10 2/ = 252.
10 a:; + 7 2/ = 61. 19 ;z; + 17 y = 154.7.
5. 41x'dly = 4.. 6. 235 £c 234?/:.= 236.
43 a; + 39 2/ = 82. 411 x + 410 ?/ . 412.
7. 52 a; 39 2/ = 13. 8. 0.5 ic  0.3 ?/ 0.021.
18 a^ + 18 3/ = 15. 0.6 X + 2 y ^ 0.332.
APPENDIX.
405
Three linear equations with three unknown quantities.
On solving the system
a^x + biT/ + Ciz = di,
a^x 4 b0 + 02Z = c?2>
a^x + b.^// + c^z = ds,
the roots are found to be
d^b^Cs + d^br^Ci 4 dsbjC^ — dJ)2C^ — djj^c^ — d^^c^
^i<^2<^3 + «'2fi?3Ci + a^diCo, — a^d^Ci — a^d^c^ — a^d^c^
X =
(^\hlCz + «2^3Cl 4 «3^lC2 — «3^2Cl — ^^a^l^S
^i^2f^3 + ciibzdx + a^bidi^ — a^b^di — a.^xdz
 aJy^d^
dih
Cl
d,M
C2
chbs
C3
aih
^1
a2 62
C2
a3&3
C3
ai
di
Cl
a2
d.
C2
as
ds
C3
tti
61
Cl
ag
62
C2
a3
63
C3
ai
W
dl
a2
&2
d2
ag
&3
ds
ai
&1
Cl
az
62
C2
as
&3
C3
ai^a^s + a^bsCi + aaZfiCg — asb^c^^ — ag^i^s — «i^3C2
It is at once seen that the same law already set forth
holds here, and that the roots may be expressed thus :
y =
It is thus seen that the roots of three linear simultaneous
equations can be written down, in the determinant form, at
sight. It then becomes merely a matter of simplifying.
Whether it is easier to solve by determinants, or to solve
by elimination through addition and subtraction, depends
largely on the size of the coefficients. If the coefficients
are small, there is usually no advantage in using deter
minants ; if they are large there is often a great gain. In
the problems on the next two pages the coefficients are not
in general large enough to make it worth while to use
determinants except for practice.
406
ELEMENTS OF ALGEBRA.
Illustrative problem. Solve the system
13x + lly \llz = 0,
17 a? + 15 2/ + 80;^ = 30.
The common denominator for ic, ?/, z is
11 9 33
2 9 33
2 9 33
19 33
13 11 71 =
2 11 71
=
2 38
= 22
1 19
17 15 80
2 15 80
04 9
04 9
= 2 . 2 (9  4 • 19) =  4 ■ 67.
(How is the second determinant obtained from the first ? the third
from the second ? and so on.)
The numerator for x is
52 9 33
26 9 33
Oil 71
= 25
11 71
= 10
20 15 80
3 3 16
17 7!
11 71
13
10(2611 13
 10 • 134.
10 134 ^
= 5
467
1771 3.117)
11 52 33
13 71
= 2
17 30 80
The numerator for y is
11 26
13 32 = 2(17322629. 132611 1532)
17 15 29
=  2 938.
. _ ^^
"We may now find z by substitution. Or the numerator for
11 9 52
2 9 26
2 26
13 11
= 2
2 11
= 22
1 11
17 15 30
2 15 15
4 15
by factoring by 2 and subtracting the second row from each of the
others. This equals 4 . 134.
4 • 1.34
.. z = =  2.
 4  67
APPENDIX. 407
EXERCISES. CLXXXI.
Solve by determinants, checking all numerical results in
the usual way.
2. 5ic3^ = 3.
2x + y = 5.
3 2/ + « = 7.5.
X . y rr ^
3 + 2^=^^
2 3^4
6. 7x3ij2z = lQ.
2x5 7j{Sz = S9.
5x + 1/ { 5z = 31.
8. 3ic + 3 2/ 4 3 2; = 144.
9x \y — z = 154.
10. a^^ + Z^'^^z+c^^^ft + Z' + c.
aa: + 6?/ + cs; = 1.
X + y + z = 0.
12. 2a;32/ + 42; = 18.
3a; + 4?/5^ = 34.
X {y + z = 0.
13. 123x + 17 2/139;s = l.
51 ic + 37 2/  97 ^ =  9.
5 a^ + 31 2/  35 s = 1.
1.
x + y = 10.
y{z = 10.
x\ z = 6.
3.
5 + f = 3.
a
z
5 + 5 = 4.
a c
5.
12x47 7/ = 109.
52/ 2.^ = 11.
4a; + 3;^ = 26.
7.
4:X{9y\z = 16.
2a; + 32/ + .^ = 4.
a; + 2/ + ^ = l
9.
^'■^o; + q^y + r^t; = s\
p^x + </^2/ + ^'^^ — "^^•
^ic + 2'?/ + r;s = s .
11.
3cc + 42/ + 2;^ = 47.
