ration is found to divide the numerator without a remainder; and consequently dividing both the numerator and denominator of the fraction 5a3+10a2b+5ab2 a3+2ab+2ab+b3

by a+b, we have the fraction

* NOTE.-The answer to Ex. 11, page 100, is misprinted, for a—x, read a2 —2ax + x2

To reduce fractions to other equivalent ones, that shall have a common denominator.

RULE I.

152. Multiply each of the numerators, separately, into all the denominators, except its own, for the new numerators, and all the denominators together for the common denominator.

It is necessary to remark, that, if there are whole or mixed quantities, they must be reduced to improper fractions, and then proceed according to the rule.

4XcXa=4ac common denominator;

Hence the fractions required are

3a2c

20ab

and

4ac 4ac

4cx

4ac

2x+1

2a2

Ex. 2. Reduce

and

to a common

denominator.

(2x+1)Xx=2x2+x) 2a2×3b=6a2b S

new numerator;

3b Xx=3bx common denominator;

Hence the fractions required are

2x2+x

and

, 36x

153. Find the least common multiple of all the denominators of the given fractions, (Art. 147), and it will be the common denominator required.

Divide the denominator by the denominator of each fraction, separately, and multiply the quotient by the respective numerators, and the products will be the numerators of the fractions required.

Or, as 4ax (the least common multiple) is the denominator of one of the fractions, it is only ne

3a2b

cessary to reduce the fraction to an equivalent

x2

fraction required.

These rules appear evident from (Art. 118).

For, let

α C "e

be the proposed fractions;

then

b' ď ƒ' adf cbf edb

bdf' bdf bdf'

are fractions of the same

value with the former, having the common denomi

adf_a cbf C
; and
bdƒ b' bdf d

3a2b y

4cx3 2x

common denominator.

and

edb e

bdf f