## Basic College Mathematics (10th Edition)

$\frac{4}{45}$
$\frac{2}{3}(\frac{4}{5}-\frac{2}{3})$ (First solve Bracket) = $\frac{2}{3}(\frac{4\times3}{5\times3}-\frac{2\times5}{3\times5})$ (LCM of 5 & 3 is 15) = $\frac{2}{3}(\frac{12}{15}-\frac{10}{15})$ = $\frac{2}{3}(\frac{12-10}{15})$ (subtraction of fractions with same denomiator) = $\frac{2}{3}(\frac{2}{15})$ = $\frac{2}{3}\times\frac{2}{15}$ = $\frac{4}{45}$