## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

$C=\dfrac{5}{9}(F-32)$
$\bf{\text{Solution Outline:}}$ 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}