ff ice Copy

OFFICE Cb
 Presented to the
library of the
UNIVERSITY OF TORONTO
by
Gopp, Clark Pitman Ltd.
m
 :

ALGEBRA FOR BEGINNERS.

BY

JAMES LOUDON, M.A.

PROFESSOR OF MATHEMATICS AND NATURAL PHILOSOPHY,
UNIVERSITY COLLEGE, TORONTO.

TORONTO
COPP, CLARK & CO., FRONT STREET,
1876.
Entered according to Act of the Parliament of Canada, in the year one
thousand eight hundred and seventy-six, by James Loudon, M.A.,
Toronto, Ontario, in the Office of the Minister of Agriculture.
 PREFACE.

Beginners, for whom this book is intended, are recom-

mended to work according to the rules printed in italics, and
to omit, on a first reading, the articles in small type. Those
who desire only a practical -acquaintance with the methods
of working, may also omit Chapters X and XXI.
The author will feel obliged to teachers for any suggestions
they may have to communicate.

September, 1875.
 CONTENTS.
PAGE.
I. Introduction i

II. Addition..... 8
III. Subtraction 16
IV. The use of Double Signs and Brackets 20
V. Multiplication 23
VI. Division 30
VII. Examples involving the application of the first -

four Rules 39
VIII. Simple Equations 42
IX. Problems 49
X. Particular results in Multiplication and Division 55
XI. Involution and Evolution 61
XII. The Highest Common Measure 72
XIII. The Lowest Common Multiple 84
XIV. Fractions 88
XV. Simple Equations, continued 105
XVI. Problems, continued 108
XVII. Quadratic Equations 113
XVIII. Problems 119
XIX. Simultaneous Equations 123
XX. Problems 129
XXI. Exponential Notation 133
Answers 141
 — ;

ALGEBRA.

CHAPTEE I.

INTBOD UG TION.
1. The operations of addition and subtraction, which in
Arithmetic are stated in words, are denoted in Algebra by
the signs + and — respectively. ,

Thus, add together 12, 5, and 6, is expressed 12 + 5 + 6 ;

from 15 take 9, is expressed 15 — 9 add together 3, \, 5J, and

;

from the sum take \ and 3|, is expressed 3 + i + 5j— i— 3^-;

and so on.

2. The sign + is called the plus sign, and the sign — the
minus sign.
Thus, 24+2—15, is read 24 plus 2 minus 15.

Exeecise I.

Employ plus and minus signs to express the following

operations in Algebraical language :

(1.) Add together 56, 10 and 15; 10, 12 and f; 2, land f

5. __§. 11 Prirj 1

(2.) From 29 take 15 ; from 2j take lh ; from 2 5 take 1 •

6.

(3.) From the sum of 11, 35 and 6 ; take 17; from the sum
of 81 and 75 take 69 and 42.
B
 ;

2 INTRODUCTION.

(4.) To the difference between 15 and 7 add the sum of

8 and 9.

(5.) To the difference between 28 and 16 add the difference

between 10 and 4.

3. The signs + and — are also used to denote that the

quantities before which they are written are, respectively,
io be added to and subtracted from some quantity not neces-
sarily expressed.

Thus, +4 number 4 to be added to some number

denotes a
—5 denotes a number 5 to be subtracted from some number.

4. Quantities to be added are called positive quantities,

and quantities to be subtracted negative quantities.
Thus, + 2, + 7, + i, are positive quantities; and —3, — 1,
— f, are negative quantities.
5. As an illustration of positive and negative quantities,
we may take the following examples :—
(i.) A man has a certain amount of cash in handhe owes ;

$150 to one man, and $280 to another and there are owing
;

to him the several sums of $100, $210, and $120. Now, in

order to determine what the man is worth, these several
sums are to be considered in connection with the cash in
hand; $150 and $280 are evidently to be subtracted, and $100,
$210, and $120, to be added. The former, therefore, may be
denoted by -150, -280, and the latter by +100, +210,
+ 120. In other words, in the process of finding out how a
man's business stands, the sign + may be used to denote his
assets, and the sign — his liabilities.

(ii.) The mercury in a thermometer rises and falls in con-

sequence of changes in the temperature, and the amounts of

these variations are expressed in degrees. In order to deter-
mine the reading of the thermometer, some of these variations
must be added to, and others subtracted from, the reading
before the variations took place. Suppose, for example, that
during the day there take place a rise of 5°, a fall of 7°, a
 ,
:

INTROD UCTION. 3

rise of 12°,and a fall of 8° 5° and 12° are to be added to,

;

and 7° and 8° subtracted from, the reading of the morning.

The former, therefore, may be denoted by +5°, + 12°, and
the latter by —7°, —8°. If the degrees are measured from a
zero point, distances above may therefore be denoted by +
and distances below by — Thus +20° means 20° above
.

zero, and —5° means 5° below zero.

6. In like manner the signs + and — may generally be

employed to denote the two relations of contrariety which

magnitudes of the same kind may bear to one another in
some defined respect. Thus, if money received be denoted
by + money paid away will be denoted by — if distances
, ;

walked in one direction be denoted by + distances walked ,

in the contrary direction will be denoted by — if + denote ;

games won, — will denote games lost and so on. ;

Exercise II.

(1.) A owes B $60, B owes C $30, and C owes A $20;

employ the signs + and — and liabilities
to denote the assets
of A, B, C.

(2.) B and C owe A $20 each, C and A owe B $30 each,

and A and B owe C $40 each express in Algebraical lan-
;

guage the assets and liabilities of A, B, C.

(3.) A
pays B $10, B pays C $7, and C pays $4 express A ;

Algebraically the amounts paid and received by A, B, C.

(4.) Denote the following variations in the thermometer

a fall of 2°, a rise of 5°, a fall of 3°.

(5.) Denote that the thermometer rises and falls alternately

1° per hour for 5 hours.
(6.) Denote that the thermometer falls and rises alternately
2° per hour for 4 hours.
(7.) Denote the following readings: 25° above zero, 7°
below zero.

7. In Algebra the numerical values of quantities are denoted

B
4 INTRODUCTION.

by (1) figures, as in Arithmetic; (2) the letters of the

alphabet, either alone or in combination.
Thus, if there be three points, A, B, C, situated in that
order on a right line, the distances between A and B, and

between B and C, if unknown or variable, may be denoted by

a and b feet, respectively and, consequently, the distance
;

between A and C will be denoted by a -f b feet. If C lie

between A and B, and a and b denote, as before, the lengths
AB, BC, the length AC will be denoted by a— b feet.
Again, if x denote the length, and y the breadth, in feet, of
a room, the dimensions of a room 3 feet longer and 5 feet
narrower will be &+3 and y— 5 feet, respectively.

8. In Arithmetical operations which are performed with

figure symbols alone, all mention of the unit is suppressed,
the results of these operations being true, whatever be the
unit in view. Thus, 10 and 5 added together make 15,
whether the suppressed unit be a pound, a gallon, or an inch.
Algebraical symbols have a still greater generality. Not only
is the unit suppressed, but the number of units is not assigned,
as in Arithmetic. Thus a may represent any number referred
to any unit. Operations performed in Algebraical symbols
will, therefore, give results which are true for any numerical
values which may be assigned to the symbols.

9. The product of symbols which denote numbers, is repre-

sented by writing them down in a horizontal line one after
another, in any order, with or without the multiplication
sign x , or dot . , between them.
Thus ab, ba, a>b, ha a x b } b x a, all denote the product of a
and b ; abc, a>b-c, axbxc, the product of a, b } c ; and
so on.
Figure symbols are written first in order; thus, 3a, 5ab,
6abc, fa. "When there are two figure symbols, the sign x
only must be used between them.
Thus, the product of 5 and 6a is denoted by 5 x 6a, and
not by 5 -6a, or 56a, whose values are five decimal six times a,
and fifty-six times a, respectively.
 INTROD UCTION. s

When there are three or more figure symbols, either the

sign x or • must be used.
Thus the product of 2, 5 and 7 is denoted by 2-5 -7, or
2x5x7.
10. The symbol a 2 stands for aa ; a5 for aaa ; a4 for aaaa ;

and so on. a2 is read a squared, or a to the power of 2; a3 is

read a cubed, or a to the power of 3 ; a4 is read a to the power of
4 ; and so on.
11. The power to which a letter is raised is called the
index or exponent of that power.
Thus, 1 is the index or exponent of a ; 2 of a2 ; 3 of a3,
4 of a\
12. The quotient of one quantity, a, divided by another, b,

is denoted by either of the forms, a -f- b or

b
Thus the quotient of 2a divided by Sbc is denoted by
2a-^3kor by|^.
36c

13. Any object or result of an Algebraical operation is

called an Algebraical quantity or expression.

be observed that the numerical value of an ex-

14. It is to
pression which is not fractional in form may be a" fraction,

and the numerical value of an expression which is fractional

in form may be an integer.
Thus, if to a we assign the value 2, and to b the value i,

the value of ^ will be 2 -f- \, or 4 ; whilst if the values of a

and b be I and i, respectively, the value of ab will be J xi,

or \, and the value of | will be i -f- i, or 2.

15. The symbol = stands for is or are equal to, and is

written between the quantities whose equality it is desired to
express.
Thus a = 3 denotes that the value of a is 3.
 : —
6 INTRODUCTION.

Exercise III.

If a = l, h-% c = 3, d = 4:, e=5, x = Q, find the values of the

following expressions :

(1.) a + 2b + 3c + 4d. (2.) 13a-46 + 5c. (3.) db + de.

(4.) ab + de— ex. (5.) 4a&c, bbod, 6dex.

(6.) a 2
+ F- + c 2
. (7.) 4a 2
+ 36 2 - c 2 . (8.) 5 a d + 46 3 - cU

/o \ a ^ /in\ c a cc /11 N ab ad bx
(9.) -
+? (10.) 3- ?+? (11.) 5-
e+
^--.

^ j
oF2 ""3?5
+ 8^ T
2 4,

(13.) If b = c = d=a, find what each of the following be-

comes in terms of a
2bca } Sa2 b, 2a2 + 3abc + 4:b 2cd.

16. When two or more quantities are multiplied together,

each is said to be a factor of the product.

Thus, a and b are factors of ab ; 3 and a 2 are factors of 3a 2 ;

and 5, b} and c are factors of 5bc.

17. One factor of a quantity is said to be a coefficient of the

remaining factor, and is said to be a literal or a numerical
coefficient according as it involves letters or not.
Thus, in 3x and %a2 the numerical coefficients of x and a 2
are 3 and f respectively in ax2 and 3cd the literal coefficients
,
;

of x 2 and d are a and 3c, respectively.

Also, since x = 1 x x, the coefficient of x in the quantity x

is 1.

The sign + or — when it precedes a quantity is also a sign

of the coefficient.

Thus, the coefficient of x in + 3x is +3, of x2 in — 6ax2 is

—5a, and of dz2 in — \edz 2

is — §c.
 INTRODUCTION. 7

In the case of fractional numerical coefficients the letter

symbols are sometimes written with the numerator thus, ;

