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atb

b
Ex. 7. Add and together.

a-b
a+b

2a +262 Ans.

(12-62 a-tEx. 8. Add 2, and

together. at

4ar Ans.

aa - 2

-3 Ex. 9. Add 2x + and 30+ together.

3

4

10x -17 Ans. 5x to

12
72
Ex. 10. Add 4x, and 2+= together.

9
5

442 Ans. 4.c +-2+

45

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20

50

Es. 11. Add 5w and -4x together. 7 9

17x Ans. x-to

63 Ex. 12. It is required to find the sum of 20, and

Ans. 2a +2+

2

a

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a

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ac

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To subtract one fractional quantity from another.

2

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.

a-6

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с

a-6

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-()

C

с

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 sign to the fractional quantity it becomes

b

; to the fractional quanrata

22 +

22 ta tity it becomes

:+ y

y

y

-6 (Art. 128); to the fractional quantity

5 ax-6

-b becomes

;

to the mixed

5 3a-t6

3a +6 quantity 5x

it becomes y

y 3a +6 -5x + ; and to the mixed quantity --- 30+

Y 2

2.

2it becomes

30 to

3a

ax

it

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-(

5

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C

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numerators, 50X5=25

35 4.2

is the differ35

5X7–35 com, denom. ence required.

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+

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50 X 3b=15bc common denominator. Whence,

5cr-5cy 6ab-12bx 5cx-5cy
15bc 15bc

15bc 12bx--6ab_5cx—-5cy +12bx —6ab

is the difference 15bc

15bc required.

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

2a-44 43 - 2a it becomes

;

then it only 50

50

5c 4x2a remains to add

and

2y 50

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

(-X
Ex. 3. From 2ab + subtract 2ab
ata

a+ Here prefixing the negative sign to the quantity 2ab

-2ab + a to?

ato ;

hence the difference of the proposed fraca+ tions is equivalent to the sum of 2ab+ä

and a to a

;
but the sum of the fractional parts

2aa +- 232
and is

: Therefore the differat

a -X2

222 +- 2002 2aa +-2.1 ence required is 2ab-2ab +

al

a2 x2

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we have – (2ab

a 7)

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a

C

-2ab to

a

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10x-9

32-5 Ex. 4. From

subtract 15

7 Here (10x-9) X 7=70x--637

numerators. (3x-5) X15=45x75

15 X7=105 common denominator.

70x - 63 45x 75 Therefore,

105

105 70x-63—45x +-75 25x+12.

is the fraction requi105

105 red.

a 6

4ah Ex. 5. From subtract

Ans. atb

-62 1 1

20 Ex. 6. From subtract

Ans. ata

22. 4x + 2

2x -- 3

4x +3 Ex, 7. From subtract . Ans. 3

3.0

a+6

a

2 a

a-X

X-a

с

Ex. 8. From 3x +7 subtract x

cx+-6x-ab Ans. 2+

bc 2x +7

3x +a? Ex. 9. Subtract from

8

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

246

3 Ex. 10. Subtract 4x

from 5.0+ 5

3

11x ----19 Ans. **

15 ata

al Ex. 11. Subtract

from at ala-2)

a(a + x)

4% Ans. a

a -32

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X-2

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

3x - 3a

Ans. 5

5 57 -2

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

7

6

168 +23 Ans.

42

3 VI. MULTIPLICATION AND

DIVISION OF ALGEBRAIC 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 :

Thus, xc="6, and *5=+*; or, which is the same, making an improper fraction of the integral quantity, and then proceeding according to the

200 5 10x rule, we havex

Х b b'

1 b Hence, if a fraction be multiplied by its denominator, the product is the numerator; thus, 7 xb=

a

C

ac

1

a

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