5x3y + 7z = 4:l.
7x2y5z = 24..
Pa
408 ELEMENTS OF ALGEBRA.
VIII» GRAPHIC REPRESENTATION OF LINEAR
EQUATIONS (p. 202).
In the annexed figure the two lines XX^ and YY\ inter
secting at right angles at 0, are called rectangular axes.
A segment, OA, on OX is called
the abscissa of any point, as Pi,
on a perpendicular to XX' at A.
A segment, OB, on Z is called
the ordinate of any point, as Pj,
X on a perpendicular to FF' at P.
The abscissa and ordinate to
gether are called the coordinates
of the point, the abscissa always
being named first.
Abscissas to the right of are called positive, those to
the left negative. Ordinates above are called positive,
those below negative.
E.g.^ in the figure the coordinates of Pi are 3, 2 ; those of P^ are
— 4, 1 ; those of P3 are — 2, — 3 ; those of P4 are 2, — 2 ; those of
A are 3, ; those of are 0, 0. The ordinate of any point on XX'
is evidently 0, and the abscissa of any point on YY' is also.
Hence, when the axes are given a point in their plane is
fixed when its coordinates are known. Conversely, when
a point is fixed its coordinates with respect to any given
axes are evidently fixed also.
A point, as Pi, is designated by its coordinates.
Thus, Pi is designated by (3, 2), P2 by ( 4, 1), P3 by ( 2,  3),
and P4 by (2,  2).
If the coordinates are unknown, they are designated by x and ?/,
the point being designated by the symbol (x, y). That is, if it is
desired to speak of two general points, as we speak of two unknown
quantities in algebra, they may be designated either as Pi, P2, or as
(«1, Vl), (X2, ?/2).
. . APPENDIX. 409
EXERCISES. CLXXXII.
In each exercise draw a pair of rectangular axes and
take I inch as the unit of measure for laying off the coor
dinates.
1. Represent the points (2, 5), (— 4, — 7).
2. Also (5, 0), (0, 5).
3. Also (0, 0), (2, 2), ( 4,  4). Join these. Do they,
or do they not, lie in the same straight line ?
4. Similarly for the points (3,0), (0,3), (3,0), (0,3).
5. What kind of a figure is formed by joining, in order,
the points (2, 4), ( 2, 4), ( 2,  4), (2,  4)?
6. Also the points (6, 3), (3, 3), (3,  5)?
The graph of an equation. The equation y = x — 1 is
satisfied if
x = 1, 0, 1, 2, . . .
while 7/ = 2, 1, 0, 1, •••.
The points (—1,2), (0,1), (1, 0), • • • may, therefore,
be thought of as lying on a line representing this equation.
Hence, in the figure the line
MN is considered the graphic
representation of the equatioil
y = x — l. Such a graphic rep
resentation is called the graph
of the equation. ^ '
The word locus is sometimes
used instead of graph. Locus M
is a Latin word meaning jyZace,
and the line is the place where the points are found.
Strictly, therefore, it is the graph of the equation and the
locus of the points which we have.
410 ELEMENTS OF ALGEBRA.
In a case like that of y = x — 1, y is a, function of x.
Hence, the abscissas represent the variable x, and the ordi
nates represent the function.
A simple equation containing two unknown quantities can
always be represented graphically by a straight line.
This is the reason why it is called a linear equation, a
term which has, however, been extended to include all
equations of the first degree.
Hence, it is necessary to locate only
/ two points to determine the graph of a
simple equation.
The easiest plan usually consists in
letting X = and finding the corres
ponding value of y ; then letting y =
and finding the corresponding value of x.
E.g., to draw the graph of the equation 2x — Sy =7. Ifx = 0,
then 2/ = — I ; if y = 0, then x = . Hence, draw a line through
(0, — I) and (I, 0), as in the figure.
Since the line represents the equation it is evident that
the coordinates of any point on the graph satisfy the equa
tion. That is,
a single linear equation in
volving two unknown quan ^
tities has an infinite number
of roots.
a single straight line has an I
infinite number of points.
EXERCISES. CLXXXIII.
Draw the graphs of the following equations :
1. X — y = 0. 2. X + y = 0.
3. 2xy = 8. 4. 7x4:y = 10.
5. 2x\3y = 5. 6. 16x{2y = 7.
APPENDIX.
411
It is also apparent that although a single
linear equation has an in
finite number of roots, two
linear equations involving
two unknown quantities have
in general but one common
pair of roots.
For example, the two equa
tions
(a) x^2y = S
(b) 2x + y = l
have the graphs a. and b.
The two equations have
the common pair of roots
whose values are
x = 2, 2/ = 3.
straight line has an infinite
number of points, two straight
lines have in general but one
common point.
The two graphs have the
common point whose coordi
nates are
x = 2, ?/ = 3.
Hence, two linear equations involving two unknown
quantities can be solved by means of graphs, although
this is not advisable in practice.
EXERCISES. CLXXXIV.
Draw the graphs of the following pairs of equations and
show that the intersections represent the solutions.
