atb and a, 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 2y—3xy 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 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 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 : 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 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 to the least |