jmc

By AM-HM, $$\frac{2x+3y}{2}\ge \frac{2}{\frac{1}{2x}+\frac{1}{3y}}.$$Then $$\frac{1}{2x}+\frac{1}{3y}\ge \frac{4}{2x+3y}=\frac{4}{5}.$$Multiplying both sides by 6, we get $$\frac{3}{x}+\frac{2}{y}\ge \frac{24}{5}.$$Equality occurs when $2x=3y=\frac{5}{2},$ or $x=\frac{5}{4}$ and $y=\frac{5}{6},$ so the minimum value is $\boxed{\frac{24}{5}}.$