1. X \y = 0.
X — y = 0.
3. 3x^5y = 12.
x + y = 2.
2. 5x{2y = 16.
3 X — y = 3.
4. 5x \7y=ll.
7x^oy = l.
412
ELEMENTS OF ALGEBRA.
Discussion of Solutions.
While in general
two linear equations involv
ing two unknown quantities
have a single common pair of
roots, they may not, for they
may be inconsistent or they
may be indeterminate.
E.g., the equations
2x + Sy = 6
2 X + 3 2/ = 4
have evidently no common pair
of roots, since that would make
6 = 4. Hence, they are called
inconsistent.
Also the equations
? + ^ = l
2 3
Sx + 2y = 6
have no determinate solution, for
the members of the second are
merely six times those of the first.
They are, therefore, equivalent
equations, and reduce to a single
indeterminate equation.
two straight lines in a iihine
have a single common point,
they may not, for they may
be parallel or they may coin
E.g., the graphs of
2x + 3y = 6
2 X + 3 2/ = 4
are parallel.
^<^
Also the graphs of
2 3
3 X + 2 ?/ =
coincide.
It is a mistake quite often made by students to think
that it is possible to solve anj/ two equations like
8 cc + 4 ?/ = 5.
They may not be simultaneous, as in this case.
APPENDIX. 413
EXERCISES. CLXXXV.
Discuss the following systems of equations, solving if
possible and drawing graphs in all cases.
2, X \ 7/ = 4:.
X — y = 6.
4. x\3y = 6.
X = 4:.
6. 6x\10y = 10.
5 3
8. 7x + 35y = 15.
3x^15y = S.
10. Sx12y = 2.
3 2 12"
12. 1.02 a; 0.01 2/ = 20.1.
0.2£c0.l2/ = l.
In the same way it may happen that three equations in
volving three unknown quantities may be inconsistent or
indeterminate.
Illustrative problems. 1. Solve the following system :
1. 9ic + 62/ + 3^ = 30.
2. 6£cf4^ + 2^ = 20.
3. x + 2y + 3z = 14.
Equations 1 and 2 are easily seen to be equivalent. Hence, there
are only two independent equations, involving three unknown quan
tities, and they are indeterminate.
1.
x\y = l.
x — y = 1.
3.
2/ + 3ic = 6.
2/ = 4.
5.
2x + 3y = ^.
2^3 •
7.
X 1.52/ = 10.
2x3y = B.
9.
10 a; + 6?/ = 5.
"i + yl.
3^5 6
11.
6 a; + 0.8?/ = 10
3 a; + 0.4 2/ = 6.
414 ELEMENTS OF ALGEBRA.
2. Solve the following system :
1. 6x + liy + 2z = Ul
2. dx\2y + Sz = 22.
3. x\2y\3z = 14:.
Equations 1 and 2 are easily seen to be equivalent. Hence, there
are only two independent equations, involving three unknown quan
tities. But these two are determinate as to x, for subtracting 3 from
2 we have 8 x = 8, .. x = 1. But y and z are indeterminate. That is,
these two equations are inconsistent except for x = 1.
3. Solve the following system :
1. 9x{6y\3z = 30.
2. 6cc + 4y + 2^ = 30.
3. x\2y\Sz = U.
Equations 1 and 2 are easily seen to be inconsistent ; for if the
members of 1 are multiplied byf, Qx + 4:y + 2z = 20 instead of 30.
EXERCISES. CLXXXVI.
Discuss the following systems of equations with respect
to their being indeterminate or inconsistent.
1. x^2y2z = 0. 2. 5x + 3 y  z =  17.
2x — y + z = 10. 6 .r + 4 ?/ = — 14.
Sx{yz = 10. x + y + z = 3.
3. 6a; + 97/ + 12s! = 5. 4. lx + lly + Az = 22.
2x + 3y + 4.z = 32. 2x + 3y { 4:Z = 9.
3x\2yz = 8. 5x + 2yz = 6.
5. lOx + 5y — 15 z = 5. 6. 2x — 3y — 4:Z = — 5.
2x + y3z = 0. x + 6y + 3z =^10.
3x{2 y + z = Q. x + 5y + 3z = 9.
APPENDIX.
415
IX. GRAPHS OF QUADRATIC EQUATIONS (p. 296).
The student has already learned in Appendix VIII how
graphically to represent a simple equation. Furthermore
he has learned that to every point on the graph corresponds
one root and only one of the equation, and vice versa, a
"onetoone correspondence" between points and roots.
He has also learned that
as, in general,
two straight lines in a plane
have one common point and
only one,
We shall now consider the graphs
of equations of degrees above the
first.
so, in general,
two linear equatio7is have
one common root and only
one.
Y
Illustrative problems. 1. Required
the graph of the equation
a;2 + 7/2 ^ 10.
