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a+b

- b Ex. 7. Add and

together.
a-6
atb

2aa +262 Ans.

2-62 a ta Ex. 8. Add 2, and

together. at

40x Ans.

aa 302 2

2x --3 Ex. 9. Add 20+ and 3x +

together. 3

4

10x -17 Ans. 5x +

12 72 Es. 10. Add 4.0, ģ and 2+together.

44% Ans. 4x4-2+

45 20

50 Ex. 11. Add 52 and ---4x together. 7 9

17 Ans. +

63 Ex. 12. It is required to find the sum of 2a,

X2 and

Ans. 2a +2+

a2

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To subtract one fractional quantity from another.

RULE.

155. Reduce the fractions to a common denominator, if necessary, and then subtract the numerators from each other, and under the difference write the common denominator, and it will give the difference of the fractions required.

с

a

-(

-(-)

с

с

с

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-(

v2 +)=+

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9

axb

Or, enclose the fractional quantity to be subtracted in a parenthesis; then, prefixing the negative sign, and performing the operation, observing the same remarks and rules as in addition, the result will be the difference required.

The reason of this is evident; because, adding a negative quantity is equivalent to subtracting a positive one (Art. 63): thus, prefixing the negative

a-b sign to the fractional quantity

it becomes -b b

'; to the fractional quan+

a ? ta tity

it becomes y

y

y

-b (Art. 128); to the fractional quantity

it

5 ax

; to the mixed 5

5 3a +b

3a +b quantity 5x

Y 3a+b -5x + ; and to the mixed quantity - 30+

Y 2

2

2-1 it becomes

=3a

с 2 =3a+ с

3x 520 Ex. 1. Subtract from

5 Here 30 X 7=21x)

25x - 212 numerators, 53 X5=25x

35

4.2 5X7=35 com, denom.

is the differ

35 ence required.

becomes -(
-(-43-4)

:)
it becomes - (3:37)

"

-(-3a+:-)

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3b

2a-428 1-Y Ex. 2. Subtract from

50 Here (2a --4x) x 3b=6ab-12bx

numerators. (-y) x 5c= 50. - 5cy S

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50 X 36=15bc common denominator.
5cx 5cy 6ab-12bx

5cx 5cy Whence,

+
15bc
15bc

15bc 12bx--6ab_5cx-5cy +126x —Gab

is the difference 15bc

15bc required.

Or, by prefixing the negative sign to the quantity 20-40

2a-4

42-2a it becomes

; then it only 50

50

50 4xremains to add

and

together, as in ad50

36 dition, and the result will be the same as above.

aX

(-X Ex. 3. From 2ab +

subtract 2ab a to

at Here prefixing the negative sign to the quantity 2ab

we have - 2abatæ

hence the difference of the proposed fracat tions is equivalent to the sum of 2ab + and

at - 2ab to

; but the sum of the fractional parts a

at 2a2 +- 2x2 and is

: Therefore the differatoa

ad X2
2aa -+- 2002

2a +-2.1 ence required is 2ab-2ab+

al 22 ad X2

a

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- (2aba * )=–2ab+

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a

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atx

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a

[graphic]

15

10x-_-9

38-5 Ex. 4. From

subtract

7 Here (10x –9) X 7=70x—637

numerators. (3x ---5) X 15=45x—75

15X7=105 common denominator.

70x - 63 450-75 Therefore,

105

105 70-63-45%-+-75 25x+12

is the fraction requi105

105 red.

4ah Ex. 5. From subtract

Ans. a+b

a?-62 1 1

20 Ex. 6. From subtract

Ans. at

a? - 22. 4x+2

23 -- 3

4x2 +3 Ex. 7. From subtract Ans. 3

3x

a+

a-6

a

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Ex. 8. From 3x+; subtract x

.

с

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b

cx-t-bx - ab Ans. 2x+

bc 2.0 +7 30? ta? Ex. 9. Subtract from

8

36 24x2 +8a2-6bx - 216 Ans.

246 2x - 3

-2 Ex. 10. Subtract 4x

from 50+ 5

3

11-19 Ans. **

15 a tox Ex. 11. Subtract

from atala

a(at)

4% Ans. a

a ??

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a

Ex. 12. Required the difference of 3r and 3a +12x

3х — за

Ans. 5

5 50-2

4x +5 Ex. 13. From 2x -+- subtract 3x

7

6

168 +23 Ans.

42

3 VI. MULTIPLICATION AND DIVISION OF ALGE

BRAIC FRACTIONS.

To multiply fractional quantities logether.

RULE.

156. Multiply their numerators together for a new numerator, and their denominators together for a new denominator; reduce the resulting fraction to its lowest terms, and it will be the product of the fractions required.

It has been already observed, (Art. 119), that when a fraction is to be multiplied by a whole quantity, the numerator is multiplied by that quantity, and the denominator is retained :

100 Thus, 6 xc="C, and ** 5=%"; or, which is =

X5 b b

b the same, making an improper fraction of the integral quantity, and then proceeding according to the

2x 5 10. rule, we have xi b b'

b. Hence, if a fraction be multiplied by its denominator, the product is the numerator; thus, g xb=

a

200

a

с

ac

and 7 *

1

1

a

a

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