a , %x 6a2 b •

= a 6"
= *X = ~~
a
%
2 i' >
~~~T
>

18. Quantities are said to be like or unlike according as they

involve the same or different combinations of letters.
Thus, + 5a, —la are like quantities and so ; also are + §x 2y,
—5x2y; —2a, —a2 are unlike quantities.

Exekoise IY.

(1.) Name the coefficient of x in 2x, Sax, &x, hex, 4:a

2
bx }
\ax.

(2.) Name the coefficient of a in — a + 3a,

} —fa, + 2a&,
— 5ax 2
.

(3.) Name the coefficient of x 2 in +x2 —x 2 Sax2 +%cdx2

, , , .

(4.) Name the coefficient of xyin —xy, +3a 2

xy, —\axyz.

(5.) Name the numerical coefficients in ~, _~, -j-??'

5x

(6.) Name the numerical coefficients in x, —y, —2x2 —3yz, ,

+§«, -lie.
(7.) Name the like quantities among 2x, ax, x, 3cx2 .

(8.) Name the like quantities among —a 2

, + 2a #, + fa2
2
,

—a2x.
(9.) Name the like quantities among —3a?x, ax2 abx2 , ,

+ 2a 2
a?, — ax2 .
 ( 8 )

CHAPTEE II.

ADDITION.

I. Figure Symbols.

19. In order to explain the meaning of Algebraical addition,

we shall, in the first instance, suppose the numerical values
of the quantities to be represented by figure symbols, as in
Arithmetic. We shall consider in order the addition of

(i.) Positive Quantities.

(ii.) Negative Quantities.

(iii.) Positive and Negative Quantities.

20. (i.) Positive Quantities.

We have seen that a positive quantity, as + 5, representsa
quantity 5 to be some number which may or may not
added to

be known or expressed. The sum of any number of positive

quantities is denoted by writing them in a row with their signs
between them, or by a positive quantity whose numerical value is

their Arithmetical sum.

Thus the sum of +4, + 2, and +10 is +4 + 2 + 10,or +16;

that is to say, the addition of 4 and 2 and 10 to a number

is equivalent to the addition of 16 to that number.

In like manner, the sum of +2, + §, +f and +5 is

+2+|+|+5 =+8ft.
The Algebraical statement
+5 + 6+J+3i = +14f
 ADDITION. 9

may therefore be read the addition of 5,6, ^ and 3 J to a number

is equivalent to the addition of 14f to that number.

21. (ii.) Negative Quantities.

Since —4 denotes
a quantity 4 to be subtracted from some
number, when there are several negative quantities, as —4,
—7, —8, denoting that they are all to be subtracted from
some number, the operation may be denoted by writing them
in a row, thus —4—7—8, or by a negative quantity, —19,
whose numerical value is their Arithmetical sum that is to ;

In Algebra —4— 7— 8, or —19, is called the sum oj —4, —7,

7

and —8. Thus the sum of —1, —10, — §, and — f is

It follows therefore that the Algebraical sum of any number

of negative quantities is a negative quantity whose numerical
value is their Arithmetical sum.
It will be observed that, instead of saying to subtract 12, we
may say in Algebra to add —12. The Algebraical statement
-2-7-10 =-19
may therefore be read, in Arithmetical language, the subtrac-
tion of 2, 7, and 10 from a number is equivalent to the subtrac-
tion of Id ;
Algebraical language, the addition of
or, in 2, — — 7,
and — 10 to a number is equivalent to the addition of —19.
As an illustration of the foregoing phraseology we may
take the following example: Suppose a man's gains to be
denoted by -f , and his losses by — ; then the statement

-200-60-500 = -760
may be read, if a dollar is the unit understood, the sum of a
loss of 200 dollars, a loss of 60 dollars, and a loss of 500 dollars
is equivalent to a loss of 760 dollars. It may also be read, the
subtraction of a gain of 200 dollars, a gain of 60 dollars, and a
gain of 500 dollars is equivalent to the subtraction of a gain
of 760 dollars.
lO ADDITION.

22. (iii.) Positive and Negative Quantities.

Since + 5 denotes a number 5 to be added, and —2a number
2 to be subtracted, the performance of both these operations
may be denoted by +5—2; and since the performance of
these two operations is equivalent to adding 3, we may write

+ 5-2 = +3.
In Algebra +5— 2, or +3, is called the sum of +5 and
-2.
Thus, +7-5, or +2, is the sum of +7 and -5 ; +f -i,
or +i, is the sum of +t and —
Again, since to add 2 to a number and then subtract 5
is equivalent to subtracting 3, we may write

+ 2-5 = -3;
that is, in Algebraical language, to add +2 and —5 to a
number is equivalent to adding —3.
The statement
+ 7-5+2= +4
may therefore be read, in Arithmetical language, to add 7 to,

then subtract 5 from, and finally add 2 to a number is equi-

valent to adding 4 ;
or, in Algebraical language, the sum of
+ 7, —5, and +2 is + 4.
So also the statement
-3 + 10-15= -8
may be read, in Arithmetical language, to subtract 3 from,
then add 10 to, and finally subtract 15 from a number is equi-
valent to subtracting 8 ;
or, in Algebraical language, the sum
of —3, +10, and —15 is equal to —8.

As an illustration of the foregoing phraseology, we may

again take the case of a man's gains and losses. Thus the
statement
+ 10-8=+2
may be read, if a dollar is the unit understood, a gain of
10 dollars and a loss of 8 dollars are equivalent to a gain of
2 dollars. So also the statement
 —
ADDITION. n
-25 + 20- -5
may be read a loss of 25 dollars and a gain of 20 dollars are
equivalent to a loss of 5 dollars.

23. From the preceding cases we can deduce the follow-

ing rule for finding the sum of any positive and negative
numbers.

(i.) When the signs are all alike— Find their Arithmetical
sum, and prefix the common sign.

(ii.) "When the signs are Find the numerical

different
difference betiveen the Arithmetical sum of the positives and
the Arithmetical sum of the negatives, and prefix the sign of the
numerically greater sum.

Examples.

(1.) The sum of + 4, +3, + i, and +7 is

+4+3+^+7= + 14i.
(2.) The sum of -5, -12, -f, and -3 is

-5-12-1-3= -20f.
(3.) The sum of +4,-2, -3, +5, and +7 is
+4-2-3 + 5 + 7- +4+5 + 7-2-3
= +16-5
= +11.
Here +16 is the sum of the positives, and —5 of the
negatives ; 11 is the numerical difference between these sums,
and has the sign of the numerically greater + 16.
(4.) The sum of + i, -2, -i, +1, and -f is

4-i-2-i + l-|= +i + l-2-J-f

= A_
_I_
4 '
1_3
4
= -2.
Here is the sum of the positives, and
+f of the —^
negatives 2 is the numerical difference between these sums,
;

and has the sign of the numerically greater — 1^3 .

 —
12 ADDITION.

EXEECISE V.
Find the sum of
(1.) + 2, + 5, +18. (2.) +i +3, +1.
(3.) - 8, -13, - 7. (4.) -f, -1, -f.
(5.) +18,-13. (6.) -26, +20.
(7.) +*, -f (8.) -f, +1, (9.) +2-5, -3-2.
(10.) +2, -3, +12. (11.) 3, +4, -6, +7.
(12.) +5, -8, -12, +3. (13.) +J, -1, +i, -f
(14.) -f, -i, +2, -i, + fc
(15.) +2-58, -3-26, +1*089, -0'067.

II. Letter Symbols.

24. (i.) Like Quantities.
"We shall now show how to find the sum of quantities
whose numerical values are represented by letters. When
these quantities are like quantities, their sum is obtained by
the rule

The sum of any number of like quantities is a like quantity

vjhose coefficient is the sum of the several coefficients.

Examples,

(1.) The sum of +2a, + 5a, and +10a is

+ 2a + 5a + 10a= + 17a.
Here + 17 is the sum of +2, + 5, and + 10.

(2.) The sum of -3c, -10c, and -12c is

-3c-10c-12c=-25c.
Here —25 is the sum of —3, —10, and —12.
(3.) The sum of + Sx, — 12x, and +lx is
+ 8x— 12cc + 7^=+3^.
Here +3 is the sum of +8, —12, and +7.
(4.) The sum of — Kb, + 12x, and —5x is
— lOx + 12x—5x=—3x.
Here -3 is the sum of -10, +12, and -5.
 ADDITION. n

Exercise VI.
Find the sum of
(1.) +a, + 2a. (2.) + 3a, +5a, + 7a. (3.) -a, -4a
(4.) -2^ -6a, -5a. (5.) + 5x2 -3a2
, .

(6.) -a2 + 4a
,
2
. (7.) + 4a, -7a. (8.) -*2c, + 5c, + 7c.

(9.) -10c, + 8c, -3c. +x2 -7x2 +3x2 -x2

(10.) , , , .

(11.) -2a&, -fllafc, +a&, -3a&. (12.) -ia, + §a.

2
(13.) -f-fa , -fa 2 . (14) -2a, + £a, -a.
(15.) +a, —fa, +|a, —2a.

25. (ii.) Unlike Quantities.

The sum of any number of unlike quantities is denoted by
writing them in a row in any order, with their proper signs
between them ; and each quantity is called a term of the sum.
Thus the sum of -f2a and + 3b is +2a + 36; of + 3x and
— 6y is +3cc— 5y; of — \a2 and — 2& 2 is — \a2 — 2b 2 of + 2a, ;

+ 3b, and — 5c is + 2% + 3b — 5c. The terms of — 2x + 5y are

— 2x and +5?/; and of — 6a + 7&— 3c are —6a, + 7b, and
-3c.

26. If a quantity contains no parts connected by the sign

+ or — it is called a mononomial.
Thus + 2x, — 3ab, + 6x ^, are mononomials.
2

27. When a quantity consists of two terms it is called a

binomial expression; when it consists of three terms it is

called a trinomial expression ; and generally when it consists

of several terms it is called a polynomial, or multinomial
expression. t
Thus +2a— 3b is a binomial, and +a— 26 + 3c a trinomial
expression.

28. The sign of a mononomial, or of the first term of a

polynomial, if it is positive, is generally omitted.
Thus 2x 2 stands for + 2cc2 and a— + c
, Z> for +a— b + c.

Like terms when they occur must be added together.

29.
The operation may be conducted by arranging the several
 ADDITION.

quantities in rows under each other, so that like terms shall

stand in the same column.

Examples,

(1.) Find the sum of 2a + 3& and 5a —2b.

2a + 3b

7a + b.

Here 7a is the sum of 2a and 5a, and -f b of + 3b and -*-2b.

(2.) Find the sum of xy— 6x and x— Sxy.

—6x + xy
x—Sxy
— 5x—7xy.
Here — 5x is the sum of — 6x and x, and —7xy of +xy
and — Sxy.
(3.) Find the sum of 3 + x + xy— 8x 2 ,
— 3xy + 2— Gx, and
4:xy + x2 + 1.
3+ cc —8.x 2
-f- cc?/

2— 6&— 3cc*/
1 +4cc?/ + 2 cc

6— 5£c + 2cc?/— 7cc 2 .

Here the sum of 1, 2, 3 in the first column is 6; of +x,

— 6x in the second is — 5cc; of + xy, Sxy, +4txy in the
third is +2xy ; and of — 8x +x in the fourth is —7x
2
,
2 2
.

(4.) Add together \a— \b— lc, J& + Jc + ia, ^c— |a—

f^a— Tzb + ^c.

Here T ^a is the sum of \a, \a, and — \a ; — Ty> and +
8

are the sums of the quantities in the second and third

columns, respectively.
 ADDITION, 15

EXEECISE YII.
Find the sum of

(1.) 2a, -3b. (2.) -x, +32/. (3.) ~2x, -3y, -z.
(4.) 3a, 2x, -5y. (5.) 4, -a, 2y. (6.) a 2, -6 2 ,

(7.) a, -2b, 3c, -a 7

. (8.) 2y, -z, 1.

(9.) a -36, 3a + 6. (10.) -2a2 + 6c, 3a 2 -5Z>c.

(11.) 3a -56, -4a + 26, 5a -66.
(12.) a -36, 26 -5c, 4c -3a.
(13.) 4tf— 3y + 2z, —3x + y—4:Z, x—ty + z.
(14.) a-x + 3, 5a + 2x-5, -2a + 7.
(15.) 2a -7, 5a + 4, -6x + 3.
(16.) a-26 + 3c, 6-2c + 3a, c-2a + 36.
(17.) cc-2?/ + 3z-l, 2x + 3-4:z, 5y-2 + 7x.

(18.) a 2 -f 2acc + a 2 2x2 -2a £2 -2aa; + a 2

,
2
, .

(19.) 3a + a 2 6-2a6 2 -f 6 3 3a& -2a 6-{-a3 a?b-ah*+3h\

8
,
2 2
,

(20.) + a +
&-
J +
f
 —
( 16 )

CHAPTEE III.

SUBTRACTION.

30. The Algebraical difference between one quantity and

another is the quantity which added Algebraically to the
latter will produce the former.
Thus the difference between 2 and —4 is the quantity
which added to —4 will produce 2; the difference between
—ha and 2a is the quantity which added to 2a will produce

—5a; and the difference between 3x and 2x— 5 is the quan-

tity which added to 2x—5 will produce 3x.

31. The quantity to be diminished is called the minuend,

and the quantity to be subtracted the subtrahend.

32. The difference between two quantities is found by the

rule

Add the first quantity to the second with its sign or signs
changed.

The reason for this rule will appear from the following

Examples.

(1.) From 5 take 3.

Here the difference is the sum of 5 and —3=5—3=2,

because 2 added to 3 makes 5.

(2.) From 7 take -4.

Here the difference is 7+4=11, because the sum of 11 and

-4=11-4=7.
 x

SUBTRACTION. 17

(3.) From -6 take -4.

Hero the difference = —6+4= —2, because the sum of
-2 and -4= -2-4= -6.
(4.) From -8 take 5.

The difference =—8— 5 =—13, because the sum of —13

and5=-13 + 5=-8.
(5.) From 2a take —3a.
The difference =2a + 3a=5a, because the sum of 5a and
— 3a=5a— 3a=2a.
(6.) From — 5x take 4.

The difference =— 5x— 4, because the sum of — 5x— 4 and

4 is — 5x.
(7.) From 2a take -3a + 26.
The difference = the sum- of

2a and 3a-2Z>=2a + 3a-26=5a-2&,

because the sum of 5a— 2b and —3a + 26 is 2a.

The operation of changing signs and adding may be per-

formed mentally, and the difference exhibited as in the fol-
lowing examples, in which the minuend and subtrahend are
written in rows with the difference underneath. In this
arrangement the first row is equal to the sum of the second
and third rows.

(8.) From 3x* + y take x 2 -5?/.

3x 2 + ?/
x 2 — 5?/
2x 2 + 6?/
Hero 2x2 is the sum of 3x 2 and — 2
; and + $y of +y and

(9.) From 5a+3&-c take a- 5+ 3c.

5a + 36— c
a— 5 + 3c
4a + 46— 4c
0
i8 SUBTRACTION.

Here 4a is the sum of 5a and —a; -f 4& of -f 36 and +b;

and — 4c of — c and —3c.

33. From the foregoing examples it appears that, in Alge-

braical language, to subtract a positive quantity is equivalent
to adding a negative ; and to subtract a negative quantity is

equivalent to adding a positive.

This phraseology may be illustrated by taking the case of
a man's gains and losses to be denoted by + and — , respec-
tively. Thus, to subtract a gain of 10 dollars is equivalent to
adding a loss of 10 dollars; and to subtract a loss of 25 dollars
is equivalent to adding a gain of 25 dollars.

Moreover, if a man gains a dollars and loses b dollars, we

say, in Arithmetical language, either that he gains a—b dollars,
if a is greater than b, or loses b— a dollars, if b is greater than
a. Either of these phrases may be employed indifferently if
we agree that a gain of — c dollars means a loss of c dollars,
and that a loss of — c dollars means a gain of c dollars.
Thus if a man gains 10 dollars and loses 5 dollars, we may
either say that he gained 10—5, or 5 he lost
dollars, or that
5 — 10, or —5 dollars. Again, if he gains 8 dollars and loses
12 dollars, we may either say that he gained 8—12, or —4
dollars, or that he lost 12—8, or 4 dollars.

Exercise VIII.

(1.) From 1 take -3.

(2.) From 1 take -1.
(3.) From -5 take 4.

(4.) From 12 take 15.

(5.) From -3 take -8.
(6.) From -8 take -5.
(7.) From 4'56 take -6*04
(8.) From -101 take 2 35.
(9.) From -4 32 take -216.
(10.) From -1*089 take -0123.
 SUBTRACTION. 19

(11.) From 2a take 3a.

(12.) From — 5a> take 2x.
(13.) From 8a2 take -2a 2 .

(14.) From -3c take -5c.

(15.) From 2a take ~
A
(160 From -a take 2a 2
2
.

(17.) From 5x 2 take -6x 2 .

(18.) From la take 4a-&.

(19.) From a + cc take a— sc.

(20.) From 5a -2s + 3 take 2a-cc-l,

(21.) From 3a 2 -4aZ> + 2 take 3a 2 + ab~b\
/>

(22.) From ace— 4&y + 3cz take 2ax\by~ cz*

(23.) —
From ia + b-1 take 2a- * + 3.
(24.) From 12x 2 -5# + l take 7^-16a2 + l.
(25.) From ~+y+| take
J+g-*
(28.) Fromfa + &-fc take a + hb—Jc.

o2
 — ;

( 20 )

CHAPTER IV.

THE USE OF DOUBLE SIGNS AND OF DBA CKETS.

34. The operations of addition and subtraction of positive

and negative quantities may also be denoted by the use of +
for the former operation and — for the latter.
Thus instead of saying add together 2a and —3b, t?g may
employ the notation 2.H 3b, the equivalent of which is
of course 2:t— 3b. So also the sum of —5a 2 + 3b, and —2c
,

may be written — 5a 2 + + 3b-\ 2c, which is equivalent to

— 5a 2 + 35 — 2s and the difference between 5cc and —ly may
;

be expressed hx ly, the equivalent of which is 5x + 7y.

According to this notation, therefore, 2a2 H 35+ + 2c
means ihe sum cf2ci 2 —3b, and + 2z; — 5x + +8y+ — 1 the
,

sum of —hx, -rSy, and —1; 7a— +23 the difference between
7a and + 2>; — 2x 5 the difference between — 2x and —5;
4a 2 7 £Ae difference between 4a 2 and —7.

35. When any of the quantities before which the double

more terms than one, it must
signs are to be used contains
be enclosed in a bracket; thus + (2a— 3b), + {—x + 4),
- (x-b), - (-2 + a 2 + x).
Thus a + (3b— c) denotes the sum of a and 3b— c 2x ;

+ (4?/-2) + (-22 + 1) the sumiy-2, and -22 + 1

of 2x,
2a 2 — b— (3a2 +4) the difference between 2a 2 — b and 3a 2 + 4;
and 2a + (6—1) — (c+4) the sum of 2a and b—1 less c + 4.

Exercise IX.
Eetaining the given quantities, denote by using the double
signs and brackets (when necessary) in the following opera-
tions :
 — + , + —
USE OF DOUBLE SIGNS AND OF BRACKETS, 21

(1.) The sum -I; 3x,

of 2x 2 } -5; 2a, -35, + 4c.
(2.) From 2a take -5a.
(3.) From — 6 take + 5x.

(4.) From the sum of 2a and —35 take +7.

(5.) From the sum of 5 and + x take —3a.

(6.) The sum of 5a and 5—4.

(7.) The sum of —a and —5-1-5.

(8.) The sum of a— 4 and 25— c.

(9.) The sum of x and 2y + 5 less 2.

2

(10.) The sum of a -I and 35 + 5 less -3c.

(11.) The sum of x, 2x -l and-3x -8.
2 2
}

(12.) The dilference between 4a 2 and 5

2
— c.

(13.) The difference between a 2 + 4 and -25 + 3.

(14.) The difference between 2a— 5 and a2 — 2a + 3.
(15.) From the sum of a + 5 + c and a— 5— c take —a
+ 25-3^.

36. Double signs may be equivalently replaced by single

ones by the rule :

Like signs produce + and unlike signs —

, ; that is

+ + = -- = +,

Thus a+ + 5=a + 5 ; 2x a=2x + a ; 3+— 4c=3— 4c;

c —h2a=c— 2a.

37. Expressions may be cleared of brackets by the rule :

The sign + before a bracket does not change the signs within,
whilst the sign — changes every sign within.

Thus 4 + (5 - c) = 4 + 5 - c,
2a + (— cc + c— 2^)=2a— cc + c— 2d,
4a2 - 1 - (2 b 2 + c) = 4a 2 - 1 - 25 2 - c,
3 # - ( - 4?/ + 5 ) = 3# + 4?/ - 5
a— (#—3) + 4— (— 3# x)=x—y + s + 4 3y —as.
 —
22 USE OF DOUBLE SIGNS AND OF BRACKETS.

Exercise X.

Ecplaco the double signs by single ones in the expres-

sions :
—
(1.) 2a++3&+-c. (2.) ab-+bc c.

(3.) x2 + -da* -1. (4.) 5x3 + -3a2 7a- +8.

Clear of brackets :

(5.) 8a-(6 + c). (6.) 8a-(5-c).

(7.) 8»-(-»+ 3c). (8.) 22-l + (&-5) + c.
(9.) + 5-(2-%) + 8.
a; (10.) a + (&-c)-(a-c).

(11.) 3^ -l-(-x+4) + 2a-(a2 -5).

2
 .

C 23 )

CHAPTEE V.

MULTIPLICATION.

38. When it is desired to denote the operation of multiply-

ing several expressions together so as to exhibit the various
factors, we enclose each in a bracket and write them together
in a row in any order,
Thus ( + 2a) (—36) denotes the product of + 2a and — 3b ;

(2a -1) (-b) the product of 2a -1 and -b; (x2 -3) (2a; + 5)
the product of cc
2
—3 and2sc + 5; and (x—1) (cc + 2) (2x— 5)
the product of x + 2, and 2x—5.

39. Each of the quantities so enclosed in brackets is called

a factor of the product.
Thus —2a, a 2 —l, and 2a— 3 are the factors of (—2a) (a2 — 1)
(2a -3).

40. When a factor is mononomial, it is called a simple

factor ; otherwise a compound factor.
Thus in the preceding example —2a is a simple factor, and
a2 — 1 and 2a— 3 compound factors.

41. In the case of a simple factor the bracket may be

omitted (i.) if the simple factor is written in the first place;
(ii.) if the sign of the simple factor is not expressed.

Thus we generally write —2a (x—1), x(x2 — 2), (a—b)x,

;
(a—b) (c—d)x.
If the sign of a simple factor is expressed, the bracket
may be replaced by a multiplication sign, x
 : — . .

24 MUL TIPLICA TJOiV.

Thus -2* x +3 j = -2a( + 33),
7

3x x —5?/ = 3^( — 5#),

(a-1) x - 2a2 = (a-1) (-2a 2 ).

Exercise XT.
Express in Algebraical language, retaining the given factors
and using brackets when necessary :

(1) The products of a— 1, 2a — 3 — 2 + a, — 3— a

2
;
2
;
x—5 }

-2x + 7.
(2.) The products of —2a2 , Z>
2
— 1; a2 — 1, —3a; 5x,
-a + 3. 2

(3.) The products of J, as— 1; &/2s-3; 2

|, a ~5. —
(4.) The products of — 5x, cc—1, .t + 2; a 2 — 4, +5x, 2a + 3.
(5.) The products of + 8x, —hy, %y—l —7a, c/&— 3, + 80. ;

42. The mode of performing the operation of multiplication

whereby products are expressed as mononomials or poly-
nomials will now be explained. It is convenient to make
three cases

I. The Multiplication of Simple Factors.

II. The Multiplication of a Simple and a Compound
Factor.
III. The Multiplication of Compound Factors.
43. I. The product of two simple factors is obtained by
the following rule :—
(i.) The sign of the 'product is obtained by the ride of signs:
like signs produce + , and unlike signs —
Thus the signs of + 2a ( + 33), -2a (-33), + 2a (-3b),
—2a ( + 3">) are, respectively, + + — —
, , ,

(ii.) The numerical of the product is the product of

coefficient

the numerical of the factors.

coefficients
Thus the numerical value of tlio coefficient of the product
of -2a and + 33c is 2x3 = 6.
(iii.) The literal part of the product is the product of the
literal parts of the factors (9).
 MUL TIPLICA TION.
Thus tlio literal part of the product of —ox and A.y
2
z is
xy 2 z.

(iv.) The p oditct of two powers of the same letter is a power

whose index is the sum of the ii dices of the factors.

Thus a • a 2 =a 1+2 =a 3 ,
a2 ' a s =a 2+3 =a 5 ,
a4 - a7 ==a4+7 =an .

Examples*

(1.) The product of + 2a and -35= + -2 x 3a5=-Ga&.

(2.) The product of — oa and 2cfc — 5 x 2ac£= — lOud.
(3.) The product of —ha and + 3// =- + j x3a5 2 =-|ci&2 2
.

(4.) -a&(-3c) = 3a&c=+3a&c.

(5.) -ia (-h^)=^+i -|a 5=-ia2 5.
2 .
2

-2a;( + 35c2)= -+2x3cc cc =-Gx

2 3 •
(6.) .

(7.) _6a (-3a

3 4
) = ~-6x3 a 3
- a 4 =+18a 7 .

(8.) + 2a2 fo (-5a5V)^ + -2 x 5

3
a • a2 - 6 - Z>
2
• c
3
• c
4
=
~10aW.

Exercise XII.
Find the product of
(1.) +37, -25; -a, +5c; -2a2 -35; ,
5cc, — G?/.
(2.) 2*&, -7c 2
;
-4a 2
,
+5fo; -2^,8^; -6, -8a.
(3.) — lx, By; |a, -2b; —\x —\y\
}
2a 2 , — |/
(4.) 2xy 2 , —3x2y ; — ax y, 2
—dxy*; \ab 2 c } — |a 2
Z>c
3
.

W a __2a 2
5* ~3~ '
a& a 2 5 3
5"'
T ;
_2xy 2 3x2y 3
"8""
~5~' J
__axy 2 __5x2
T' T'
44. II. T7*e product of a simple and a compound factor is the

Algebraical sum of the products of the simple factor and the

several terms of the compound factor.
 — 4 5

26 MUL TlPLICA TION.

Thus the product of 2a and 35 — oc-hc? is the sum of the
products of 2a, 3b; 2a, —5c; and 2a + d; and , is therefore
equal to Qab— 10ac + 2ad.

The work may be arranged as in the following

Examples.

(1.) Find the product of —3x and 2?/

2
— 4cc?/— 5.
2?/
2 — 4x?/—
-3x
— 6xy + 12x 4- 15cc.
2 2
?/

(2.) Find the product of 2a — $x + \ and 2

—%xy.
2^_i x + i
HN*|y

(3.) Clear of brackets the expression —2a (3a 2 — 5a 4-1).

As this expression denotes product of the —2a and
3a2 — 5a 4-1, it is equivalent to —6a3 4- 10a 2 — 2a.

Exercise XIII.
Find the product of
(1.) 4a-3&4-c, -2x; 3a?-2x + l,4y; 2a5-3c, -d
(2.) x 2 -2x-5, 3x; 2a 2 -3a4-7, -a3 x 2 -ax + 2a 2 , -4acc. ;

(3.) 2x + y—3z,2xyz; 7a 2 b-ab 2 + ab, -4a&.

(4.) a + £b— fc, 2a&; 2a 2 — 3a 4-4, — ia; fee 2 — acc4-ia2 , Jaa.

(5.) -15; ^-l+^ilax.

Clear of brackets :

(6.) 5(x 2 -x4-4); -2(a-a&4-3); -a(a 2 -2aa> + l).

(7.) (a— x)x; (a — b + c)c; (— 1 + afr— 3a )5a&. 2

2
(8.) |?(6a - 9a 4- 12) ;-
2
(- 10 4- 2x- 15a ). 3 -

45. III. The product of two compound factors is the Alge-

braical sum of the products of one factor and the several terms
of the other.
 MUL TIPLICA T10N.
Thus the product of 2a— 3b and 4c + 5i is the sum of the
7
products 2a— 3.6, 4c and 2a— 3b, + 50 and is, therefore, equal ,

to 8«c-12&c + 10arf-15fcc?.
The work may be conveniently arranged as in the following

Examples.
(1.) Multiply 2z2 -Bx + 5 by 4a-7.
2^2 -3x + 5
4a; -7

-Iix 2 + 2Ix-35
8x -26x2 + 41cc-35.
3

Here 8& — 12x 4,x and 2& — 3x-f 5;

3 2 2
-f 20x is the product of
— 14^ + 21x— 35
2
is and 2x 2 —3x + 5; and
the product of —7
the sum of these two partial products, which are arranged so
that like terms stand in the same column, is the product
required.