'•' y = ± vlO — x2, .. by giving x va
rious values (noticing that x^ ^ 10 for real values of y) we have corres
ponding values of y as follows :
x = ± VIo, ± 3, ± VS, ± Vt, ± VO, ± V5, ±2, ± V2, ±1, 0.
y= 0, ± 1, ± V2, ± Vs, ±2, ± V5, ± Vo, ± Vs, ±.3, ± VlO.
Taking the approximate square roots, and laying off the abscissas
and ordinates as indicated, and then connecting the successive points,
the graph is the circumference of a circle.
So, in general, the graph of every equation of the forno x^ + y^ = k^
is the circumference of a circle.
2. Required the graph of the equation 2 a;^ f 3 y"^ = 10.
■•• y = ± iV6(5 — x^), .. by giving x various values (noticing that
X ^ 5 for real values of y) we have corresponding values of y as fol
lows:
±V5, ±2, ±V3, ±V2, ±1, ±0.
0, ±iV6, ±fV3, ±V2, ±fV6, iiVyo.
416
ELEMENTS OF ALGEBRA.
Taking the approximate square roots, and
laying off the abscissas and ordinates as indi
cated, and then connecting the successive points,
the graph is a curve known as an ellipse.
So, in general, the graph of every equation of
the form ax^ + by2 = c, where a, b, c are posi
tive, is an ellipse.
3. Required the graph cf the equation 2x^ — Sy^ = 10.
'.' y = ±k "v^C (ic2 — 5), .. by giving x various values (noticing that
x2 <^ 5 for real values of y) we have corresponding values of y as fol
lows:
X = ± V5, ± Ve, ± Vt, ± Vs,
±iV6, ±fV3, ±V2,
y
0,
±3,
Y
± Vio,
Taking the approximate
square roots, and laying off
the abscissas and ordinates as
indicated, and then connect
ing the successive points the X
graph is the curve known as
the hyperbola.
So, in general, the graph
of every equation of the form
ax2 — by2 = c, where a, b, c, are positive, is an hyperbola.
4. Required the graph of the equation t/'^ = Sx.
'■' y — ±i "^2 X, .. by giving x various values (noticing tl^at x <^
for real values of y) we have corresponding values for y as follows :
x = 0, 1, 2, 3, 4, 5, 6, •••.
y = 0, ±iV2, ±1, ±iV6, ±V2, iiVio, ±V3, ....
Taking the approximate square roots, and
laying off the abscissas and ordinates as indi
cated, and connecting the successive points, the
graph is a curve known as the parabola.
So, in general, the graph of every equation
of the form y2 = ax is a parabola.
The ellipse, hyperbola, and parabola are curves formed by cutting
into a right circular cone, and hence are called conic sections.
APPENDIX. 417
5. Required the graph of the equation
cc^ + 3 ^2  a;  3 = ?/.
Giving X various values, we have corres
ponding values of y as follows :
x = 4,_3,_2,l, 0,1,2,3, 4,.
y= 15, 0, 3, 0, 3,0,15,48, 105,. ■..
The curve is seen to be one which can be
cut by a straight line in three points, and this is the general charac
teristic of graphs of cubic equations.
EXERCISES. CLXXXVII.
1. Required the graph of the equation x'^\y^ — '26. In
how many points at the most could a straight line cut this
curve ?
2. Similarly for y^ = 18 aj.
3. Similarly for 2x'^ + 5i/=^ 10.
4. Similarly for 2x''^i/ = 10.
5. In how many points, at the most, can a straight line
cut the graph of a quadratic equation, judging by the results
of exs. 14 ?
6. Required the graphs of the equations
x^ + y' = 13,
drawn with respect to the same axes. What are the abcis
sas and ordinates of the points of intersection of tl^e two
graphs ? How do these compare with the common roots
of the two equations ?
7. Similarly for the equations
418
ELEMENTS OF ALGEBRA.
8. Similarly for the equations
f 2x^ = 80.
9. Required the graph of the equation y = x^ — 9x,
In how many points does this curve cut the X axis ?
10. Required the graph of the equation
1/ = x^ — 5 x^ { 9 x^ — 5 X.
In how many points could a straight line cut this curve ?
Graphic representation of the roots. In the equation
y = x^ — X — 2
we have the following corresponding values :
3,
2,
1,
0,
1, 2, 3, 4, .
0,
4,
0,
2,
2, 0, 4, 10, .
The graph is shown in the annexed figure."
When y = 0, x = — 1, or 2. That is, the
values of x, in the equation x^
2 = 0,
"^■^""X are the abscissas of the intersection of the^
graph with the X axis.
Similarly, any equation f(x) = can be
solved by writing it f(x) = y and plotting it.
The abscissas of the intersections of the graph with the X,
axis will then be the roots of the equation.
Imaginary roots show themselves by a curve which does
not reach the X axis.
JE.g., in studying the equation x^ — 5 x^ ^ 8 x — & = 0, let
y =f(x). Then we have the following corresponding values :
x =  1, 0, 1, 1.5, 2, 3, 4.
y = 20, 6, 2, 1.875, 2, 0, 10.
