Elementary Algebra

Published by Cengage Learning
ISBN 10: 1285194055
ISBN 13: 978-1-28519-405-9

Chapter 9 - Roots and Radicals - Chapters 1-9 Cumulative Review Problem Set - Page 432: 12



Work Step by Step

Using the $LCD= (x-2)(x+3) $, the given expression, $ \dfrac{3}{x-2}-\dfrac{4}{x+3} ,$ simplifies to \begin{array}{l}\require{cancel} \dfrac{(x+3)(3)-(x-2)(4)}{(x-2)(x+3)} \\\\= \dfrac{3x+9-4x+8}{(x-2)(x+3)} \\\\= \dfrac{-x+17}{(x-2)(x+3)} .\end{array}
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