It will be observed that in the foregoing process we work
from left to right, and not from right to left, as in the cor-
responding Arithmetical operation.

(2.) Multiply 2a-b by c-3d,

2a-b
c—3d
2ac—bc
-6ad + 3bd
2ac — be— Qad -f 3bd t

In this case, as there are no like terms in the partial pro-

ducts, the second is placed entirely to the right of the first.

(3.) Multiply l-2x + 3x by 4x-5x + 2. 2 2

It will be found most convenient in this and similar

examples to arrange the given factors according to ascending
or descending powers of x that is, so that the exponents of
;

the successive terms shall continually increase or decrease.

In the former arrangement the numeral stands first, in the
latter last.
 —

2$ MULTIPLICA TION.

In the present case let them be arranged in the order

2
l-2a + 3x2 , 2 + 4cc-5x .

.1— 2x + 3x-
2 + das— 5x 2

+ 4x— 8x2 + 12x 3

-5x2 + 10x3 -15s4
2 ^ 2
+ 22x3 -15zc4 .

(4.) Multiply 2a 2 -a& + 5 2 by a 2 + a&-36 3 .

2a — a& + 5
2 2

a 2 + a&-3£2

+ 2a &-a & 2 -t-ak3 3 2

-6a 62 + 3aZ>3 -3&4 2

2a* + a 6 - 6a*b + 4a& - 3b\

3 2 3

Here the factors are arranged according to descending

powers of a and ascending powers of 6.

Exercise XIV.
Multiply together

(1.) 2x-3, x+4; 4x + 5, 2— 3sc, 1 + —«+

(2.) x 2 -2, 2^-1; + 3x 2 l + 2 ;

(3.) 2a 2 — a + 4, 3a— 2 1— a + a 2 1 + a; 1-fa + a 2 1— a.

; , ,

(4.) a + &, ra— a + 5— c, m + 2/? 2??i— w, 2w + ^.

; ;

(5.) xy + 2 cc?/ — x
tu
, + x?/ y 2 x — 2?/.
2
; cc ,
2

(6.) 2a 2 -5a + l, a2 + 3a-4; a + 2&-3c, a-26 + 3c.

(7.) 3a + 25-c, a-r26 + 3c; 3x 2 + 2x?/ -f 2 x*-2zy + 3y2 ?/ , .

(8.) x — I, x 2 — % a +
1
a— -J; 2a — a-fj.
; 7r,

cc -£x + l, 2x-£; 3x -fx + i 3a;- i.

2 2
(9.)

AO.) »+|-2,|-8y; ~-2x + i, 3x-|,

^-.i x a2 2a ... a2 . 2a 1
 —

MULTIPLICA TION. 20;

(12.) x2 + if—xy + x+y— l,x + y — 1.

(13.) a 2 + V1 + c 2 — &c col — a&, a 4- Z> + c.
4.6. The product of three expressions is found by multiplying
the product of two of them by the third.

Examples.

(1.) Multiply together 2x } —3x y, — %xy z*

2 2

Hero 2x(— 3x 2y) = —6x 3y ; and — Gx 3 ?/(— %xyh)= + %x4ysz.

(2.) Multiply together cc— 1, cc--2, x—3.
x-1
x-2
x 2 —x
-2x + 2
~2
x 2 -3x +

x3 —3x 2 + 2jc

cc
3
— 6x + llcc— 6.
2

Exekcise XV.
Multiply together

(1.) -3a 2 , + 2a2 b, - 5aZ>2 ;

ja, ~ -Jcc
2
, + fx4 ;
-8a2 ?/,
2
z, asys.
-f?/
(2.) -2x, 3xy, 4x2 -5?/; 2a&, -^a2 , 2a 2 -3a6 + l.
(3.) 2x-3, 4x + l, 05-2.
(4.) x 2 — x + 1, x— 1, + 1. cc

(5.) as
2
-f 2ax -f a2 , x 2 — 2ax + a2, cc
4
+ 2a x2 + a\
2
 ( 30 )

CHAFTEK VI.

DIVISION.

47. Division being the inverse of multiplication, it follows

that, when two factors are multiplied together, either factor
will be the quotient of the product divided by the other.
Thus, since + 2b) = —2ab, +a is the quotient of —2ab
divided by — 2b, and —2b is the quotient of — 2ab divided by
+a.
Again, since — 2x 2 (x 8 — 4x + 3) =
— 2,r 5 + 8x3 — 6x2, it follows
that x 3 — 4x + 3 is the quotient of — 2x 5 + 8x 3 — 6x 2 divided
by — 2x 2 and — 2x 2 is the quotient of —2x 6 -\-8x 3 —6x 2 divided
,

by x 3 — 4x + 3.

48. When it is desired to denote that one quantity is to

be divided by another, they are enclosed in brackets and
written in a row with the sign ~~ between them or the ;

second quantity is written below the first with a line between

them.
Thus (-2x2)-f-( + 3x), or ~~ f
denotes that -2x 2 is to

be divided by + 3x; (3x 2 -2x + 5) -f-

?
(x-4), or ^l^±^ }

that 3x 2 —2x-f 5 is to be divided by x— 4.

49. The first or upper quantity is called the dividend , and
the second or lower the divisor.

50. The bracket is generally omitted in the case of a

mononomial.
Thus (-2x) ^
(-3?/) is written -2x~-3z/; (2z 2 -l)
~(4x) is written (2x 2 -l) -f. 4x ;
(-3x3) ~ (2x-l) is
written — 3x 3
~ (2x — 1).
 : : . . — :
;

DIVISION. 3*

Exercise XVI.
Eetaining the given quantities and employing brackets only
when necessary, express in Algebraical language
(1.) Divide 2a-5 by -3a; 4a 2 -3a + l by 3a-4.
(2.) Divide 2a by -3b; -x 2 by +2x; 3x by 2a.
(3.) Divide Ax2 by 2x— 5; —ax 2 by x—a.
The mode of performing the operation of division
51.
whereby quotients are expressed as mononomials or polyno-
mials will now be explained. We shall consider in order
three cases
I. W hen the Dividend and Divisor are Mononomials.
II. When the Divisor only is a Mononomial.
III. When the Dividend and Divisor are Polynomials.
52. I. The quotient of one mononomial divided by another

is obtained by the following rule

(i.) The sign of the quotient is obtained by the ride of signs
like signs produce + , and unlike signs —
Thus the signs of + 2ab~3cd, -3x2 ~-2x, 4a2 -f--~2a,
—5x 2y-$- + xy }
are, respectively, +, +, — —
,

This rule follows from the rule of signs in multiplication

thus, since + a ( + b) —+ab, it follows that
±^- = +b.
+a
So also from the equivalent forms —a( + b) = —ab —a (~-b)
}

= + ab, +a (—b) =—ab, we deduce

—ab _ ^ —ab _
—a '
—a 3
+a
(ii.) The numerical coefficient, without regard to sign, of the
quotient is obtained by dividing the numerical coefficient of the
dividend by the numerical coefficient of the divisor.
Thus the numerical coefficient of 12a2 -~ 3a is 12—3=4, of
2x 3 -f- 3x is f, and of £a 2 -f- fa is \ -f §=}; -

(iii.) The literal part of the quotient is obtained by dividing

the literal part of the dividend by the literal part of the divisor
(12).
 —
32 DIVISION,

Thus the literal part of tlic quotient of 2a 2 ~Z>c is ~. In

be
applying this rule the fractional form of expressing the
quotient should always be used.

(iv.) The quotient of two powers of the same letter is a power

whose index is the difference of the indices of dividend and
divisor.

The reason for this rule will he evident from the following
examples :

a3
From (10) — = aaa = act =cr =crn_i ? 1 *

a a
5
a aaaaa
-=aa = a'=a°
cr acta
a 7 aaaaaaa
-=aaa = a3 =»7 " 4 :

aaaa

So likewise

a — ay«_9^ — rr* 4.

a1

53. Since the quotient of any quantity divided by itself is

1, this rule can be applied when the indices of the dividend
and divisor are equal, if the zero power of a letter is considered

eiual
1
to unity. Thus ~=1 ;
and, by the rule, %=za2 ~2 =a° :

cr cr

therefore a° = l. Whenever, therefore, by applying the pre-

n
ceding rule, we get such symbols as x°, y°, c we must ,

replace each of them by 1.

Examples.

(1.) -12-5- 4-6«-V=-2;

+*-r— >- j- v.

(2.) -6aJ-86 = -#.

 —
DIVISION. 33

(4.) 10a 3 -5a2 = -2a3 " 2 = -2a.

(5.) -4a5 ^
-7a 2 =+-fra5 - 2 = +fa3 .

(6.) 2a W -f- 3a5c - fa

2 2 " 1 3 " 1 5 ~2
6 c = fa&2c3 .

Exercise XVII.
Divide

(1.) -16 by +4; 20 by -4; -fby-f; +5 by -fr

(2.) -a by +2x; 3a2 by -26; -6x2/ by -3a.
2
(3.) by -|; bv
Dy bv
y -?-
I 3' 7 3*
(4.) 2a 5 by a2 ; 3a 4 by 3a; 8x 6 by 2a4 .

(5.) -4a 6by2a&;

2
lOaW by 2a6c.
(6.) ax 2 y 4 by — 3axy; —3a xy2 bj 2
—5axy.

54. II. The quotient of a polynomial divided by a mono-

nomial the Algebraical sum of the quotients of the several
is

terms of the former divided by the latter.

——
6x 2 + 8x divided by — 2x
3
Thus the quotient
sum of the quotients 3x 3 -i
of 3cc
— 6x2 ~ 2x, +8x-i
2x, — is the
2x;
and is therefore equal to — fx2 + 3o5— 4.

The work may be arranged as in the following

Examples.

(1.) Divide 8x*-&x 2

+2x by -2x.
— 2a ) 8x -4x + 2a
3 2
;

— 4x + 2.a5.—l.2

(2.) Divide 5a3 6-10a2 Z> 3 + 2a& by 5ab.

5ab)5a»b- 10a 2 3
Z> + 2aj>
a2 - 2a& 2 +|
(3.) Collect coefficients of x in 2ax—*3x.
3>
 — —
34 DIVISION.

As this means that 2acc— 3ce is to be expressed as the pro-

duct of two factors one of which shall be x, we have merely
to divide the given quantity by x to get the other factor.

Thus 2ax-3x=(2a-3)x.
(4.) Collect coefficients of x2 in 5ax 2 — bx + x
2 2
.

By dividing the given quantity by x2 we get

Sax2 — bx 2 + x2 = (5a b + 1) x 2 .

EXEECISE XVIII.
Divide

(1.) 10a-156 + 20by -5; -4aa + 12-8a2 by -4.

(2.) 4a 2cc-3a 2 + a3 by-a2 12x3 -6x2 + 9x by Sx.
;

(3.) 3a - 12a + 15a2 by -3a 2x sy-6x 2y 2 + 8xy* by

4 3 2
;
2xy.
(4.) 2a Z>c-3a6 2 c + a6c2 by abc.
2

(5.) 20a2 5c 2 -15aZ> 2 c 3 + 5aZ>by -5a&.

(6.) Collect coefficients of x in ax—bx, 2ax—cx + x,

(7.) Collect coefficients of xy in kxy—axy, Sx2y—xy2 .

55. Since by (54) a *<*

*>y (44) K*-l)=g-l*
rfc
it follows that 2^
o
and i(x— 1) are equivalent forms. So
likewise

^=i(2a^3%

56. III. When the dividend and divisor are polynomials,

the quotient is obtained by the following rule :

(i.) Arrange both dividend and divisor according to ascending

or descending powers of some letter.

(ii.) The first term of the quotient is found by dividing the

leading term of the dividend by the leading term of the divisor.

The product of the divisor and the first term of the quotient is
subtracted from the dividend, giving the first difference.
 1

DIVISION. 35

(iii.) The second term of the quotient is found by dividing the

leading term of the first difference by the leading term of the

divisor. The product of the divisor and the second term of the
quotient is subtracted from the first difference, giving the second

difference in the process.

And so on until the last difference is zero.

Examples.

(1.) Divide 6® 3 -5® -3x + 2 by 2 + 3®.

2

3x + 2) 6x 3 - 5x 2 - 3x + 2 (2x2 - 3x +
6x 3 + 4-x2
"~-9®2 -3®+2
— 9cc — 6x
2

~~3x + 2
3x + 2

Here the divisor is first arranged, like the dividend, ac-

cording to descending powers of x. The first term of the
quotient 2x 2 is obtained by dividing 6x3 , the leading term of
the dividend, by 3a?, the leading term of the divisor. The
product of the divisor 3x + 2 and 2x 2 , which is 6cc3 +4&2 , is
then subtracted from the dividend, giving — 9cc 2 — 3# + 2, the
first difference. The second term of the quotient —3® is
obtained by dividing — 9x 2, the leading term of the first
difference, by 3x, the leading term of the divisor. The pro-
duct of the divisor and —3®, which is — 9sc2 — 6x, is then
subtracted from the first difference, giving 3x + 2, the second
difference. The third term of the quotient +1 is obtained
! by dividing 3x, the leading term of the second difference, by
j
3x, the leading term of the divisor. The product of the
II
divisor and + 1, which is 3x + 2, is then subtracted from the
second difference, giving the last difference zero ; and thus
the process ends.
The latter terms of the differences need not be expressed
I
until the corresponding like terms in the partial products
are to be subtracted from them; thus in the following ex-
d 2
36 DIVISION.

ample —17a; + 6 is not expressed in the first difference, nor

+6 in the second.

(2.) Divide 6a; + 5a;3 -f 6a; - 17a; + 6 by 2x-l.

4 2

2x - 1) 6x 4
+ 5a; 3 + 6x - 17 x + 6 (3a)3 + 4a; 2 + 5a; - 6
2

6a4 - 3a3
8x3 + 6x2
8a;3 — 4a; 2

10^1703
10a;
2
- 5a;
-12a; + 6
-12a; + 6

(3.) Divide l-2x3 + a; 6 by l-2a; + a;2 .

1 - 2x + a; 2) 1 - 2a; 3 + x (1 + 2x + 3x 2 + 2a; 3 +
6
a;
4

l-2a; +a; 2
2a;— a; 2 — 2x 3
2a; - 4 + 2a;
a;
2 3

3
3a;2 -4a;

2a;
3
-3a; 4
3
2a; -4a 4 + 2a; 5
;

a;
4 — 2a; + 5
a;
6

x*—2x 5 + x 6

Here +xQ is not expressed until we reach the last dif-

ference.

(4.) Divide 2a5 -6a 3 b + 13a?b 2 --6ab 3 -3ai b by 2a-3b.

Here we shall arrange the dividend and divisor according
to descending powers of a.

2a - 3b) 2a* - 3a 4 b - 6a3 b + 13a2 b 2 - 6a¥ (a4 - 3a 2 b + 2ab 2

2a 5 -3a4 b
-6a36 + 13a2 6 2
-6a3 5 + 9a2 62
MW-Sab 3

2 2
4ta b -6ab3
 1 a a

DIVISION. 37

EXEECISE XIX.
Divide

(1.) x 2 -7x + 12 by x-3; and 3x2 + 7x + 2 by x + 2.

(2.) x 2 — 4:x— 5 by cc— 5; and 4x 2 — 9 by 2cc + 3.
(3.) 6x 2
-5x-6 by 2x-3 ; and 9x 3 -18x 2 + 26x-24 by
3x-4.
(4.) x3 — 4x 2 + 5x— 2 by x2 — 3x + 2; and x4 + x 2 +l by
X2 + X+l.
(5.) 4x4 -15x 3 -t-14x 2 --6x + l by 4x2 -3x + l.
(6.) x 4 — lbyx— 1; and x 5 + l by x + 1.
(7.) x 2 —2xy + y 2 by x— y and xd + y3 by x + y.
;

(8.) a4 + a2 6 2 + Z>
4
by a2 -ah + 2 .

(9.) 20x 2 + 9^-12x-18?/ 2 + 92/ by 4cc-3y.

(10.) x 5 — 2ax 4 + 3a 2 x 3 — 3a3x 2 + 2a 4 x — 5
by x s — ax2 + 2
x — a3 .
-

(11.) x —x y—xy + y* by x + x?/ +

i s 3 2
?/
2
.

fa x-^a x + fa x +^acc ~x —fa^ +

4 3 4 5
2 2 3
(12.) by fa 3 Jcc
8
.

57. When the division is not exact, the last difference is

called the remainder.In this case the product of the quotient
and divisor added to the remainder will be equal to the
dividend.

Example.

Find the quotient and remainder in dividing 10x 3 + 7x 2

-8x-2by2x + 3.
2x + 3) 10x3 + 7x 2 - 8x -2 (5x 2 -4x + 2
10x 3 + 15x 2_
~-8x2 -8x
-8x 2 -12x
4x^2
4x + 6
-8
Quotient = 5x2 — 4x + 2.
Eemainder - —8.
38 DIVISION.

Exercise XX.

Tind the quotient and remainder in dividing

(1.) 4a?-4a? + 8a* + 2by 2x + L

(2.) x2 +a 2 byx + a.
(3.) cc — a by x + a.
3 3

2a; -2xH9a by 2a + cc+l.

5 3 3
(4.)

(5.) 2x5 +2a; 4 + 5<c 3 by xs +x2 +x+l.

 ( 39 )

CHAPTEE VII.

EXAMPLES INVOLVING THE APPLICATION OF TEE

FIBST FOUR EULES.

58. In the following examples some of the given quantities

are expressed by letter symbols, and the object of the exercises
is to express in like manner other quantities which by the
conditions of the question are related to the former. When
a doubt exists as to the manner of solving a question, it will
be well to substitute numbers for letters in order to see what
operations ought to be performed in the given symbols.

59. The sign .

*
. will be used to mean hence, or therefore, and
the sign v since, or because.
Examples,

(1.) I buy goods for 2a+3&— c dollars, and sell them for
4:a—b + 2c dollars what do I gain ?
;

4a— h + 2c
2a + Sb-c
2a-46 + 3c
.•.the gain, which is the selling price less the cost price, is
2a— 4&-l-3c dollars.

(2.) A man has 3x2 + 7x + 2 dollars and spends x + 2 of them

per day ; how long will his money last ?

cc+2)3xH7a; + 2(3^ + l
Sx 2 +6x

cc+2
 £

40 EXAMPLES OF THE FIRST FOUR RULES.

. the required number of days, which is equal to the num-
•
.

ber of times the amount of his daily expenses is contained in

the amount he possesses, is 3x + 1.
(3.) A man walks
x miles in y hours at what rate per hour :

does he walk how far will he walk in 5 hours and how long
; ;

will he be in walking 12 miles ?

V he walks x miles in y hours,

. „ „ ~ „ „ 1 hour,
y
and 1 mile in ^ hours.
x
Again, Y he walks - miles in 1 hour,
y

:. „ » - » „ 5 hours;
y
and v he takes ^ hours to walk 1 mile,
x

„ „ — x
» » „ 12 miles.

The answers are, therefore,

x
- miles, —
5x
miles, and —x
12?/

y y
hours.

Exercise XXI.

(1.) A man x+a, and x—2a miles in the same

walks x }

direction ;how far does he walk altogether ?

(2.) A man has 100 dollars, and owes 50— x dollars; what
is he worth ?
(3.) A man walks a + b miles and returns a— b miles; how
far is he from the starting point ?
(4.) What is the area of a room x + yfeet long and x—y feet
broad ?
(5.) A man walks x miles at a miles an hour ; how long is

he on the road ?
(6.) A has x dollars, B 50y cents, and C 75z cents ; how
many dollars have A, B, and C together ?
 EXAMPLES OF THE FIRST FOUR RULES. 41

(7.) A has x pounds, B has y shillings, and C z pence ; how

many pounds have A, B, and C together ?

(8.) A spends a dollars in x days ; in how many days will

he spend 10 dollars ?
(9.) How many square yards in a floor which is a feet by
x feet ?

(10.) What is the cost in dollars of painting a floor x + y feet

by cc— y feet at x2 +y 2 cents per square foot?
(11.) A owns a acres, B b acres, and G 5 acres less than one-
and one-third of C's together
half of A's ; what is the whole
amount possessed by A, B, and C ?
(12.) A owns a + b acres, B a— b acres, C half as much as A,
and D half as much as B how much more
; do A and C own
than B and D?
(13.) A walks a miles in t hours, and B half as far again in
the same time ; how far will B walk in 10 hours ?
(14.) A walks 10 miles in x hours ; how long will he be in
walking a miles ?
(15.) A
spends at the rate of x dollars a day for a days, a
dollar a less for twice that time, and a dollar a day
day
more for three times that time how much does he spend;

altogether?
 CHAPTER VIII.

SIMPLE EQUATIONS.

60. An equation is the statement of the equality of different

quantities, and these quantities are called the equation's
members or sides.

Thus 2cc + 3=7 is an equation whose sides are 2.r + 3 and

7 , and cc
2
— 5& + 6=0 is an equation whose sides are cc 2 — 5^ + 6
and 0.
61. An identity is the statement of the equality of two like
or different forms of the same quantity.
Thus 2a + b=2a + b, 2x + 3x=5x, cc
2
-5a) + 6=(cc-2)C^~3),
are identities.

62. In the case of an identity the equality holds for all

values of the quantities involved, whereas in an equation the
equality does not exist except for a limited number of values
of the quantities involved.
Thus the statement x2 + 2x + l= (x + 1)2 holds no matter
what x is but hx — 3=7 holds only when x=2, and x 2 + 6= 5a?
;
|
only when x=2, or cc=3.

63. A symbol to which a particular value or values must

be assigned in order that the statement contained in an equa-
tion may be true is called an unknown quantity.
Thus the unknown quantity in 5cc— 6=9 is x, and in
2y
2
-y=8 is y.
The letters a, b, c, I, m, n, p, q, r are generally used to I

denote quantities which are supposed to be known, and x, y } z

those which are for the time unknown.
 — ;

SIMPLE EQUATIONS, 43

64. Quantities which on being substituted for the unknown

reduce the equation to an identity are said to satisfy the
equation, and are called its roots.

Thus 5 a root of 2sc— 3=7, because 5 when substituted

is
for x reduces the equation to the identity 10—3=7. So 2
and 3 are the roots of x 2 + 6=5x, because when either is sub-
stituted for x the equation is satisfied.

65. The determination of the root or roots is called the

solution of the equation.

66. An equation is said to be reduced to its simplest form

when its members consist of a series of mononomials involving
positive integral powers only of the unknown.
Thus 5^-8=0, x2 -5x + 8=0, 2x 2 + 6cc=7, xs -Qx2 =7x-8
are in their simplest forms.

67. Equations when reduced to their simplest forms are

classed according to their order or degree,

68. Simple equation s, or those of the first degree, are those

in which the highest power of the unknown quantity is the
first; as, for example, 2cc=5, 5cc— 8=0, 3x— 7=0.

69. Quadratic equations, or those of the second degree, are

those in which the highest power of the unknown quantity
is thesecond; as, for example, cc
2
— 2cc + 8=0,
2& 2 =9,
4x2 -3=10cc.
70. Equations of the third and fourth degrees are called
cubic and biquadratic equations, respectively thus + 2cc=10 ;

is a cubic, and x 4 —2x 3 =10x~ 5 a biquadratic.

proved in works on the Theory of Equations that

71. It is
the numberof the roots of an equation is equal to its degree
so that a simple equation has one root, a quadratic two roots,
a cubic three roots, and so on.

72. In order to solve an equation it is generally necessary

to reduce it by one or both of the following processes :

I. Transposition of Teems.
II. Clearing of Fractions.
44 SIMPLE EQUATIONS.

These operations will be illustrated by applying them in

order to the solution of simple equations.

I. Transposition of Terms.
an equation contains no fractions it may be solved by
73. If
transposition of terms, which consists in talcing the unknown
quantities to one side of the equation and the known to the
other, the signs of the quantities which are so transposed being
changed.
Thus, if the equation is 4cc + 5=10, by subtracting 5 from
each side we get
4^ + 5-5=10-5,
or 4cc=5;
and so any quantity may be transposed from one side to the
other by changing its sign.

Examples.
(1.) Solve 5^ + 15=25.

Transposing + 15 we get
5^=25-15=10.
The value of x is then found by dividing both sides by its

coefficient 5.
.\ x=2.
(2.) Solve 8#-4=2x + 20.
Transposing —4, 8x=2as + 20 + 4.
Transposing 2x, Sx — 2x = 20 + 4 ;

6x=24.
;.x =4.

(3.) Solve 10 + 2(6^-1) =32-3(^-4).

Clearing of brackets,
10 + 12a;-2=32-3cc + 12.
Transposing 10, —2, — 3x,
12£ + 3a=32 + 12-10 + 2;
15^=36.
 b ;

SIMPLE EQUATIONS. 45

(4.) Solve 3(aj

a
+ 2a?) + 13 = 3^-7+4(3aj-l).
Clearing of brackets,

Sx2 + 6x + 13 = 3^2 - 7 + 12a; -4.

2
Transposing +13, 3a? ,
+12x,
3x2 -3o:2 + 6x-12^ = -13-7-4;
- 6oj = -24.
Dividing by— 6,
x = 4z.

When, as in this case, the same quantity is common to

both sides, it may be struck out without actually transposing
thus
5a;
2
- 6x + 7 = 8a + 5a;2 - 10
becomes —6x + 7 = 8x—10.
(5.) Solve ax + b = c.
Transposing + ax—c—b.
Dividing by a, x= c-^.
(6.) Solve ax + b=cx + d.
Transposing + b, cx,
ax—cx = d—b.
Collecting coefficients of x,
(a—c)x = d—b.
dividing by a— c,
x= d-b
a—c
.

(7.) Solve a(x— b) = b(x + a)-~ c.

Clearing of brackets,
ax—ab = bx + ab~c.
Transposing — ab, bx,
ax—bx = ab + ab—c.
Collecting coefficients of x,
(a—b)x = 2ab—c.
Dividing by a— 9

2ab — c
 — ; ;

46 SIMPLE EQUATIONS.

Exercise XXII.

(1.) 34-a5=5. (2.) a5-6=4. (3.) ^ + 5 = 12.

(4.) 05 + 9=4. (5.) 2®-l = 3. (6.) 5o;+4 = 29.
(7.) 4-3o5 = 5. (8.) l-x = (j. (9.) 3=6-2o5.
(10.) 2x + 3=o5 + 5. (11.) 505-^-2 = 2x + 7.
(12.) 05 + 4 = 18-405-4. (13.) 2x + 3 = 3o5-4.
(14.) 16-2o;=:46-5o5. (15.) 3(cc-l) + 4 = 4(4-o;>
(16.) 5-3(4-2o5) + 4(3-4a5)=0.
(17.) 05-l-2(o5-2) + 3(^-3)-6.
(18.) 5(o5-5) + 2(o^-3)-(x-l)=9.
(19.) 2(o?-2)-3(o5-3)+4(o;-4)-5(o -5) = 0. !

(20.) 4(o5-ll)-7(o5-12) = 6-(o5-8).

(21.) 05=2a-05. (22.) 2a-3x=8a-5x.
(23.) x-Zb=2a-x. (24.) a + x-b = a + b.
(25.) ax—ab—ac=0. (26.) ax—a = b—bx.
(27.) ax—a3 =bx — b 3 . (28.) a(o5— b) = c(x— a).

II. Clearing op Fractions.

74. If an equation contains fractions, it may be reduced to
a form capable of solution by transposition, by multiplying
both sides of the equation by the L. C. M. of all the denominators
of the fractions.
In the following examples numerical denominators only
will be considered.

= L
l-g g~
(1.) Solve

Multiplying by 30, the l.c.m. of 2, 3, 5,

15:r lOx = 60; — 30.

Transposing, 15o5 — lOo; —6x= —30
-05= -30.
.-. o; = 30.

/0
(2.)
N a
Solve _+^
x—1 ,
2^ + 3
__,
= 6x + 19
Multiplying by 24, the l.c.m. of 2, 3, 8,

12(o5 - 1) + 8(2x + 3) = 3(6a; + 19)

 1

SIMPLE EQUATIONS. 47

whence on clearing of brackets and transposing we get

05=4:5.

X + 3 = i(2a> + 3),
It must be observed that ~^\(x-Y\ ??£
A o

and ®^~?==±(6x + 19); and therefore the brackets must be

8
supplied in the first step since the numerators become bino-
mial factors.

(3.) Solve 3-^=1 +

A
^ o
= 0.

Multiplying by 6, the l.o.m. of 2 and 3,

18-3(x-l)+2(a + 2)=0;
whence on clearing of brackets and transposing we get
o;=25.

EXEKCISE XXIII.

(1.) |— 2=3. (2.) |+|=7. (3.) |-g+_|=m

x/
(4.) ^ 4
+ ^=10.
6
(5.)
y
5=§ + «=20-2^.
2 3 2

(6.) |+?=?=4--?=?. (7.) ** + 2=*-K* + l).

(8.) 2(a?-l)-4(2aJ-9) = 4(17-2aj).

(9.) §^=4- 5^. (10.) K*-4)+K«-6)=aB-liB.

(11.) 3-fcc=l-^(7oj-18).
(12.) 6(x - 1) - 1 = f (5 - 2x) + i(x + 1) + 4oj.

(13.) *+f£^-^-9*=0.
(14.) 4x+2Jx +-(-~-) =^+lf.

/1^\ -+ ^ - — 2
£C : 305 i__05 —
T
,

(15.) g-.

8-2as 6a: 5 _3(2» + 6) 2*

4s SIMPLE EQUATIONS.

/iv\ 3x — l__13 — sc__7a; ll.x + 33 n

'
K }
~T~ ~T~ ~3~
6

(ib.) --+.
5
_+__ g -,,
/i q \ ^__2:c_L 3a?__4i; 2(^ — 3) _^
^ ;
2 "3 T 5" 9
'

(20.) | (*-8)-?^-g=a
 —
( 49 )

CHAPTER IX.

PROBLEMS.

75. When a question is assigned for solution the unknown

quantity, or number, is generally involved in the various
conditions which are proposed for its determination. The
