Intermediate Algebra (6th Edition)

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

Chapter 6 - Sections 6.1-6.5 - Integrated Review - Expressions and Equations Containing Rational Expressions - Page 380: 20

Answer

$\dfrac{-a-8}{4a(a-2)}$

Work Step by Step

Factoring the given expression, $ \dfrac{3}{4a-8}-\dfrac{a+2}{a^2-2a} ,$ results to \begin{array}{l}\require{cancel} \dfrac{3}{4(a-2)}-\dfrac{a+2}{a(a-2)} .\end{array} Using the $LCD= 4a(a-2) $, the expression above simplifies to \begin{array}{l} \dfrac{a(3)-4(a+2)}{4a(a-2)} \\\\= \dfrac{3a-4a-8}{4a(a-2)} \\\\= \dfrac{-a-8}{4a(a-2)} .\end{array}
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