Intermediate Algebra (6th Edition)

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

Chapter 9 - Cumulative Review: 15

Answer

$x=\left\{ -1,\dfrac{1}{2} \right\}$

Work Step by Step

The expression $ \dfrac{2x}{2x-1}+\dfrac{1}{x}=\dfrac{1}{2x-1} $ simplifies to \begin{array}{l} x(2x-1)\left( \dfrac{2x}{2x-1}+\dfrac{1}{x} \right)=\left(\dfrac{1}{2x-1}\right)x(2x-1) \\\\ x(2x)+1(2x-1)=1(x) \\\\ 2x^2+2x-1=x \\\\ 2x^2+x-1=0 \\\\ (2x-1)(x+1)=0 \\\\ x=\left\{ -1,\dfrac{1}{2} \right\} .\end{array}
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