expression of these conditions in Algebraical language leads
to an equation, the solution of which will be the solution of
the question.

76. In some cases, although there are more unknowns than

one, they are related to each other in such a manner that
when one determined the others become immediately
is

known. In such cases the unknowns can be expressed in

terms of one unknown.
Thus, if the sum of two unknowns is equal to 8, we may
denote one of them by x and the other by 8— x, or one of
them by 4 + £c and the other by 4:-~x; if the greater of two
unknowns exceeds the less by 3, the former may be denoted
by x and the latter by x— 3 if there be two numbers of which
} ;

one exceeds 4 times the other by 7, the former may be denoted

by 4:x + 7, and the latter by x.
In like manner, if there be three unknowns, of which the
first exceeds the second by 3, and the second exceeds the
third by 5, the first may be denoted by a?, the second by ic—3,
and the third bj x—8.
77. The following examples will illustrate the method of
solving problems by means of simple equations :

(1.) What number exceeds its fifth part by 20 ?

Let x be the required number.
E
 :

PROBLEMS.

Then its fifth part = % ; and by the condition of the

5
question
4=20.
5
x=25.

(2.) The sum of two numbers is 71, and their difference 43 :

find them.

Let x be the greater number.

Then cc— 43 is the less and since their : sum in 71, we have
a + cc— 43=71.
,\ cc=57, the greater;
and x— 43=14, the less.

This question may also be solved as follows

Let x be one number, the greater suppose.

Then 71— x is the less and since their difference
; is 43, we
have
£C -(7l-^)=43.
^=57,
and 71— cc=14.
(3.) A boy is one-third the age of his father, and has a
brother one-sixth of his own age; the ages of all three
amount to 50 years. Find the age of each.
Let the boy's age =x years.
Then the father's age =3x years,

And the brother's age —% years.

o
And by the condition of the question

x + 3cc + ^=50.
o
.\ 8 = 12,
3cc=36,

Fractions may be avoided by supposing the ages of boy,

father, and brother to be 6x, 18x, x years, respectively.
 :

PROBLEMS. 5i

(4.) A and B start from two places, 90 miles apart, at the

same moment, A walking 4 miles per hour, and B 5 when ;

will they meet, and how far will each have walked ?

Let the time of meeting be x hours after starting.

Then A will have walked 4cc miles, and B 5x miles ; and
since the sum of these two distances is 90 miles,

4a + 5a =90.
x=lQ.
»\ 4a =40, and 5a =50, are the distances in miles walked by
A and B, respectively.

(5.) How much tea at 90 cents per lb. must be mixed with
50 lbs. at $1*20, that tbe mixture may be sold at $1*10?
Let x = the number of lbs. at 90 cents, the value of which
will be *90a dollars.

Then, since there will be a + 50 lbs. in the mixture, its

Value will be 1*10 (a + 50) dollars and since the value of the
;

50 lbs. at $1*20 is 60 dollars, we have

9Ob+6O=1-IO(«&+60):
Multiplying by 100,
90a + 6000=110(a + 50).
.\ a=25.

Exercise XXIV.

(1.) Divide 25 into two such parts that 6 times the greater
exceeds twice the less by 70.

(2.) Divide 135 into two parts such that one shall be f the
other.

(3.) The sum of two numbers is 37 and their difference 3

find them.

(4.) A fish weighed 71bs. and half its weight ; how much
did it weigh ?
(5.) At a meeting 43 members were present, and the motion
was carried by 9 how many voted on each side ?
:

e 2
52 PROBLEMS.

(6.) Divide 326 into two parts, such that f of the one shall
be equal to the other diminished by 7.

(7.) What is the number whose 4th and 5th parts added
together make 2J ?
(8.) Forty-two years hence a boy will be 7 times as old as he
was 6 years ago how old is he ?
:

(9.) A father is 57 years old, his son 13 : when will the

father be 3 times as old as his son ?

(10.) I have made 164 runs at cricket this season in 12

innings: how many must I make in my next innings to
average 14?

(11.) My grandfather told me 10 years ago that he was 7

times as old as myself; I am now 18: how old is my
grandfather ?

(12.) If in a theatre f of the seats are in the pit, in the

lower gallery, \ in the upper, and there are 50 reserved seats,
how many are there altogether ?

\ of our men by sickness, and 210 killed

(13.) After losing
and wounded, the regiment was reduced by i how many :

men did the regiment originally contain ?

(14.) In a certain examination f of a boy's marks were gained

by translation, | by mathematics, and by Latin prose he :

also obtained 1 mark for French. How many marks did he

obtain for each subject ?

(15.) Two men receive the same sum but if one were to
;

receive 15 shillings more, and the other 9 shillings less, the

one would receive 3 times as much as the other. "What sum
did they receive ?

(16.) A and B begin trade, A with 3 times as much stock

as B. They each gain £50, and then 3 times A's stock is
exactly equal to 7 times B's. What were their original stocks?

(17.) One-tenth of a rod is coloured red, one- twentieth

orange, one-thirtieth yellow, one-fortieth green, one-fiftieth
 PROBLEMS. 53

blue, one-sixtieth indigo, and the remainder, which is 302

inches long, white : what is its length ?

Find three numbers whose sum is 37, such that the

(18.)
greater exceeds the second by 7,
and the second exceeds the
third by 3.
(19.) Find a number such that if 5, 11, and 17 be suc-
cessively subtracted from it, the sum of the third, fourth,
and sixth parts of the respective results shall be equal to 19.

How much wheat at 44s. a quarter must be mixed

(20.)
with 120 quarters at 60s. that the mixture may be sold for
50s. a quarter ?

(21.) How many lbs. of tea at 2s. 6d. per lb. must be mixed
with 18 lbs. at 5s. per lb. that the mixture may be sold for
4s. per lb. ?

(22.) How much sugar at 4^d. per lb. must be mixed with
50 lbs. at 6£d. per lb., that the mixture may be worth 5d.

per lb. ?
A bag contains a certain number of sovereigns, twice
(23.)
as many shillings, and three times as many pence and the ;

whole sum is £267 find the number of sovereigns, shillings,

;

and pence.
(24.) I wish to divide £5 4s. into the same number of
crowns, florins, and shillings ; how many coins must I have
of each sort ?

(25.) A
person gets an income of £550 a year from a
capital of £13,000, part of which produces 5 per cent, and
part 4 per cent. what are the amounts producing 5 and 4 per
:

cent., respectively ?

(26.) I invest £800, partly at 4i per cent., and partly at 5£

per cent. ;
my income is £39 10s. : what are the sums invested
at 4| and h\ per cent., respectively ?

(27.) A
garrison consists of 2600 men, of whom there are
9 times as many infantry and 3 times as many artillery as
there are cavalry how many men are there of each ?
:

(28.) My grandfather's age is 5 times my own ; if I had

 :;

54 PROBLEMS.

been born 100 years ago, I should have been born 15 years
before my grandfather how old am I ?
:

(29.) There is a number consisting of two figures of which

the figure in the unit's place 3 times that in the ten's;
is

if 36 be added, the sum is expressed by the digits reversed

what is the number ?

(30.)A miner works for 6 weeks (exclusive of Sundays),

his wages being at the rate of 24s. per week, but he is to
forfeit Is. besides his pay for each day that he is absent at ;

the end of the time he receives 4 guineas how many days :

was he absent ?
(31.) A contractor finds that if he pays his workmen
2s. 6d. per day, he will gain 10s. per day on the job if he ;

pays them 3s. a day, he will lose 18s. how many workmen :

are there, and what does the contractor receive per day?
An officer on drawing up his men in a solid square
(32.)
findshe has 34 men to spare, but increasing the side by 1
man he wants 39 to make up the square how many men :

had he ?
(33.) If the mean velocity of a cannon-ball at effective
ranges is 1430 feet per second, and that of sound 1100 feet,
how from a fort who hears the report of a gun
far is a soldier
9
To of a second after he is hit ?
(34.) An army in a defeat loses one-sixth of its number in
killed and wounded, and 4000 prisoners. It is reinforced by
3000 men ; but retreats, losing a fourth of its number in
doing so. There remain 18,000 men. What was the
original force ?

(35.) Suppose the distance between London and Edinburgh

is 360 miles, and that one traveller starts from Edinburgh
and travels at the rate of 10 miles an hour, while another
starts at the same time from London and travels at the rate of
8 miles an hour it is required to know where they will meet.
:

There are two places 154 miles apart, from which

(36.)
two persons start at the same time with a design to meet
one travels at the rate of 3 miles in 2 hours, and the other at
the rate of 5 miles in 4 hours when will they meet ? :
 ( 55 )

CHAPTER X.

PABTICVLAB RESULTS IN MULTIPLICATION AND

DIVISION

78. Theee are several results in multiplication and division

which should be committed to memory, as they enable us to
dispense with the labour of performing the operations. The
following cases occur most frequently.

I. Since by actual multiplication

(a + b) 2 =a + b + 2ab,
2 2

la-b) 2 =a + b2 -2ab,
2

(a + b-c) 2 =a + b + c2 + 2ab-2ac-2bc,
2 2

&c. = &c.

we can hence write down the square of a polynomial by the

rule:

The square of a polynomial is equal to tlie sum of the squares

of the several terms and twice the sum of the products of every
tivo terms.

Thus in the last example a2, +b2} + c2, are, respectively, the
squares of a, +b, —c; +2ab is twice the product of a and
+ — 2ac is twice the product of a and c, and 2bc is twice — —
the product of +b and — c.

In taking the products of the terms, two and two, it will

be found most convenient to take in order the products of
the first term and every term that follows it, then the pro-
ducts of the second term and every term that follows it, and
so on, .if there be more terms than three.
 PARTICULAR RESULTS IN

Examples.

(1.) (a + 2x)
2
=a2 + 4a + 4aa.
2

Here + 4ax is twice the product of a and + 2cc.

(2.) (2a-5a)2 =4a 2 + 25a 2 -20a*.
Here + 25cc2 is the square of —bx, and —20ax is twice the
product of 2a and — 5x.
(3.) (2z 2 -3x + 4)2 =4x4 + 9cc2 + 16-12cc3 + 16a2 -24a
=4cc 4 - 12x 3 + 25cc 2 - 24cc + 16.
Here — 12sc 3
is twice the product of 2x 2 and — 3x, + 16sc 2
of
2cc
2
and
4, and — 24cc of — 3x and
+4. Like terms are added
together and the terms are arranged according to descending
powers of x.
(4.) 99 2 =(100-1) 2 =10000 + 1-200=9801.

Exercise XXV.
Write down the squares of
(1.) x— 1, x + a, cc— 5, x + 3.
(2.) 2a; + 1, to-l, 2x + 3, 3x-2.
(3.) x2 -a, 2xy + l, Zx -2a, ax2 -kb.2

(4.) a;— 2x + 3y— z, x—2y—5z, 2x-- 4?/ + l.

2a 2 + a + 3, 3a 2 -4a + l, a -2a-4.
2
(5.)

(6.) Find the squares of 49, 98 and 995.

79. II. Since (a + b) (a—b) =a —b

2 2
, it follows that the pro-
duct of the sum and difference of tivo quantities is equal to the
difference of their squares.

Thus (2^ + 3*/) (2^-3*/) =±x 2 -9y2 ;

0 +l)0
2 2
-l) =a -l; 4

2 2 4 2
(5cc +4?/) (5x -4?/) =25cc -16*/ ;
(2x3 + a 4 ) (2x 3 -a ) =4;x -a s
4 6
;

501x499 =(500 + 1) (500-1)

=500 -l 2

=249999.
 MULTIPLICATION\AND DIVISION. 57

Exercise XXVI.
Write down the products of
(1.) x— 1, + 1;cc a + 3,a—3; 2 + x,2—x.
(2.) 2x + l, 2a?— 1; 5a + 2, 5a-2; 4x + a, 4cc-a.
(3.) a 2 + x, a2 -cc; a 3 + l, a 3 -l; a 5 + x 2 , a 5 -x2 .

2 2
(4) 3a + 26, 3a -26; 4a3 + 2x 2, 4a 3 -2x- 2 ;
7a 4 -5a 3 ,
7a4 + 5a3 .

(5.) Find the products of 48, 52; 95, 105; 695, 705.

80. III. Since by actual multiplication

(o? + b 2 -ab) ( a + =0? + b 3
,

( a a
+ &2 + a&) a -6)
( =a3 -b 3
,

it follows that £Ae swm of ^e squares less the product of two

quantities multiplied by their sum is equal to the sum of their
aubes.
In the latter identity the two quantities are a and — b;
the sum of their squares less their product is, therefore,
a2 + b 2 ab=a2 + b 2 + ab; and, since the cube of — b is — b 3
,

the sum of their cubes is a 3 —b 3 .

Examples.

(1.) (x -cc + l)
2
0 + 1) =cc 3
+ l.
Here the two quantities are x and 1.

(2.) (x + x + l) (cc-,1) =cc -l.

2 3

Here the two quantities are x and — 1.

(3.) (4x2 -2x + l) (2« + l)=8x 3 + l.
Here the two quantities are 2x and 1, the cubes of which
are Sx 3 and 1.

(4.) (x*-a 2 x 2 + a 4 ) (x 2 + a 2 ) =x + a 6 G
.

Here the two quantities are x 2 and a 2 } the cubes of which

are x 6 and a 6 .

(5.) (4x 4 + 6x 2 ?/ + 9?/ 2 ) (2x 2 -3?/) =8x 6 -27?/ 3 .

2
Here the two quantities are 2x and —3y } the cubes of
which are 8x 6 and —21 y 3 .
58 PARTICULAR RESULTS IN

Exercise XXVII.
Write down the products of
(1.) m — mn + n m + n; p +pq + q ,p—q.
2 2
,
2 2

(2.) m — m + 1, m+1; 1 + q + q 1— q,
2 2
,

(3.) x2 -3x + 9,x + 3; a 2 +±a + 16, a -4.

(4) ±a -2a + l, 2a + l; 16x + ±ax + a , ±x-a.
2 2 2

(5.) 4a 2 -6a& + 96 2 2a + 3Z>; 9x 2 + 15x?/ + 25?/ 2 , 3aj-5y.

,

(6.) a;
4
+ as
2
+ 1, cc
2 —1 ; a?
6 -— a 2 x 3 + a*, x 3 + a2 .

81. IV. By actual division it can be shown that the sum of

any the same odd powers of two quantities is exactly divisible by
the sum of the quantities.

x
-±i
Thus, =i,
x+y
t±yl= X*- Xy + y\
x+y
x ~^~y
= — x y + x y — xy
cc
4 3 2 2 5
+ w4 ,

&c.=&c.
It will be observed that the signs of the quotient are
alternately + and — and , that the successive powers of x
are in descending whilst those of y are in ascending order.

82. V. The difference of any the same odd powers of two

quantities is exactly divisible by the difference between the
quantities.

Thus, —
x—y
V- =1,

^J^JL=x2 + xy + y 2
x-y ,

- ~y =xi + x3y + x2y + xy + y* 2 s

x—y 9

&c.=&c.
 x y .

MULTIPLICATION AND DIVISION 59

Here the signs of the quotient are all +

It may be noted that this case is included in the preceding
(81) by supposing the two quantities to be x and —y. Thus
the sum of the cubes of x and —y is x 3 —y 3, which is exactly
divisible by their sum x—y. In fact the formulas of (82)
are deducible from those of (81) by substituting ~y fov y in
the latter.

83. VI. The difference between any the same even powers of
tivo quantities is exactly divisible by the sum of the quantities
and also by their difference.

Thus (i.)
K }
~~zt=x-y,
Ji
v+y
4
oc —y*_
' •x 6 —x y-\-xyl —i
l

x+y
x —y 6 Q
x 5 — x*y + x 3y 2 — x2y 3 + xy* — y 5 ,
x+y
&C.=&C

(ii.) ^= x + y 3
x-y
x^—y 4-

x3 + 2
y + xy
2
+ 3

x-y ,

x —y 6 G
x 5 + x*y + x 3y 2 + x2y 3 + xy* + y 6 ,
x-y
&c.=&c.

It willbe observed that when the divisor is cc— y, the signs

of the quotient are all + and when the divisor
; is x + y the
signs of the quotient are alternately + and — . It should
also be noted that the formulas (ii.) are deducible from (i.) by
substituting —y for y in the latter.

Exercise XXVIII.

"Write down the quotients of

(1.) a 3 + 1 and x 5 + l divided by cc+1.

6o PARTICULAR RESULTS.

(2.) x 3 — 1 and x 5 —l divided by x— 1.

(3.) a;
2
—1 and cc
4
—1 divided by os + l.
(4.) x2 —1 and as
4
—1 divided by 1.

(5.) 4a 2 -96 2 divided by 2a + 3b.

(6.) 9x 6 -4a 2 divided by 3x 3 -2a.

(7.) |a 4 -x 6 divided by ha 2 + x 3 .

(8.) Find what the quotient of x 3 + y 3 divided by cc + y

becomes when (i.) x=2a, y—3b; (ii.) x=a 2 } y—2.
(9.) Find what the quotient of x 3 —y 3 divided by x—y
becomes when (i.) x=3a, y=b (ii.) x=2a2, y=3b. ;
 ( 61 )

CHAPTER XI.

INVOLUTION AND EVOLUTION.

f
84. The process by which the powers of quantities are
Impressed as mononomials is called Involution. The powers
If polynomials when so expressed are said to be developed, or
mjoanded.

85. We have already explained the notation for denoting

he powers of a single symbol, as a, x, y. In all other cases
he power of a quantity is denoted by enclosing it in brackets
kith the number indicating the power above and to the right
|f the bracket.
I Thus (—2a) 3 denotes the third power of —2a; (a 2 b) 2 the
iquare of a 2 b; (a 3 bc 5 y the fourth power of a 3 bc 5 (a— b) 3 the ;

tube of a— b; (x 2 — 2x-\- 3) 5 the fifth power of x 2 —2x-\-3.

86. The same notation is used for denoting powers of

J

lowers of a quantity, brackets of different shapes being em-

lloyed when necessary.
Thus (a3 ) 2 denotes the square of a3 {(—%xy) 2 } 3 the ;

{(sc — 5) the fourth power of (cc 2 — 5) 3

2 2 3 4
lube of (—2xy) ; }
<

Exercise XXIX.

j
Eetaining the given quantities, denote

(1.) The cubes of —a, 2x, 3xy 2 %fb 2 c. ,

(2.) The squares of 2a— 1, a— & + 1, a?

3 — 1.
!
(3.) The squares of (x )\ (-2a) , (4aa) 5 , (3a2 &c4 ) 3
3 3
.

(4.) The cubes of (a-6) 2 , (a2 -!) 4 , (x2 -3x +2) 2 .

62 1NV0L UT10N AND E VOL UTION.

(5.) The squares of the cubes of x } —2x, a?b, x—a,

x 2 —ax + l.
(6.) The cubes 2
of the squares of ~~a 3 x , <&x— 1, cc
8
~a 3
.

87. A power of a power of a quantity is expressed as a

power of that quantity according to the rule
m n mn
(a ) =:a .

Thus, (a 2
y=a 2
-a -a
2 2
=a 6
;

3
(a y=za 3
*a
3
=^a (i

(a 3 ) 4 =a 3 'a 3 -a 3 -a 3 =:a 12 ;
(a±y=a12 ;

(a 5 ) 3 =a 15 ;

88. So also
(am bPc*ys£safm 'VM 'cz n 9
&c.

Thus. (a b) 3 ^a 2 ha 2 b-a2 b^a 6 b 3

2
;

(aW) ~a<W 3 2
;

(ab 3 c 5 y=2a 4 b l2 c 20 .

89. A rule has already been given in Art. 78 for expanding

the square of any polynomial. The expansion may also be
effected as in the following examples, in which the various j

parts are arranged in rows. In the first row occurs the square

of the first term of the given quantity ; in the second row the
product of twice the first term added to the second and the
second; in the third row the product of twice the first term
added to twice the second term added to the third term and the
third ; and so on*

Examples,

(1.) (a + 5) =a
2 s
-.2

+ (2a + &)6 ssr&c.

(2.) (b-cy~b 2

+ (2&-c) (-c)=&c.
 INVOLUTION AND EVOLUTION. 63

(3.) (a+b + c) 2 =a2

+ (2a + b)b
+ (2a + 2b+c)c~&c.
(4.) (a-h-cf=a 2

+ (2a -2b- c)( - c) = &c.

(5.) (a -b + c -d) 2 =(a2f
2 2

2
+ (2a -&)(-&)
+ (2«2 -2Z> + c )c2 2

+ (2a2 - 26 + 2c2 - d)(- cO=&c,

Exeecise XXX.
Express as powers or products of powers

(1.) O) 2 3
,
(2x2) 3 , (a 3) 3 , (2x3) 8 , (3a 2 ) 4 .

(2.) (ax2 )2 } (A ) 3 2
,
(a2 *; 3 ?/ 4 ) 2 .

(3.) (abc
2 2 2
)\ (a bc f } (2aW)3 .

(4.) (ccyz 4 ) 5 , (a& 4 c7 ) 6 .

Expand
(5.) (a; + l)2 (2x-3) 2 (*2 -5)
, ,
2
,
(cc -2a2) 2
3
.

(6.) (x + 2x + 3) 2 (x 2 -3x + 4) 2
2
, ,
(2x3 -x2 + 5) 2 .

90. Higher powers of polynomials are developed by the

Binomial and Multinomial Theorems, the explanation of which
may be found in more advanced works.

91. The process by which the roots of quantities are deter-

mined is called Evolution,

92. The wth root of a quantity is denoted by writing the

quantity under the sign the line above being y ,

sometimes replaced by brackets enclosing the given quantity-

Thus, V 2a denotes the square root of 2ci ;

*I 5a 2 „ cube „ 5a2 ;

4/^2T3 „ fourth „ a 2 + 3;
^ # (a»-2a + 3) denotes the fifth root of a -2a + 3. 2
64 INVOL UTION AND E VOL UTION.

93. The mth root of the wth root of a quantity is denoted

by writing the quantity under the sign y/ % .

Thus, vVsa denotes the cube root of </2a ;

V''$W2 „ fourth „ ^5^";

a/Vcc 5 —x + 1
8
„ cube „ vV— £c
3
+l.

94. The mth root of the ^th root of a quantity is expressed

as a root of that quantity by the rule

V
a*/ nf
V« =
mn
va.
I

95. The reason of this rule will appear from the following case t—
Let \/ fj a~x. Then on cubing both sides of this equation we have
tj a = x z .

Squaring a = x6 .

Extracting the sixth root,

Exercise XXXI.

detaining the given quantities, denote

(1.) The square roots of 2x, ax 2 or— 1, x 2 — 3x4-4.

,

(2.) The cube roots of —a: 6

,
dco
3
,
a-b, (a 3 —Sa +4).

(3.) The square roots of the cube roots of Sax, x— 1,

x6 + l.
(4.) The fourth roots of the cube roots of 2, 3a? — 1,
2a 4 -a 2 + 3.

Express as roots of the quantities under the double sign

(5.) VVa, VJF^x, VV'Sx, y/il^a.

 b

INVOLUTION AND EVOLUTION. 65

(6,) • VZIx±*-l, V^a -6a;* +

8
7.

96. Since the square of a quantity is equal to the square of

the same quantity with its sign or signs changed, it follows
that there will be two square roots (if there be any), one
being derived from the other by a change of signs.
=
Thus, since ( + ct) 2 =(— a) 2 + a2 , it follows that the square
root of + a is + a or —a. These two roots may be repre-
2

sented by the symbol ±a (read plus or minus a) ; so that we

have A Ja 2 =±a; Jx'
A
L
=±x 2
; aJ9x 2 =±3x.

Again, since by Art. 78

(a~b) 2 =(-a + hf=2a 2 --%ab + h 2 ,

(a~b + c) 2 =(-a + b-c)2 =a2 Vb 2 + c2 -2ab + 2ac-2bc5

it follows that

J a — 2ab-\-b t=za—
2 2
}
or —a-\-b ;
~±{a-b);
*J {p? i- b
2
+ c2 — 2ab -f 2ac -- 26c) =a—b + c, or — a + b—c
= ±(a—b + c).
In the following examples we shall only determine that
square root of a polynomial whose leading term is -h, the
other being derivable by a mere change of signs.

97. Since Vx^m = xn\ it follows that

Jltf^x Jx^~x jjx^x*;
2
,
z
}

where it will be observed the index of the root is one-half the

index of the given power.

98. The square root of a polynomial can generally be found

by the following rule.
(i.) Arrange the given quantity according to ascending or
descending powers of some letter.

(ii„) The first term of the root is the square root of the leading
term of the given quafitity, from which its square is subtracted^

leaving the first difference,

(iii.) The first divisor is twice the first term of the root added
F
 1 1

66 1NV0L UTION AND E VOL UT10N.

to the second term. The second term is the quotient of the

leading term of the first difference divided by the leading term
of the first divisor. The product of the first divisor and the
second term of the root is subtracted from the first difference,
leaving the second difference.

(iv.) The second divisor is twice the sum of the first and
second terms of the root added to the third term. The third
term is the quotient of the leading term of the second difference
divided by the leading term of the second divisor. The product
of the second divisor and the third term of the root is then sub"
traded from the second difference, leaving the third difference.

The process is thus continued until the difference is zero.

Examples.

(1.) Find the square root of 9a;

2
— 12a; +4.
9#2 -12a;+4(3a;-2
2
9a;

6a;-2)-12a;+4

Here the first term 3x is the square root of the leading

term of the given quantity, from which its square 9a; 2 is sub-
tracted, leaving — 12a; +4. The leading term of the first
divisor is 2 x 3x=6x. This is divided into — 12x, the leading
term of the first difference, giving —2, the second term of
the root, which is also the second term of the first divisor.
The product of 6a;— 2 and —2 is subtracted from the first
difference, leaving remainder zero. The root is therefore
3a;-2.

(2.) Find the square root of 4a;— 12a; + 5a; -h6a;4-l.

4 3 2

4a;
4
- 12x + 5a;
3 2
+ 6a; + 1 (2a; - 3a; -
2

4
4a?

4« 2 - 3a?) - 12a;3 + 5a; 2 + 6x + l

-12 x3 + 9a;2
4a;
2
— 6a;— 1)— 4a; + 6a; + 2

— 4a; + 6a: + 1
2
 INVOLUTION AND EVOLUTION 67

The first two terms, 2x2 and —3 x are found as in Ex. 1. j

The first two terms of the second divisor = 2 (2x 2 — 3x)

= 4:x 2 —6x, the leading term of which is divided into —dec 2 ,
the leading term of the second difference, giving 1, the third —
term of the root, which is also the third term of the second
