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 2 - Equations, Inequalities, and Problem Solving - 2.1 Solving Equations - 2.1 Exercise Set - Page 87: 58



Work Step by Step

Multiplying both sides by $ \dfrac{4}{3} $, then the solution to the given equation, $ \dfrac{3}{4}x=27 $, is \begin{array}{l}\require{cancel} \dfrac{4}{3}\cdot\dfrac{3}{4}x=27\cdot\dfrac{4}{3} \\\\ \dfrac{\cancel{4}}{\cancel{3}}\cdot\dfrac{\cancel{3}}{\cancel{4}}x=\cancel{3}\cdot9\cdot\dfrac{4}{\cancel{3}} \\\\ x=36 .\end{array}
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