## Prealgebra (7th Edition)

$-\frac{2}{9}$
When $x=-\frac{1}{3}$, y=$\frac{2}{5}$,and $z=\frac{5}{6}$ $x^{2}-yz$ $=(-\frac{1}{3})^{2}-(\frac{2}{5}\times\frac{5}{6})$ $=[(-\frac{1}{3})\times(-\frac{1}{3})]-(\frac{1}{1}\times\frac{1}{3})$ $=(\frac{1}{3}\times\frac{1}{3})-\frac{1}{3}$ $=\frac{1}{9}-\frac{1}{3}$ $=\frac{1}{9}-\frac{3}{9}$ $=-\frac{2}{9}$