Math 017 Materials With Exercises

c) Numerator: one term, 3ab (with three factors 3, a and b). Denominator: two terms, ab (with factors a and b) and ─ 22(with factors ─1 and a ). ALL terms in the numerator AND in the denominator have a common factor, a. We cantherefore divide both the numerator AND the denominator by a.20 x is NOT a factor in the denominator, but it can be viewed as a factor in the numerator. We can NOT cancel x, because itis not a factor in the denominator.217x IS a factor in BOTH the numerator and the denominator, therefore we can cancel x. The result isxy22 a IS a factor in BOTH the numerator and denominator, thus we can cancel it. The result: a x .23 ab IS a factor in the denominator, but NOT in the denominator, therefore we cannot cancel it.1 c24 a) ; 3xy b)3 4f)25 a)h)m)Lesson 6a( b c)21x y132 x1y x3; a g) 20x2 ;b)y z3; 2ab c)12b;2a d)xy 3 ; 5y e)43(a b)51 x h) c( b e); b i) ; ( a b)j) not possible7c) 4 5xd)i) not possible j)n) 111y1x2 o) 2 3u1e)v p)113xf); 5 x2 4 3a b 4ba g) 1k) not possible l) x 3y 2z13x 3 6xyza1 b) and d)2 Yes.3 No. (All three are unlike)42 26x y 2x 2 y5 xy0.3yx2563 ab225(ab ) 2 a 22b 77y 3 x 2 xy 3 x 2x 3 y2 y 2 xxbayyxxy78 a) x2 5 3a b a b 2 a5 b3 3( ab ) b3 22(ab ) a2 b) not possible c) 0 d) not possible e) st or 2tsh) 6 hmv i)3 z3 5xy j) m 3 n9 k) not possible l)a3 b 52 f)2 22ab9 a) ( 3 4)x xb) 1 2 1 x x c) 2 3 1 x x 3 7 21 1122 22d) ( 0.3 0.5) x 0.2x 7 2 53 1 2 3 1e) x x x 9 5 45 5 3 10 610 a) x b) 5ac) x d) x e) 7ab11 a) 3x 8x 2x 8y 13x 8yb) 5a 7a 3b 2b 12a 5bc) 2ab 4ba 3ab 2 1 3ab1 f) x f) not possible g) not possible h) 1.2x 2 y22ac g) not possible124Answers to Exercises