Rationalize the denominator of $\frac{\sqrt{8}+\sqrt{3}}{\sqrt{2}+\sqrt{3}}$. Express your answer in simplest form.
Multiplying top and bottom by the conjugate, we have $\frac{\sqrt{8}+\sqrt{3}}{\sqrt{2}+\sqrt{3}} = \frac{(\sqrt{8}+\sqrt{3})(\sqrt{2}-\sqrt{3})}{(\sqrt{2}+\sqrt{3})(\sqrt{2}-\sqrt{3})}$. Simplifying, we obtain $\frac{\sqrt{16}-\sqrt{24}+\sqrt{6}-\sqrt{9}}{\sqrt{4}-\sqrt{9}} = \frac{1-\sqrt{6}}{-1} = \boxed{\sqrt{6}-1}$.