(navigation image)
Home American Libraries | Canadian Libraries | Universal Library | Community Texts | Project Gutenberg | Children's Library | Biodiversity Heritage Library | Additional Collections
Search: Advanced Search
Anonymous User (login or join us)
Upload
See other formats

Full text of "Elements of algebra"

fiKiii 



IN MEMORIAM 
FLORIAN CAJORI 




Digitized by the Internet Arciiive 

in 2008 witii funding from 

IVIicrosoft Corporation 



littp://www.archive.org/details/elementsofalgebrOObemaricli 



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 
text-book 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 text-books 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 one-to-one 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 text-book, 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 
2-3. 

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. 1-9. 

^ 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. 10-17. 



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 many-termed. 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) 2-3V7H-1. (f) 2x^-?>x'-^x-V\. 

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 number-symbol 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^-100-2,. (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#-j-a2-^ + 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. Two-thirds 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, "Two-thirds 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 ? 

fx-H0 = 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 two-thirds 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. .-. |-x-5 + 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=^-1-7. 

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. 

«Ai-20^ ^ .. . 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 one-third 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. .-. x-0.02x-5 = 117.50. 

5. .-. 0.98x = 122.60. (Why?) 

6. .-. X = 125. (Why ?) 
Check. 125-0.02of 125-5 = 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^^x-iof 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". 



o-f 3 or 

O-f-2 " 
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 ? 
of-f? 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, a-b, 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'^ + {2a-b + 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 + 3-4 + 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 c-d. (f ) (^ + *) (p + d). 

(g) 2 + a^H^ H-a + 5. (h)2axb-^dxc-a. 

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) (b-ay = b^-2ba + a^ 

(c) (b^-a^)/(b-a)=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)(a-b) = a^-P. 

(b) (a-by = a'' + P-2ab. 

(c) (a - by = a^ - b^ - 3ab(a - b). 

6. Show that a a = 3, b = 2, c = 1, 

(a) (a-]-by-c'^ = (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^ thi-ee 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 2-11). 
weight, a 4-lb. 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 thi-ee 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 + i-15=- 
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 = 9-5; 

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 + 4a6-562 

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 
4a2-5a6 + 262 4 _ lo + 8 = 2 

3a2 + 4ab-5b2 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) 3ic3-7a;2 + 2ic-13 (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%'' -j-J^ 

(e) 18^2^,^ + 4a -3Z>« (f) 2xy-y''-^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) F-Q. (b) P - P. (c) Q-E. 

(d) Q-P. .- - ^ - .„. _ _ 

(g) F-Q-E. 



(b) F-E. (c) Q-E. 

(e) F-{-Q-E. {i) F + E-Q. 
(h) E-Q-F. (i) Q-F-E. 



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^ 


-3-1+1 


= -3 


4a2-3a&-362 


4-3-3 


= -2 


2a2_2a6- &2 


2-2-1 


= - 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)P-g. (h)Q-R. (G)R-P. (d)Q-F. 
(e)Ii-Q. (i)F-R. (g)F-^Q + B. (h)P-{.Q-E. 



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 6-c + a) = 2a-3& + c-a = a-36 + 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'' + 3xy-y^). 

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 — a-b+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^ -{2ab-b''- 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 -{-x-y)- 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.pq-2q'' + 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 0-Si 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 fii-st. 

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 •c-d) = (a-b) ■ (c-d) = (a'b-cy'd = b- a- (d-c)= • • • , 
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'-h--h 


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{x-i-y) = {x + y) + {x + y) = 2x + 2y. 



II. When the monomial is a positive fraction, as in the 

0- 

X l + lH-l + ---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^-i-2c''d). 

5. -5a[-3a + 2(a-2)]. 

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){x-y). 

3. {x^ - y^) {x + y). 4. {p^ + q'') {x^ - 3 /). 

7. (a3 + a2 + a + l)(a-l). 

8. {bx^-x + l){3x^-x-2). 

9. {2x + 3y -z){2x-3y + 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 



xy-i + 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 

2-4 = 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 ^ah-h'' + 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. 2-3 = 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. 1-13, 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. {x-yy. 11. {x -{- yf. 

12. (x-ijy. 13. (x + y)(x-y). 

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. 9-13 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 = AP-2DP-\-BH, 

.-. {X -y)^^x^ -2xy + y^. 
Expressions of the form x -\- y, x — y, are 
called conjugates of each other. 



(x-y)' 



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{x-y) +y{x-y) = {x + y){x- y). And 
•.• GG + FB = AC -AF, 

.-. (.T + 2/) (X - ?/) = X2 - 2/2. 



x-y ^^"--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. 1-18. 

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. (2cc2-l)3. 

8. (2 x"" - yf. ■ 9. [a-(b + c)^. 

10. (2cc2 + l)(2cc2_l). 11. (a2 + 3)(a2_3). 

12. (a-]-b-ab)(a + b-^ ab). 13. [(a + b) {a - b)J. 

14. {a''-\-2ab-{-b'){a''-2ab + b''). 15. (x'' + y') (x^ - y^) . 

16. 42^. 17. 49-51. 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^ ; 

«"•" " a-a-a-- • 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. 1-12. 
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. (x-2y)2 = x2-4x2/ + 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?/2-32x?/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. 1-20. 

1. (20 + 1)2. 2. (x'--3 7/y. 

3. (x + 3yy. 4. (2x-7yy. 

5. (x"" + 2/")*. 6. (a + b + cy. 

7. (^x-^yy. 8. (-x-3yy. 

9. (2x^-3yy. 10. (a -h 2 b -hey. 

11. (-a-b-cy. 12. (a^-h2ab + by. 

15. (a^o-b'^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 (§§ 128-133) 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.= 8-36 + 64-27=: -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. (1-ay. 

3. (x^ — yy. 4. (x — yy\ 

5. (a-2Z')2. 6. (2CC + 2/')'- 

7. {x^y-Zy^y. 8. (a; + 3/ - «)2. 



MULTIPLICATION. 59 

11. {a-h-cy. 12. (2a -^ly. 

13. {a-b + 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. 3-T7-- 

— 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, ^^!-^ ^^^^ = 2a-h. 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 + bY-3(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') 

(2x-iy -{■^{2x-iy-{2x-iy 
^^' -(2x-iy 

- 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 x-y + 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 
1-1)1+0 + 0-1 

1 — 1 Check. Let x = 2. 

1 (2 - 1) (8 - 1) = 4 + 2 + 1 

^-^ or 1 . 7 = 7. 

1 

1-1 X2 + X + 1. 



66 ELEMENTS OF ALGEBRA. 

EXERCISES. XXX. 

Perform the divisions indicated in exs. 1-14. 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 4-8aS6-14a4Z>2 4.Q,3^3 + 6a2^,4by a3-3«^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*4-xy + y^ by ?/^ — cc?/ + cc^. (Rearrange the divisor.) 

Perform the divisions indicated in exs. 15-31 by using 
detached coefficients, checking as above. 

15. £cs - 5 £c2 _ 3000 by x-h. 

16. 16x^-81?/* by 2x4-3?/. 

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 4-3 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 a-y'' by {x — yf. 

24. 26 rx2 _^ 4 ^3 _ 3 ^4 ^ ^5 _ 92 ^_^ 55 ^^ «^-3a4-ll. 

25. 24 7M*-14m3-9m2-84 + 43m by 7 - 3 /?^ + 4 7/2,2. 

26. ic«-3a;^-5cc5 + 2£c^ + 5a;^ + 4x2 + 2 by .t« + 2x-1. 

27. 3 m« + 7 m^ - 12 «^^ + 2 m^ - 3 wt^ _|_ 13 ^^^ - 6 by m^ 
+ 3 m - 2. 

28. a;^-x^-2x^ + 5iz;^-5iz;^ + 8x2 + 6ic-12 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;^4-3ic^ + 3x8 + 3x2 + 2a34-l by x^ 
+ x^ + x^ + x^ + X + 1. 

