Intermediate Algebra (6th Edition)

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

Chapter 7 - Cumulative Review - Page 473: 34b

Answer

$\dfrac{3x^2-21x}{(x+3)(x-3)}$

Work Step by Step

Using the $LCD= (x+3)(x-3) ,$ the given expression, $ \dfrac{5x}{x+3}-\dfrac{2x}{x-3} ,$ simplifies to \begin{array}{l}\require{cancel} \dfrac{(x-3)(5x)-(x+3)(2x)}{(x+3)(x-3)} \\\\= \dfrac{5x^2-15x-2x^2-6x}{(x+3)(x-3)} \\\\= \dfrac{3x^2-21x}{(x+3)(x-3)} .\end{array}
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