divisor. The product of 4x2 — Qx — 1 and —1 is then sub-
tracted from the second difference, leaving remainder zero.
The root is therefore 2sc
2
— 3#— 1.
The terms of the differences need not be expressed
latter
until the corresponding terms in the partial products are to
be subtracted thus in the foregoing example $x + 1 might
;

have been omitted from the first difference.

(3.) Find the square root of cc4 — 4:X 3y + 10x 2y 2 — 12xy 3 + 9y\
x*-4:x 3y + 10x 2y 2 - I2xy* + 9/ (x 2 -2xy + 3y 2
4
cc

2x 2 -2xy)-4:x 5y + 10x 2y 2
—4:X 3y-{- A.x
2
y
2

2x -4ay + 3y 2 2
) 6ctV-lW + V
6^y~12^ 3
+% 4

In this example the given quantity is arranged according

to descending powers of x, and the first two terms only of
the first difference are expressed.

99. The reason for the rule given in the preceding Article will
appear from the following method of considering the last example.
The given quantity is there seen to be equal to
a*
~-4:X 5 y+4x2y 2
+ 6x y 2 2
-12xy z -{-9y 4 '

that is, to
(x 2 ) 2
+ (2x 2 ~2xy)(-2xy)
+ (2x2 -4xy+3y ) 3y* 2

Which by Art. 89 is equal to (x 2 - 2xy+3y 2 )2 i

Now, since a?
4
= (# 2 2
) , the first term, x 2 , of the root is the square root
of a?
4
, the leading term of the given quantity. Also since — 4cX 8
y as
2x 2 (—2xy\ it follows that the second term ~-2xy is the quotient of
p 2
 i

INVOLUTION AND EVOLUTION

— 4# 3 ?/, the leading term of the first difference, divided by 2x 2 the
,

y — 2x (?>y
2 2 2 2
leading term of the first divisor. Again, since <ox it '),

2
follows that the third term 3y is the quotient of 6x 2 y 2 the leading ,

term of the second difference, divided by 2x 2 the leading term of the

,

second divisor.

100. When the process for extracting the square root is

applied to a quantity which is not an exact square, a result

is obtained the square of which added to the last difference is

equal to the given quantity.

Example.

Find three terms of the square root of 1— 2sc.

1— 2x (I—as— 2~
1

2-a;)--2a>
-2x+x2
2-2x-%- )-*2
— x* + ar + -r
4
3 &
2
The square root is, therefore, 1— and remainder
A

«~x3 ~^. Hence ( 1~«j-^Y-x3 -^ = 1-2^

4 \ 2 4

Exercise XXXII.

Find the square roots of

(1.) 4a 4 & 2 Toxhf, 81xy.

, (2.) I6x 2 + AOx + 25.

(3.) 36x -36x + 9.

2
(4.) l + 6x + 9x 2 >

(5.) ^ 2
+ | + TV (6.) aM*+f.
 INVOLUTION AND EVOLUTION. 69

(7.) 4^ 2 -^ + T i¥ . (8.) 4.x

2
-12xy + 9y 2 .

(9.) x* + ±x* + 6x + 4;x + l.2

(10,) x* + 2xs + '3x2 + 2x + l.
(11.) .x
4
— 4x 2/+6cc — 4cc?/ 4-?/ 4
3 2
2/
2 3
.

(12.) 4cr;
6
-4x - llx* + 14x + 5a - 12sc + 4.
5 3 2

(13.) Extract to three terms the square root of 1 + cc.

The method of extracting the square root of a numerical

101.
quantity is founded on the Algebraical process, as Avill appear by

comparing the examples given below. We shall first show how the
number of figures in the root is determined by dividing the given
quantity into periods.

Since ^1 = 1, V^ = 10 V 10000 =
0 > 100, ^0000000 = 1000, &c,
it follows that the square root of a number between 1 and 100, that is,

containing 1 or 2 figures, lies between

and 10, and therefore contains 1

1 figure ; the square root of a number between 100 and 10000, that is
containing 3 or 4 figures, lies between 10 and 100, and therefore con-
tains 2 figures ; so likewise the square root of a number containing
5 or 6 figures contains 3 figures ; and so on. If, therefore, we divide
a number into periods of 2 figures each, beginning at the units, the
number of such periods, whether complete or not, will be the number
of figures in the root.

In Arithmetic if the root is a-f6-fc+ &c,

2a is called the first trial-divisor,
2a -f- 26 „ second „
2a + 26+ 2c „ third „
and so on.

Instead of obtaining the divisors as in the previous examples, we may

form them as below, where it will be observed that the sum of a
divisor and its last term or digit gives the next trial-divisor.

Examples.

(1.) a a 2
+2a5+6 2 +2ac+26c+c2 (a+b+o

2a+b ) 2a6+6 2 -f 2ac+26c-f-c 2

6 2a6+6 2
2a+26+c ) 2ac+2bc-\-c 2
2ac+2bc-\-c 2
 —
7o INVOL UTION AND E VOL UTION.

300 10W29 (300 + 20+7

300 9 00 00

600 + 20 ) 1 69 29
20 124 00
600+40+7 ) 45 29
45 29

In the numerical example, since the given number contains 3 periods,

the root will contain 3 figures. The leading figure of the root,
which is also the number of hundreds, will be 3, since the given
number lies between 90000 = 300 2 and 160000 = 400 2
The second .

figure of the root, which is also the number of tens, is obtained by

dividing the first remainder 16929 by the first trial-divisor 600. The
third figure of the root, which is the number of units, is obtained
by dividing the second remainder 4529 by the second trial-divisor
640.

Omitting all unnecessary figures, we may arrange and describe the

work as follows :

3 10'69'29 (327
3 9

62 )169
2 124

647 )4529
4529

The leading figure of the root is 3, the square of which is the

greatest square number under 10, the first period ; the square 'of 3 is

subtracted from the first period, and to the remainder is annexed

the second period 69 to form the first dividend 169. The first

figure of the root is doubled to give the first trial-divisor 6, the

division of which into 16 indicates the second figure of the root.

The second figure of the root is annexed to the trial-divisor to form

the first divisor 62, which is multiplied by the second figure of

the root, and the product is subtracted from 169. To the remainder
is annexed the third period to form the second dividend 4529.
Under the divisor 62 is written its last digit, and the sum forms
the second trial-divisor 64. The third and last figure of the root
is 7, because when annexed to the trial-divisor to form the
 INVOL UTION AND E VOL UTION. 71

second divisor, the product of 64-7 and 7 is equal to 4529, the last
dividend.

(2.) 2 5'83'51'23'36 (24156

2 4
~44 183
4 176
~~751
481
1 481

4825 27023
5 24125

48306 289836
289836
 ( 72 )

CHAPTEE XII.

THE HIGHEST COMMON MEASUBK

102. A quantity is said to be of so many dimensions, in any

letter, as are indicated by the index of the highest power of
that letter involved in it.

Thus 3x2 —2cc+4is of 2 dimensions in x 3y* + <2y— 5 is of ;

4 dimensions in y and ax 3 — bx 2 + c is of 3 dimensions in x.

;

103. A whole expression, or quantity, is one which involves

no fractional forms.
Thus 3x2 —bxy, 2x3 — 3cc + 4, are whole expressions, as are
,

also all positive and negative integers.

104. When two or more whole expressions are multiplied

together, each is said to be a measure of the product, and the
product is said to be a multiple of each factor.
Thus 1, 4, a, and b are measures of 4a5; 1, 3, x and x + 1
5, x x—1, and y —l are measures
2
are measures of 3x 2 + Sx ; 2
,

2
of 5x2 (x-1) (?/ -l).

105. It must be carefully observed that the terms measure

and multiple are to be used only in connection with whole
expressions. In order, therefore, to obtain a multiple of a
quantity it must be multiplied by another whole quantity ; and
to obtain a measure of a quantity it must be divided by a
whole quantity, the quotient also being a whole quantity.
Thus the terms measure and multiple cannot be used in
connection with \ax 2 , x 2 — ? + 3, because they involve fractions;
A
whilst 1, 3, a, x, x 2, and x—1 are measures of 3ax 2 (x— 1),
 1

THE HIGHEST COMMON MEASURE.

because the quotient of the latter divided by each of the

former is a whole quantity.

106; The highest measure of a quantity is the quantity

divided by +1 or — 1, that divisor being taken which will
make the first term of the quotient positive.
Thus the highest measure of — 4cc2 is 4cc 2 of 5cc— 7 , is 5cc— 7,
and of — 2cc + 2
cc— 3 is 2cc
2
— + 3.
cc

107. The lowest multiple of a quantity is the quantity mul-

tiplied by +1 or —1, that multiplier being taken which will
make the first term of the product positive.
Thus the lowest multiple of 2cc— 3 is 2cc— 3, and of
— + cc— 5 is
cc
3
cc
3
— + 5.
cc

108. When
one quantity is a measure of two or more others,
it is common measure of those quantities.
said to be a
Thus 2x is a common measure of 4x 2 and 2x 2 — 4cc, and cc—
is a common measure of 2cc— 2, cc 2 — 2x + 1, and cc 2 — 1.

109. The highest common measure of two or more quantities

is the common measure of highest dimensions and greatest
numerical coefficient or coefficients.
Thus the common measures of 4x3 and ftx
2
y are
1, 2, x, 2cc, 2x\
of which the last is called the highest common measure
(h.c.m.) ; the common measures of 4(x 2 — 1) and 6(cc— l) 2
are
1, 2, 2(^-1),
of which the last is the h.c.m.

110. We shall consider the process for finding the h.c.m.

in the three following cases, namely — I. When one of the
quantities is a mononomial. II. When the two quantities are
polynomials, neither of which has a mononomial measure.
III.
T
W
hen the two quantities are polynomials, one or both
of which have mononomial measures.
111. I. (i.) When the given quantities consist of two or
more mononomials, their h.c.m. is the product of the g.c.m.
 — ;

74 THE HIGHEST COMMON MEASURE.

of the numerical coefficients and the highest power or powers

common to the several given quantities.

Examples.

(1.) Find the h.c.m. of I8a¥x3 and 15aW.

Here the g.c.m. of 18 and 15 is 3;

the highest power of a common to both is a ;

99 99
b
U 99 39
b2 1-
U

and there is no power of x common to both

/. H.O.M. is dab 2 .

(2.) Find the h.o.m. of 12x y 3z*, 16x3y 2z3, and 28x *yz\
2

The g.cm. of 12, 16, and 28 is 4;

the highest power of x common to the three quantities is x2 ;

99 99 V 99 93 2/ >

99 99 & Jf 9f Z3 j

h.o.m. is &x2yz\

(ii.) The h.c.m. of a mononomial and a polynonr'*

1
112.
is the h.c.m. of the mononomial and the h.c.m. of the several
terms of the polynomial and may be fonnd by the following
:

rule:

Express the polynomial as a product one factor of which is the

H.C.M. of its several terms
: the h.c.m. of this simple factor and
the given mononomial will be the h.c.m. required.

Examples.

(1.) Find the h.c.m. of Qab 3 and 8a 2 b 3 -12a 3 b 2 .

Here 8a 2 b 3 -l<2a 3 b 2 =4:a 2 b 2 (2b-3a), where 4a 2 b 2 is the h.c.m.

of 8a2 b3 and 12a 3 2 and the h.c.m. of 4a% 2 and 6a& 3 is 2ab 2 ,
Z> ;

the h.c.m. required.

(2.) Find the h.c.m. of 15xy 2z3 and 1Q# 3 ^V— 15a%V
+ 20x2y 2z 3 .

Here lOxhfz 2 - \hx 2yH 2 + 20x 2y 2 z 3 = 5x 2y 2z 2 ('2x +By 4 4z),

where 5x 2 y 2 z 2 is the h.c.m. of 10x 3y 2 z 2 15a; 2 ?/V, and 20x 2y 2z 3
, ;
 THE HIGHEST COMMON MEASURE. 75

and the h.c.m.

2 3
of 5x 2y 2z 2 and 15xy z is 5xy 2z2 , the h.c.m.
required.

Exekcise XXXIII.

Find the h.c.m. of

(1.) 12ab 2 and 16a 2 b 2 . (2.) 15a 2 b and 20ah 2 .

(3.) 9axy and SQxyz. (4.) 9ax 2 y 2 and 15a 2 xz.

(5.) 42a 2 & 4 2/ and 35a 3cc 2 ?/ 4 . (6.) ab 2 cu2 v and 3a 2 buv 2w.
4 3 2 2 3
(7.) 8a 5, 12a ^ , and 16a Z> .

(8.) 3Ctey, 426ary, and l&asY.

(9.) 4a& 2 and 12a2 bx-8ab 2y.
2 4
(10.) 10aZ> c and 30a3 5 3 +45a2 Z> .

(11.) WaPxy and ItiaVcx-lOVcy.

(12.) Suv 2 w and 12w3 w 3 — 24w2 3 + 36^V. w
113. II. "When two polynomials, neither of which contains
a mononomial measure (other than unity), involve powers of
a single letter, their h.c.m. can be obtained by the following
rule :—
(i.) Having arranged the given quantities according to descend-
ing poivers, choose that one which is not of lower dimensions than
the other as divisor.

(ii.) Divide this into the other multiplied by the least number

which will make its leading term a multiple of the leading term
of the divisor. When this number is unity, actual multiplication
may be dispensed with.
(iii.) Divide the first difference by the highest mononomial

measure contained in it. When this measure is + 1, actual

division may be dispensed with.

(iv.) Bepeat the steps (i.), (ii.), (iii.), with respect to this last
quotient (or difference) and the first divisor; and so on, until there
is no difference.

The last divisor will be the h.c.m. required.

76 THE HIGHEST COMMON MEASURE,

It will be observed that no fractions occur in the process,

and that the leading signs of all divisors are made positive.

Examples.

2
(L) Find the hx.m. of 2x2 -7cc + 5 and 3cc -7cc+4.

2oc
2
-7a; + 5)6x 2 -14x+ 8 (3
6cc -21x + 15
2

7)7a>- 7
a- 1

Here since the dimensions of the two given quantities are

equal, either onemay be made the divisor. 2cc2 — 7x + 5 being
taken as divisor, 3x 2 — 7cc + 4 is multiplied by 2 in order to
make the leading term 6x 2 a multiple of the leading term 2cc2
of the divisor. The first difference 7x— 7 is divided by its
highest mononomial measure 7.

In the next step cc— 1 and 2x 2 —7x + 5 are to be treated in

the same manner as the given quantities.

fic-1) 2x 2 -7x + 5(2x

2x2 -2x
-5) -5x+5
03 —1
The leading term of 2x2 — 7x + 5 is a multiple of the leading
term of x— 1, and therefore the multiplication of the former
by 1 is omitted. The difference — 5# + 5 is divided by its
highest mononomial measure —5.

In the next step the quotient x— 1 and divisor 1 are to

be treated as the previous quotient and divisor were.

x-l)x-l(X
0-1
 :

THE HIGHEST COMMON MEASURE. 77

The process thus terminates and the h.c.m. is the last divisor
as-1.

Whenever as in the second step the difference is a multiple

of the divisor, the division may be continued and the work of

the last step avoided. Thus

2cc
2 — 2x

—5a? + 5

The whole work may be arranged as follows

3x2 -7x+ 4
2
2x2 -7x + 5) 6x 2 -Ux + 8(3
6x2 -2Ls + 15

,7 )7s-~7
x~ l)2.x 2 -7o:+ 5(2a-*5
2a;
2
-2a
-5£c + 5
H.C.M. ^03— 1. — 5cc + 5
(2.) Find the h.c.m. of a 2 +2jc— 3 and x2 +5x+6.

x 2
+2x-S)x + 5x + 6(I
2

x 2 + 2x-3
3) 3a; + 9

^ + 3)a 2 + 2a-3(cc--l
a: + 3a;
2

—x—3
H.C.M. —a: + 3. —x—3
Here the multiplication of x 2 + 5x + § by unity is unneces-
sary. The other steps are similar to Ex. 1.

(3.) Find the h.c.m. of 2x s -7x-2 and 2x 2 ~x-6.

 6

78 THE HIGHEST COMMON MEASURE.

2x2 -x-6)2x 3 -7x-2 (x, 1

2a; — tc — 6x
8 2

x2 —x—2
2
2x 2 -2x-4:
2x 2 — x—Q
-l)-^+2
~~x~^2) 2x 2 -x-6 (2x+3
2x 2 —4a>
3x— 6
H.c.M.=cc— 2. 3.x—

Here 2x 2 —x—6 used as divisor in the second step, the

is

dimensions of the first difference being 2. The partial

quotients x, 1 of 2x 3 — lx— 2 and 2x2 — 2x— 4 divided by
2x 2 — x— 6 are separated by a comma to distinguish them
from parts of an ordinary quotient.

(4.) Find the h.o*m. of

4x2 + 3cc-10 and 4a;

8
+7a;2 -3sc-15.

4x +3a;-10)4x + 7cc - 3a; -15

2 3 2
0+1
4£3 + 3£ 2 -l(k
4x 2 + 7cc-15
'

2
4a; -}-3a;-10

4a;-5)4^ 2 + 3^-10(a;-f2

&c-10
H c.M.^4cc-5,
t 8x-10

In this example there is no necessity to introduce or

suppress any mononomial factors.

114* The process of the foregoing examples will frequently

enable us to find the h.c.m. of polynomials involving powers
of several letters, as in the following
 THE HIGHEST COMMON MEASURE. 79

Example.

Find the h.c.m. of 2x 2

+xy—3y2 and 3x2 —4:xy+y2 .

Sx 2
*-4:xy-\-y 2

2
W + xy-Sy 2
) 6x 2 -8xy + 2y 2 (3
e>x + 3xy- 9y
2 2

~- lly)-llxy + ll y 2
x~~y) 2x 2
+ xy — By 2 (2x + Sy
2x 2 —2xy
3xy—3y 2
H.c.M. 3^2/ — 3?/ 2
Here the mononomial factor —11?/ is suppressed.

115. The reason for the rule in Art. 113 will appear from the follow*
ing proposition and its application in the next Art.

When one quantity is a measure of two others, it will measure the sum
and difference of any multiples of them*

Let the quantities be A, B, C ; and let A measure B and (7, so that

B=mA, C~nA. where m and n are whole quantities.

Take any multiples pB, qC of B,C, where p and q are any whole
quantities whatsoever. Then, since pB —pmA, qG—qnA 1

pBdcqC -pmA±LqnA — (pmdbqri)A,

pBzkqC
1 .
£
.*. . .
zzpmzkqn ;,

that is, A is a measure of pB^zqC, the sum or difference of any

multiples of B and C, because the quotient pm^kqn is a whole
quantity.

Thus, 2x 2 , which is a measure of Qx z and $x 2 y, will measure

GxX-2a)-8x\/(-3x), 6x*-hSx2 y, 6x*-8x 2 y(4;xy), &c.

116. Suppose, now, that A and B denote two polynomials (as in

Art. 113), neither of which contains a mononomial measure other than
unity ; and let the dimensions of A be not greater than the dimensions
of B, Divide A into B multiplied by a mononomial w hole quantity a,
which makes its first term a multiple of the first term of A ; and
So THE HIGHEST COMMON MEASURE.
divide the difference C by the highest mononomial measure which it

contains, and let the quotient be ZX

B
a

A)aB(b
bA

c)Q

D
Mow, C being equal to aB—bA, or the difference of two multiples
of A and B, is a multiple of all the common measures of A and B, and
therefore of their H.C.M.
Again, every of C and A is a measure of C+bA, or
common measure
aB, and therefore of B, because A has no mononomial measure.
Hence the H.C.M. of A and B is the same as the h.c.m* of A and (7,

which is the same as the H.C.M. of A and Z), because A has no mono-
nomial measure.
The problem is thus reduced to finding the H.C.M. of A and D>
These two quantities, A and D, are then treated in precisely the
same manner as A and B and the
; process is continued until it

terminates as follows, when the last divisor, P (suppose), is a measure

of the last dividend Q.
P)Q(r
rP

The problem is thus finally reduced to finding the H.C.M. of P and

Q. This is evidently P.
Hence the last divisor in the above process will be the H.C.M.
required.

Exercise XXXIV.
Find the h.c.m. of

(1.) Sx 2 + 2x- 21 and c

+ 13x-6.5cc
2

(2.) 2x + x-3 and 3x -4a + l.

2 2

(3.) a 2 -5£ + 6anda 2 -6a + 9.

(4) x 2 +10x + 2I and z 2 -2£-15.
 :

THE HIGHEST COMMON MEASURE. Si

(5.) 2x 2 + cc-15 and 2x 2 -19a; + 35.

(6.) x 2 -4x + 3 and 4x3 -9x 2 - 15^ + 18.
(7.) cc
2
f lOx + 25 and xz + 15a> 2 + 75cc + 125.
(8.) x3 -6cc 2 -fllaj-6 and x 3 ~cc 2 -14x-{-24.
(9.) a>
3
-3x2 -9a;-f27 and 3x 3 -a; 2 -27aj + 9.
(10.) 3x 2
-22x + 32 and cc
3
-llas2 -f 32as-28.
(11.) 7a:
2
-12a; + 5 and 2x 3 + x 2 -8x + 5.
(12.) 5x 3 + 2x2 -15x-6 and 7x3 —4x2 -21x + 12.
3
(13.) 2cc -f 9x
2
+ 4a;- 15 and 4a;3 + Sx 2 + 3x + 20.
(14.) a 3 -6x 2 +llcc-6 and x4 -2x 3 -13x2 -f-14a;+24.
(15.) x4 -2x 2 + l and ce
4
-4a; 3 + 6a; 2 -4a; + l.
as — 5xy + 6y
2 2
(16.) x 2 -\-xy— 12y 2 and .

(17.) 2x 2 + 3^+^ 2 and 3x2 + 2^-2/ 2 .

(18.) as
3
+ 2
as ?/ + xy-{-y2 and as
4 — ?/ .
2

(19.) 5cc
2
+26^+33?/ and7^ + 19^-62/2
2 2
.

(20.) 3as
4
-a%2 -2?/ 4
and 2a;
4
+ 3afy-2a;y-3a;2/s .

117. III. The e.c.m. oftwo polynomials involving mono-

nomial measures is found as follows

Express each polynomial as the product of a mononomial and

a polynomial which contains no mononomial measure. Then
the H.C.M. required is the product of the H.C.M. of the mono-
nomial factors and the H.C.M. of the polynomial factors.

Example.
Find the h.c.m. of
Sx^y + 12x3y -f- Ax 2y and 6a% 2 — 6a%2 — 12xy2 .

Here the given quantities are equal to

2 2
4:x y(%x + 3x + l) and 6xy (x2 -x-2).
2

The h.c.m. of 4:x2y and 6xy 2 is 2xy ; and the h.c.m. of

2as
2
+ 3as + l and as
2
— as— 2 is + 1. as

the h.c.m. required is 2xy(x + 1).

118. The h.c.m. of three polynomials is the h.c.m. of any

one and of the h.c.m. of the other two.
 ; ; ;

82 THE HIGHEST COMMON MEASURE.

Example.
4
Find the + 2£2 -3, and 2cc4 + 2cc3 + 3x + 3.
h.c.m. of cc^-l, cc

The h.c.m. of — 1 and + 2cc2 — 3 is cc2 — 1; and the h.c.m.

cc
4
cc
4

of x 2 — 1 and 2x4 + 2x3 + 3x + 3 is x + 1, the h.c.m. required.

Exekcisb XXXV.
Find the h.c.m. of

(1.) 12ax4 -27ax 2 and 2a2x*+a?x2 --3a2x,

(2.) 10cc
2
+40cc + 30and4« 3 -16x2 -84x.

(3.) 2x -6x3 -4x2 and3x4 -3x3 --12cc.

5

(4.) 2x +x—3, x 2 —l, and cc2 +4cc— 5.

2

(5.) 6x 2 -x-2, 21cc 2 -17;c + 2, and 15x 2 + 5x-10.

119. When all the component factors of the given quan-

tities are known or can be determined, the h.c.m. may be
found by the rule of Art. 111.

Examples.

(1.) Find the h.c.m. of

4(>-l)2 0+2) s and 6(cc-l) 3 (x+2).

The g.c.m. of 4 and 6 is 2
the highest power of cc— 1 common to both is (cc— l) 2 ;
x+2 „ „ x+2.
the h.c.m. required is 2(cc— l) 2 (x + 2).

(2.) Find the h.c.m. of

2
%a2x(x*-l), IZaxXx 4 -!), and 2Qax(x -l).

By Art. 80, 8a 2 x(x 3 - 1) = $a 2 x(x- l)(x 2 + x + 1)

by Art. 79, I<2axXx*-I) = mx%x 2
-I)(x*+l)
= 12ax2 (x- l)(x + l)(x2 + 1)
20ax(x 2 ~l) = 20ax(x + 1)0-1).
 THE HIGHEST COMMON MEASURE. 83

Now the g.c.m. of 8, 12, and 20 is 4;

the highest power of a common to the three quantities is a;

j> ?? ^ a » » ^I
55 55 X—\ 5> jj ,?
X — 1,
and the other factors x + l,x 2 + l, x 2 -\-x + l are not common.
,\ the h.c.m. required is 4ax(cc— 1).

Exeecisb XXXVI.
Find the h.c.m. of

(1.) 2a(x— a) 2 and 3ax(x + a)(x— a).

(2.) 6a 2 (x + 2)(x-3) and 8a^(cc— 3)(cc-h8).
(3.) ax2 — 2a& + a and 2a2x 2 — 2a 2 .

(4.) x2 -l 3 x3 +l and }
cc
4
-l.
(5.) x + 2, x2 -4andx 3 + 8.
3
(6.) 3cc -81, cc
2
-6a+9, and 2x*-18x.

g2
 —
( 84 )

OHAPTEK XIII.

THE LOWEST COMMON MULTIPLE.

120. When one quantity is a multiple of two or more

others, it is said to be a common multiple of those quantities.

Thus 12a; 2 is a common multiple of 2x and 3x 2 ; and

15x(x—-1) of 3, 5, 15sc and x— 1.
121. The lowest common multiple (l.c.m.) of two or more
quantities is the common multiple of lowest dimensions and
least numerical coefficient or coefficients.

Thus of the following common multiples of 2x and Sx 2,

namely,
6a 2 , 12x 2 , W,
6x 3, 12x s , I8x\
6x\ 12x\ 18x\ &g.,

thq first 6x2 is called the lowest.

122. The l.c.m. of two quantities is found by the following

rule :

(i.) If they contain no common measure except unity, their

L.C.M. is their product.
Thus, the l.c.m. of 4tx and lab is 28abx.

If they contain a common measure, their l.c.m. is equal

(ii.)

to one of the given quantities multiplied by the quotient of the

other divided by their h.c.m.

It will be generally found most convenient to express the

l.c.m. as the product of several factors.
 —
THE LOWEST COMMON MULTIPLE. 83

Examples.

(1.) Find the l.c.m. of 6x 2y and 9xy 2 .

The h.o.m. of these quantities is 3xy.

:. by the rule, l.c.m. =^ 6xy

x 9xy 2 =lSxhj 2 .

(2.) Find the l.c.m. of 2cc 2 -7^ + 5 and 3x 2 -7x + 4:.

The H.C.M. is found to be as— 1; and since -

as—l
= 2^-5, the l.cm. will be (2cc— 5) (3a 2 -7a;-f 4).

123. The following is the proof of the rule given in the preceding
Article :

Let the two quantities be denoted by A and B, and their h.c.m. by

C ; and let A = aC, B = bC, where a and b are whole expressions which
have no common measure except unity.

Then, since the L.C.M. of a and b is ab, the L.C.M. of aC ( = ^-) and
ri/
bC(=B)
i r>. .

is
n aC.-bC = AB = A
abO=
7
—— r,
—
B A
~.B=-.A.

124. The l.c.m. of three quantities is the L.C.M. of any one

and the L.C.M. of the remaining two.

The l.c.m. of four quantities is the L.C.M. of any one and the
L.C.M. of the remaining three.
And so on.
'

Example.

Find the l.c.m. of 3x 2y 2z }

ftxyz
2
, and 10x2yz s .

The l.c.m. of 3x2y 2 z and 6xyz2 is 6x y 2 2 2

z ; and the l.c.m. of
§x2y2 z 2 and 10x2 yz 3 is 30x 2y 2z 3 .

125. When
the component factors of the given quan-
all
tities are known
or can be found, their l.c.m. may also be
obtained by multiplying the l.c.m. of the numerical factors by
the highest poiver or powers of the several factors that, occur in
the given quantities.
86 THE LOWEST COMMON MULTIPLE.

Examples.

(1.) Find the l.c.m. of 6x 2yz 2, 4:X

2
y
3
z, and Sx^z.