32. Divide the product of {x -V){x- 2){x - 3)(x - 4) 
by 2(4-3x) + 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 + 2-2 + 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^ + x-l, 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) = x-y, 
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) -i-f{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 do-sym- 
metry, or, where no misunderstanding will result, simply 
symmetry. 

EXERCISES. XXXIII. 

Select from exs. 1-13 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\b-e) + b\c-a)-hc\a-b). 

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. 14-20 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 Y-x"^ + 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^-2x-36 hj x-2. 

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 + 5-3+0-2+ 7 
-7 1-7 14-77 539 -3759 

1-2 11-77 537 ; - 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;i0-33a;« + l by a; - 1. 

5. a;* + 2a;2-3a;-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^ + 3x-y/a--i^? 
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) = 
x-y. 

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'^ + 2xy-y^ 
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 = 2-3. 

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;2-10a;. 4. m« + 14m8 + 49. 



FACTORS. 81 

5. 121ic2-22£c + l. 6. 4a;2 4-42/(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^ + 2xz-j-2yz-{- 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)(x-y) = x^- y"- (§ 69, 3), it follows that 
expressions in the form of x^ — y"^ can be factored. 

5 • - 3. 



5-3 -1. 



3 1. 



E.g., x'-2-16 = (x + 4)(x-4). 


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;2-92/'. 

5. 16a;* + 4a;22^2_^^4 6 81 ic* + 9 a;^ + 1. 

7. a;2 4-2a;2/ + 2/'-^'. 8. (cc + y)^ - (i» - 2/)^- 

9. a'' + h'' - x"" -1+2 ah + 2x. 

I 10. a2-|-2a^> + &2_(a;2_2iC2/ + 2/^). 

n 11. 4a2-f-4a-3(=4a2_^4a + l-4). 



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){y-2x) = 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)(-X-l)(l-X) 
= (-x2-l)(X+l)(l-x). 

Check. •Letx = 2. Then 

16-l=5.31 = 5-3--l=-5-3- -1. 

EXERCISES. XL. 

Factor the following, giving the various forms of the 
results and checking each. 

1. l-a\ 2. x^-1. 

3. 16 -£C*. 4. a^ — h\ 

5. lQx^-^ly\ 6. 2 + ^* -2^2^. 

7. 121-f £c2-22£c. 8. z^ + 2-2z^V2. 

9. a;i<>-26a;« + 168. 10. a'' - c'' -^r b"" + 2 ah. 

11. 16£c*-f-8a;2 + i_252/^ 12. -x''-l^x^-{-2^x^-lx. 

13. 121 x^ -f 121 2/2 - 9 - 242 xy. 

14. 4.x''-\-l-t/-2ijz-z^ + 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 

= (3x2-2?/)3. C/iecA;. 1 = 18. 

x8 x2?/2 xy* y« /x\3 o{^V{y'^\ .o(^\{y'^\^ /v'^V 

S~'~T^~6r-Y7 = \2)~^\2)\'3)'^^\2)\B) ~\s) 

_/X_^\3 

~ V2 3/ 

Check. Let X = 2, y = 3. Then l_9 + 27-27=-8 = (l- 3)3. 



EXERCISES. XLI. 

Factor the following expressions : 
1. l-Sx + Sx'^-x^ 2. a^-3a^ + 3a-l. 

x^-^- 3x^ + 3x^-1. 4. 27x3-27a;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 + ^>)H-l. 



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^ + 2n-l. 



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 = 5-8, 

and - 40 = 5 • - 8, 

x2-3x-40 = (x + 5)(x-8). Check. -42 = 6.-7. 

EXERCISES. XLIII. 

Factor the following expressions : 

1. x^ + Zx + 2. 2. x''-x-2. 

3. ic^ + a;2_12. 4. x^-^x + Q. 

5. cc2-4iz;-165. 6. j^^-j^-GOO. 

7. a2-aa-130. 8. a;2_4^_21. 

9. ^2- 11 a -60. 10. x'-4.x^-4.5. 

11. 4a;2_|_8£c-45. 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 = 7-5. 
Consider also 8 x^ + 7 x — 3. 
Here 7 = 9-2, 

and 6 - 3 = - 18 = 9 • - 2. 

6x2 4-7x-3 = 6x2 + 9x-2x-3 

' =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. 


12P--P-1, 




4a;2_4cc-3. 


4. 


3 ^2 + 8 a + 4. 




600 a'^ — a- 1. 


6. 


9cc2-17cc-2. 




Ux^-5x-l. 


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 ab-S4:b\ 


15 


16xyz^-{-S9xi/z-27. 


16. 


30x^-4:1 xz-15z\ 



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^ — 4x4-3 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. x8-19i»-30. 2. x^-Sx-2. 

3. m* — 2 mn^ + n^. 4. a^ — a^ — a — 2. 

5. a^-a-2 -\-2a^ 6. a;« + 9 x^ + 20 cc + 12. 

7. a^-6a^-{-lla-6. 8. a^ + 8a^ -112 a + 256. 

9. a^'-a^-lBa-h 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. 81-87. 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_23x4-20. 

17. ccy 4- 2 ccy + xy. 

L9. a^-lba%'' + ^b\ 

II. ah At y'^ — ay — hy. 
13. £c* - 8 icy + 16 ^/. 
\h. (a + by + (a-by. 

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;4-12. 

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)2-5cc-29. 

26. 16 cc* - 28 ic2y2 _|_ ^4^ 

28. 7x3 + 96a;2-103£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^a-2 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;2-14a; + 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;2-4)?/-2a^y/2_ 

58. 121 a* - 795 ^2^2 _^ 9 Z-^ 

59. (a-4)2-4(a-4) + 4. 

60. (a; -5)2 -8(0; -5)+ 12. 

61. x^{x — 2y) — y^{ij — 2x). 

62. l-(a-Z.)-110(a-J)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, " 2-2 = 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 x-2 = 0, " x = 2, " 2-2 = 0; 

id if X - 3 = 0, " X = 3, " 3-3 = 0. 

EXERCISES. XLVII. 

Solve the following equations : 

1. a;2-l = 0. 2. ic2 + 287 =48a;. 

3. 2x^ + 2 = Bx. 4. 6a;2-13x + 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. 2ic3-67cc2 + 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^¥ = 2-2-aa-b'b-b-b, 

V4a2&4^ V(2 ■ab-b)-{2-a-b-b) = ±2-a-b-b 

= ±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 + 4-6- 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. 1-6. 

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. 7-12. 

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;S-40ic2_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., 4-12 + 9 |2-3 

4 

4 -12 + 9 ±(2x''-3y) 

4-3 -12 + 9 



I 



FACTORS. 97 

2. Kequired the square root of 

Root =±(a-62 +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/+7i-2a-62 -2a62-f54 ^ ^^ 

2/ =2 a-2 62 4 ac -4 62c+4 c2 contains 2/n + n2 
2/+n=2a-2&2-j-2c 4ac-4b2c+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 = 1-2+1 + 4-4 + 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. 1-16. 

1. x^ + 2x^-x + i. 

2. l + 8a + 22a2 4-24a» + 9a^ 

3. 9(a^-iy-12(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^ + 6bc-10ca~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()bH-4.0cH-\-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<^4-a; 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. 
Kequii-ed 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. 1-6. 
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 a62-63 contains 




+n2 


+ n2 


3/2n+3/n2+n3 


12 a2 


-6ab 


12a2-6a6 


-12a26+6a62-63 = 




+ 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. 1-6. 

1. 8a8-36a26+-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 hundred-millions 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. 1-4. 

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. 1-3. 

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^, 

a-6 " " " {a-Hf " 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(6-a)(6 + o), • -4(a-6)(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^—x-2. 

Here a;2 - 3x + 2 = (x - 2) (x - 1), or (2 - x) (1 - x), 
and x2-x-2 = (x-2)(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. x3-x2-2x = x(x-2)(x + 1). 

3. ix2 + |x-3 =i(x-2)(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'^-x-Q>, 2-5x-Sx''. 

7. 2x^ — xy — y^, 4: x^ -\- 10 xy -\- 4: y\ 

8. 6a^ + 19ab-7b% 2 a^ -\- ab - 21 b\ 

9. 4a2(et8-68)^ ^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' 
+ 2X-11L 

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. 