APPENDIX.
419
The curve does not reach the X axis between 1 and 2.
In solvmg the equation the roots are found to be 1 + i,
1 — i, and 3.
The fact that complex roots enter in pairs
is readily understood by a study of the graph.
E.g., consider the equation ic^ — 4 = 0.
Let f{x) = y. We then have the following
corresponding values :
x= 0, ±1, ±2, ±3, ....
2/4, 3, 0, 5, ....
If, now, we make each y 3 units greater, i.e., if we lift
the curve 3 units (or, what is the same thing, lower the X
axis 3 units) x will equal ± 1 when y = 0.
I.e., the roots will approach each other.
This can be done by making the equation
x''l=y.
If we lift the curve another unit (or lower
the X axis 4 units), making the equation
x^ = y, X will have only the double root
when y = 0; i.e., the two roots are now equal.
If we lift the curve another unit (or lower the X axis 5
units), making the equation x^\.l = y, x will have the
imaginary value ± i when y = 0, the two imaginaries enter
ing together. In other words, complex roots (of which pure
imaginaries form a special class) enter in pairs.
Roots of simultaneous equations. We have seen that two
linear equations, involving but two unknown quantities, can
be solved by finding the point of intersection of their graphs.
Similarly, if we have two equations like
I.
11.
x' + 3i/ = 2S,
2x^y^ = 7,
420
ELEMENTS OF ALGEBRA.
the coordinates of the common points of their graphs repre
sent the common roots of the equations.
From I, we have y = ±^ V3(28 — x^).
Hence, if
x= 0, ± 1, ± 2, ±3, ± 4, ± 5, ± V28,
then 2/ = ±fV2l, ±3, ±2V2, ±iV57, ±2, ±1, 0.
The graph is marked I in the annexed figure. It is an
ellipse.
From II, we have 3/ = ± V2 x^ + 7.
Hereifcc= 0, ± 1, ± 2, ± 3, ± 4, ••.,
then 7/ = ± V7, ±3, ± Vl5, ± 5, ± V39, ■ • •.
The graph is marked II in the annexed figure. It is an
hyperbola, a twobranched figure.
The common roots are
x = l, 1,1,1.
y = 3, 3, 3, 3.
From the preceding figure we see confirmed the fact
already mentioned, that two simultaneous quadratic equa
tions involving two unknown quantities have, in general,
four roots, the two curves intersecting in four points.
Two of the points may coincide, as in Fig. 1, giving a
double root, or there may be two double roots, as in Fig. 2,
APPENDIX.
421
or two of the roots may be imaginary, as in Fig. 3, or both
pairs of roots may be imaginary.
Fio. 1.
Fig. 2.
Fig. 3.
From similar considerations it may be inferred that there
are 6 roots common to two simultaneous equations of which
one is a quadratic and the other a cubic. In general, if one
equation is of the Mth degree and the other of the fith,
there are m?i roots.
EXERCISES. CLXXXVIII.
Represent graphically the following sets of simultaneous
equations, and find at least one value of x and one of i/
common to the two.
1. x^ + y^ = S.
2. x'^y = 7.
y^ = 2x.
x + y' = ll.
3. a; + 37/ = 10.
4. 2.^2^7/2 = 19.
x'h5 = 2y.
x^y^ = 8.
5. 2x' + y^ = 3.
ic2 _ 3 7/2 = _ 2.
6. x^ + x^ — x — 3 = y.
x + y = 5.
7. a;2 + 7/2 = 5.
y = x^ + 3x^4:X\2.
TABLE OF BIOGRAPHIES.
The simple equation was known to the Egyptians, its solution
appearing in the oldest deciphered mathematical work extant, the
papyrus of Ahnies. The quadratic equation was solved by the
Greeks, and indeterminate equations formed a considerable portion
of the works of Diophantus. The Hindus, Persians, and Arabs next
took up the science and made considerable progress in the study of
equations and series. The Arabs gave to algebra its name.
The sixteenth century saw a great revival of learning in general
and of algebra in particular. The cubic and quartic equations were
now solved.
The seventeenth century saw modern symbolism established, thus
forming elementary algebra as it is known today.
The following table contains the names mentioned in this work,
together with a few others prominent in the history of algebra. The
notes are chiefly from those prepared by the authors for their transla
tion of Fink's " History of Mathematics" (Chicago, The Open Court
Publishing Co., 1900), to which reference is made for a more complete
account of the development of the science.
Abel, Niels Henrik. Born at Pindoe, Norway, August 6, 1802 ; died
April 6, 1829. Proved the impossibility of the algebraic solution
of the quintic equation.
Ahmes. An Egyptian scribe. Lived about —1700. Wrote the earliest
deciphered mathematical manuscript extant, on arithmetic, alge
bra, and mensuration.
Al Khowarazmi, Abu Jafar Mohammed ibn Musa. First part of ninth
century. Native of Khwarazm (Khiva). Arab mathematician
and astronomer. The title of his work gave the name to algebra.
