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: 35



Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $ F=\dfrac{9}{5}C+32 $ for $ C ,$ 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} F=\dfrac{9}{5}C+32 \\\\ F-32=\dfrac{9}{5}C \\\\ 5(F-32)=5\left( \dfrac{9}{5}C \right) \\\\ 5(F-32)=9C \\\\ \dfrac{5(F-32)}{9}=\dfrac{9C}{9} \\\\ \dfrac{5}{9}(F-32)=C \\\\ C=\dfrac{5}{9}(F-32) .\end{array}
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