^2-4, x^ + 7x2 + 100. 


4. 


cc^ + 1, x^ + ax^ + ax + 1. 



5. x^ - 3 X 4- 2, x2 - 9 X + 14. 

6. x2-9x + 14, 2x^-5x2-441. 

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, x2-l, 293xS-200x'^ + 7x8-50x2-25x-25. 
[12. 1-x^ x^-1, x^-l^Zx-Zx", 247x2-240x-7. 
[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» 

+ 204x2-27. 



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 -i-1. 

x4-x3+2x2-ic+l|a:44- x^+2x'^+ x + l[l 
x4- x^+2x^- x + 1 
2a;l 2x3 +2x 

x2 +I|x4-x3+2x2-ic+llx2-a;+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. 

6x3-13x2+ 9x-2 

2 

4x3-12x2 + llx-3|l2x3-26x2 + 18x-4[3 
12x3--36x2 + 33x-9 
5 |l0x2-15x + 5 

~ 2x2- 3x+l|4x3-12x2+llx-3 |2x-3 
' " 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 + 2x2-x- 1 and2x3 + 9x2 + 9x + 7. 



3+3 + 


2- 


1 - 


1 
2 
2|3 




2 + 9 + 9 + 7|6+ 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 + 


l|2 + 9 + 9 + 7|2 + 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^ + 2x-4., x^-3x-{-2. 

3. x^ + 4., x^-2x^'{-x^ + 2x-2. 

4. x^-4:0x + 63, x^-7x^-h63x-Sl. 



5. x^ + y^ 



2/^ x' + xV + xy + y\ 



6. x«(6a: + l)-£c, 4£c»-2cc(3£c + 2) + 3. 

7. £c4-15£c2 4-28x-12, 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£c5-llic* + 8iK»-2ic2. 

3. 9a2-4^>2 + 4^>c-c2, 2^ ^ c" -\-Z ah -3hc -3ac. 

4. {a - b) {a^ - c2) -(a-c) {a^ - b^), a^ - b', ab - b^ 
— ac -\- be. 

5. cc3-10(a;2 + 3)+ 31a!, ^^^(x - 11) + 2(19 :b - 20), 
a;3_9cr2 + 26£c-24. 

6. a*52 + 4 a^b^ + 3 a%'' -4.ab^-4. b\ a^'b + 3 a%^ - a'^h^ 

7. 3a^-7 ab + 2b^ + bac-6bc + 2c% 12 a^ - 19 ab 
+ 5b'' + llac-llbc + 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. 15-2x-x2= -(x-3)(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+ x-45 
2x3+ 8x2-3x-27 



2 1 4 x2 + 4 X - 


-18 


2x2 + 2x- 


- 9|2x3 + 8x2-3x-27|x + 3 




2 x3 + 2 x2 - 9 X 




6x2 + 6x-27 




6x2 + 6x-27 



:- (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. 1-15. 

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, 10cc2-3 + 13a;. 

5. x^ + 2ax + a% x^ + ab -^ (a + b)x. 

6. Gx^ + llx^'-dx + l, 2a;2 + 3x-2. 

7. x^-x^ + x^ — x-4.,x^-x^ + 2x-8. 

8. x^ + l + 3{x^ + x), x^ + l -^4:{x^ + x)-{-6x^ 

9. 3x^-15ax^ + a^x-5a^, 6£c* - 25aV - 9 a*. 

10. x^ + 20x + 91,35-2x-x'',x^ + 6x^-6x'' + 6x-7. 

11. 2a;^-2x3-a;2-4a;-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(K-7, x^ + 1, 

15. 4^2(3 ^^2) -(27 a + 18), 12a3- a(8a + 27) + 18, 
6(3a-2) + 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'-2x-l, 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^B-10. 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 

|ig|ultiple. 

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 lai-ger. 



■ 



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, 36X-84 (§ 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 
x-1 ' 



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'{i-o)-!'Hc~a)-e^a-i ) 
^ ^ (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 

_ 21aj-10-9a;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- TV-r.- 10. 



11. -^Z "Ta - 12. 

nx*y — nx^y^ 



3ic2-26x + 35 

a« + g^ + 3 g - 5 
^2 _ 4 <^ + 3 

Q>x'' + lxy-^y'' 
Q> x^ + 11 xy-ir^y^ 

x^ -x^-lx + ^ ^ 
a;4 + 2ic« + 2cc-l' 

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^ + 2a-16' 

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 


2-12 a^ 


ic^ + 4a;2 


+ 5 a; + 20 


1 


-6^2 


(1 + axy 


'■-(a-^xy 


x'-5x' 


'-{-7X-8 


2^3 -5a 


;2_^4iC-l 


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^{c-a) + c^{a - b) 
abc (a — b)(b — c) (c — a) 

(a — b) (b — c) (c — a) 
a\b -c) + b^{c-a) + 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. 1-10 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 + 2ic-l 




3x-l 




^x'-^xy-^y^ 




x + y 




x^ -\-^ x^ -^Vlx 


+ 8 



3. — Vr-i 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)'^ 

Kx-y (X + 2/) (X - 2/) ' 
x-y {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^ + 12xy-5y^ 6x^ -11 xy -]- 4^y\ 
^' 8x^-xy-10y^' 2x^-5xy + 2y^' 

X — y X -\- y 2 x^y'^ 



10. 



11, 



y{x-y) x^ + y{x + y) x^ — y^ 
2ic2 + 3cc-4 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 4-8 g + 5 

* a2-9a + 18' a2 + a-12' 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 x-2^ (x-l)(x-2) 

X + 1 X2 - 1 

X x + 3 X -2 _x + (x + l)(x + 3)-(x - l)(x -2) 

■ * X2 - 1 X-1 ~X+ 1 ~ X2- 1 

_x + x2 + 4x + 8-x2 + 3x-2 

~ X2-1 

8x + 1 



X2-1 

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 ^^V-J^-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 ^ 2-x , x-2 



32/^2 g^2^ • -j^_^2 ' \^x-2x^ 

-1,1 x X 2xy 

x + y x — y ' x-\-y x — y x'^ — if 



FRACTIONS. 



125 



g + l Q^ + 2 a-1 
' a + 2 a-{-3 a-{-2' 

' {x-yf x + y x-y 



9. 



10. 



5x^-1x^-^x^ + 11 x-1 

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 a-b "* a^ - b^ 



L3. 



+ 



+ 



{a-b){a-c) Q)-c)(b-a) (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 


= b---d-- 
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 

_ {x-h){x-Z){x-l){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 a-b 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 + 7x4-12 ^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. 1-8 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, 


' x-1. 


5. 


x^-x^ + x^-x + 1, ' 


' x + 1. 


6. 


a« - b% 


' a^ + bK 


7. 


x^-j-xy-^ y\ 


' x^-xy + y\ 


8. 


{a-b){b-c){c-a), ^ 


i (a + b){b-\-i 



Reduce the following to fractional forms, as in § 158, 
checking each result : 



9. 4. a 



6ab-2 
Sb 



11. CC^ + iC + 1 + 



x-1 
13. 1 + a + a2 



10. a + b-i 

a — b 

12. ^8_3^_ 3a;(3-a;) 

x-2 



a^ + 
14. x^ -{- 2 xy -{- 2/ - 



a-1 

(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. 1-6 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^ 

■ 4ic2-f 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 4-5 

2. To find the product of - + - and 

a a 

/a . 6\/a h\ /a\^ /b\'^ 



{i-i)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;2-2aic + 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 = q-T--- ■ 

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^ ' x-y 






1. 


27 
8(x2-2/2) 


Sx _x - y 27 
x-y 3 X 8 {X -{- y) {X - y) 




§161 


o 




{x-y)27 




§156 




Sx-S{x-i-y){x-y) 


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 = 

8-3 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 ^ n-MTT-^ 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 



25x-29 



\5-4:X 2-x) (5-4ic)(2-a;) 

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 _ a-b \ 
• ^2-^.2 a'^^b'') ''\a-b a + bj 

( la-\3 
\ a -3b 



3. . ■--13^ ^2^^-5^_^, 



b ^ 3b-a ^J ' a-3b 

x'^-Q>xy + 9i/ ^ f x^ — 9y^ ^ x^ -{- xy - Q y^ \ 
• x^-^xy + 4.y^ ' \x'' + 4:y^^ x''-xy-Q»y^)' 

10. f^^+_3^y 3.-^^/_l i_\ 

\3a-b 3a-bJ 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' oj-b ^ 

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. 

