The problem asks us to simplify the expression $\frac{4y}{\sqrt{3} + \sqrt{2}}$ by rationalizing the denominator.

1. Problem Description

The problem asks us to simplify the expression $\frac{4y}{\sqrt{3} + \sqrt{2}}$ by rationalizing the denominator.

2. Solution Steps

To rationalize the denominator, we multiply both the numerator and the denominator by the conjugate of the denominator. The conjugate of $\sqrt{3} + \sqrt{2}$ is $\sqrt{3} - \sqrt{2}$.

$\frac{4y}{\sqrt{3} + \sqrt{2}} = \frac{4y}{\sqrt{3} + \sqrt{2}} \cdot \frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} - \sqrt{2}}$

$= \frac{4y(\sqrt{3} - \sqrt{2})}{(\sqrt{3} + \sqrt{2})(\sqrt{3} - \sqrt{2})}$

The denominator is now in the form $(a+b)(a-b)$, which simplifies to $a^2 - b^2$. Therefore, the denominator becomes $(\sqrt{3})^2 - (\sqrt{2})^2 = 3 - 2 = 1$.

$\frac{4y(\sqrt{3} - \sqrt{2})}{1} = 4y(\sqrt{3} - \sqrt{2})$

$= 4y\sqrt{3} - 4y\sqrt{2}$

3. Final Answer

$4y\sqrt{3} - 4y\sqrt{2}$