Intermediate Algebra (6th Edition)

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

Chapter 6 - Review - Page 404: 65

Answer

$x=\dfrac{3}{2}$

Work Step by Step

Factoring the given expression, $ \dfrac{1}{x-2}-\dfrac{3x}{x^2-4}=\dfrac{2}{x+2} ,$ results to \begin{array}{l}\require{cancel} \dfrac{1}{x-2}-\dfrac{3x}{(x+2)(x-2)}=\dfrac{2}{x+2} .\end{array} Multiplying both sides by the $LCD= (x+2)(x-2) $, the value of the variable that satisfies the equation above, $ \dfrac{1}{x-2}-\dfrac{3x}{(x+2)(x-2)}=\dfrac{2}{x+2} ,$ is \begin{array}{l}\require{cancel} (x+2)(1)-1(3x)=(x-2)(2) \\\\ x+2-3x=2x-4 \\\\ x-3x-2x=-4-2 \\\\ -4x=-6 \\\\ x=\dfrac{-6}{-4} \\\\ x=\dfrac{3}{2} .\end{array}
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