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 - Test: Chapter 2 - Page 151: 31



Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $ c=\dfrac{2cd}{a-d}, $ for $ d ,$ use the properties of equality to isolate the variable. $\bf{\text{Solution Details:}}$ Using the properties of equality to isolate the variable, the equation above is equivalent to \begin{array}{l}\require{cancel} c=\dfrac{2cd}{a-d} \\\\ (a-d)\cdot c=(a-d)\cdot \dfrac{2cd}{a-d} \\\\ (a-d)c=2cd \\\\ ac-dc=2cd \\\\ -dc-2cd=-ac \\\\ d(-c-2c)=-ac \\\\ d=\dfrac{-ac}{-c-2c} \\\\ d=\dfrac{-ac}{-3c} \\\\ d=\dfrac{\cancel-a\cancel c}{\cancel-3\cancel c} \\\\ d=\dfrac{a}{3} .\end{array}
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