Intermediate Algebra (6th Edition)

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

Chapter 6 - Test - Page 406: 3



Work Step by Step

The given expression, $ \dfrac{7x-21}{24-8x} ,$ simplifies to \begin{array}{l}\require{cancel} \dfrac{7(x-3)}{8(3-x)} \\\\= \dfrac{7(x-3)}{-8(x-3)} \\\\= \dfrac{7(\cancel{x-3})}{-8(\cancel{x-3})} \\\\= \dfrac{7}{-8} \\\\= -\dfrac{7}{8} .\end{array}
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