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atb

and a,

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

by a+b, we have the fraction a3 + 2ab + 2ab +63

5a: +10a2b +- 5abo in its lowest terms; that is, =5a +-5ab; +-2a 6 + 2ab? +63

=atab+62 : atb

5a+-5ab Hence,

is the fraction in its lowest a' tab +62 terms. Ex. 7. Reduce 14x2y3 -- 21x y?

to its lowest 70%y

2y3xy terms.

Ans. 5113 – 1722 +34.0 Ex. 8. Reduce

to its lowest 17:05

332 terms.

Ans.

24 - 6 Ex. 9. Reduce

to its lowest terms, -63

1 Ans.

a2 + ab +-62

to taxa ta Ex. 10. Reduce

to its lowest 254 tax3-a

22 -- axta terms.

Ans.

22 -a?

7a---23ab +662 Ex. 11. Reduce

to its 5a-18a2b+11ab2-663

7a-26 lowest terms.

Ans.

5a? --3ab+-26

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a

3

Ex. 12. Reduce

to its lowest terms, ao bo

a+62 Ans.

a' taoba +64

ya-X Ex. 13. Reduce

to its lowest yoy2x−yx2 + x3

y2 + x2 Cerms.

Ans.

y-3 al-20'6+-2ab-63 Ex. 14. Reduce

to its lowa' ta'63 +64

a-6

3

est terms.

Ans.

a2 + ab +63 Q3.

-3a2x + 3axa -1 Ex, 15. Reduce

to its low. a_22

a? — 2ax to me est terms.

Ans.

ato a +2ba36a Ex. 16. Reduce

to its lowest 2a4-3ba3--562 a

Q-26 + 362 terms.

Ans.

2a2---3ba-562 6x5 +15ax 4c23 -10ac2x3 Ex. 17. Reduce 9ax3-27acx -6ac2x +18ac

2x3 + 5ax to its lowest terms.

Ans.

3r-9ac

a-2axx2 Ex. 18. Reduce

to its lowest al-a--ax+23

1 terms.

Ans.

ata

* Note.The answer to Ex. 11, page 100, is misprinted, for a -», read a? —2ar+r.

ai-624 Ex. 19. Reduce

to its lowest terms. a + 2ab +62

@_ab Ans.

atb

CASE IV.

To reduce fractions to other equivalent ones, that shall

hare 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.

3a

56 Ex. 1. Reduce

and – to a common deno

4' minator.

3a XcXa=3ac
56 X 4 Xa=20ab new numerators;

3 XCX4=4cx

с

a

4Xc Xa=4ac common denominator;

3ac 20ab Hence the fractions required are

and

4ac' ac' 4cx 4ac

2x +1

2a2 Ex. 2. Reduce

and

to a common

denominator.

(2x + 1) Xx=2x+x? 2a2 X 36=6a2 b )

new numerator :

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36 Xx=3b2 common denominator;

2x2 te Ilence the fractions required are

and

3be 62' 3bx 3 50

3r? Es. 3. Reduce

and a-t- to a com4' 3

5 mon denominator.

302 5a +- 3r TIere at

5

5 3X3 X5=45

5x X4 X5=1000 new numerator; (5a +32°) X4X3=60a+36x2

and

4X3 X5=60 common denominator ;

45 100x llence the fractions required are

60 60 B0a + 36 xa

60

2

RULE. II.

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.

to the least com

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3a2b 5ab Ex. 4. Reduce and

4a x2 mon denominator.

Here 4axa is the least common multiple of zi? and 40x2 ;

4αα2 then X 3a'b=4a x 3a'b=12a3b

new numera4ax2

rators. and x 5ab=5ab 4ax?

12a36 5ab Hence

and

are the fractions required. 4ax 2 - 4ax Or, as 4ax? (the least common multiple) is the denominator of one of the fractions, it is only ne

3a2 b cessary to reduce the fraction

to an equivalent

22 one whose denominator shall be 1ax; hence, 4αα2

3a2b

4a 3a2b X 4a 12a3b

X 22

is the X2 40 202 + 4a 4axa fraction required.

These rules appear evident from (Art. 118). For, let ūū f?

be the proposed fractions ; then adfcbf edb

are fractions of the same bdfbdfbdf value with the former, having the common denomiadf _ a cbf

edb nator bdf. Since

; and bdfbdf

bdff

-=4a, and

a

с

e

.

с

e

bdf

and

За?ь

y Ex. 5. Reduce

40x3 2x common denominator.

522
8ac?

to the least

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