The and 8 is 24 the highest power of x
l.c.m. of 6, 4, ;

which occurs amongst the factors of the given quantities is

and the highest powers of y and z are, respectively, y 3
;

and z 2 .

4 3 2
Therefore the l.c.m. required is 24:X y z .

(2.) Find the l.c.m. of l&ab(a—b), 21a(a + b)(a— b), and

3§b 2 (a + b).

The l.c.m. of 15, 21, and 35 is 105 the highest powers of ;

a, b,a—b, a + b, which occur amongst the given quantities,

are, respectively, a, b a—b, a + b.
2
,

Therefore the l.c.m. required is 105a£ 2 (a— b)(a + b).

(3.) Find the l.c.m. of x 2 — 1, x 3 — 1, and x 3 +l.

Here x 2 -l = (x + l)(x-l);
x 3 -l = (x-l)(x 2 + x + l);
x3 + l = (x + IXx 2 --x + V).

the l.cm. = (x + 1Xx-1Xx + x + 1)(x -x + 1)

2 2

= (x2 -lXx* + x2 + l).

Exeeoise XXXVII.

Find the l.c.m. of

(1.) Sabx, Ibxy. (2.) Qa*xy, Uax2y. (3.) ab 2, be 2, ca\

2
(4.) Sa 2
bc, 12ab% 2A.abe .

(5.) mcu 2
,
IQcav 2 , 20abiv 2 , 40aW.
(6.) x 2 -7x + 12,x 2 -x-6. (7.) 2x 2 ---5x-3,4a2 +4a;+l.
2
(8.) 3x -llx + 6, 2x -7cc+3.
2

(9.) x 3 -iax 2 + 5a2x-2a

3
, a3 -2a2a--4a8.
2
(10.) 8(^-1), 120-1) .
 THE LOWEST COMMON MULTIPLE. 87

(11.) cc
2
-l,(a-l) 2,(a + l)2 .

(12.) x\x+y), xy(x—y), y\x+y).

(13.) a* + a3 b, ab-b2 a?-b 2
, .

(14.) 2x(x2 +x + l), 3x3 -S, 4#?-~4u

p + q ,p —q ,p + q
2 2 2 2 3 3
(15.) .

(16.) f-l,^-l,^-L
(17.) (a-&) (a-c), (6-c) (6-a), (c-a) (c-6).
(18.) %a2b(a-b), 12a&(&-a), 3(a2 -6 2), &W(5 2 -a2).
 —
( 88 )

CHAPTEE XIV.

FBACTIONS.

126. When one quantity is not exactly divisible by another,

the quotient is represented by writing them in the form of a
fraction.
o
Thus, —
+o
= denotes the quotient of —2 divided by +3;

—3a?
the quotient of -2a divided by -3x;
x
^
or— dec + 4
the

quotient of x— 1 divided by cc
2
— 3a; +4.
127. A fraction is not altered in value by multiplying
or dividing the numerator and denominator by the same
quantity.
—2 -2x3 _ -6 +6 + 6x-4 _ -24
Thus,
+3 - +3x3 - +9 ; __ 8
- _ 8x _ 4 - +32 ;

-20 _ -204 5 — _ +4 -4 _ -4x | -f .

-25 ~ -25^4—5 ~ +5 ;
+5 ~ + 5xf ~ +y'
a ac — acc
&
~~
&c
~~ — '

a? -J Q-l) + _ Ov
2 -!
2a- 3 (>-3) (a> + l) ~ 2a 2 -a-3*

128. The statement in the preceding Article depends on the two

following propositions :

I. The numerical value of a fraction is unaltered by multiplying or

dividing its numerator and denominator by the same quantity.
(i.) Let a, 6, m, be integers. Then, since ? denotes a of the b parte

into which a unit is divided, it follows that

 FRACTIONS. 89

"x»=™
O 0
.
'

. . . (1)

TTW
0
=-
mb
.... (2)

a _ma /(rv

Before considering the case where a, 6, m are fractional, we shall

show how the operations of multiplication and division of numerical
fractions must be performed.

Let a, b y c, d, m be integers. Then, since ^=m, it follows that

c
a mc a ma
b c b b

But by
ma mac
(3),
1 be

. a
"b' ~c~Tc~
mc_mac
* ' *
^
^ ;

Hence, if — be multiplied by an integer in a fractional form, the

b

productis a fraction whose numerator is the product of the numerators

and denominator the product of the denominators. If, now, we wish

to find the product of - and - where , - is not equal to an integer, the

d b d
operation of multiplication must be in accordance with this rule ; for
any application of the term multiplication to cases where its primary
meaning (which is repetition) does not apply, must not be inconsistent
with the cases where it does apply.

. a c __ac
'
"V'dTbd ^

Again, since division is the inverse of multiplication, both in its

primary and extended applications, it follows from (5) that

bd d b

acd , /ox

ac d .
 FRACTIONS.

Also
bd b d

= 2*2
abd
by
y (3)w
=o-- by (5)

Hence it follows that the quotient of the fraction

^ divided by

—
d
is equal to the product of
be
~ and — ; that is,

ct^c _a d__ad
K '
6
'
d~b '~c~bc ' '

(ii.) Now, let a, &, m be fractional, which will include the case
where some of them may be integers.

Let a=-, b=-,m=2

y u d
x
Then °=^^by(6);
b z yz

mr
And '=£_J?=±by(5)
mb c z cz
d u du x

cxdu ,
/CN

=2^ by (3)
yz

m a __ma
b mb"

II. If, in the fraction a and 6 denote positive and negative

quantities, the sign of the fraction depends on the signs of the nu-
merator and denominator. These signs will be either like or unlike,
and on multiplying or dividing by a positive or negative quantity they
will still be like or unlike, and the sign of the fraction will therefore
remain unchanged.
 1 +

FRACTIONS. 91

For example, multiplying by — 1,

-i-? = —- = -jr~ a positive fraction ,

f 3 —3
-
3
;

—
_l_3
-

-4 +4 4'
—
= 3 = — -3 a negative
S fraction. ,

129. A fraction which involves fractional coefficients in the

terms of the numerator and denominator can always be
reduced to one whose numerator and denominator are whole
expressions by multiplying the numerator and denominator
by the l.c.m. of the several denominators.
2
Thus, o 2^ s , i
is reduced by multiplying numerator

and denominator by 12, the L.O.M. of 2, 4, and 6, to the

equivalent from _ 9x+2 .

24x2

130. A fraction is said to be in lowest terms when its

numerator and denominator contain no common measure,
except unity. Hence a fraction is reduced to lowest terms by
dividing its numerator and denominator by their H.C.M.

Examples.
Sa 2 b 3x
Eeduce
(1.)
YM&y to lowes^ terms.
The h.c.m. of Sa 2 b 3 x and 12a s b 2y is 4a2 & 2 .

Sa 2 b 3 x _ 2fo
12a 3 b 2y ~~ Say
'
'

x2 —
(2.) Eeduce 3^-^ to lowest terms.
x
The H.C.M. = x + 1.
_ x—1
x 2 —l
4
* '
+ 1 ~ X2 — X-\-l
£C
3

%x 2 + 3x 2
(3.) Eeduce 2
— 5% 2 to lowest terms.
2a?

The h.o.m. = 2x—l.

2^-3^-2 sc+2
2o>
2
-5cc + 2 ~ a>-2
 —
92 FRACTIONS.

Exercise XXXVIII.
Keduce to lowest terms
h
(2.)^. (i.)-^ -
(Djjft
4:5ax 2'
acy '
8xy 2 z^' 2'
a b—ab2
~

w 9a& 3
6a 2 £ 2
-18^ ^
*
;
x+1
2 -!*
a;
/
K7 %J
n ^—2
x 2 -£

m
&s x ^-7^+10
C9.)
|eJ-
n9A}
do-)

x +7x + 12
2
3 ,:i3-
K J K
x 2 -x-2' x2 -x-20'

1 ;
3+8x-3a2 ' U ;
x3 -39x + 70*

fl*nJ
cc
3
— 6a3 — 9 ,-in\ 12x 2 —15x + 3
K
xT+3x 3 -9x-9' K }
6x 3 -6x 2 + 2x-2'
r-jn-N
_ x 4 —4:X 2y 2 /inx x 3 — 3cc 2 ?/ + 3xy 2 — y 3
x 3 —6x 2y + 12xy 2 —8y 3 ' '
x 3 —x 2y—xy 2 + y 3

131. Fractions are said to be like or unlike, according as thej

have the same or different denominators.

Thus
a;
-
cc
are like fractions, as are also -f^,
cr— 1 or— 1
; ^
3*/
.

2a
=j- are unlike fractions.
ox
132. Unlike fractions are reduced to like fractions by mul-
tiplying the numerator and denominator of each fraction by the
quotient of the L.C.M. of the several denominators divided by
its denominator.

The common denominator will accordingly be the l.o.m.

of the several denominators.

Examples.
2a 3 c
CI.) Eeduce j^, to like fractions.

The l.o.m. of 3b and 4c? in L2bd.

 ; — —
' ; 1

FRACTIONS. 93

The multiplier for the first fraction is ~^f=4:d;

ob
. 2a 8ad

IQbd
The multiplier for the second fraction is ~~ = 3b

(2.) Eeduce ——1

a~b
,
b—a
2
to like fractions.

The l.c.m. of a— b and b—a is a—b 3 the quotient of which

divided by b—a is —1

b— a a—b'

(3.)
v J Eeduce -i_ to like fractions.
x—V x2 + x + l x 3 —l

The l.c.m. of the denominators is x3 — 1, the quotients of

which divided by x—1, x2 + x + l, x 3 —l are, respectively,
x2 -\-x + l, x—1, 1.
1 _x2 + x + l

x— 1 = x2 —2^ + l
9
x2 + x + l x 3 -l
x—2x _x—2x
2 2

x —l x 3 —l
3

Exeecise XXXIX.
Eeduce to like fractions :

4
a 6
(2.) \,\.
a* ab
(3.)
x
1,
xy
-

xyz
.

//In
(4:.)
2
a
J
—a
/k \
(5.) —1
acc
,
— xy
2
,
3
axy
/£
(6.)
\ c&

cc—-1
,
2c&
—o —
cr—
,

/i7 v 1 2 x 2 3 x /-Q v a 1— c&

 1 4 —1
94 FRACTIONS.

nn^ 4 8 1
g;n _L_ _J^_
CiU,;
ra* v ;
a^-^+l ' *+l *3 +l*'

2 3 1
(12-)
(x + 2) (as-1) '
(2-as) '
0-2) (s+2)"

rm
±0
^
L___ L. _ _J:
-''

(c _a) '
(a-6) (c-6) '
(a-c) (&-c)
1 1
aiv a(a— (a;— a) '
b(b—a) (x—b) '
a&x
&)

Addition and Subtraction.

133. The operations of addition and subtraction of frac-

tions may be denoted by the signs + and respectively. —
Thus, the sum of T,
-~
2
— o>
and o — k>
may be denoted by

__a -2b x-2 .

x-1 iSx-5'
x
and the difference between . f
3ar-— 1
<mc2
oa;—
—
^ ^, by
4cc _ cc—

Thus also +^ denotes the fraction % to be added to, and

b b

—- the fraction ? to be subtracted from some quantity not

b b
expressed.

134. Sums and differences of fractions when expressed as

single fractions are said to be simplified, the operation being
performed according to the three following rules :

The sum of any number of like fractions is a Wee fraction

(i.)

whose numerator is the sum of their numerators.

m1
1 iius, — 4
-r
=-
—x ,
,2x— 3 _4z—x-\-2x— 3__x + l-.
- -t-
X — 1 £C — 1
—CC — 1 03 — 1 CC —

(ii.) The difference between tivo like fractions is a like fraction

ivhose numerator is the difference between their numerators.
 FRACTIONS. 95

Examples.

5x-4:__ 6 _5a;-4-6_5x-10
(1.)
2x-3 2cc-3 2x-3 2aj-3*

(2.) l'
X2 — 1 cc
2
—1 ~cc
2 —1 CC
2
—
It will be observed that the — before the second fraction
changes the sign of — 2cc.

n 3cc
2
— + 5___£c — 6cc— 8_3x — + 5—(cc
cc
2 2
cc
2
—6cc— 8)

_ 3x 2 -^ + 5- + 6cc + 8
2
cc

a3 + l
_ 2s 2 + 5s + 13
[

x3 + l
In this case the — before the second fraction changes the
signs of all the terms of x 2 —6x—8.

It appears from the preceding rules that

c c

and conversely ;in other words, the subtraction of a fraction

is equivalent to the addition of a like fraction, whose numerator
is the numerator of the former with its sign or signs changed.

Thus, _^+g
—
= + -«' + fc-8
X6 — 1 X6 1

— 2cc + 5_ __ 2cc- 5
3x 2 -7~ 3x 2 -7'

(iii.) Addition and subtraction of unlike fractions are per-

formed by reducing the unlike to like fractions and proceeding

as above (i.), (ii.).

Examples.

l+^ + ^
-
(1.) Simplify
h
 — a

96 FRACTIONS.

Since the l.o.m. of be, ca, ab is abc, this sum equal to

abc
is

a+b+c
~~
abc abc a&c a&c

1 2^_ _1_
1+0 + I- a ~ 1-a2
'

The l.c.m. of the denominators is 1— 2

.

1 1 1+a 2a
2a 1—a
" l+a
+ 1-a "1-a2 ~ 1-a* + 1-a2 ~ 1-a2
l + a— 2&
= 1—a +l^a 2

_ 2-2a
~ 1-a2
2
~~
1 +a

(3.) Simplify
^-y - 2
-

The denominators (x + y)2 (x—y) 2 =

Ill
l.c.m. of the is

(x 2
-y )2 2
.

(x+y) 2 y
2
-x 2
(x—y)2
(x—y) 2y
2
—x2 (x+yf
~ (x -y ) ~ (x -y )
2 ~ (x -y )
2 2 2 2 2 2 2 2

__ Q-?/) - + ^ - Q+ 2
2/
2 2
2/)
2

(a 2 -*/ 2 ) 2

_ x —2xy + y —y + x —x —2xy—y
2 2 2 2 2 2

(x 2 -y2)2
2 2
x —4:xy—y
= (x2 -y2f~'

Exercise XL.

Simplify
a b ,~ v x 1 ,n x a? 1 2
-
a-) 6 +r; <
2 -)
s*V 3
< -> 3+2«+$?-
 C " "

FRACTIONS. 97

1 2 2 3a cc-1 x-3
^'\x"3x2 '
W« + xr ^' 1^
tf>.;
2x a2
2a-3& Sa-2Z> a-36 /ON a + 5 a-6
rrTN
< 7 ') — S
^ .

a+& a— & ; 2a 1

^ & e2
a + a + & + a 2 + a&
*
^ 2 __3_ 2a~3
a;""2cc-l~4a; 2 -l

^ 13 -^ + a;

1
+ l + ^-^Tl" W 2(«--l).""2(a? + l>^?'
nsS 1 a?

2a)-%""2^4-42/
+ cc
2
-42/2
"

_2 _1_ *+3 o

^ x 2 + xy + y2 x 2 —xy + y 2 x^ + x^+y*

^ 1 q \
a
^—1
a; -a; + l
1 +1 1
+ aj + l + {»*+aJ + l + a-l
2 2 1
( 19 0 (a-lXa; + 2)">- 1)(<b-2)
+ (a-2)0* + 2) *

1^0 (a ~&)(&_ c),+(a^ c )(c^&)'

/Ol N _
^ _ C
*
(cc— b)(x—c)~*~ (x—c)(x— a)~*" (x>—a)(x--b)
a 2 + ^2 + c 2
*

(as— a) (x—b) (x— c)

sc\c\ \ 95™"* W X"~~l) X **™"

a 2 — be b
2
—ac c
2
—ab
t^O + (6~aX6-c) + (?^(7-^)
( C5 -&)(a-c)

135. A m^ecZ quantity is the sum of a whole expression and

a fraction ;
as, for example,
b 9 _ 3x n x—1
c ar-f-5 x2 + 2
 —
93 FRACTIONS.

136. A mixed quantity may be expressed as a fraction by

considering a whole expression as a fraction whose denominator
is unity ; and conversely, a fraction may be expressed as a
mixed quantity when part of its numerator is a multiple of its
denominator.

Examples.

„ 12x1 2ax + l
T + -=-^-'
N O
2*+-=
(J.)

.
-I
1 x—1 1 x2 m
(2.) *-i+5+i=-r +x + l=xTl

W x2 + i
v + 1- X2 + 1 x -a.2 + 1

It will be observed here that —x2 +l=—(x 2

—l)=z — —j

(4.) Express — ^
— as a mixed quantity.

On dividing 2cc
2 — 3cc + 4: by cc + 1 we get quotient 2x~5 and
remainder 9.

2J5!=^±4 =
... 2z-5+-A_.
cc +1 ce +1

(5.) Express — as a mixed quantity.

x y 2
cc — a?-f-l
In this case the quotient is x + 2, and remainder —cc + 1.

. a3 + a2 -2cc + 3_ .
9j_ -x + 1
" x*^x+i

CC
2
— +1 £C
'

Here the + before the fraction is changed to — by changing

at thesame time the signs of all the terms in the numerator
 — —3
FRACTIONS. 99

Exercise XLI.
Simplify—

(1.)
f-+5.
(2.) j-L (3.) 2-1
SC CI LI

(4)y
x
(5.) ??-o+L
x
(6.) 2+^.
x-

(7.) S-^-1 . (80 4-^^

(9.) * +8. (10.) (ll.)2-^|.

4b-1-^±£. x 2 + xtj + y 2 - W^t.

(

(12.) (13.)
x+6 x—y
Express as mixed quantities

a *- 1
CM.) ^±5. (i5.) . (16.)^=M.

/lr7 s
(17.)
2a;
2
— +3
cc /in
(18.)
\
—
4— 3a;
g-
+ 6cc2

(19.) (20.) ^=f . (21.) i+i 23

.

or— -1 cc— 1 1 + a;
/0 o\ a;
3 — 3a; + 2 2
/0Q n 6a;
2 — 4a; + 5 s a;
5 —x + 5
< 22 ->
n?=g— (23 } '

2*^+1 •
/0/(
(24)
wr-
Multiplication.

137. To denote that two or more fractions, or a fraction and

a mononomial whole quantity, are to be multiplied together,
they are written in a row with the multiplication sign x , or
(dot), between them.
-

Thus ^x- 7,
or % •
4 denotes that % is to be multiplied by
o a o d b

x—1 —3 2
—— and — — 2
c
-7;
a
.

— —x —
n-
cr —1 *
2a;
„
6
-
-, ,

denotes the product of

, « a;— 1
0
cr
2a;
-
x3 ~-l'

-a "^x— 3acc denotes the product of

^a ~^-
and — 3ax>
h2
 —a
IOO FRACTIONS,

Sums and differences when they are factors must bo

enclosed in brackets.
a
Thus, Q
X cP—ax + x*) denotes the product of -s^-, and

a?—ax + x2 ;
-?L-(x + -) denotes the product
x + a\ xl
of— —
x\a
and

x + ®; 2 denotes the product of 4=f and

X X — 1\2 05 05
2
/ —1
05
2

2%
2 x x

138. Multiplication of fractions is performed according to

the following rule :

The product of any number of fractions is a fraction whose

numerator is the product of their numerators and denominator
the product of their denominators.

A whole quantity is to be considered as a fraction whose

denominator is unity and sums and differences of fractions
;

and mixed quantities must first be reduced to fractions.

Examples.

v x2 — 2
t
ax _(x 2 —a 2 )ax_x—a
2ax x-\-a 2ax(x + a) 2

common to any numerator and any denominator

Factors
may be struck out. Thus ax which is common to the first
denominator and second numerator, and x + a which is com-
mon to the first numerator and second denominator, may
be struck out before multiplying, and the result will be in
lowest terms. When possible, therefore, the component
factors of the several numerators and denominators should
be obtained.

4a 3 — 4acc 2 bc + bx _£a(a + x) (a— x)

t
b(c + x)
^ J 3bc 2 -3bx 2 "
aF^ax ~ "3T(c + 05) (c - x) '
a(a-x)
_ £(a + x)
3(c-a>)'
 &—
FRACTIONS., 101

Here a,a—x, b, and c + x are struck out, being common to

the numerators and denominators.

-49) =
x (x + 7)(x-7)
0+6)0 + 7)0
2
(3.)
0 + 6)0 + 7)' 1
_x (x-l)
+6 £C

2cc -4\ cc-l_3cc -3-2x + 4 x-l

2 2 2

(4) (S-**=1\
x 2 -l x2 + l
x2 + l x-l
"0+1)0-1) a2 +l
1
"aj+1'

Exebcise XLXL
Simplify

(2-)f
be
.
- -4
ca ab
2
/o>, c a a2 ^
2/2
2 "
zx2 '
xy 2
' (4.) 1— " l +#
*

a3 — b z a-\-b
(5.) (6.)
a— ab a3 + 6 3 *
a— b"
a2 —ab + b2 9
as b—ab 3 a 3 —6 s
(8.)

1 a? +l a2 +l x 2 -l x 3 -l x4 - l
(9.) (10.)
cc +1 '
a2 + l '
a 3 + l* cc
3
+l "
x2 + l '
X Q -V

(ii.) fi+i+lW (12.) JEJ*-»\.

O 2/ 2/ a + x\a xj
x2 + x V( x
(14.)
x 2 + y 2 \as— 2/ cc + 2//
(15 0
(^^aT+62 ~5)^^6•

3 as.) (s±»+^±gy«jt».
05
8
+ 2/
 ; '

102 FRACTIONS.

Division.

139. To denote that one fraction is to be divided by another,

they are written in a row with the sign -f between them. -

The same notation is employed when the dividend or divisor

is a mononomial whole quantity.

Sums and differences when they are the object of division

must be enclosed in brackets.
2 x 2 x
Thus,' --^- denotes that - is to be divided by
J
-
x a x a'

ia • Q„ *? o •

f-H>-l) „ \ „ „ 2a-l;

&C .
• /
f o
2
3
3cc\
*\ x n Sx
y "
\ a / " "2/ .a '

(ic
2 — 5)~— „ „ x
2
5 „ „
—
The same thing may also be denoted by writing the
quantities which are the objects of division in the form of
a fraction.

2x a

Thus, f ra -^(2^-l),

140. "When the product of two quantities is unity, each is

said to be the reciprocal of the other.

Ttas
lows that
'
1
• 1 • != 1 •
£a •
fer=Mt foi-
 — ;

FRACTIONS.

- is the reciprocal of a , and a of -

a a
a b b a m

k-1 2x-3 2a;-3 g?--l

ts 9
2a;-3 " x-1 " a-l " 2x-3*

141. Diyision of fractions is performed according to the

following rule :

The quotient of one fraction divided by another is the product

of the former and the reciprocal of the latter.
A whole quantity is to be considered as a fraction whose
denominator is unity and sums and differences of fractions
;

and mixed quantities must first be reduced to fractions.

Examples.

4za b
2
2ab2 _ 4a 2 b 15x 2y _ (5ax
a) hxy 2 '
lbx 2y bxy 2
m

2ab 2 by
a 2
_^ a 2
+ ax _ a2 a 2 -\-ax-\-x2
(2.)
a8 —x 3 '
a 2 -tax+x 2 6
a' —x 3
a 2 + ax
_ a2 o
a 2 -{-ax + x2
(a — x) (a 2
4- ax + x2) a(a + x)
a
<r-— or

bx-\-ay . bx—ay
<3
->(MMH)= ab *
ab
__bx + ay ab
ab bx—ay
_bx + ay
bx—ay

\ar xy y j
2
\ar xy y*/x3 + i
2 2
_y —xy-\-x
x2y 2 (x + y) (x 2 —xy + y2 )
1
x2y 2 (x+yY
 5

lOf FRACTIONS.

Exercise XLXII.
Simplify-

v 2ab . Sx 4a2ccs . Sax

T 105° 5F27 ~ 5%
; "
1 ; V
6££ 2

(3.) JL . 1 . (4.)

x a 3 —b 3 m
a—b ,n x a4 — 4
. a2 —ah

(1 1\ . 1 /a+5 . « 3 -^3

a+i> a—i> a+aj a— a;

a— p a-i-p

y L_jL,L
x y ' <W 11"
cc— 2/ a; -4-2/ a3 +£&3

(
^
m)0 '
0.
X X+X X
oa)SZ2JO.
X X Xy
X
4+5 6+G 3+10 10+12

t
 - h

( io 5 )

CHAPTEE XV.
SIMPLE EQUATIONS (continued).

142. We shall give in this chapter some examples of equa-

tions involving fractions with literal denominators. Such
equations may be cleared of fractions by the rule already
given in Art. 74. In some cases before applying this rule
it will be found more advantageous to simplify parts of the

equation separately.

Examples.

Solve
cc—-a x—b
(1.)
b ~ a
Multiplying by ab, the l.c.m. of the denominators, we get

(x — a)a = (x— b)b.

Clearing of brackets and transposing,
cix — bx — 0?— 2
.

Collecting coefficients of x,

(a—b)x = a2 —b 2.
Dividing by a— 5,
a 2 -b2
a—b

Solve
3x-l 4cc-2
(2.)

Multiplying by the l.c.m. 6 (2x-l) (3x-2),

6(3a-l)(3x-2) . 6(4* 2)(2x - 1) := (2x - l)(3x - 2),
 — 5 5 5 *

io6 SIMPLE EQUATIONS.

Clearing of brackets,
54x 2 - 54a; + 12 - 48x 2 + 48 X _ 1 2 = 6x 2 - 7x + 2.
Transposing,
54a2 -48x 2 - 6x 2 - 54x + 48a; +7^=2-12+ 12,
.\ cc=2.

a?— 1_&— 2_#— 4_cc---5

(3 ) Solve
^ x—2 x—3 x—5
Simplifying the sides separately,

( £C „3)^( -2)2 _ (x~4)(a;~6)~Ccc-5) 2

a;

Clearing the numerators of brackets,

-1 -1
(aj-2)(aj-3) (a-5)(a>-6)'
Multiplying by the l.c.m. (cc—2)(^— 3)(cc— 5)(a;— 6), clearing
of brackets, and solving,

£C=4.

Exekoise XLIY.
Solve

n v 12 1 _*a /9 n 42 _ 35

/on 16
3x— 4
_ 27
5cc— 6*
m + *
y
45
2cc 3
_ 57 5*
4cc—
,r v
1 ;
cc— l_7cc— 21
^2 ~ 7a; -26 '
W 2%— 6_23^7
^ \

3x~=$ ~
cs—

5cc— 3_2x_+3_^ 6% + 13_2cc_ 3^ + 5

/•i-i v eg— _ i ,
cc— /-jqx a;— 14 _ 2%— 29__ 1
^ ;
a;
~ 2^-5' * *
;
x "2^20
25'

^ (2a+3) (x-5)
2 :=
3
(3a;-2) (a-ll)
'
 SIMPLE EQUATIONS. 107

v = 0.
dec
2
—! 2cc+l

x x — 1 __x — 3 5c — 5*
4-
(15.)
a;— 1 a;— 2 as— 4 a;—

(16.)
.« -|=l-f=? = l. (17.) ?-"=c.
b—x 5— ft; 4— X'
a b

/to \ x—a
r— +
a;— &
= a b
5 + «=£ + *.
— abba
.
,

(le.) (19.)

(20.) 5=| + 5=2=1. (21.) ii_-^=&2_ a2

ax bx

X '~'b ^ x+a x+h

Xm~ a
(22) + ^
x—b x—a x — b x—a*
 ( ioS )

CHAPTEE XVI.

PROBLEMS (continued).

143. We shall give in this chapter some examples whicl

are more difficult than those in Chapter IX.

Examples.

(1.) If A can perform a given work in 60 days, and B in 4C

days, in how many days will A and B, working together, ba
able to perform it?

Let to denote the work to be done, and x the required

number of days. Then
w .

amount of work done by A in one day=gg'

» » -° » ""40

» AandB „ =~ *

•'•
www
S
= +
60 40

^
Divide by
1.
111
Multiply by 120x, 120 = 2a + 3a>.
,\ cc = 24.

(2.) At what time between 5 and 6 is the minute hand of

a watch 5 minute divisions behind the hour hand ?
 PROBLEMS. 109

Let x = the number of minute divisions between the hour

hand and then 5— £c=the number of minute divisions
5;
between the minute hand and 5. But the number of minute
divisions between 12 and 5 is 25 therefore the number of;

minute divisions between 12 and the minute hand is 25—

(5-x)=20 + x.
The hour hand thus moves over x minute divisions while
the minute hand moves over 20 + x; and since the latter
moves 12 times faster than the former, it follows that

Hence the required time is 12cc=21 T9T minutes past 5.

(3.) A grocer bought 200 lbs. of tea and 1000 lbs. of sugar,
the price of the sugar being ^ of that of the tea. He sold the
tea at a profit of 40 per cent., and the sugar at a loss of 2^
per cent., gaining on the whole $45 "50. What were his buy-
ing and selling prices ?
Let the cost price of the sugar per lb. =x dollars.

Then „ „ 1000 lbs. of sugar -lOOOa dollars.

„ „ 200 „ tea =1200x „

40
The profit on the tea = jqq- 1200cc=480a;

2i
The loss „ sugar = jqq- 1000& == 2ox

480x-25x=45-50
X-'10

.\ the buying price of sugar is 10 cents, and the selling

price 91 cents per the buying price of tea is 60 cents,
lb. ;

and the selling price 84 cents per lb.

Exekcise XLY.

(1.) I arrange 1024 men 8 deep in a hollow square : how

many men will there be in each outer face ?
no PROBLEMS.

(2.) A regiment containing 700 men is formed into a hollow

square 5 ranks deep how many men are there in the front
:

rank?
(3.) A man has a number of cents which he tries to
arrange in the form of a square on the first attempt he has
;

130 over when he increases the side of the square by 3 cents

;

he has only 31 over. How many cents has he ?

(4.) On a side of cricket consisting of 11 men, one-third
more were bowled than run out, and 3 times as many run out
as stumped two were caught out. How many were bowled
;

and run out, respectively ?

(5.) Water expands 10 per cent, when it turns to ice. How
much per cent, does ice contract when it turns to water ?
(6.) A manufacturer adds to the cost price of goods 20 per
cent, of it to give the selling price ;
afterwards, to effect a
rapid he deducts from the selling price of each article a
sale,
discount of 10 per cent., and then obtains on each article a
profit of 8 shillings. What was the cost price of each article ?

(7.) A
person invests £14,970 in the purchase of 3 per
90 and 3i per cents, at 97. His total income being
cents, at
£500, how much of each stock did he buy ?
(8.) A and B join capital for a commercial enterprise, B
contributing £250 more than A. If their profits amount to
10 per cent, on their joint capital-, B's share of them is 12 per
cent, on A's capital. How much does each contribute ?

*(9.) In a concert room 800 persons are seated on benches

of equal length. If there were 20 fewer benches, it would be
necessary that two persons more should sit on each bench.
Find the number of benches.
*(10.) A man travelled
105 miles, and then found that if he
had not travelled so by 2 miles an hour, he would have
fast
been 6 hours longer in performing the journey. Determine
his rate of travelling.

(11.) An express train running from London to Wake-

field (a distance of 180 miles) travels half as fast again as an

* These questions belong to Exercise XLVIII.

 PROBLEMS. in

ordinary train, and performs the distance in two hours less

time find the rates of travelling.
;

(12.) A much work as B, B can do half as

can do half as
much and together they can complete a piece of work
as 0,
in 24 days; in what time could each alone complete the