X2-2/2 



E.g., to simplify — 



y1 

Multiplying both terms by xy2, this equals 
y (g2 - y^) _ y (x + y) 
x(x-y) ~ 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 

a-b 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 

\x-y 



_^x-y\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 — ■ 



1-x 



FRACTIONS. 139 



3. Simplify ^ 



1 



4. Simplify — 



X 



.+ 1 



5. Simplify 



6. Simplify 



7. Simplify 



x-1 
1 



x^ 



x+ 1 



X -\-l 

X + y 



x + y -\ 



x-y + 



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 

(a-b) 



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 ; 



x-2 x-2 

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 : 



x-lji 



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_ ^ {a-bY^{b-cY^{c-aY 

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 



1-a 1 + a 1 + a'^ (1 - «) (1 + a2^ 

2a^ 



(1 + a) (1 + a^) 1-6^4 

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^{f-b^) 

= 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) 9x74-3ic* + lli«V5-4/5 + 4/25iz;2. 

(b) 1 +- 4/cc + 20/£c8 + 25/^* + lO/ic^ + 24/^^ + 16/a^«. 

(c) 178/7-20£c/72/ + 9?/V16£c2+4a;V497/2-15y/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) l-3cc/2 + 3icV2-5:rV4 + 3icV4-3icV8 + 
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) (x-2)(x-3)=0. 

(c) (x + l)(x + 2)=0. (d) (x + 2)(x-3)=0. 

3. What is meant by an equation being satisfied ? What 
values of x satisfy these equations ? 

(a) (x + ^)(3x-2)=0. 

(b) (2£c-l)(2ic + 3)=0. 

(c) x(x-l)(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. 5x-2 = 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 = 8-f3ic. 6. 11 -x = 2x-l. 
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=2x-5, 

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 
x-1 = 6 
are multiplied by x — 2, we have 

(X - 1) (X - 2) = 5 (X - 2), 
or x2-8x + 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. cc-2 = 4ic + l " 3. 

5. x-5 = 5x- 21 " X. 

6. x^-\-x =(1 — xy " x. 

7. 3x-4. = 4:X-3 " 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 

.^.^ x-lx — 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^' 






x-2. 


3. 


hl*h' 






X. 


4. 


X^ 1 






X-1. 


x-1 x-1 


5. 


x^_6x + 5 

x^-5x + 4. 


5 




x^-5x + 4.. 


6. 


x^-5x + 6 = 







1/(03-2). 


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. 1-4, 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. (2-x){5-x) = x\ 6. 3(a;-l) = 4(a; + l). 

7. 9x-3(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;(aj-2)=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 : -^r-r — I — 7-=— = 8. 

lo IT 

1. The l.c.d. of the fractions is 15 • 17. 

2. Multiplying both members by 15-17, by ax. 6, 

17 a; -61 -|- 15 a -l- 105 = 15- 17 -8. 

3. Subtracting —51-1- 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 _| = |. 







7X-12 




8. 


Transposing and uniting terms, 








7x- 


-4) 
12 


I 


Dividing by 2 and multiplying by 7 


x-12 


I 


14 X 


-24 = 16x- 


30. 


4. 

f 


Adding 24- 15 X, 


- X = - 6. 




h 


Multiplying by - 1, 


x = Q. 




Check. H-ir% = l. 







SIMPLE EQUATIONS. 155 

3. An equation which should be cleared of fractions part 

^ . 3a; + 7 2x-4: 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, 

xa-x-x2+x + 2 x«-15x + 56-x«+ 15X-54 



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. 
■— 



(x-2)(x-l) (x-6)(x- 


-7) 




5 . 2 _ 2 






(x-2)(x-l) (x-6)(x-7) 




8. Dividing by 2 and cleai'ing of fractions, 






xa_l3x + 42=x2-3x + 2. 






4. .-. -10x= -40. 




(Why ?) 


6. .-. x=4. 




(Why ?) 


4 5 -4 -5 

Check. = 

2 3-2-3 







156 ELEMENTS OF ALGEBRA. 

5. An equation in which the fractions should be reduced 
to mixed expressions before clearing : 

5a;-8 6x-U _ x-S 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, 

x2-7x + 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 x-3 


3. 


50 12 49 
Ax^ X lO' 


4. 2^1 = 2.4-15. 
X — o 


5. 


2 + 3-^^ 4 


6. 0.5a:4-0.25rr = 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 
x-b ~ {2x- by ' 

1 + i^ i^ + 1 





^»iC 


a 








iC 


^«^ 


3b-x 









X 


+a ' 


2b + x 






6 


x + 7 
12 


3x- 
4.x- 


4 
3 


+ 


X 

2* 


iC 


-2 


X-4: 


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 


^x-2 
3 


8 




9 


15 


a; + 3 


2x 


+ 6 


ra^ + 2 


lx + 5 


5 


X — 


6 8-5a; 



23. 

6 4 12 

op; 2a;-5 , 6x + 3 ^ 35 

25. _^ + ___ = 5x--. 

2a; + 3 6a; + 22 _ 3a; + 17 
5 15 ~5(l-a;)" 

a — c> X a -\-h x 

28 ^ + 46^ + ^* 4a7 + a + 2^ > ^ ' 
a; + a + ^ iz;-[_(t — ^ ""* 

29. 1 2 3 4 



-1 x-2 x-3 x-4. 

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(l-4^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 

\x-l^-^x-l = ^% + \^x-\l-^x. 

13.r-10 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; H-1 _lf a; + 1 

3 ~2y~~Y~ 



x-2 



\{'-'-^) 



+ 



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. .-. x-5 = l+x-2- 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, 
x-2 = l+x-5 + 2Va 



l. Then, isolating the radical Vx — 5, by subtracting 1 + x — 5 

dividing by 2, 

1 = Vx-6. 

5. .•. \ z= X — 5, whence x = 6. 



Check. V6-2 _ 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 + 5Vx-7 = 4 Vx -4. 

2. Squaring 

X + 1 + 25x - 175 + 10 Vx2-6x-7 = 16x - 64. 

3. .-. X - 11 = - Vx2 - 6 X - 7. (Why ?) 
4..-. x2-22x+ 121 =x2-6x-7. (Why?) 
5. .-. x = 8. (Why?) 
Check. V9-4V4 + 5V1 =3-8 + 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^+5-2 V2x-l = l.* 

6. 4 ^x + 2 - Vx + 7 - 5 Vic-1. 



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 52-45 and 66-37 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 + 2-6i%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. .-. x-30%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 one-fourth 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 coal-pit 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 1-second pen- 
dulum; of a 2-seconds 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 right-angled 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 bui-dens." 
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 Al-Khowarazmi, 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 '^ 

(^■' b-a 

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 



3a-2\a + 3[a-2(a-a-2x)^\ = 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. Zx-2y = 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. 9x-62/ = 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. 3x-2y = l. 




3. ••• 


12x + Oy = 123, multiplying 1 by 3, 




4. and 


12x-82/ = 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. 


6x-5y: 


-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,x-3y = 26., 


4. 


5x + 2y = l. 




18x-{-8y = ll. 


6. 


l.Tx-{-l.ly = 13, 




1.3cc-0.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 30x-2?/ = 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. 


x-y-l = 0. 


6. 


ax + hy = c. 




2x + y-29 = 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^^^. 




llx-y = 104.. 




4|2/-^ = 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. .-. 


Q-ix + 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/H-l 



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 



x-y 



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 


-x-y = 


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^-^2xy-h288- 


6 2/^ 


= 2x 


+ 3 


2/- 


131, 


2^ + 13-22/ 




5a;- 


-42/: 


^22. 









