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

$L=W-\dfrac{NR-Nr}{400}$
$\bf{\text{Solution Outline:}}$ 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}