work?
(13.) Three persons can together complete a piece of work
in 60 days and it is found that the first does f of what the
;

second does, and the second f of what the third does in what :

time could each alone complete the work ?

(14.) What is the first time after 7 o'clock when the hour
and minute hands of a watch are exactly opposite ?

(15.) The hour is between 2 and 3 o'clock, and the minute

hand is in advance of the hour hand by 14^ minute spaces of
the dial. "What o'clock is it ?

(16.) At what time between 3 and 4

o'clock is one hand of
a watch exactly in the direction of the other hand produced ?

(17.) The hands of a watch are at right angles to each

other at 3 o'clock : when are they next at right angles ?

(18.) How much

water must be mixed with 60 gallons of
spirit which £1 per gallon, that on selling the mixture at
cost
22s. per gallon a gain of £17 may be made ?

(19.) How much

water must be mixed with 80 gallons of
spirit bought at on selling the mix-
15s. per gallon, so that
ture at 12s. per gallon there may be a profit of 10 per cent,
on the outlay ?
(20.) If 16 oz. of sea-water contain 0*8 oz. of salt, how
much pure water must be added that 16 oz. of the mixture
may contain only 0*1 oz. of salt ?

(21.) Ihave a bar of metal containing 80 per cent, pure

gold, which weighs 30 grains how much must I add to this
:

of metal containing 90 per cent, pure gold, in order that the

mixture may contain 87 per cent. ?

(22.) How much silver must I add to 2 lbs. 6oz. of an

 ;

112 PROBLEMS.

alloy of silver and gold containing 91*7 per cent, of pure

gold, in order that the mixture may contain 84 per cent, of
gold?
(23.) A
person started at a certain pace to walk to a rail-
way off, intending to arrive at a certain time
station 3 miles
but, after walking a mile, he was detained 10 minutes, and
was in consequence obliged to walk the rest of the way a mile
an hour faster. At what pace did he start ?
(24.) A person started at the rate of 3 miles an hour to
walk to a railway station in order to catch a train, but after
he had walked \ of the distance he was detained 15 minutes,
and was obliged in consequence to walk the rest of the way
at the rate of 4 miles an hour. How far off was the station ?
(25.) A wins the 200 yard race in 28 \ seconds, B the con-
solation stakes (same distance) in 30 seconds: how many
yards ought A to give B in a handicap ?
(26.) A
wins a mile race with B in 5' 19". B runs at a
uniform pace all the way A runs at \& of B's pace for the
;

greater part of the distance, and then doubles his pace, win-
ning by a second how far did A run before changing his
:

pace?
(27.) A boy swam half a mile down a stream in 10 minutes;
without the aid of the stream it would have taken him a
quarter of an hour. What was the rate of the stream per
hour and how long would it take him to return against it ?
;

(28.) A contractorundertook to build a house in 21 days,

and engaged 15 men to do the work. But after 10 days he
found it necessary to engage 10 men more, and then he accom-
plished the work one day too. soon. How many days behind-
hand would he have been if he had not engaged the 10
additional men ?
(29.) Two crews row a match over a four-mile course one
;

pulls 42 strokes a minute, the other 38, and the latter does
the distance in 25 minutes; supposing both crews to row
uniformly, and 40 strokes of the former to be equivalent to
36 of the latter, find the position of the losing boat at the end
of the race.
 t "3 )

CHAPTER XVH.
QUADEATIC EQUATIONS.

144. We have already defined Quadratic Equations in

Art. 69. They are further called adfected or pure, according
as the term involving the first power of the unknown quan-
tity does, or does not, appear.

Thus, 4o32 — 5# + 7= 0, x 2 + 6x—3=Q, x 2 —Sx—0, are adfected

quadratics; 3x 2 — 8—0, x2 + 6=0, are pure quadratics.
145. I. Pure Quadratics are solved hy transposition of
terms and extraction of the square root.

Examples.
(1.) Solve rc
2
-4=0.
2
Transposing, cc =4.
Since the square root of a positive quantity is either + or
— , we have, extracting the square root,

x=±2.
Thus the two roots are + 2, —2.
(2.) Solve x 2 +5=^°x 2 ^m
Clearing of fractions,
3x 2 +15=10x2 -48.

Transposing and dividing by —7,

x 2 =9.
x =±3.
Thus the roots are -f3 and —3.
I
 — ;

U4 QUADRATIC EQUATIONS.

146. II. Adfected Quadeatics may be solved by one of

the following three rules :

*
(i.) When the equation is in the form of the product of two
factors, each containing the unknown, equated to zero, the solu-
tion is effected by equating to zero each factor in turn.

Examples.

(1.) Solve (a?+l) (2x-3) =0.

Since, when the product of two factors vanishes, one or

other must be zero, we have

either cc + l=0, and /. x=—l ;

or 2cc— 3=0, and ,\ cc=f.

Thus the two roots are —1 and f.

(2.) Solve x 2 - ox=Q.

r

Factoring, x (x — 5) = 0.
/. either x=0,
or x— 5=0, and cc=5.

Thus the roots are 0 and 5.

Whenever, as in this case, the terms of an equation are

by the unknown x, we can infer that one root is
divisible
zero.

(3.) Solve (2sc-5) (ax-4tb) =0.

Here either 2x— 5=0, and .% cc=f

4&
or ax—4:b=0, and £=~ #

Thus the roots are f and •

 —
Q UADRA TIC EQUA TIONS,

Exercise XL VI.
(1.) ce
2
-36=0. (2.)
2
5x =45. (3.) | =27.
(4.) 2(x 2 -7)+3(cc 2 -ll)=33.

*
(5).
jCa + 4)+K^+8)=^+l. (60 J^=ra'
2

K^ -^)= 5 ^
2
(7.)
2
x =3x. (8.) |- + 6x=0. (9.)

(10.) a;
2
-^ 0 - C 11 -) ^ +|=0.
2

(12.) x*-^=2x2 + x. (13.) (cc-3) (a-5)=0.

(14.) (® + 5)Oe-7)=0. (15.) (cc-t-l) (a + 3)=0.

(16.) (2cc-l)(3rc-4)=Q. (17.) (3a-5)(2a;+7)=0.
(18.) (5x + 6)(6cc + 7)=0. (19.) (aa;-.&)(ca> + <Q=0.
(20.) cc
2
— aa;=aa;— as
2
.

147. When the quadratic is not in a form adapted for

applying rule (i.), it may be solved by either of the following
rules :

(ii.)Having transposed the unknowns separately to one side,

make the coefficient of x2 unity by division (if necessary). Then
add the square of -J the coefficient of x, and the solution is effected
by the extraction of the square root of both sides.

148. (iii.) Having cleared the equation offractions (if neces-

and transposed the unknowns separately to one side, mul-

sary) j
tiply both sides by 4 times the coefficient of x2 and add the ,

square of the coefficient of x. The solution is then effected by the

extraction of the square root of both sides.

Examples.

(1.) Solve x 2 -12cc + 35=0.

By rule (ii.), transposing,

x2 -12x=-35.
i2
 Q UADRA TIC EQUA TIONS.

Adding the square of one-half 12,

;
2
-12x+6 2 =36-35=l.
Extracting the square root,
rc-6=±l;
that is, cc— 6=1, and ,\ x=7,
or x— 6=—l, and .*. cc=5.

Thus the roots are 7 and 5,

(2.) Solve 2a 2 +5x-3=0.

By rule (ii.), transposing and dividing by 2,

x 2 + %x=\.

Adding the square of one-half f.

Extracting the square root,

-o±7 , q

Thus the roots are £ and— 3.

(3.) Solve X 2 +pX + q=zQ.

;

By (ii.), transposing,
X 2 +pX= —q.

Adding the square of one-half

Extracting the square root,

2
--4 g
= -j)=fc yV /i>

2
 QUADRATIC EQUATIONS. 117

Thus the roots are

2 2

(4.) Solve 2x 2
+5x=3 by rule (iiL).

Multiplying by 4 x 2=8,
16x 2 +40cc=24.

Adding the square of 5,

16x 2 + 40a + 5 2 =25 + 24=49.

Extracting the square root,

4^ + 5= -1=7.
.". w— or —3.

(5.) Solve aaj

2
+ foc + c=0.
By (iii.), transposing,
ax 2 + bx=— c.
Multiplying by 4a and adding b 2,
4aV + 4abx + & =62 -4ac. 2

Extracting the square root,

2ax + b=±\f b 2 —4-ac,

. _-6±V& 2
-4ac

Thus the roots are

2a 2a

• Exbeoise XLYIE.

(1.) cc
2
-6a-!-8 = 0. (2.) x2 -4:X-5=0.
2
(3.) a +4a--21=0. (4.) 2^ 2 ~5a3 + 2=:0.

(5.) l-*2 => (6.) +

 1

QUADRATIC EQUATIONS,

(7.) 4x 2 -4x=15. (8.) 6a2 -ll*-f4=0.

(9.) (10.) *±L*%.

J.-*=12.

(11.) (12.) 1 + 1=^.

cc
2
+5 OX 2
X
21a 8 - 16
*
(13.) (14.)
3a 2 -4: * + 60 3x-5'
2-a?, I—as — 2& — 3 oc —
is
(j^g ^
3
2x + l 3x + 2

a=5-
3a-2 __ 2x-5 _
(17.) (18.)
cc-3' 2^-5 3a-2
^ + 4 _ 5 (^ + 2 ) -8. CC +2 7 _4r— X
(19.) (20.)
as— 2x

(21.) (x-3) 2 -3(x-2)(cc- -7)=21.

(22.) a2 -(a + &>+a&=0.
 ( H9 )

CHAPTEE XVIIL
PROBLEMS.

149. "When the Algebraical statement of a problem leads to

a quadratic equation, the unknown quantity will be one of
the roots. In some cases either root may be taken, but it
will generally be found that one of the roots must be rejected
as being inconsistent with the conditions of the particular
question proposed.

Examples,

(1.) A person laid out a certain sum of money in goods

which he sold again for $24, and lost as much per cent, as he
laid out. Find out how much he laid out.

Let x = number of dollars laid out.

,\£c— 24= „ „ lost.

But the loss is also x per cent, of x =^^xx=~-,

••• 24 -

fir*-
2
x -10Gb=-2400.
cc=40 or 60.

The amount laid out was, therefore, $40 or $60. Thus

both roots satisfy the conditions of the problem.

(2.) A person buys a certain number of shares for as many

dollars per share as he buys shares ; after they have risen as
many cents per share as he has shares, he sells and gains
$100. How many shares did he buy ?
120 PROBLEMS.

Let x = the number of shares bought, the price of which

at x dollars per share is x 2 dollars.

The rise being x cents or dollars, the price for which

he afterwards sells the x shares at ^+ jjjg dollars per share is

• (x + ^)x-x*=lOO;

a2 = 10000.
/, x = + 100, or -100.

As the negative root would not answer the conditions of

the problem, it must be rejected. The answer is, therefore,
100.

Exeecise XL VIII.
(1.) A rectangular room which contains 1800 square feet is

twice as long as it is broad : find its dimensions.

(2.) Divide 20 into two parts whose product shall be 91.

Find a number whose square increased by 20

(3.) is 12
times as great as the number itself.

(4.) Divide 15 into two parts such that their product shall
be 4 times their difference.

(5.) By what number must I divide 24 in order that the

sum of the divisor and quotient may be 10 ?

(6.) Find three consecutive numbers such that the square

of the greater shall be equal to the sum of the squares of the
other two.

(7.) A ladder 34 feet long just reached a window of a house,

when placed in such a position that the height of the window
above the ground exceeded the distance of the foot of the
ladder from the wall by 14 feet. Find the height of the
window.
 :

PROBLEMS. 121

(8.) A
horse is sold for £24, and the number expressing
the profit per cent, also expresses the cost price of the horse
what did he cost ?

(9.) An article is sold for £9 at a loss of as much per cent,

as it is worth. Find its value.

(10.) A and B start together for a walk of 10 miles; A

walks 1% miles an hour faster than B, and arrives li hours
sooner than he does at what rate did each walk ?
:

(11.) After selling a part of an estate, and the same part of

the remainder, I find I have left nine- tenths of the part first

sold what part did I sell at

: first ?

(12.) An uncle leaves 14,000 dollars among his nephews

and nieces, but 3 of them having died in his lifetime, the
others received 600 dollars apiece more : how many nephews
and nieces were there originally ?

(13.) A number is composed of two digits, the first of

which exceeds the second by unity, and the number itself
falls short of the sum of the squares of its digits by 26. What

is the number ?
The sides of a rectangle are 12 and 20 feet
(14.) : what is
the breadth of the border which must be added all round
that the whole area may be 384 square feet ?

One hundred and ten bushels of coals are distributed

(15.)
among a certain number of poor persons; if each had
received one bushel more, then he would have received as
many bushels as there were persons. How many persons
were there ?

(16.) A sum of £23 is divided among a certain number of

persons ; if each one had received 3 shillings more, he would
have received as many shillings as there were persons. How
many persons were there ?
A
company at an inn had £7 4s. to pay, but before
(17.)
the was settled 3 of them left the room, and then those
bill

who remained had 4s. apiece more to pay than before; of

how many did the company consist ?
122 PROBLEMS.

(18.) A person rents a certain number of acres of pasture

land for £70; lie keeps 8 acres in his own possession, and
sublets the remainder at 5s. an acre more than he gave, and
thus covers his rent and has £2 over. How many acres were
there?

(19.) An officer can form the men in his battalion into a

solid square, and also into a hollow square 12 deep ; if the
front in the latter formation exceed the front in the former
by 3, find the number of men in the battalion.
 ; —
( 123 )

CHAPTEE XIX.
SIMULTANEOUS EQUATIONS.

150. If two unknowns are to be determined, there must

be two independent equations. These equations are called
simultaneous equations, because- the same values of the un-
knowns x and y must be substituted in both equations.
Thus if
2x-y = 9,
2y-x = 3,
the only values which satisfy both these equations at the
same time are x = 7) y=5.
151. It must be borne in mind that there is an infinite
number of values which will satisfy either equation sepa-
rately.

Thus, in the equation 2x—y=9,

if x— 1, 2— y=9,
and ,\ y= — 7;
if x— 2,4:—y=9, and .". y— — 5;
if x=lQ, 20-#=9, and .\ y= 11
and so on.

152. If three unknowns are to be determined, there must

be three independent equations and generally the number of
;

unknowns must be the same as the number of independent

equations connecting them.

153. The solution of simultaneous equations is effected by

deducing from them other equations, each of which involves
one unknown. This process is called elimination, and may be
conducted according to one of the following methods :
124 SIMULTANEOUS EQUATIONS.

I. Substitution.

II. Comparison.
III. Cross Multiplication.

I. Method of Substitution.

154. This method consists in finding from one equation the

value of one unknown in terms of the other, and substituting the
value so found in the second equation, which is thereby reduced to
a simple equation in one unknown.

For convenience of reference the given equations and others

which arise in the process of solution are numbered (1), (2),
(3), &c.

Example.

Solve x+y= 3 .... (1),

2x + y = 4: .... (2).

From(l) we find y = 3-x .... (3).

Substituting this value of y in (2),

2ic + 3— 05=4.
.*. x=l.
Substituting this value of x in (3),

Thus the solution is x=l, y=2.

II. Method of Compakison.

155. This method consists in finding from each of the pro-
posed equations the value of one and the same unknown in terms
of the other, and equating the values so found.

Example.

Solve 7x-3y = 19 .... (1),

4<r + 72/-37 .... (2).

 SIMUL TANEOUS EQ UA TIONS. 125

From (1) we find

• (3);

and from (2) . (4).

Equating these values of y,

7.x-19_37~4a
*

3 7
,\ x=4z.
Substituting this value of x in (3),
28-19 o

Thus the solution is as =4, 2/ =3.

III. Method of Ceoss Multiplication.

156. This method consists in multiplying the given equations

(reduced to the form ax + by=c) by such quantities as will
render the coefficients of the same unknown numerically equal.
By adding or subtracting the equations so found, we obtain a
simple equation in one unknown.

Examples.

(1.) Solve 7x-9y = 5 . . . .

(1) ;
13cc+4y = 30 .... (2).

Multiplying (1) by 4 and (2) by 9,

28o;-362/= 20 ... .
(3),
117a + 36^=270 .... (4).

Adding (3) and (4),

145a = 290,
,\ x = 2.
Again, multiplying (1) by 13 and (2) by 7,
91a- 117^ 65 ... .
(5),
91a + 28.?/ = 210 .... (6).
126 SIMUL TANEOUS EQ UA TIONS.

Subtracting (5) from (6),

14% = 145,

Thus the solution is x = 2, y = 1.

(2.) Solve 8x + 25y = 9 . . . . (1),

12a>- 10^ = 4 .... (2).

Multiplying (1) by 2 and (2) by 5,

16^+50^=18 .... (3),

60cc-5% = 20 .... (4).

Adding (3) and (4),

76x = 38,
m \ x= %.

Again, multiplying (1) by 3 and (2) by 2,

24^ + 75^ = 27 .... (5),
24^-20?/ = 8 . . . . (6).

Subtracting (6) from (5),

9% = 19,

Thus the solution is x=h, y=i>

Exekcise XLIX.

(1.) 4cc+2/=ll, £c + 4y=14.

(2.) 2x + 3y=21, 3x + 5^=34.

(3.) 3x=23-2y, 10 + 2x=%.

(4.) ^+3 I, + 0.
3 2

° 5 + 6~2 + A 3
(5 4
"10 I*

(6.) 3a>-2y = 3(6-a0> 3(4rc-3y) = 7y.

(7.) 7(x-l) = 3(*/ + 8), ^ + 2 = ^.

 SIMULTANEOUS EQUATIONS. 127

(9.) ^6 + ^2 =& -2,5^+^i=y+8.

(10.) 2x-^-3 =^2, 2^-£^=^.
(11.) J5(x+ll)+i(2/-4)=a -7,K* + 5)-K2/-7) = 32/-cC
; .

a;-24 = + 2/)4-a;=f(22/-!») + 105.

(12.)
| + 16, |(a;

(13.) K3cc-720=K2a+2/+l), 8-K*-sO = 6.

s 2x+7y_ 2(2x-6y + l) «
a4 1

(15.) x + «/ = a, x—y = h.
(16.) aa; + (M/=a2 +&2 , c»=a.

(17.) |+|=1,. x+y=c

157. TFAe^ £Aere are three simultaneous equations containing

three unknoiuns, the solution is effected by eliminating one of the
unknowns hetiveen the first and second equations , and also be-
tween the first and third, or second and third. Two equations
are thus obtained involving tivo unknowns, which may be found
by the methods already explained. The value of the third un-
known may then be found by substitution.

Example.

Solve 2x—3y-\- z— 1 . . . (1),

3x—5y + 4z= 3 . . .
(2),
4x + 2^/-3^13 . . . (3).

Mnltiplyin g (1) by 3 and (2) by 2,

6x- 9y + Sz=3 . . . (4),

6x-lQy + 82=6 . . . (5).
128 SIMULTANEOUS EQUATIONS.

Subtracting (5) from (4),

2/-5z=-3 .... (6).

Thus x is eliminated from (1) and (2).

Again, multiplying (1) by 2,

4a-%H-2z= 2 . . . . (7),
and 4o5 + 2?/ --32=13 .... (3).

Subtracting (7) from (3),

8#-5z= 11 ... . (8).

And 2/-5^=-3 .... (6).

From (6) and (8) we find y=% 2=1.

Substituting these values of y and z in (1), (2), or (3)
we get x=3.
Thus the solution is x =3, y= % 2=1.

EXEECISE L.

(1.) a5 + 3y + 2z=ll, 2x + 3/ + 3^=14, 3x + 2?/ + 2=ll.

(2.) a + 2?/ + 3z=13, + 3t/ + 2=13, 3x + + 2z=10.
2.x ?/

(3.) 2x + 3?/-42=10, 3x-4?/ + 22=5, 4a-2y + 3z=21.

(4) 10cc-2?/ + 42=10, 3^ + 5^ + 32=20, cc+3*/-22=21.

(5.) 3x + 2y=13, 3y + 2z=%, 32 + 2x=9.

(6.) | + | + |=3, 4x + 52/ + 62=77,2 + a;=22/.

(7.) 3x-2y=6, 3y-22=5, 32-2x=-2.

(8.) l(x-l)-y=35, %-52=43, cc + + 2=30.

*/

(9.) 2/-2 + 3=0, z—x=5, x + y=6.

(10.) 2/ + 2=a, 2 + x=&, + ?/=c. £c
 .

{ 129 )

CHAPTER XX.
PROBLEMS,

158. In the following problems the various unknowns are

expressed in terms of separate and distinct symbols, and the
relations between the quantities involved then take the form
of equations. If two symbols x and y be employed, the
conditions of the problem must give two independent equa-
tions and three independent equations will be required to
;

determine three unknowns, x, y3 and z»

Examples.

(1.) A fraction becomes equal to 1 when I is added to the

numerator, and equal to -| when 4 is added to the denomi-

nator. Find the fraction.

Let - be the fraction,

V
Then by the conditions of the question,

the solution of which is x=5, #=6.

Hence the fraction is f

(2.) Find three numbers such that the sum of the first,
one-fifth the second, and one-tenth the third, shall be equal
 ;

130 PROBLEMS.

to 4; the sum of one-half the first, the second, and one-

tenth the third equal to 7 and the sum of one-half the first,
;

one-fifth the second, and the third equal to 12.

Let x— the first, y= the second, and z= the third number.

Then by the conditions of the question,

Multiply these severally by 10,

10x + 2y+ 2= 40 ... . (1),

5x + 10zj+ z= 70 ... . (2),
bx + 2^ + 10^=120 .... (3),
2x(2)-(l), 18*/ + s=100 .... (4),

(3)-(2), -8y + &= 50 ... . (5),

9x(4)-(5), 170*/=850.
.-. y=5.
Therefore from (4) z= 100 -18*/ =10
and from (1) x=2.
The required numbers are thus 2, 5, and 10.

Exekcise LI.

(1.) One of the digits of a number is greater by 5 than the

other. When the digits are inverted, the number becomes
f of the original number. Find the digits.

(2.) In a division the majority was 162, which was T3T of

the whole number of votes how many voted on each side ?
;

(3.)The sum of two digits is 9. Six times one of the

numbers they form is equal to 5 times the other number.
Find the digits.
 ;

PROBLEMS.

(4.) If the numerator and denominator of a fraction be

each increased by 3, the fraction becomes 2 if each be in- ;

creased by 11, it becomes f Find the fraction.

.

(5.) A number consists of two digits whose sum is 12, and

such that, if the digits be reversed in order, the number pro-
duced will be less by 36, Find the number.
Three towns A, B, and C are at the angles of a
(6.)
triangle.From A to C through B, the distance is 82 miles
from B to A through C, is 97 miles; and from C to B
through A, is 89 miles. Find the direct distances between
the towns.

(7.) The diameter of a five-franc piece is 37 millimetres,

and of a two-franc piece is 27 millimetres. Thirty pieces laid
in contact in a straight line measure one metre exactly. How
many of each kind are there ?

(8.) At a contested election there are two members to be

returned and three candidates, A, B, C* A obtains 2112
votes, B 1974, C 1866. Now 170 voted for B and C, 1500 for
C and A, 316 for A and B. How many plumped for A, B, C,
respectively ?

(9.) A boat goes up stream 30 miles and down stream 44

miles in 10 hours. Again, it goes up stream 40 miles and
down stream 55 miles in 13 hours. Find the rates of the
stream and boat.

(10.) At a contested election there are two members to be

returned, and three candidates, A, B, C. A obtains 1056
votes, B 987, and C 933. Now 85 voted for B and C, 744 for
B only, 98 for C only. How many voted for C and A, how
many for A and B, how many for A only ?
(11.) Seventeen gold coins, all of equal value, and as many
silver coins, all of equal value, are placed in a row at random.
A is to have one-half of the row, B the other half. A's share
is found to include seven gold coins,and the value of it is
£6. The value of B's share is £6 15s. Find the value of
each gold and silver coin.
k2
*32 PROBLEMS.

The road from A to D passes through B and C suc-

(12.)
cessively. The distance between A and B is six miles greater
than that between C and D, the distance between A and C
is of a mile short of being half as great again as that
between B and D, and the point half-way from A to D is
between B and 0 half a mile from B. Determine the dis-
tances between A and B, B and C, C and D.

(13.) Fifteen octavos and twelve duodecimo volumes are

arranged on a table, occupying the whole of it. After six of
the octavos and four of the duodecimos are removed, only f
of the table is occupied. How many duodecimos only, or
octavos only, might be arranged similarly on the table ?

Three thalers are worth \d. more than 11 francs.

(14s)
Five francs are worth \d. more than 2 florins* One thaler is
worth 2d. more than a franc and a florin together. Find the
value of each coin in English money.

(15.) Six Prussian pounds weigh i oz. more than 5 Austrian

pounds. Twenty-five Austrian pounds weigh i oz. more than
14 kilogrammes. One kilogramme weighs 1 oz. less than the
sum of the weights of a Prussian and an Austrian pound.
Find the number of ounces in each foreign measure of
Weight.

(16.) A person walks from A to B, a distance of 9j miles,

in 2 hours and 52 minutes, and returns in 2 hours and 44
minutes, his rates of walking up hill, down hill, and on the
level being 3, 31, and 3j miles an hour, respectively. Find
the length of level ground between A and B.
 ( 133 )

GHAPTEE XXI*