23. 



24. 7y + 13-5. _ 3y + 2x-16 . 

4 ^ 3 

,5y + 2x 3.T-12 + 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 = 60-5x. 
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 4-J 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. 9x-2y-z = 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 + 1-1 = 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. ic4-2/ + 2« = — 3. 

3. x-2y -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 + 2-6= -3. 
1-4 + 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 + 3x-y-\-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. to-9x=-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. 
3-2 + 2-2 = 1. 
4 + 6-1+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 


llz-7u = 


1. 




7x + 4:y-3z = 19 


4rz-7y = 


1. 




x + y = l. 


19x-3u^ 


1. 




a; + 6 « = 39 



196 ELEMENTS OF ALGEBRA. 

5. x + y = 16. 6. x-\-y-z = 132. 

z-{-x = 22. x-y + z = 65.4. 

y + z=2S. -x + y + z = - 1.2. 

7. a* • a^ + 2 == a>\ 8. a^x + h^y + Ci-t; = ^i. 

~ a^' ■ a^ + ^ = a^"^. a^x + h^y + Cg;^ = c^g. 

9. a; = 21— 4?/. 10. ic + y + ;^ = 5. 

z = 9-^x, 3x-5y + 7z = 75. 

y = 64 - 7^5:. 9ic - 11 ^ + 10 = 0. 



11. 



X y 


L8 = i8. 


a: ?/ ^ 


X y 


^ = 25. 

z 


13. 


3y-l 
4 




5.-^4;*; 
4+3 


3a; + l ^ 1 



12. 7ic — 2;t; + 3ii = 17. 
4?/ — 2« + V = 11. 
5?/-3x-2i^ = 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 

jft-hand 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 left-hand 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:. 

-i-i + 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 — f-4by 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-{-0.3y-hOAz = 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 




3x-5y + 4.z = 0.5. 




7x-{-2y-3z = 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^ +2y-0.5z = 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) 5x-2y = 17. 

(c) 5cc + 232/ = 100. 

2. Find two entirely different sets of roots of each of 
the following equations : 

(a) x-Stj-\-4:Z = 20. 

(b) 2x-^10y-z = 15. 

(c) 8x-7y-^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"" -T-a, 


a, 


(( 


a^^a, 


1, 


a 


a -7- a, 


1 

— J 
a 


ii 


1 -T-a, 


1 


a 


1 

a 


1 






— J 







206 ELEMENTS OF ALGEBRA. 

Illustrative problems. 1. Express 2"^ as a decimal frac- 
tion. 

1. 2-2 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. =x-i?/. §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 (2-2)-2. 

1. 2-2 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. 2-1 " i 

4. .-. the expression reduces to ^. 



THE THEORY OF INDICES. 207 

EXERCISES. XCVIII. 

Express exs. 1-4 without exponents. 

- ^ f- p-:' (?r- — .-■(r 

Express exs. 5-9 with, positive exponents. 

9. 2(t-3, t^-^ (-cc)-^. 

Express exs. 10-16 in the form of common fractions, with 
positive exponents for the factors. 

, .- , .. , a-^^-^' 2-^3-2 

10. -^^;;-r- • 11. 






12- ^ ^, . 13. |£c-V-^H-|iC2/V. 

14. [(£C-«)-«]-«. 15. [(1 - £C)-2(1 - £C2>)-|-1_ 

16. a-%-^cH^, a-'^b-^'c-P. 
Simplify exs. 17-20. 

2-8 4-5 e-7 68 ^-6 ^-a J-c 



• 3- 


~^'l 


-4 


7-6 


• 77' 


18. 


ar'' c-^ 


&-« 




19. 


- 


- a-^ 


■ 1 + a- 
2 3 + «- 


-3 


3«^ + l 






(- 


-a)- 






20. 


- 


a-2. 


(2-i)^ 


• 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. 1-3. 

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. 4-9. 

1 



a a -{- b a 1 

4. -T- + 



V^ ^a-b 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^ +{a-b-'^)-\ 

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. ^a--r^^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^ = X-X «'" r^-<l^ = X-X-^ = 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 a-i, 

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 (X-l -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. 
x-2 + x-i + 1 3 

g-2 _ a;-i + 1 1 

X-4 -I- X-8 + X-2 

- X-3 - X-2 - X-1 
X-2 + X-l + 1 

x-4 + x-2 +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 = x-2 + 2x-i + 1 

x-i + l|x-3 + 3x-2 4- 3x-i + 1 

X-3 + X-2 



Check. 8 -- 2 = 4. 



2x-2 + 3x-i 
2 x-2 + 2x-i 



+ 1 



x-i + 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)(x-3 _ 1) =0. 

x~5 = 2, or x~^ = 1. 

x-i = 23 = 8, or x-i = V^ = 1. 



8-1 



or X = 1, a::d these roots check. 



218 



ELEMENTS OF ALGEBRA. 



EXERCISES, C. 
Eemove the parentheses and simplify in exs. 1-8. 

1. [(-arf. 

2. (ar' + y-y. 

1 _1_ 

4. [(«'» + '*)"'-".(«»')«]'«*. 
_i _i 

6. \l(x-Y^T^~% («^^^^)*. 



7. (203-2- 2/- 2)-^ V64[(x-2/)-6]i 

Express with positive integral or fractional exponents, in 
simplest form, exs. 9-14. 

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. 15-19. 

15. 3a~^-4:a~^'2a^; ax—^y-'^-bx^'y'^. 

16. a^-a^; 3a-^-^c^-4.a-%^c-^; a;-^.(-aj^). 

17. 5-y/x^-2xhf', -a-'b^c^d-''-a'b-^c'd-\ 

18. (a;2 + 2xy + y^)- (x-' - 2 x-hf-"- + y-'). 

19. (a;-3 + 3 x-'y + 3 ic-y + y^) ■ (x-' + 2x-hy + y^). 



THE THEORY OF INDICES. 

Perform the divisions indicated in exs. 20-30. 

21. 4:abh^:2b^ck 

22. (ah^ — ah^ + 4 Jb^) : ah\ 

23. 4a;-* + 11^"' -45 by 2ic-i-3. 

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 a-4 : 5 a't, x^ : J [x-i^/'i (ccy)^]-^^-^. 

28. 16cc-8 + 6a3-2 + 5a;-i-6 by 2ic-i-l. 



219 



29. V^4 x-h/z^ : [(1 : ^12a;V^«') • '^lOSx-h/z-^y 



30. ic-^ — 2ic-* 



4ic-« + 19x-2_3l£c-i + 15 by a;" 



3 1 . rind the remainder when a;- ^ — 11 cc" ^ + 1 is divided 
byx-i-1. 



32. Also when x~^ + (a — 3) x' 
divided by x~'^ — 3. 



(b-3a)x-^-3bis 



33. Factor 2 x-^ — 9 x-^ — S x-'^ -\- 15, negative expo- 
nents being allowed in the factors. 

34. Also 6a;-« + x-2-5a;-i-2. 

35. Also 6a;-^ + 17ic-2-18a:-i-45. 

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 




n-tic, 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) 3-v/5. (c) ab^. (d) V2-V^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) 5-2*. (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 V8-2 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. (3-5V3)2. 4. V^.^.^^.-4^. 

5. V7 . -^7 • ^. 6. {-^/a-b + Va + bf. 

7. 2V2.3V3.5V6. 8. VT2I. V^.-J/14641. 

9. (2 + 8 V3)(4-5V3). 

10. 3 V2.2^3.4V^.5V^. 

11. V^/a + V^ . V^Va - V^. 

12. (V2 + V3)(2 V2-5 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 a-b^ 

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. 3-V2. 

9. V5-I. 

11. aJn'^c » . 

13. Vt + Vs. 



2. V7-V5. 

4. Va — V^. 

6. Va + ^» + c. 

8. V5-V2-V3. 

10. V5 + V7 + Vil. 

12. V2 + V7-Vli. 



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) 



2-3 



230 ELEMENTS OF ALGEBRA. 

EXERCISES. CVI. 

