Algebra: A Combined Approach (4th Edition)

Published by Pearson
ISBN 10: 0321726391
ISBN 13: 978-0-32172-639-1

Chapter 12 - Cumulative Review: 48

Answer

$a.) \ \dfrac{2a}{a-1}$ $b.) \ -\dfrac{3(a+6)}{4(a-3)}$ $c.) \ \dfrac{x+y}{(xy)^2}$

Work Step by Step

$a.) \ \dfrac{\dfrac{a}{5}}{\dfrac{a-1}{10}} = \dfrac{10a}{5(a-1)} = \dfrac{5(2a)}{5(a-1)} = \dfrac{2a}{a-1}$ $b.) \ \dfrac{\dfrac{3}{2+a}+\dfrac{6}{2-a}}{\dfrac{5}{a+2}-\dfrac{1}{a-2}} = \dfrac{\dfrac{3(2-a)+6(2+a)}{(2+a)(2-a)}}{\dfrac{5(a-2)-(a+2)}{(a+2)(a-2)}} = \dfrac{\dfrac{6-3a+12+6a}{2^2-a^2}}{\dfrac{5a-10-a-2}{a^2-2^2}} = \dfrac{\dfrac{18+3a}{4-a^2}}{\dfrac{4a-12}{a^2-4}} = \dfrac{\dfrac{18+3a}{-(a^2-4)}}{\dfrac{4a-12}{a^2-4}} = \dfrac{(a^2-4)(18+3a)}{-(a^2-4)(4a-12)} = -\dfrac{3a+18}{4a-12} = -\dfrac{3(a+6)}{4(a-3)}$ $c.) \ \dfrac{x^{-1}+y^{-1}}{xy} = \dfrac{\dfrac{1}{x}+\dfrac{1}{y}}{xy} = \dfrac{\dfrac{x+y}{xy}}{xy} = \dfrac{x+y}{(xy)^2}$
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