Intermediate Algebra (6th Edition)

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

Chapter 10 - Cumulative Review - Page 634: 32

Answer

$x=4$

Work Step by Step

The values of $x$ that satisfy the equation $ \dfrac{4}{x-2}-\dfrac{x}{x+2}=\dfrac{16}{x^2-4} $ are \begin{array}{l} \dfrac{4}{x-2}-\dfrac{x}{x+2}=\dfrac{16}{(x+2)(x-2)} \\\\ (x+2)(x-2)\left( \dfrac{4}{x-2}-\dfrac{x}{x+2} \right)=\left( \dfrac{16}{(x+2)(x-2)} \right)(x+2)(x-2) \\\\ (x+2)(4)-(x-2)(x)=16(1) \\\\ 4x+8-(x^2-2x)=16 \\\\ 4x+8-x^2+2x=16 \\\\ -x^2+6x-8=0 \\\\ x^2-6x+8=0 \\\\ (x-4)(x-2)=0 \\\\ x=\{ 2,4 \} .\end{array} Upon checking, only $ x=4 $ satisfies the given equation.
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