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: 65



Work Step by Step

Multiplying both sides by $ -\dfrac{2}{3} $, then the solution to the given equation, $ \dfrac{-3r}{2}=-\dfrac{27}{4} $, is \begin{array}{l}\require{cancel} \left(-\dfrac{2}{3}\right)\left(\dfrac{-3r}{2}\right)=\left(-\dfrac{27}{4}\right)\left(-\dfrac{2}{3}\right) \\\\ \dfrac{6r}{6}=\dfrac{54}{12} \\\\ r=\dfrac{\cancel{6}\cdot9}{\cancel{6}\cdot2} \\\\ r=\dfrac{9}{2} .\end{array}
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