Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 6 - Cumulative Review - Page 408: 48



Work Step by Step

The factored form of the given expression, $ \dfrac{5a}{a^2-4}-\dfrac{3}{2-a} ,$ is \begin{array}{l}\require{cancel} \dfrac{5a}{a^2-4}-\dfrac{3}{-(a-2)} \\\\= \dfrac{5a}{a^2-4}+\dfrac{3}{a-2} \\\\= \dfrac{5a}{(a+2)(a-2)}+\dfrac{3}{a-2} .\end{array} Using the $LCD= (a+2)(a-2) ,$ then the given expression simplifies to \begin{array}{l}\require{cancel} \dfrac{1(5a)+(a+2)(3)}{(a+2)(a-2)} \\\\= \dfrac{5a+3a+6}{(a+2)(a-2)} \\\\= \dfrac{8a+6}{(a+2)(a-2)} \\\\= \dfrac{2(4a+3)}{(a+2)(a-2)} .\end{array}
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