\(3\frac{1}{3}-2\frac{2}{5}\\\\find\ LCD(\frac{1}{3};\ \frac{2}{5})\\\\3\ and\ 5\ are\ the\ prime\ numbers,\ therefore\ LCD=3\cdot5=15\\\\\frac{1}{3}=\frac{1\cdot5}{3\cdot5}=\frac{5}{15};\ \frac{2}{5}=\frac{2\cdot3}{5\cdot3}=\frac{6}{15}\\\\3\frac{1}{3}-2\frac{2}{5}=3\frac{5}{15}-2\frac{6}{15}=2\frac{15+5}{15}-2\frac{6}{15}=2\frac{20}{15}-2\frac{6}{15}=\boxed{\frac{14}{15}}\)