Intermediate Algebra (6th Edition)

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

Chapter 6 - Review - Page 403: 13

Answer

$-\dfrac{3}{2}$

Work Step by Step

Factoring the expressions and then cancelling the common factor/s between the numerator and the denominator, the given expression, $ \dfrac{4-x}{5}\cdot \dfrac{15}{2x-8} ,$ simplifies to \begin{array}{l}\require{cancel} \dfrac{-(x-4)}{5}\cdot \dfrac{5\cdot3}{2(x-4)} \\\\= \dfrac{-(\cancel{x-4})}{\cancel{5}}\cdot \dfrac{\cancel{5}\cdot3}{2(\cancel{x-4})} \\\\= \dfrac{-3}{2} \\\\= -\dfrac{3}{2} .\end{array}
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