Squaring the equation, we get $(\sqrt{a}+\sqrt{b})^2 = 2 + \sqrt{3}$. And expanding the LHS, we get $a + b + 2\sqrt{ab} = 2 + \sqrt{3}$. Comparing rational and irrational parts, we can see $a + b = 2$ and $2\sqrt{ab} = \sqrt{3}$. From the second equation, divide by $2$ and square: $ab = \frac{3}{4}$. We can solve for $x$ and $y$ by treating them as the roots of a quadratic equation. By Vieta's Formulas, $t^2 - (x + y)t + xy = 0$. Substituting: $t ^ 2 - 2t + \frac{3}{4} = 0$. And solving we get $t = \frac{3}{2}$ and $t = \frac{1}{2}$. So $a = \frac{3}{2}$ and $b = \frac{1}{2}$.