Ampere, Andr^Marie. Born at Lyons, France, in 1775 ; died at
Marseilles in 1836. Founder of the science of electrodynamics.
423
424 TABLE OF BIOGRAPHIES.
Apollonius of Perga, in Pamphylia. Taught at Alexandria between
— 250 and — 200, in the reign of Ptolemy Philopator. Solved
.the general quadratic with the help of conies.
Aigand, Jean Robert. Born at Geneva, 1768 ; died c. 1825. Private
life unknown. One of the inventors of the present method of
geometrically representing complex numbers (1806).
Aristotle. Born at Stageira, Macedonia, — 384 ; died at Chalcis,
Eubcea, — 322. Great philosopher. Represented unknown quan
tities by letters ; suggested the theory of combinations.
Aryabhatta. Born at Pataliputra on the Upper Ganges, 476. Hindu
mathematician. Wrote on algebra, including quadratic equations,
permutations, indeterminate equations, and magic squares.
B^zout, Etienne. Born at Nemours in 1730 ; died at Paris in 1783.
Prominent in the study of symmetric functions and determinants.
Bhaskara Acharya. Born in 1114. Hindu mathematician and astrono
mer. Author of the " Lilavati " and the " Vijaganita," contain
ing the elements of arithmetic and algebra. One of the most
prominent mathematicians of his time.
Bombelli, Rafaele. Italian. Born c. 1530. His algebra (1572) sum
marized all then known on the subject, especially the cubic.
Boyle, Robert. Born in Ireland, Jan. 25, 1627 ; died Dec. 30, 1691.
Celebrated physicist.
Brahmagupta. Born in 598. Hindu mathematician. Contributed to
geometry, trigonometry, and algebra.
Briggs, Henry. Born at Warley Wood, near Halifax, Yorkshire,
February, 15601 ; died at Oxford Jan. 26, 16301. Savilian pro
fessor of geometry at Oxford. Among the first to recognize the
value of logarithms ; those with decimal base bear his name.
Burgi, Joost (Jobst). Born at Lichtensteig, St. Gall, Switzerland,
1552 ; died at Cassel in 1632. One of the first to suggest a system
of logarithms. The first to recognize the value of making the
second member of an equation zero.
Cardan, Jerome (Hieronymus, Girolamo). Born at Pa via, 1501 ; died
at Rome, 1576. Professor of mathematics at Bologna and Padua.
Mathematician, physician, astrologer. First to publish (1545) the
solution of the cubic equation.
Cataldi, Pietro Antonio. Italian mathematician, born 1548 ; died at
Bologna, 1626. Professor of mathematics at Florence, Perugia,
and Bologna. Pioneer in the use of continued fractions.
I
TABLE OF biographip:s. 425
Cauchy, Augustin Louis. Born at Paris, 1789 ; died at Sceaux, 1857.
Professor of niatliematics at Paris. One of tlie most prominent
mathematicians of his time. Contributed to the theory of deter
minants, series, and algebra in general.
Cramer, Gabriel. Born at Geneva, 1704 ; died at Bagnols, 1752.
Added to the theory of equations and revived the study of deter
minants (begun by Leibnitz).
D'Alembert, Jean le Rond. Born at Paris, 1717; died there, 1783.
Physicist, mathematician, astronomer. Contributed to the theory
of equations.
De Moivre, Abraham. Born at Vitry, Champagne, 1667 ; died at
London, 1754. Contributed to the theory of complex numbers
and of probabilities.
Descartes, Ren^, du Perron. Born at La Haye, Touraine, 1596 ; died
at Stockholm, 1650. Discoverer of analytic geometry. Contrib
uted extensively to algebra.
Diophantus of Alexandria. Lived about 275. Most prominent of
Greek algebraists, contributing to indeterminate equations.
Euclid. Lived about — 300. Taught at Alexandria in the reign of
Ptolemy Soter. The author or compiler of the most famous text
book of geometry ever vv^ritten, the "Elements," in thirteen books.
Euler, Leonhard. Born at Basel, 1707 ; died at St. Petersburg, 1783.
One of the greatest physicists, astronomers, and mathematicians of
the eighteenth century.
Ferrari, Ludovico. Born at Bologna, 1522 ; died in 1562. Solved the
biquadratic.
Ferro, Scipione del. Born at Bologna, c. 1465 ; died between Oct. 29
and Nov. 16, 1526. Professor of mathematics at Bologna. Inves
tigated the geometry based on a single setting of the compasses,
and was the first to solve the special cubic x^ + px = q.
Gauss, Karl Friedrich. Born at Brunswick, 1777 ; died at Gottingen,
1855. The greatest mathematician of modern times. Prominent
as a physicist and astronomer. The theories of numbers, of func
tions, of equations, of determinants, of complex numbers, of
hyperbolic geometry, are all largely indebted to his great genius.
Harriot, Thomas. Born at Oxford, 1560 ; died at Siou House, near
Isleworth, July 2, 1621. The most celebrated English algebraist
of his time.
