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 - 2.3 Formulas - 2.3 Exercise Set - Page 101: 55



Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $ R=r+\dfrac{400(W-L)}{N} $ for $ L ,$ 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} R=r+\dfrac{400(W-L)}{N} \\\\ N(R)=N\left( r+\dfrac{400(W-L)}{N} \right) \\\\ NR=Nr+400(W-L) \\\\ NR-Nr=400(W-L) \\\\ \dfrac{NR-Nr}{400}=\dfrac{400(W-L)}{400} \\\\ \dfrac{NR-Nr}{400}=W-L \\\\ L=W-\dfrac{NR-Nr}{400} .\end{array}
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