Intermediate Algebra (6th Edition)

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

Chapter 6 - Section 6.2 - Adding and Subtracting Rational Expressions - Exercise Set - Page 355: 80

Answer

$\dfrac{x-1}{4x}$

Work Step by Step

The given expression, $ \dfrac{x}{2x+2}\div\left( \dfrac{x}{x+1}+\dfrac{x}{x-1} \right) ,$ simplifies to \begin{array}{l}\require{cancel} \dfrac{x}{2(x+1)}\div\dfrac{(x-1)(x)+(x+1)(x)}{(x+1)(x-1)} \\\\= \dfrac{x}{2(x+1)}\div\dfrac{x^2-x+x^2+x}{(x+1)(x-1)} \\\\= \dfrac{x}{2(x+1)}\div\dfrac{2x^2}{(x+1)(x-1)} \\\\= \dfrac{x}{2(x+1)}\cdot\dfrac{(x+1)(x-1)}{2x^2} \\\\= \dfrac{\cancel{x}}{2(\cancel{x+1})}\cdot\dfrac{(\cancel{x+1})(x-1)}{2\cancel{x}\cdot x} \\\\= \dfrac{x-1}{4x} .\end{array}
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