Intermediate Algebra (6th Edition)

$\frac{7}{18}$
We are asked to evaluate the expression $\frac{y}{2x}-\frac{\sqrt x}{3y}$, where $x=9$ and $y=-2$ $\frac{-2}{2\times9}-\frac{\sqrt 9}{3\times-2}=\frac{-2}{18}-\frac{3}{-6}=-\frac{2}{18}-(-\frac{3}{6})=-\frac{2}{18}+\frac{3}{6}$ $=-\frac{2}{18}+(\frac{3}{6}\times\frac{3}{3})=-\frac{2}{18}+\frac{9}{18}=\frac{7}{18}$ We know that $\sqrt 9=3$ because $3^{2}=3\times3=9$