Perform the divisions indicated in exs. 1-16. 
1. 6:4-v'24. 2. 24 : (2 Vt - 6). 

3. ^a%^'.2^^h. 4. 15V24:3V^. 

5. 58:(8+V35). 6. 12 ^192 : 4 ■V^729. 

7. 16-V^^^':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. (3V5-8 V2):(3 V3-4V5). 

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. 
17-23. 

17. i4±4- 18. ^' 



11^ - 5^ 2-3^ + 5^ 

7 + 3V7 3_5i_2^ 

19. p=- 20. -• 

12-6 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 

V8-2 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) a-2-yf2ab^2h. (b) a-2a'-^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) a-3V^^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) 32-80 -^ + 80-v^2/^-407/v^^ + 10y/V^-2/2 

4. Extract the square roots of 

(a) 8-2 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) 2x4-2V^ 



(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^+ia-h -ia-^62 




+tV«-^6« 


.-. V5=(4 + l)^ 






=4^ + 1.4-^-1.4-* 




+ T^6-4-^ 


= 2 +i -J^ 




+ 5l^ 


_/i 1 ^\—a. 






(l+X)3-(''^^ 




= 1 +(-3)x+^^i^ 


-V,-3(-3-l)(-3-2)^, 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 = V16-2 = 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) at-2af + 5al-4ai + 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/ + 49ic-y-24a;-y + 162/*. 

13. Extract the cube root of 
aj-6 _ 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 
■^^x-i + ^\ic-i + |a7~t + 7 a;-T% + 3 a;-i + |x-tV + 1. 

17. If a^ = b% show that (t V = «^"^- 

\^ / V3-- 

V3-V2y 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 right-angled 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 clock-hands, counter-clockwise. 





' 


' 






B. 


■i-aYn 




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^), -2V-i, -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 V3-1 = 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. V-32. 13. 3V^. 14. V2.V^. 



15. - V- 16. 16. 2V-9. 17. -^V-12. 



or 




{x-2f 


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, 

V-1)=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. -5-2 V^. .6. - 3 - 3 V^^. 



7. -i + iV3-V-l. 



8. 



i-iV3.V-l. 



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., i-M.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) = 


6-4i 



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 3-2i. 

Multiply both terms of the fraction 
12 + 5i 
3-2i 
by the conjugate of the denominator. Then 

(3 + 2 i) (12 + 50 _ 26 + 39 i _ 26 + 39 i _ 

(3 + 2i)(3-2i) ~9-4(-l) ~ 13 ~ "^ *' 

5. Cube — 1 + i * Vs. 

••• (/ + n)8 =P + SPn + 3M + n\ 
.-. (-i + iiV3)3 

= -i + 3.i-iiV3+3.(-i).f.(-l) + f(-l).iiV3 

= -i + |iV3 + f-fiV3 

= 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 text-book. 
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) 5-7 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 
i-i^V3. 

2. Multiply 

(a) 3 - 4^- by 5 + 21. (b) - ^ + ^ ^ by i + ^ /. 

(c) 2 + 9t by 9 + 2/. (d) -4 + 2^ by -4-2/. 

(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 3-6 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) -5-121 

(d) _45-28^. (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, 

■i-a^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..-. 2x-l =0, or 3x-2 = 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''-2x-lb = (). 

11. cc2 + 5x-14 = 0. 

13. x^ + 19x + lS = 0. 

15. a;2-12a;-85 = 0. 

17. x''-22x-\-121 =0. 

19. ic2 _ 24 x + 143=^0. 



1 1 

6. x^ = 2{12~^x). 

8. 8ic-a;2_12 = 0. 

10. X (10 + £c) = - 21. 

12. 6a;24-7ic + 2 = 0. 

14. ic2 + 26 X -: - 120. 

16. £c (4 — ic) + 77 == 0. 

18. 3cc2-10x + 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 + 4x4-4-3=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 + 3-8t4V3 + 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. 1-16. 



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. (x-iy + 4.{x-l), 

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 + f-i) = 0, 

(X + 2) (X + 1) = 0. 

4. .-. X = — 2, or — 1. 
Check. 4-6 + 2 = 0, 1-3 + 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. .-. x2-x + ^-f = 0. 






4. ...(a;-i+iVg)(x-i-iV5) = 0. 






5. .-. » = i ± 


iVs. 




Check. i±|V5-V3±iV5 




= i±^V5-iV6±2V5 






'=^±|V5-iVl±2V5 + 5 




§245 


= |±iV5-|(l± V5) = 0. 







EXERCISES. CXVII. 

Solve the equations : 



X 25 



3. 



3. a;2-2x = -2. 
5. cc2 _ 9 ^ _ 1 ^ Q^ 

T,. x''-7x + 5 = 0. 



(-9-- 



2. 1/ 

4. x'' + 6x + 2 = 0. 

6. x2-6ic + 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 + 3-8i4V3 + 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 = -2-V3, 

(3) - (X + 2) = + V3, " - X = 2 + v'S and .-. X = - 2 - V3, 

(4) -(x + 2)=-V3, "-x= 2-V3 " .•.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; + 7-23 = 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 + 3x-18 = 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 + 5-23 = 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. 4x2-4x = 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£c2-4fa; = 304. 
7. 3ix^-4.x = 96. 8. ic2_,_i32a; = _l33l. 
9. x'' + 6x + 25 = 0. 10. 7ic2-5x- 150 = 0. 

11. 4x2_5^_pg2 = 0. 12. 4..05x^-7.2x = U76. 

13. (a; + a)2-f-2(£c + a)+l = 0. 

14. ( a; + -y_3| ^ + - 1 + 2 = 0. 



(. + ^y-3(x + l) + 2 



15. (ic2 + 2a;)2_3(ic2 + 2ic) + 2 = 0. 

16. (x^-^x-iy + 4:(x^ + x-l) + 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 

2-6 2-6 ^ ' 

= {i±^ Vl69 - 144 

Similarly, the roots of 
2 3 



4.6-6 



+ 1 = are - = 

X X 



2-2 2-2 ^ ' 



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 = - 



^±J-Vb^ 

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 1-30, and then 
simplify the results. 

1. a;2-3a; + l = 0. 2. ic^ + 6ic + 2 = 0. 

3. cc^ + dcc — 4 = 0. 4:. x^ — 5x-\-l =0. 

5. x^ + 2x + 2 = 0. 6. x2 + 2cc-24 = 0. . 

7. ic2-2ic + 3 = 0. 8. x''-5x-36 = 0. 

9. ic^ + 2 ic - 3 = 0. 10. 0-2 + 7 cc - 44 = 0. 

11. x^-5x-S6 = 0. 12. ^2 + 10 a^ + 5 = 0. 

13. x^ + 7x-\-l() = 0. 14. x^ — 4.x -12 = 0. 

15. 12ic2 + aj-6 = 0. 16. x''^ 4.x -45 = 0. 

17. cc2-7ic + 12 = 0. 18. x2-3cc-28 = 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. 6cc2-37cc + 6 = 0. 

25. ic2-2.l£c-l = 0. 26. 6x^-\-5x-56 = 0. 

27. i»2-llic-60 = 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^ -j-bx -\-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) (3x-5)(3x-7) 

2. Multiplying by i (x + 3) (x + 5) (3 x - 5) (3 x - 7), 

(3x-5)(3x-7)=(x + 3)(x+_5). Ax. 6 

3. .-. 8 x2 - 44 x + 20 = 0, (Why ?) 
or 2x2-llx + 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 
2-3 2-3 ^ ' 



= 2±iV3. 
Check. ^ H ^ + 



l±iV3 ±iV3 _l±^V3 

l-i_ 1-i 

= f TiV3± V3-f 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 1-x 4.x -1 


2. 


3 2 1 


3-x 2-x l-3ic 


3. 


2cc4-l x-j-1 x-6 
x + 1 x + 2~ x-1 


4. 


4:X X -{- 1 X -\- 5 
2ic-l 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. 


x-2a x-3b x''-6ab_ 
2a 3b 6ab 



9. V2-a; + V3 + a? - Vll + ic = 0. 

10. (1 + 2 x)^ - (3 + x)* + (2 - a:)^ = 0. 