EXPONENTIAL NOTATION.

159. Although the notation adopted in the preceding pages

the purposes of the operations herein treated
is sufficient for

of, yet found expedient, before proceeding farther, to

it is

employ another notation to express roots, powers of roots,

and their reciprocals. This notation, which consists in em-
ploying fractional exponents instead of radical signs and
integral exponents, and negative exponents instead of reci-
procal forms, possesses the great advantage of reducing to a
few uniform laws the operations of Multiplication, Division,
Involution and Evolution, with respect to powers, roots,
and powers of roots, of a quantity, and their reciprocals.

160. The exponential notation consists in writing

m
an instead of ^/ am ,

and or* „ „ i
;

Thus, according to this notation,

a2 =Va, a3 =^ 2
,
a?=ya\
-1 "3 -8
1 1 1
8'
a a3
L i 3 -

* This Chapter may be omitted by those who do not intend to read

more advanced works on Algebra.
 — ;

134 EXPONENTIAL NOTATION.

-3 2 -i 5_ -3 4
2a 5a =j a >
±a

161. When the exponential notation is employed, the

quantity is said to be raised to the power indicated by the
exponent.

Thus a2 is read a to the power ^

3
a4 a a # •

X i> » ;

x~* „ X )} » 5*

The term power in this extended Algebraical sense thus

includes the terms power, root, root of a power, reciprocal of a
power, reciprocal of a root, reciprocal of a root of a power, as
used in the ordinary or Arithmetical sense.

Exercise LIL

Express in the Arithmetical notation—

l X 2. A
(1.) a 3', a 6 , a 5 , a 2 .

(2.) x- 2 ,
x~\ x-10 .

_1 •
_ 5. _7.
(3.) m 3
, n 2
, p *.

(4.) 2ai, Sx~*, 6m" 2.

Express in the Exponential notation

(5.) aJx, %~m, ZJ~n.

W
m\ 1
ff
1
a*
1
W
1
a?

(7.) *R\ ff&, J/S?.

...Ill
 —
EXPONENTIAL NOTATION. 135

(9.) K
m % 4
n p 2
x
s

2 5 7
(10,)
7% W 2
'

162. The utility of employing the exponential notation will

be exhibited in the statement of the three following rules,
which are usually called Index Laws,
I. The product of any two powers of the same quantity is a

power whose exponent is the Algebraic sum of the exponents of the

factors.

Since the product of a quantity and its reciprocal =1, this

rule cannot be applied when the exponents of the two factors
are numerically equal and of opposite signs, unless the zero
power of a quantity he considered=l.

Examples.

(1.) a2 • a3 = a2 3 =a 6 .

x 3. 5. 3 +5 _U>
(2.) a* • =a l *.
a a ==a* a

~1
(3.) a 3
a~ =a =a 2 1 3
.

-3
(4.) cr 2 • a- 3 =a~ 2 =a' 5 .

(5.) a • a~i=a 1 ~i=a 2 .

§_ __3 5.-3. 1
(6.) a6 - a * = aQ * = a 12 .

a- 1 =a 1'1
(7.) a • =a°=L
(8.) a2 • a~i=a°=h

This law may also be expressed briefly as follows :

m
a .
n
a = am+n
where m a?i<2 n are any quantities whatsoever, positive or nega-
tive, integral or fractional , including zero ifaP =1.
 .

236 EXPONENTIAL NOTATION

163. Proofs of the preceding rule will be exhibited in the

following particular cases.

(1.) If m and n be positive integers,

am • an — aaa . . , . (m factors) x aaa . . . . (n factors),

= aaa . . . . (m-f 71 factors),

+
(2.) a* • a^ = /Ja *fa 3 y~a 2 = fya^cfi ==a% *.

Here .<fa = ^a3 , because each of them when multiplied

by itself six times produces a 3 . So also $~a = %[a 2 ; and
generally, if m f n > jp, are integers,

V « =V a
because each of these quantities when multiplied by itself np
times produces a™?.

(3.) If m, n, p, q be positive integers, then

/r

a» •
a «=V am '
V^a* = \/

= \/amv +Pn ,
by Ex. la

m q+pn

(4.) <fi.a-*J#^=a-*=a**.

Exercise MIL
JTind the products of

(1.) 2sc, So;

71

;
a 2, 4x m ; 4a™ a2" 3
.

(2.) cc% 2x±; 3ccs, 2a;2; 5»*.

n
(3.) a , a? 2; 2a*, 3a"; S>f.
 —
EXPONENTIAL NOTATION 137

3
(4) 2a , a~* ;
a~\ 3a3 ;
5a, 6a~ 2 .

(5.) a"2, a" 3; 2a*, a, of ^

(6.) a*, a'l; a 2 ", a~^.

n
(7.) a 3, a" 3 ; a", a~ ;
2a, da- 1 ; ma2,na .2
164. II. When one power of a quantity is divided by another^
the quotient is a power whose exponent is the Algebraic difference
between the exponents of the dividend and divisor.

Since the quotient is=I when the dividend and divisor

are the same, this rule cannot be applied in the case of equal
powers unless the zero power of a quantity be considered =1.

Examples.

(1.) ^=a«=A (2.) 4=a*~

a3
W.
2
0
(3.)
a 2-3
=a
-1
•
(4.) —
a3
a ?
=a3+4 =a7 .
a

(5.)
a 3=^i +
t=aA, (6.) ^=a*-6=a0 =l.

This rule may also be expressed briefly as follows :

am m ~n
-~=a
an

where m and n are any quantities whatsoever y positive or nega*

tive } integral or fractional, including zero if a°=l.

165. The proof of this rule rests on that of the preceding.

~n
For since a m . an = a m, it follows that
am
 EXPONENTIAL NOTATION.

Also, since \=a~ n , the second rule may be considered to

be included under the first. Thus

am
—=zam a~ n = am " n. .

Exekcise LIV.
Divide
(1.) a2m bjam ;
a5n bja\
2 ! 2
3 3 3
(2.) a by a ; a by a .

(3.) a>* by a; a>* by a*

2 -1 3 2
(4.) as by £c ; cc by a?" .

2 2 6
(5.) or by or 1
; x~ by a?' .

8 3 2 1
(6.) a: by of ; a by or .

n 2n
(7.) x~ bjx- ; x* by x~*.

166. III. The power of a power of a quantity is a power

whose exponent is product of the numbers expressing those
the
powers. In other words,
(am n =amn )

where m and n are any quantities whatsoever, positive or nega*

tive, integral or fractional, including zero ifsb°=l.

Examples.

(1.) 0 2 3
)
= a6 ;
(a 5y = a 10 .

(2.) (a
3
f=a 3 ;
(a
3
f = a\
4
(a* )i=a,i; (a*) =a*.
(3.)
1 2
(4.) (a- ) ^ 2
;
(a- 2 ) 3 = a- 6 .
1
(5.) (a- )'
2
-a2 ;
(a~*)-*=a12 .

(6.) (aV = a~"s ;

 — — —
EXPONENTIAL NOTATION 139

167. Proofs of the preceding rule will be exhibited in the

following particular cases :

(1.) 0) 2 3
=a2 • a2 • a2 =a 6 =a Zx2
11 =a*i+i =a
.

1 SL

(2.) (a 3 ) 2 =a3 '

a3 3 s

(3.) If n be a positive integer,

(a m) n =a m am to n factors,
• . .

. ^m+m+ ....ton terms

=amn .

(4.) If p and # be positive integers,

p np

because each of these quantities when raised to the qth power

produces anP.

Exercise LY.
Express the following as powers of a :

(1) (O 3
; (« )
2 4
;
(a 3)3.
(2.) (a- 1) 2 ; (a- 2) 3 ;
(a" 3) 4 -

2)- 3 -
(3.) (a ;
(a- 2)" 3 ;
(a- 3)" 4 .

(*) (a*) 5 ;
(a!)*
2 3 3 2 5 4
(5.) (a*)a ;
(a') 3 ;
(a*) 3 .

(6.)
2 _o2 „4 _ 5
s) 3
; (« 3)
168. The Index Laws (I., II., III.) are thus seen to be true
on the assumption that

°>~ P =~, a°=l,

a~n=lfa™,

for all values of m and n3 positive or negative, integral or

fractional, including zero.
Instead, however, of treating the subject of the Index Laws
and notation as in the preceding Articles, we may proceed as
follows :<

If ra and n be positive integers, it may be proved, as is done

in Arts. 163, 165, 167, that
140 EXPONENTIAL NOTATION

I. am • a n =am+n .

II. =am - n m being greater than n.

,

III. (am )
n
=a mn
.

These laws, which are thus proved to hold in the particular

case where m and n are positive integers, are then assumed to
be true when m and n are any quantities whatsoever, posi-
tive or negative, integral or fractional, including zero ; and
from this extension of these laws we deduce that
m n ,
1

o» must=V am , a-*=a? and a°=l. 9

Thus, by I., $ ' ai=a.

But aJoT. jsja =a.

By. EL,

But

a+==tfc?.
By I., a? - a°=as .

But a3 x l=a3 .

.\ a°=l.
By L, a3 • a~ 3 =a0 =L
But «»xL3=L

169. From the remarks of this chapter it will be thus seen

that,although there is no absolute necessity for using such
3 _2
symbols as a , a ,a 3 , still their introduction gives rise to a
41

uniformity in certain Algebraical processes and the number ;

of rules which otherwise would be required to meet the

different cases that arise in these operations thus becomes
largely reduced.
 ;

( Hi )

ANSWERS.

(1.) 56 + 104-15; 10 + 12+f; 2+f+f

(2.) 29-15; 2J— 14; 2-5-1-6.
(3.) 11 + 35 + 6-17; 81 + 75-69-42,
(4.) 15-7+8 + 9.
(5.) 28-16 + 10-4

1L

(1.) +20; B's +60, -30; C's +30, -20.

A's -60,
(2.) A's+20, +20, -30,-40; B's -20, +30, +30, -40;
C's -20, -30, +40, +40.

(3.) A -10, +4; B -7, +10; G -4, +7.

(4.) -2°, +5°, -3°.
(5.) +1°, -1°, +1°, -1°, +1°.
(6.) -2°, +2°, -2°, +2°.
(7.) +25°, -7°.

ni.

(1.) 30. (2.) 20. (83 22.

(4.) 22. (5.) 24, 120, 0. (6.) 14.

(7.) 7. (8.) 16. (9.) »

H2 ANSWERS.

(10.) I. f (11.) |*. (12.) f§|§.

(13.) 2a3 , 3a3 , 2a 2 + 3a3 +4a4 .

IV.

(1.) % 3a, a
Aa% la.he,

(2.) -1, +3, + 2Z>, -5a2 .

(3.) +1, -1, -3a, + fca\

(4.) -1, + 3a 2 -faz. , (5.) i, |, -i, + f, -f,
(6.) 1, -1, -2, -3, + i -f. (7.) 2x, x.

(8.) -a2 + fa2 + 2a2 -a2a.

, ;
<r,

(9.) — 3a2x +2a2 cc; ax2 —ax2

} , .

V*

(1.) +25. (2.) +4^. (3.) -28,

(4.) -2ft. (5.) +5. (6.) -6.
(7 .) +A . (8.) -A* (9.) -0-7,
(10.) + 1L (11.) +2. (12.) -12.
(13.) (14.) +1A. (15.) +0*342.

VI.

(1.) +3a. (2.) + 15a. (3.) -5a.

(4.) -13a. (5.) +2a 2 . (6.) +3a2
(7.) -3a. (8.) +Wc. (9.) -5c.
(10.) -4a2 . (11.) + 7a&. (12.) +ia.
(13.) -&a>\ (14.) -fa. (15.) -ffa,

VII.

(1.) 2a-36. (2.) -a>+%. (3.) -2cc-3y~s.

(4.) 3a + 2cc-5*/. (5.) 4-a + 2y.
(6.) a 2 -52 + i. (7.) a-26 + 3c-a\
(8.) —x + 2y-z + l, (9.) 4a-2Z>.
 ANSWERS. 143

(10.) a2 -46c. (11.) 4a-92>.

(12,) -2a-&--c, (13.) 2a-6#-z.

(14) 4a + a + 5. (15.) 5a -4a.
(16.) 2a + 26 + 2c. (17.) lOx + Sy-z.
(18.) Ax2 . (19.) 4a3 +46 8 . (20.) a+

(21.) (22.) 2a+f 5-fc.

vux
(1.) 4. (2.) 2. (3.) -9.
(4.) -3. (5.) 5. (6.) -3.
(7.) 10-6. (8.) -3-39- (9.) -216.
(10.) -0-960. (11.) -a. (12.) -7x.
2
(13.) 10a . (14.) 2c. (15.) 2a-|.

(16.) -3a 2
. (17.) lb 2
. (18.) 3a + b.
(19.) 2a. (20.) 3a-a+4. -5a& + 2&2
(21.) ,

(22.) -aa-5fy+4cz. (23.) -6a + + a-4.

Z>

-7a*+28a?-5».
(24.) (25.)
|+|+^
(26.) Ja+4*-4c

IX.

(1.) +-l; 3a+-4y+~5; 2a+-35++4c

2a 2
i

(2.) 2a- -5a. (3.) -6-+5a.

(4.) 2a+-3&-+7. (5.) 5++a 3a.

(6.) 5a + (Z>-4). (7.) -a + (-5 + 5).

(8.) a-4 + (26-c).

2
(9.) a + (2*/ + 5)-z.

(10.) a -l + (36 + 5)--3c.

(11.) a + (2a 2 -l) + (-3a2 -8). 2
(12.) 4a2-(2> -c),

(13.) a
2
+ 4- (-26 + 3). (14.) 2a-5-(a -2a-f 3).
2

(15.) a + 5 + c + (a— 5~c)— (— a + 26— 3c).

 x

144 ANSWERS.

X.

(1.) 2a+36-c. (2.) a&— 6c +o.

(3.) X2 -4:X-1. (4.) 5* 3 -3a 2 +7»-8.
(5.) 8a-&-c. (6.) 8a-Z> + c.
(7.) 8a+26-3c. (8.) 2a-6 + 6 + c.
(9.) x + 11+%. (10.) 6. (11.) 2x 2 +3x.

XI.

(1.) (a-l)(2a2 -3); (-2+a)(-3-a2); (a-5) (-2z+7) :

(2.) -2a2(Z> 2 -l); (a 2 -l)(~3a); 5<-a 2 + 3).

2
(3.) l(x-l); K^-3); -f(* -5).
(4.) -5<*-l)(a; + 2); (^-4)( + 5cc)(2a+3).
(5.) +8<-5*/)(^~.l); -7a(a&-3)(+8&).

XIL
(1.) -6c^; -5ac; +6a 2
5; -30^
(2.) -Uabc2 ;
-20a2 6c; -16a^s; +48a.
(3.) -f^; -f^; +±xy; -fa2 &.
(4) -6x 3
?/
3
;
+3a^ 6
; —|aW.
5
a 3
a 3 4
Z> 3cc
3
?/ abx sy 2
(5.) 5
3 '
15 12

XIII.

(1.) -8ax + 66x— 2ca?; 12^-8^+4?/; -2a6d+3ccZ.

(2.) 3x 3 -6x 2 -15cc;-2a 5 + 3a4 -7a3 -4aa3 +4a2a2 -8a3a;. ;

2
(3.)
2 2
4:x yz + 2xy z-6xyz -28a3 & 2 +4a263 -4<x2 & 2 ; .

(4.)
2
%a b + ab -%abc;
2 ~ a + %a 2 -~2a; ±ax*-ha2x2 + %a?x.
3

(5.) -12a + 10a&-15; -^ax 2 -\%ax^\a2x2 .

(6.) 5x 2 -5a; + 20; -2a + 2a&-6; -a3 +2a2a-a.

(7.) aas— 2 ac—bc + c 2 — 5a6 + 5a2 6 2 — 15a3 5.
; ;

(8.) 4a3 -6a2 + 8a; + 2a 2 -fce 3 + 3a; 5 .

 xw ;

ANSWERS.

XIV.
2a; + 5a-12; -4:X -x + 5; 2-x-3x
2 2 2
(1.) .

(2.) 2x ~£c -4x + 2; 2-2x + 3x -3x 3+3x2 -x3 -x5

3 2 2 3
; .

(3.) 6a -7a 2 + 14a-8; 1 + a

3
1-a3 3
; .

(4.) am— an + dm hn; am + ~bm — cm+ 2<m + 2bn — 2<m

—
4m2 — 2
.

(5.) x2y 2 — x +x y—x y —2xy—2xy + 2y

4
",
z 3 2 2 2 3
.

(6.) 2a4 +a8 --22a2 + 23a-4; a2 -4&2 + 126c-9c2 .

(7.) 3a2 -4a& + 8ae-#2 + 86c-3c2 ;

3a4 -4afy + 6ay +4x?/ 8
+ 3*/4 .

(8.) ^-fx2 + i; a2 + ia-i; 2a2 -±

(9.) 2x 3
—
|x 2
+fix-|; Q^-i^+fa;-.^.

do.) l^-l-IVe,; ^^+ff*H

(11.) ia4 ~|a2 + |a-L (12.) x3 +3^-2a;-h2/ 3 -2*/ + L
(13.) a -3abc + b
3 3
+ c\
XV.
(1.) +30a5 63 ; + 136 ^y^- TV» 7
;

(2.) -24a^+30xV; -2a 5

&+3a4 &2 -a8&.
8x3 -26a2 -17a + 6. (4.) x*-x +x-l.
3
(3.)

aj -2a x + a
8 4 4 8
(5.) .

XYI.

(1.) (2a-5)-f— 3a; (4a 2

-3a + l)-4-(3a-4>
(2.) 2a~-3&; -£c2-f- + 2a; 3z-f-2a.

(3.) 4^~(2^-5); -^ -f-(^~a).

2

XVII.

(L) _ 4; -5; +r> -V . (2.) -|

2

; A
146 ANSWERS.

2
4a; .

(5.) -2a; 5a&V. (6.) -^ 3

;

XVIII.

(1.) -2a+3&-4; aa-3 + 2a2 .

(2.) -4a+3-a; 4cc2 -2a + 3.

(3.) -a2 +4a-5; a?-3a;y+4^ 2 .

(4.) 2a-36+c. (5.) -4ac2 + 3&c8 -l.

(6.) (»-&)», (2a-c + l>. (7.) (4-a)asy, (3x-y)xy,

XIX.
(1.) a>-4; 3a; + 1. (2.) a? + l; 2a>-8.

(3.) 3x+2; 3a;

2
-2#+6.
(4.) a?— 1; a?-cc+l. (5.) &?-8x+l.
(6.) cc
3
+ cc2 + cc + 1 ; as
4
— as
3 2
-f- as — + 1.
a?

(7.) cb— 2/ ;
x2 -xy+y*. (8.) a 2 +a& + 6 2 .

(9.) 5x + 62/-3. (10.) x2 -ax + a2 .

(11.) aj
2
-2^+^ 2
. (12.) face -2a;
2
.

XX.
(1.) 2a*-8»+¥, -I. (2.) cc-a, 2a2.

a; -ax + a -2a3 a?-cc+4, -3&-4.

2 2
(3.) , .
(4.)

(5.) 2a 4-3, -5a 2 -3a;-3,

2

XXL
(1.) 3x—a miles. (2.) 50 + x dollars.
(3.) 21 miles. (4.) x 2 —y 2 square feel

(5.) - hours. (6.)*+!+!-

l(te
(8.)
 '

ANSWERS, 147

(9.) -- < 10 ->

9 iro" loo'

(11.) ^+^-5.
A 0
(12.) 36 acres.

(13.) miles. (14.) ~ hours.

(15.) 6ax+a dollars.

XXIL
(1.) 2. (2.) 10. (3.) 7.

(4.) -5. (5.) 2. (6.) 5.

(7.) -f. (8.) -5. (9-) I-

/II ) Q
(10.) 2. (11.) 6. (12.) 2.

(14:} 10
(17}
V-1 - 8
') v ' (18.) 6i

(20.) 13. (21.) a.

('23} a4-b (24.) 2&.

(9,6 } X.
{AD.) 1 (27.) a2 + ah + b 2 .

sn*c\ x (lb OjC

(28.) .

a—c

(1.) 12. (2.) 12. (3.) 2A.

(4.) 30. (5.) 23*. (6.) 6A-

(7.) 6. (8.) 3. (9.) 3.

(10.) 9. (11.) 4. (12.) 5.

(13.) 3i. (14.) -2. (15.) 33.

(16.) 24. (17.) 2. (18.) 2.

(19.) 120. (20.) 13.

XXIV.
(1.) 15 and 10. (2.) 60 and 75. (3.) 20 and 17.

(4.) 14 lbs. (5.) .26, 17. (6.) 181 and 145.

l2
148 ANSWERS.

(7.) 5. (8.) 14 years. (9.) In 9 years.

(10.) 18. (11.) 66 years. (12.) 400.
(13.) 700. (14.) 30 for translation,
5 for mathematics.
4 for Latin prose.
(15.) 21 shillings. (16.) A's £300, B's £100.
(17.) 400 inches. (18.) 18, 11 and 8.

(19.) 35. (20.) 200 quarters. (21.) 12 lbs.

(22.) 150 lbs. (23.) 240 sovereigns, (24.) 13.

480 shillings,
720 pence.
(25.) £3000 at 5 per cent., (26.) £450 at 4£ per cent.
£10,000 at 4 per cent. £350 at 5h per cent.

(27.) 1800 infantry, (28.) 17 years.

600 artillery,
200 cayalry. (29.) 26. (30.) 12.

(31.) 56 workmen; 150 shillings. (32.) 1330.

(33.) 4290 feet. (34.) 30,000 men.

(35.) 200 miles from Edinburgh. (36.) In 56 hours.

XXV.
2
(1.) x -2x + l, x + 2ax + a2
2
, x2 -10x + 25, x 2 + 6x + 9.
(2.) 4a2 +4cc + l, 9x 2 -6x + l, 4cc
2
+ 12cc + 9, 9cc
2
-12^+4.
(3.) x*-2ax2 + a2 &x2y2 + 4:xy +
3 1, 9x 4 - 12ax + 4a aV
2 2
,

-8a&x 2 + 16&2 .

2—
(4.) x + y + z
2 2
2xy + 2xz — 2yz, 4x 2 + 9y 2 +z 2 + 12xy— &xz
-6yz, x 2 + 4:y 2 + 25z 2 -4:xy-10xz + 20yz, 4x 2 + 16?/ 2 + l--16^
+4cc— Sy.
(5.) 4a + 4a + 13a +
4 3 2
6a + 9, 9a 4 - 24a 3 + 22a 2 - 8a + I,
a4
-4a 3
-4a 2
+ 16a + 16.

(6.) 2401, 9604, 990025.

XXVI.
(1.)
2
cd — 1; a2 -9;;4-a 2 .

(2.) 4a2 -l; 25a 2 -4; 16x2 -a2 .

 a

ANSWERS. H9

(3.) a4 - 2
;
a 6 -l; a10 -a4 .

(4.) 9a -46 16a 6 -4a 4 49a 8 -25a6

4 2
; .
;

(5.) 2496; 9975; 489975.

xxvn.
(1.) m + n p —q
3 3
;
3 3
.

(2.) m + l; 1-a
3 3
.

(3.) a + 27; a - 64.

3 3

(4.) Sa + 1; 64a -a
3 3 3
.

(5.) 8a +27Z>3
27a - 125/; 3
;
3

(6.) a 6 -l; a 9 +a 6 .

XXYIII.
(1.) a 2 -a + l; a4 -a 3 +a 2 -a + L
(2.) a 2 + a+l; a 4 + a 3 + a 2 +a + L
(3.) a-1; a 3 -a 2 +a-l.
(4.) a + 1; a 3 + a2 +a + l.
(5.) 2a -35.
(6.) 3a 3 + 2a.
(7.) £a 2 -a 3 .

(8.) 4a 2 — 6a6 + 9Z>

2
; a4 -2a2 +4.
(9.) 9a 2 + 3a&-fZ> 2 ; 4a 4 + 6a 2 b + W.

XXIX.
(1.) (-a) 3 (2a) 3 (3a?/2) 3 (2a4Z>V) 3
, , , .

(2.) (2a-l) 2 (a-6 + 1) 2 (a 3 — I) 2 , ,

.

(3.) {(a 3) 4 }
2
,
{(-2a) 3 }
2
, { (4aa) 5 } \ {
(3a 2 k)
4 3
} \
(4.) { (a-lf (x*-iy } j (x
} 3,
2_ 3a + 2) 2 j 3>
{
3 ,

3 2
(5.) (a ) ,
3
{(-2a) } 2
{(a%y\\ {(a-a) 3 } 2 , ,

{(a 2 -aa + l)3 }

2
.

{(-a) 2 {(a8 -a8) 2

3
(6.) } ,
{(a 2 ) 2 }
3
, { (4a-l) 2 }
»
}
3
.

XXX.
(1.) a 6 8x 6 a 9 8a9 81a8
, , , , .

(2.) a 2a 4 ,
a4 a 6 ,
a 4 a 6 ?/ 8 .
 ANSWERS.

(3.) aW, aW, 8aW 2

.

(4.) x10y15 z20 a6 6 24c42

, .

(5.) x 2 +2x + l }
4cc
2
-12a + 9, cc
4
-10a; 2 + 25, a 6 -4A; 3 +4a4.
4 4
(6.) + 4a + 10cc + 12^ + 9,
cc
3 2
cc -6ic 3 + 17^ 2 -24x + 16,
4a 6
-4a + a4 + 20cc3 - 10x2 + 25.
5

XXXI.

(l.) v^> j^zri, jxt-Sx+l.

(2.)
^"^ 6
, .^3?, fftti ^(a -3a + 4).
3

(3.) a/^t, x +l 6
.

(4.) a/ ^2, V ^3^, a/ ^2a -c* 4 2

+ 3.
(5.) fa ^^/3x~ ™fm }
\

(6.) yS 4
^!, ^2x 3 -5, */a>
8
-6a; 4 + 7.

xxxn.
(1.) 2a26, 5xy s , 9x 2y\ (2.) 4a + 5.
(3.) 6x-3. (4.) 1+ 3^. (5.) cc+i,

(6.) ®-f. (7.) 2z- TV (8.) 2*-3*/.

(9.) x2 + 2x + l. (10.) cc
2
+ a + l.
(11.) x 2 -2xy + y 2 . (12.) 2a 3 -x 2 -3x+2.
2

(13.) 1+|-| ; remainder g-g.

XXXIII.

(1.) 4a& 2 . (2.) hob. (3.) 9a;y. (4.) 3aa.

(5.) 7a 2x 2y. (6.) afrtM/. (7.) 4a 2 2>. (8.) 6x 2y\

2 2
(9.) 4a&. (10.) bob . (11.) 2b . (12.) 4ww.

XXXIY.
(1.) x+S. (2.) a-l. (3.) a-3.
(4.) x + 3. (5.) 2® -5. (6.) *-3.
 ANSWERS.

(7.) a2 +10a+25. (8.) x2 -5x+6. (9.) x2 -9.

(10.) x-2. (11.) x-I (12.) a 2 -3.
(13.) 2^ + 5. (14.) x-2i (15.) cc
2
-2x + l.
(16.) x-3y. (17.) a+y. (18.) a 2 **/.
2 2
(19.) x+3y. (20.) a; -2/ .

XXXY.
(1.) acc(2a+3). (2.) 2(^+3). (3.) a(a-2).
(4.) (5.) &»-2.

XXXYI.
(1.) a(x-a). (2.) 2a(a?-3). (3.) a(x-l).
(4.) a+L (5.) a + 2.' (6.) a-3.

XXXVII.
(1.) 6afoy. (2.) 24a 2 a%. (3.) aW ?

24aW. (5.) 240a 5 cWw

2 2 2
(4.) .

(6.) (x -7 + 12)(cc + 2) = (^ -^-6)(x-4).

2 2

(7.) (2x -5x-3) (2x + 1) = (4x +4a + 1) (cc-3).

2 2

(8.) (3x 2
-Ilx + 6) (2a-- 1) - (2x 2
-7x + 3) (3a^2).
(9.) (x - 4aa + 5a a - 2a ) (x + 2aa + 2a ) = (x - 2a?x
3 2 2 2 3 2 s

-±a )(x -2ax+a ).

3 2 2
(10.) 24(a; + l)(^-l)
2
.

(12.) xhj (x -y )
2 2 2 2 2
(11.) (a-l) (x + l) . f

(13.) a &(a -& ). (14.) 12(x -l)(x +x + l).

3 2 2 2 2

(15.) (^ + 2 )(P -2 )(P -M + ^ ).

2 2 2 2 2

(16.) (p*-l)(jp*+jp»+l). (17.) (6-c)(c-o)(a-6).

(18.) 24a2 6 2 (a2 ~6 2).

XXXVIII.

CLy-fr m%. (B.)|-.

 5 — c 1 3

ANSWERS.
1
X ± x+y
(7-)
xT+l'
(8.) (9.)
x2 + xy + y2
I iQ
(10.) (11.) (12.)
X+ V cc—5
3+ft x—10 cc—
(13.)
3— a?" (14.)
x 2 -7x + lQ'
(15.)
ft
2
- 3"
3(4ft-l) x\x + 2y)
(16.) 2
(17.) (18.)
2(3ft +I)' ft + 2/'

XXXIX.
n n b 2a rcy v Z> a .
Q x 2?/z 3z 4
ab ab a*b a'b xyz xyz xyz

a a accy aajy aft?/

1
•
a(ft j> 1)

~ft
2
-l ' ft
2a
2
-l*
^ +3 ft

(ft + l)(ft + 3) '(ft

2(ft + 1)

+ l)(ft + 3)
2 a?— 3 /q> a a—

Hfn 4(ft
2
— 1) 3ft (ft— 1) ft
1 ; 2 2 2
ft(ft -l) '
ft(ft -l) '
ft(ft -l)'
^-.^ft 2 — 1 x2 — x + 1 3ft
K } 9 ?
x*+l ~~^+T~ ft
3
+ l*
nQ x 2(s-2) 3(ft + 2)
'
(ft-l)(ft-2)(ft + 2) '
(a;-l)(cc-2)(aj + 2)

ft-1
'

(ft-l)(ft-2)(ft + 2)

/-.ox c— a—
^ 10 ^ *
(b- c)(c--a)(a--b) '
(&-c)(c-a)(a-6)

(&— c)(c— a) (a— 6)*

. v bx(x— b) aft (a— x)
abx(a — b) (x — a)
'
^ *' (ft b) '
abx(a— b)(x— a) (ft— 5)
(a — 5)(ft— a)(ft— 5)
afrr (a - — a)
fc) (ft &) (ft
 — x a 1

ANSWERS.

XL.
n
(L)

W
(A.\
\ a?+b
-sr-
Sx ~ 2
2
/Q v
(2)
vmw
ax+2

$x 2
(3 °

+ 1 2 ax~l
—
4a2cc + 6a+6
12^~" •

~dx^~' 4^
x2 —3x+6
W 2a;
2 '
1
/tt s
;
6£— 17a
60
'

. 2a2 +25 2 /onJ^ / n 1

(8 10

«±*. ~ 2 1N 2 *S
(11.) (12.) (13.)
v
a
v
' cc(4cc — 1)
2
.
* v
cc
4
+ x2 + T

( M0 -27-^-
x (x2 —
2
1)
(15.) -V.
a— 2?/
(16.) ?|±|.
4
cc —
+% 4
n nx 2a(2x + l)

(19.) _—
(a>-l)(a>-2)(a + 2)'
/"^ —_ .
v V
(20.)
(a-6)(a-c)
"

(21.)- (22.) 0. (23.) 0.

(cc a) (x — Z>) (cc — c)
XLI.
3+£b *— 2
2a— x
a.) . (2.) (3.)
a2 "

a+ax— 3a— ax+x

(4-) (5.)
X X
2x*+x-l 2cc
2
+l
(6.) ' (7.)
x2 x2 •

ix s
-x +2x-3 2
1 + Sx
(8.) (9.) *
xs x— 1
a—a?+ab a + 3&
(10.) (11.) '

a—b a+6
4a2 +9x-4 x z —xy
(12.) " (13.)
x+3 x-y
 ANSWERS.

(14) l+t.
a
(15.) a-±. < 16 ->
BT L
(17.) 2a-l + 3 (18.) 1-1+2,.
x'

(19.) 1 + 3^+5 (20.) 5-

— 1'
03

2x
(21.) 4- (22.) ,-3 + 3£|.
1+x'
3- x-2 ^- ^t^T 5
(23.) (24.) .
sr+1

XLn.
4a
(1.) (2.) 1. (3.)
5x*

- a+b a 2 +«5 + 5 2
(4.) (5.) (6.) '

1-<b2 a 2 —ab + b 2
a?+ab+W (8^ (« + ^) («
2
+ a&+& ) 2

(7.) <
(a+by a?b\a-b)
1 (b " 1)2
(9.) 3
(10.)
V ;
aj +l" (a^-sc + l) 2

(11.) 2/a+za3+a)y. (12.) cc-a.

a4 -64
(13.) (14.) .
ab x-y
(15.) (16.) 1.
a(a2 -ab+b2)

(17.) _^L 0 (18.)

+
03 2/ (a-6) 2

XLIII.
4ax 2y
{ } (2) (3.)
3x 2y'
a—x a2 +62
(4) (5.) ' (6.)
a + x' a?—ab + b 2
a+x 2
(7.) (8.) *

a («-6) 2
 . . a .

ANSWERS, 155

x 2 + 2xy—y 2
(10.) (11.)
x*+y 2
(13.) If. (14) A-
YT TT7"
A1j±V.

(1.) 10. (2.) 8. (3.) 12.

(4.) 6. (5.) 8. (6.) 2.

(7.) -f. (8.) -1. (9.) -f.

(10.) 20. (11) v°- (12.) 15.

(13-)ff. (14) -f. (15.) 3.

,1 rr \ die
(16.) f (17.) 5;
b—
.

(19.) a+b. (20.) a.

ft2+
a+bf
(22.)
v J

XLV.
(1.) 40. (2.) 40. (3.) 355.

(4.) 4 bowled, 3 run out. (5.) 9^. (6.) £5.

(7.) £9150 of 3 per cents., £5820 of 3i per cents.

(8.) A £1250, B £1500. (9.) 100.

(10.) 7 miles an hoiir. (11.) 45 and 30 miles an hour.

(12.) C 42 days, B 84 days, A 168 days.
(13.) 240, 180, 144 days. (14.) 5^- minutes past 7.

(15.) 26^ minutes past 2. (16.) 49T1T minutes past 3.

(17.) 32-^ minutes past 3. (18.) 10 gallons.

(19.) 30 gallons. (20.) 112 oz.

(21.) 70 grains. (22.) 21 oz.

(23.) 3 miles an hour. (24.) U miles.
(25.) 10 yards in 200, (26.) 1430 yards.
(27.) 1 mile ; half-an-hour. (28.) 5f
(29.) ^ of a mile behind.
156 ANSWERS.

XLVI.
(1.) 6, -6. (2.) 3, -3. (3.) 9, -9.
(4.) 4, -4. (5.) 2, -2. (6.) V26, -V26.
(7.) 0, 3. (8.) 0, -12. (9.) 0, 19.

(10.) 0, t (11.) 0, -A- (12.) 0, -f.

(13.) 3, 5. (14) -5, 7. (15.) -1, -3.
(16.) *,f. (17.) I, (18.) -I, -h
(19.) *, -1 (20.) 0, a.

XL VII.
a.) % 4, (2.) 5, -1. (3.) 3, -7.
(4.) 2, J. (5.) i, -2. (6.) i, *.

(7.) f, -f. (8.) i, f. (9.) 6, -4.

(10.) |, f (11.) %h (12.) « -10.
(13.) 2, -A- (14.) 14, -10. (15.) 4, -1.
> s.
(16.) -1, -f. (17.) 4,4. (18.) 1, 5

(19.) 1,2. (20.) 3, H

(21.) 6,f. (22.) a, I.

XLYIII.

(1.) 60 ft. by 30 ft. (2.) 13 and 7.

(3.) 2 or 10. (4.) 12 and 3.

(5.) 4 or 6. (6.) 5, 4, 3 ; or -1, 0, 1.

(7.) 30 ft. (8.) £20. (9.) £90 or £10.

(10.) A's rate 4 miles ; B's 2 3 miles an hour.
(11.) f. (12.) 10. (13.) 87.

(14.) 2 ft. (15.) 11. (16.) 23.

(17.) 12. (18.) 56. (19.) 1296.

 a

ANSWERS.

XLIX.

ao 2,3. (2.) 3,5. (3.) 5, 4

(4) 6, 12. (5.) 30, 12. (6.) 4 3.

(7.) 4 -1. (8.) 3, 5. (9.) 2, 3.

(10.) 2, 3. (11.) 9, 4. (12.) 60, 40.

(13.) 13,3. (14) 4 1.

a ah— he
W ~2-
(16.) a, I. (17.) |=2-c ,
a—b

L.

(1.) 2, 1, 3. (2.) 1,3,2. (3.) 5,4,3.

(4.) 3, 4, -3. (5.) 3, 2, 1. (6.) 4, 5, 6.

(7.) 4,3,2. (8.) 19,7,4. (9.) 2,4,7.

b + c— c+a— b a+b—c
(10.)
2

LI.

(1.) 72. (2.) 378 and 216.

(3.) 5 and 4. (4.) V3 ' (5.) 84.

(6.) A to £ 37 miles, B to C 45 miles, C to A 52 miles.

(7.) 19 five-franc pieces, 11 two-franc pieces.
(8.) 296 for A, 1488 for B, 196 for C.
(9.) The stream 3 miles an hour; the boat 8 miles an
hour.
(10.) 148 for A, 750 for C and A, 158 for A and B.
(11.) The gold coins are half-sovereigns, the silver coins
are crowns.
(12.) A to B 11£ miles, B to C 7 miles, C to D 5| miles.
(13.) 24 octavos or 32 duodecimos.
(14.) A thaler = 2s. 11c?. ; a franc = %d. ; a florin =
Is. 114c?.
 .

i 58 ANSWERS.

(15.) A Prussian pound = 16£ oz. ; an Austrian pound

191 oz. ; a kilogramme = 35£ oz.

(16.) 3h miles.
LII.
1 1
(1.) J* Ha, tf* Ja\ 9 (2.)
\h
1

—
tKi \Aj

(3.) ^7=, -7==, (^) 2#a, a,

-1
(5.) cc
2",
(gi) a; ,
cr 2 , a' 5 , a~ 8 .
3. 4- _1 _i _JL
(7.) cc
2
,
cc 3
,
x5 . (8.) x 2
, x 3
, x 5 .

(9.) 2m" 1
, dn~ 2
, lOp" 3
. (10.) 2af 2 , 5af * 7afl

MIL
(1.) to"* 1
2wj-l
;
4cc
w +2
3n
;
4cc
5re
3ttl
. (2.) 2.^; ; 30 A
(3.) x 2
; 6aT* ;
rc
T . (4.) 2a; 3a" 1 ; 30a" 1 .

1 __1 2 J> « 5n
a 6 2a 12 a 3. a 6
a 2
a 8
(5.) ; ; (6.) ; ; .

(7.) 1; 1; 6; ron.

LIY.

(1.) aM ;
a251 . (2.) at ; at (3.) ^. ^
3 5 3
(4.) cc ;
cc . (5.) x; cc .
(6.) cc; a?f.
Sn
2
(7.) cc»; SB .

LY.
(1.) a 12
;
a 8
;
a 9
. (2.) a~ 2 ; a" 6 ; a" 12 .

6 6 12 9 10
(3.) a" ;
a ;
a . (4) a 2 ;
a ; a .