2B. Solve the system of equations

$$
\left\{\begin{array}{l}
\sqrt{x+y}-\sqrt{x-y}=\sqrt{y} \\
x^{2}-y^{2}=9
\end{array}\right.
$$
Solution. The domain of the system is the set of all pairs $(x, y)$ for which $x \geq y \geq 0$. We square the first equation and obtain the equivalent system

$$
\left\{\begin{array}{l}
x+y-2 \sqrt{x+y} \sqrt{x-y}+x-y=y \\
x^{2}-y^{2}=9
\end{array}\right.
$$

or equivalently

$$
\left\{\begin{array}{l}
2 x-2 \sqrt{(x+y)(x-y)}=y \\
x^{2}-y^{2}=9
\end{array}\right.
$$

By substituting the second equation into the first, we get

$$
\left\{\begin{array}{l}
2 x-2 \cdot 3=y \\
x^{2}-y^{2}=9
\end{array}\right.
$$

From this, we have

$$
\left\{\begin{array}{l}
2 x-2 \cdot 3=y \\
x^{2}-(2 x-6)^{2}=9
\end{array}\right.
$$

or equivalently

$$
\left\{\begin{array}{l}
2 x-2 \cdot 3=y \\
x^{2}-8 x+15=0
\end{array}\right.
$$

The solutions to the quadratic equation are $x_{1}=5$ and $x_{2}=3$. Then, $y_{1}=4$ and $y_{2}=0$. The solutions to the system are $(5,4)$ and $(3,0)$, since they are points in the domain.