Intermediate Algebra (6th Edition)

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

Chapter 6 - Review - Page 404: 31

Answer

$\dfrac{-7x-6}{5(x-3)(x+3)}$

Work Step by Step

Factoring the given expression, $ \dfrac{x}{9-x^2}-\dfrac{2}{5x-15} ,$ results to \begin{array}{l}\require{cancel} \dfrac{x}{-(x^2-9)}-\dfrac{2}{5x-15} \\\\= \dfrac{x}{-(x-3)(x+3)}-\dfrac{2}{5(x-3)} \\\\= \dfrac{-x}{(x-3)(x+3)}-\dfrac{2}{5(x-3)} .\end{array} Using the $LCD= 5(x-3)(x+3) $, the expression above simplifies to \begin{array}{l}\require{cancel} \dfrac{5(-x)-(x+3)(2)}{5(x-3)(x+3)} \\\\= \dfrac{-5x-2x-6}{5(x-3)(x+3)} \\\\= \dfrac{-7x-6}{5(x-3)(x+3)} .\end{array}
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