6-\-5x __ 3x-4 5-7 X _ :§! ^ ^ 
^^' 4(5-x) 5(5+x) ~^25-a;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. r-g^ + 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 — T-g = 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 



Y-a^2-a^' 



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 1-10 ? 



1. 5 a;2 + 1 = 0. 

3. x"" - X + 1 = 0. 

5. Sx^-\-x + 7 = 0. 

7. 7a;2-a;-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'' + 2x-6. 

17. 6 ^2 _^ 7 a: -3. 

19. 2x^-^3x-4:, 

21. 40x2 + 34^ + 6. 

23. 80x2 + 70^ + 60. 

25. 65x2-263x-42. 



12. 2x2 + 7x + 3. 

14. 2ic2-5x + 3. 

16. 55x^-27x-\-2. 

18. llx2-23ic + 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 



2x-b 
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. cc2-6a; + 8 = 0. Q. x^-x-2 = 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, .-. x-i-ii = 0. 

2. V x=:i-ii, .-. 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, 

16x2-16x4-5 = 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 leai-ned 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 + 16--0. 

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(- V5-I ±i VlO-2 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 

(x-2 - 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. 

x2-9= 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. 7x-4.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^ + 5x-l = - ^ -• 

x^ + 5x + l 

9. (ic2 + 3)2 + (x^ -I- 3) _ 42 = 0. 

10. x-(a + b)x^-2a(a-b) = 0. 

11. (x^-\-2x-\-3)(x^ + 2x-h6) = -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:c-4-2 = 
1. This may be arranged 



0. 



2x2 + 3x-4-3V2x2 + 3x-4 + 2 = 0. 

2. Let y = V2x2 + 3x-4. 

3. Then y^ -Sy + 2 = 0. 

4. .-. (y-2){y-l) = 0. 

5. .-. y = 2, or 1. 

6. .-. 2x2 + 3x-4 = 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. .-. (Vx-2)(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+(x-4.y = 5. 


5. 


x'^-\-5x-W = 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 + 2V2-ic- 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^ + l-2x(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., x5-l=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(--^5-l+ ^ VlO - 2 VS), 
x, = l(~^5-l-i 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;2-29cc + 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. 2-5 



9 . r-fi-x . 9K 



2. 3"':8P = (3ii)'. 

4. 32^ -9^ = 27^' -3. 

6. a'^:{a'=f = {ay. 

8. (3-f-3- = 27i^ 

10. 2-6^* + '^ = 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 


= L25-0.45. 


-1.25= -0.8-0.45. 



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. One-fourth 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 right-angled triangle is 10 
and one of the sides is 2 in. longer than the other. 

jquired the lengths of the sides. 

18. In a right-angled 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 right-angled 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 o-f 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 mid-points of the semi-diagonals; 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 

x-i 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 2x-l 

^ + 19 Y_ 3(2-^ 



11 ; 2 + x 

10. Solve 18 (x + iy{x + 2)2 = 8 (x - 3)2(ic + l)^. 

11. Solve {x -3y-3{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. 
x-y = -3. 

19. (7 + ^) (6 + 2/) =80. 
x + y = 5. 

21. a;2 + 7/2 = 500. 



3. 



x-y 



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. 
jx-y-50 = 0. 

10. 3a; + 4?/-8 = 10. 
x^ — y^ = — 5. 

12. ia;2+ 1^/2-60 = 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;-0.25y-3 = 0. 

20. 0.01 0^2 + 400?/ -25 = 0. 
0.5x + y-10 = 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(?/2-3)±3V2/2-3 + 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 + y-3 = 0. 



1. 


x + y^ = 11. 




3. 


2x^-i-3x-t/ = 


: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 + 3x-4y + 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, -i-iiVl5, 

and y = 2, 8, -i + izVl5, -i-iiVi5, 

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'^^xy-y'' = h. 

x^ — 2xy + y^ = 0. 

3. Bx^ -\-4txy — y^ = 0. 

x^ + X + y = 5. 

4. x^ -\- xy + X — y = — 2. 

2x^ — xy-y^ = 0. 

5. x^ + 3xy + 3x~y = 2. 

x^ + 2xy-3y^ = 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^3x-y = 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. 

JB-gr., 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 + f|-Tl8=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+3u-2u2 2-5w + 6d2 


Reducing, 






8. 




18 v2 _ 19 u 4- 1 = 0, 


or 




{18u-l)(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 v-2 _ 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;2-5xy = 18. 
2/' 2/ 


7. 


3xy + y^-lS = 


0. 


8. 


x'^-xy + y^ = 21. 




4.x''-{-xy-7 = 


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 + 4x2/4-57/2-48 = 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^ + u-67 = 0, 
or (6w + 19)(tt-3) = 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 
x-y = 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. 

x-y = -\. 

Here x = y — \; hence, 

(y-l)3 + (y-l)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 . 

1-1 = 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. 
x2-2xy + 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 io-2iViT 



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. 
x2-2x2/ + ?/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?/=i-4. 

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 




(x-2)(5x-2) = 0. 


7. .-. 


X = 2, or f. 




2/ = - 1, or H. 




z = 5, or 6^^. 


Check for the second set of values. 




l + H-10||=-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 two-figure 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 two-figure 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 right-angled 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)^ = y-l- 



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 


-10V3|A- 


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. 12x-2x + 3>18x-2 + 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. 3-3 = 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 5xH-2>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 sea-water 1.025. 
To what depth will a cubic foot of ice sink in sea-water ? 

4. From ex. 3, how much of an iceberg 500 ft. high would 
show above water, the cross-section 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. 

^ 14-30 

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 ? one-third 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 H-e + •• 
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, •■u-o.t- i J.- a -\- b -\- G a — b -\- c ^ 

9. Which IS the greater ratio, ; — — ? or ; ? 

a — b — c a -i-b — 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 3a4-5c — 6e 

11. It J = - = - = k, prove that k = 



b d f "-'^^ ^"-^ 3b + 5d-6f 

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 water-wheel) 
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 cross-section, 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 
cross-section. 

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 1-sec. pendulum 
being 39.2 in., what is the length of a 2-sec. 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 1-sec. 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 = cH-d: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 

x2-3x + 1 ~ic2 + 4x-2' 
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 =25x-75. 

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. =±Vl2-16i 

3. = ± 2 V3-4i 



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 + 2a6-8^>2:5a2 + 4a^'-12^»2 = x:5a-6^>, 
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 -\- by-j- 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 = tn-i-\-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 + {n-l)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. {x-y)^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-\-{n-l)d. 


l^-id±V{a-id)^ + 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 + {n-l)d]. 
s = i{l-\-a) + {l2-a^)/2d. 
s = in{a-^l). 
s = in[2l~{n-l)dl 


dnl 
dns 
dls 
n I s 


a 


a = l-{n-l)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 = {l-a)/{n-l). 
d = 2(s-an)/(w2 - n). 
d = {l^ -a^)/{2s-l-a). 
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 + (n-l)(-5) = 30-6w. 

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 = tn-ir — 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 = 

r-1 

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(i-|y \ 
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 
.8-73 = a(15-a)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){2y-S) = 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 + {r-l)s]/r. 


3 


a 71 s 


Z(s-0«-' -a(s-a)"-i=0. 


4 


r n s 




l = {r-l)srn-^/{rn-l). 


5 


ar n 




s = a(r«-l)/(r-l). 


6 

7 


an 

a n I 


s 


s = (W-a)/(r-l). 

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(r-l)/(r--l). 


11 


rls 


a 


a = rl-{r - l)s. 


12 


n I s 




Z (s _/)«-!_ a (s-a)«-i = 0. 


13 


a n I 




r= {l/a)^>^^. 


14 


an s 




r^-sr/a + {s-a)/a = 0. 


15 


a I s 




r = {s-a)/{s-l). 


16 


n I s 




rn _ sr^-'^/{s -l) + l/{s-l) = 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 (s-a)- 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 + /,,4-T^^ + ---. 

(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 arithmetico-geometric. 
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» 

•■■'"(1-X)2~''(1-X)" 

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. 

