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

$r=\dfrac{9}{2}$
Multiplying both sides by $-\dfrac{2}{3}$, then the solution to the given equation, $\dfrac{-3r}{2}=-\dfrac{27}{4}$, is \begin{array}{l}\require{cancel} \left(-\dfrac{2}{3}\right)\left(\dfrac{-3r}{2}\right)=\left(-\dfrac{27}{4}\right)\left(-\dfrac{2}{3}\right) \\\\ \dfrac{6r}{6}=\dfrac{54}{12} \\\\ r=\dfrac{\cancel{6}\cdot9}{\cancel{6}\cdot2} \\\\ r=\dfrac{9}{2} .\end{array}