426 TABLE OF BIOGRAPHIES.
Horner, William George. Born in 1786 ; died at Bath, Sept. 22, 1837.
Chiefly known for his method of approximating the real roots
of a numerical equation (1819).
Lagrange, Joseph Louis, Comte. Born at Turin, Jan. 25, 1736; died
at Paris, April 10, 1813. One of the foremost mathematicians of
his time. Contributed extensively to the calculus of variations,
theory of numbers, determinants, and theory of equations.
Maclaurin, Colin. Born at Kilmodan, Argyllshire, 1698 ; died at York,
June 14, 1746. Professor of mathematics at Edinburgh. Contrib
uted to the study of conies and series.
Metrodorus. Lived about 325, Author of many algebraic problems.
Napier, John. Born at Merchiston, then a suburb of Edinburgh, 1550 ;
died there in 1617. Inventor of logarithms.
Newton, Sir Isaac. Born at Woolsthorpe, Lincolnshire, Dec. 25, 1642,
(0. S.); died at Kensington, March 20, 1727. Lucasian professor
of mathematics at Cambridge (1669). The world's greatest mathe
matical physicist. Invented fluxional calculus (c. 1666). Contrib
uted extensively to the theories of series, equations, curves, etc.
Ohm, Georg Simon. Born at Erlangen, Germany, in 1781 ; died July
6; 1854. Celebrated physicist.
Pascal, Blaise. Born at Clermont, 1623 ; died at Paris, 1662. Physi
cist, philosopher, mathematician. Contributed to the theory 6t
numbers, probabilities, and geometry.
Recorde, Robert. Born at Tenby, Wales, 1510 ; died in prison, at
London, 1558. Professor of mathematics and rhetoric at Oxford.
Introduced the sign = for equality.
Tartaglia, Nicolo. (Nicholas the stammerer. Real name, Nicolo Fon
tana.) Born at Brescia, c. 1500; died at Venice, c. 1557. Physi
cist and arithmetician ; known for his work on cubic equations.
Taylor, Brook. Born at Edmonton, 1685 ; died at London, 1731.
Physicist and mathematician. Known for his work in series.
Vifete (Vieta), Frangois. Born at FontenayleComte, 1540 ; died at
Paris, 1603. The foremost algebraist of his time.
Volta, Alessandro. Born at Como, Italy, Feb. 18, 1745 ; died March
5, 1827. Celebrated physicist.
Wallis, John. Born at Ashford, 1616 ; died at Oxford, 1703. Savilian
professor of geometry at Oxford. Suggested (1685) the modem
graphic interpretation of the imaginary.
TABLE OF ETYMOLOGIES.
The following table will serve to make more clear to students the
meaning of many words used and defined in elementary algebra,
i., Latin. G., Greek. dim., diminutive.
Abscissa. L. cut off.
Absolute. L. absolutus, ab, from,
+ solvere^ loosen. That is, com
plete, unrestricted.
Abstract. L. abs, away, + tra
here, draw.
Add. L. ad, to, + dere, for dare,
put, place.
Affected. L. ad, to, +facere, do,
make; i.e., to act upon, influ
ence. Hence compounded ; an
equation of several degrees.
Aggregation. L, ad, to, f gre
gare, collect into a flock, ^from
grex, flock.
Algebra. Arabic, al, the, \jabr,
redintegration, consolidation.
The title of Al Khowarazmi's
work (see Table of Biographies)
was Him aljabr wa'' I muqa
balah, the science of redintegra
tion and equation ; of this long
title only aljabr survives.
Alternation. L. alter, other.
Antecedent. L. ante, before, 
cedere, go.
Antilogarithm. L. and G. anti,
against, opposite to, + loga
rithm. See Logarithm. The
number standing opposite to the
logarithm.
Arithmetic. G. arithmos, num
ber.
Ascend. L. ad, to, + scandere,
climb.
Associative. L. ad, to, fsociare,
join.
Axiom. G. axioma, that which
is thought fit, a requisite.
Binomial. L. bi, two, + nomen,
name.
Characteristic. G. charakterizein,
designate; from G. character,
an instrument for graving, from
charassein, to scratch.
Circulate. From circle. L. dim.
of circus, a ring, G. kirkos or
krikos, a circle, ring.
Commutative. L. com, inten
sive, I mutare, change.
Comparison. L. com, together
with, + par, equal.
Complement. L. complementum,
that which fills ; from com,
intensive, + plere, fill.
427
428
TABLE OF ETYMOLOGIES.
Complete. See Complement.
Complex. L. com., together, +
plectere, weave.
Composition. L. com, together,
4 ponere, place.
Cjompound. Same etymology as
Composition.
Consequent. L. con, together,
+ sequi, follow.
Constant. L. con, together, f
stare, stand.
Continued. L. con, together, +
tenere, hold.
Corollary. L. corollarium, a gift,
money paid for a garland of
flowers, from corolla, dim. of
corona, a crown.
Cube. G. cuhos, a die, a cube.