10-i=i, .-.logO.l =-1. 10-2= — , .-.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 
10-10*? 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 103-3932 =2473, .-.log 2473 = 3.3932. 

... 103.3991!. 101=102.8932^ .-. 102-8932 =247.3, .. log 247.3 = 2.3932. 

Similarly, ioi.3982 =24.73, .-.log 24.73 = 1.3932. 

*" '"^ 100-3932 =2.473, .-.log 2.473=0.3932. 

100-3932-1=0.2473, .-.log 0.2473=0.3932-1. 

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 right-hand 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 1-10, 


To find colog 41.5. 

9. 


9 9 10 


log 41. 5= 1. 


6 18 


colog 41. 5 = 8. 


3 8 2 0-10 


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(317-92) = 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 

317-92 _„,, log(6178. 0.13) = 2.9048 

OD.OZ. 



6178-0.13 ' ■ 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 . 10-6 = 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 . 10-8 = 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= 9.3979-10 

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)(7i-2)---(n-r- 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(n-l)(n-2)...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 = 1.2-3 = 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)---3-2-l _ nl 



(n-r){n-r-l)...3-21 (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 n-n-n- • -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 9-figure numbers are possible, all zeros 
being excluded. 

7. From ex. 6, how many 10-figure 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(n-l)(n-2)-.-(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)(/i-2).••(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 {n-r)\, giving 

n{n-l){n-2)---{n-r + l){n-r){n-r-l)-.-^.2-l 
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| = ^'^'^"^•^ . 

3.21-2.1 



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+iX-lx2 + .... 



2. Expand to 4 terms {a — 2 b)-^. 

1. V (x + ?/)« = x'* + nx'^-^y + ^^^~ ' x'^-^y^ 

n{n-l){n-2) ^ „ „ 
2-3 . ^ ^ 

2. .-. (a-2 6)-3 = a-'^ + (-3)a-4(-2 6) 

= a-3 + 6 a-46 + 24 a-''^b^ + 80 a-%^ + 



3. Expand to 3 terms (1 + x) *. 
As above, (1 + x)-i = 1 + {- i)x + (~ ^) (" ^ " ^) 3.2 4. 
= 1 ~ix-h^%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{n-V) {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(fc-l) 



afc-162 + 



A; (fc - 1) (A; - 2) 



2.3 

k{k-\) • 



a*-263 + 



a^+i + (fc + i)a% 4- ^A±il^a^-i62 



(fc + l)fc(fc-l) ^,_ 



2-3 



■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^ 


H-2a' 





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 

1-3+3+4 
1-2 1 



1 — 2 + 1 ; 5 rem. 

Check. Let x = 2. Then (8 -12 + 6 + 4- 5) ^1=4-4 + 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 CYCLO-SYMMETRY (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. E-g-, 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 non-symmetric 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) cyclo-symmetry, as seems best. 

1. (x^J^i/){x^-xY + y^)- 

2. (2x + y-z){2x-ij + 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\b-c)-irb'(c-a)-\-c\a-b). 

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, 

%(a-b) = {a-b) + {b-c) + {G-a), 
instead of 

{a-b) + {b-a) + {b-c) + {c-b) + (c-a) + {a - c) ; 
and 

%a{b + c-2 af = a{b + c - 2 af +-b{c + a - 2bY 
■\-c{a + b-2cy. 



APPENDIX. 383 

EXERCISES. CLXXIII. 

Expand the expressions in exs. 1-8. 

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. 9-14 ? 

9. ^a(b + c). 10. ^a'^(b-c). 

11. :Sa'(^>' -c«). 12. %a%\a-b). 

13. 2a'(a-b + 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{b-c)^0. 16. (:^ay-^a^ = 2:^db. 

17. (^ay=^a^-{-:$2ab. 18. ^[(^ay - ^a^2 = %ab. 

a •■■ d a •■■e 

19. ■%(a-b)(a + b -c)^0. 

20. '$(a-by=3(a-b)(b-c)(c-a). 

21. %a\b -c) = -{a-b){b-c){c- a). 

22. %ab (a — b) = —(a — b)(b — c) (c — a). 

23. (Sa) (^ab) - aZ-c = (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 + 1-2 ^. 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 

^(^c-2a-Sby. 

10. Show that (— a + b + c){a — b + c)(a -\-b — c) 

= '%a^(b + c)-^a^-2abc. 

11. Show that %{a-b) = 0. 

12. Show that {a-[-b -[-c){— a + b -\-c)(a — b -\-c) 

{a + b-c) = ^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 
c-a. (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\y-z). 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. ^{a-b){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^)-j-z (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. {x-ay{b -c) + {x-by{c-a) + {x-cy{a-b). 

28. {a + b){a- by ^ (b + c) {b - cy + ((; + a) (c - ay. 

29. 2 (a - Z*) (a2 + Z*^)^ ^-.g.^ (^^ _ ^^ (^^2 _^ ^2^ 

+ {b-c) (b^ + c2) + (c - «) (c^ + 6^2). 

30. (x4-2/ + ^)'-(^ + 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 Z-x + 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, .'.A-B = 0. 

2. .-. C(A-B) = 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) + b-r(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)^b-F{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 =2-7 — 5-3, 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^ — a-fi^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 hii-her 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 l-7-0-t-4.6.0-f3.0-2-l-6-0-4-2-0-3-7-0 = 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^ + (0-2 + ^^^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(5-7) 



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' 


= {a-b)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, 
nx-ly = U. 



Herex^'^^-^U '^ - ^ L 16(- 28 - 11) ^ 8,^^ ^ 

2(-49-55) -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£c-302/ = 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 



- a-Jy^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 


= 2-2 


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 


= 2-5 


11 71 


= 10 


20 15 80 




3 3 16 





17 7! 

11 71 

13 



10(26-11 -13 

- 10 • 134. 

-10 134 ^ 
= 5 

-4-67 



17-71 -3.11-7) 



11 52 33 




13 71 


= 2 


17 30 80 





The numerator for y is 

11 26 

13 32 = 2(17-32-26-29. 13-26-11 -15-32) 

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 


= 2-2 


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. 5ic-3^ = 3. 

2x + y = 5. 
3 2/ + « = 7.5. 

X . y rr ^ 

3 + 2-^=^-^- 

2 3^4 

6. 7x-3ij-2z = lQ. 
2x-5 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. 


12x4-7 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. 




5x-3y + 7z = 4:l. 




7x-2y-5z = 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. 2x-y = 8. 4. 7x-4: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. Sx-12y = 2. 

3 2 12" 

12. 1.02 a; -0.01 2/ = 20.1. 
0.2£c-0.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£c-f-4^ + 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. 




2x-3y = B. 


9. 


10 a; + 6?/ = 5. 




"i + y-l. 

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-^2y-2z = 0. 2. 5x + 3 y - z = - 17. 

2x — y + z = 10. 6 .-r + 4 ?/ = — 14. 

Sx-{-y-z = 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-\-2y-z = 8. 5x + 2y-z = 6. 

5. lOx + 5y — 15 z = 5. 6. 2x — 3y — 4:Z = — 5. 
2x + y-3z = 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 
"one-to-one 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. 1-4 ? 

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 two-branched 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 to-day. 

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 electro-dynamics. 
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, 1560-1 ; died at Oxford Jan. 26, 1630-1. 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 Fontenay-le-Comte, 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 al-jabr wa'' I muqa- 
balah, the science of redintegra- 
tion and equation ; of this long 
title only al-jabr 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, -f-sociare, 
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. 



14 DAY USE 

RETURN TO DESK FROM WHICH BORROWED 

LOAN DEPT. 

This book is due on the last date stamped below, or 

on the date to which renewed. 

Renewed books are subjea to immediate recall. 



^'^efc'er/H/ 




REC'D LD 




DEC] i'63-5PN 














































1 


T T^ 01 A Kn-^ Q '«i General Library 



Kt^^ ^ 



^^ 


.<'<^' 


911350 




Cl^ 


/r3 


t 


37 


THE UNIVERSITY OF CALIFORNIA LIBRARY 




[ililviM