Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter R - Elementary Algebra Review - R.2 Equations, Inequalities, and Problem Solving - R.2 Exercise Set - Page 947: 30



Work Step by Step

Multiplying both sides by the $LCD, 6 ,$ the value of the variable that satisfies the given expression, $ \dfrac{2}{3}\left( \dfrac{1}{2}-x \right)+\dfrac{5}{6}=\dfrac{3}{2}\left( \dfrac{2}{3}x+1 \right) ,$ is \begin{array}{l}\require{cancel} 2\cdot2\left( \dfrac{1}{2}-x \right)+1(5)=3\cdot3\left( \dfrac{2}{3}x+1 \right) \\\\ 4\left( \dfrac{1}{2}-x \right)+5=9\left( \dfrac{2}{3}x+1 \right) \\\\ \left( \dfrac{4}{2}-4x \right)+5=\dfrac{18}{3}x+9(1) \\\\ 2-4x+5=6x+9 \\\\ -4x-6x=9-2-5 \\\\ -10x=2 \\\\ x=\dfrac{2}{-10} \\\\ x=-\dfrac{1}{5} .\end{array}
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