Decimal. L. decem, ten.
Deduce. L. de, down, away, +
ducere, lead.
Define. L. de, +finire, limit,
settle, define.
Degree. L. de, down, + gradus,
step.
Denominator. L. namer, from
de, + nominare, name, from
nomen, name.
Descend. L. de, down, + scan
dere, climb.
Detach. Ital. des, privative, +
tacher, fasten.
Determinant. L. de, + termi
nare, bound, limit.
Determine. See Determinant.
Discriminant. L. dis, apart, +
cernere = G. krinein, separate.
Distribute. L. dis, apart, + tri
buere, give.
Divide. L. di, for dis, apart, 
videre, see.
Domain, L.dommmm, dominion,
from dominus, lord.
Eliminate. L. e, out, + limen, a
threshold. To turn out of doors.
Equal, L. aequalis, equal, from
aequus, plain.
Equation. See Equal.
Evolution. L. e, out,  volvere,
roll. To unfold the root.
Exponent. L. ex, out, + ponere,
put ; i.e., to set forth, indicate.
Extraneous. L. extra, outside.
Extreme. L. extremus, superla
tive of exter, outer.
Factor. L. a doer, from facere,
do.
Fraction, L. fractus, broken,
from, frangere, break.
Function. L, functus, performed,
from, fungi, perform.
Graph. G. graphein, write.
Homogeneous. G. homos, the
same, + genos, race.
Identical. L, idem, the same.
Imaginary. L. imago, an image.
Indeterminate, L, in, privative,
4 determinate. See Determi
nants.
Index, L, indicare, point out,
show.
Infinite, L, in, not, +Jinitus,
bounded.
Inspection. L. in, on, in, at, +
specere, look.
TABLE OF ETYMOLOGIES.
429
Integer. L. in, privative, + tan
gere, toucli; i.e., untouched,
whole, sound.
Inverse. L. in, on, toward, +
vertere, turn.
Involution. L. in, in, + volvere,
roll. To roll the root into a
power.
Limit. L. limes (limit), a cross
path, boundary.
Linear. L. linea, line.
Literal. L. littera, litera, a letter.
Logarithm. G. logos, proportion,
ratio, + arithmos, number.
Mantissa. L. an addition.
Maximum. L. greatest, superla
tive of magnus, great.
Mean. L. medius, middle.
Minimum. L. least.
Minuend. L. minuere, lessen.
Monomial. G. monos, single,
+ L. nomen, name.
Multiple. L. multus, many, +
plus, like English fold, from
plicare, fold.
Negative. L. ne, not, + que, a
generalizing suffix.
Notation. L. notatio, a marking,
from nota, a mark, a sign.
Number. L. numerus, number.
Numerator. L. numberer.
Operation. L. opus, work.
Ordinate. L. ordo (ordin), a row.
7t. Initial of G. periphereia, pe
riphery, circumference.
Polynomial. G. polus, many, +
L. nomen, name.
Positive. L. positivus, settled,
from ponere, put.
Power. L. posse, to be able.
Problem. G. problema, a question
proposed for solution ; from
pro, before,  ballein, throw.
Product. L. pro, forth, + du
cere, lead.
Proportion. L. pro, for, before,
+ portio, a share.
Proposition. L. pro, before, +
ponere, place.
Pure. L. purus, clean.
Quadratic. L. quadratus, a
square, from quattuor, four.
Quantity. L. quantus, how much,
from quam, how.
Quartic. L. quattuor, four.
Quotient. L. quot, how many.
Radical. L. radix, root.
Ratio. L. a reckoning, calcula
tion, from reri, think, estimate.
Rational. L. ratio. See Ratio.
Real. L. realis, belonging to the
thing itself, from res, thing.
Reciprocal. L. re, back, + ad
jective formative cus.
Reduce. L. re, back, + ducere,
lead.
Remainder. L. re, behind, back,
I manere, remain.
Root. L. and G. radix, a root.
Series. L. a row.
Similar. L. similis, like.
Simplify. L. simplex, simple.
430
TABLE OF ETYMOLOGIES.
Simultaneous, L. simultim, at
the same time, from simul,
together.
Solution. L. solvere, loose.
Square. L. quadra, a square,
from quattuor, four.
Substitute. L. sub, under, +
statuere, set up.
Subtract. L. sub, +trahere, draw.
Subtrahend. See Subtract.
Sum. L. sumnia, highest part.
Surd. L. surdus, deaf. A mis
translation of the G. alogos,
which does not mean stupid
(hence deaf), but inexpressible.
Symbol. G. symbolos, a mark,
from syn, together, + ballein,
put.
Symmetry. G. syn, together, +
metron, measure.
Theorem. G. theorema, a sight,
a principle contemplated.
Transpose. L. trans, over, +
ponere, place.
Trinomial. L. tres (tri), three,
+ nomen, name.
Vary, variation. L. varius, dif
ferent.
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THE UNIVERSITY OF CALIFORNIA LIBRARY
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