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

$d=\dfrac{a}{3}$
$\bf{\text{Solution Outline:}}$ 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}