Solve the following system of equations:

$$
\begin{aligned}
& x+2 \sqrt{y}=2 \\
& 2 \sqrt{x}+y=2
\end{aligned}
$$
Solve the following system of equations:

$$
x+2 \sqrt{y}=2
$$

$$
2 \sqrt{x}+y=2
$$

Solution. Subtract (2) from (1), and apply the identity $a^{2}-b^{2}=(a+b)(a-b)$ to the difference $(x-y)$:

$$
\begin{gathered}
x-y=2(\sqrt{y}-\sqrt{x})=0 \\
(\sqrt{x}+\sqrt{y})(\sqrt{x}-\sqrt{y})-2(\sqrt{x}-\sqrt{y})=(\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y}-2)=0
\end{gathered}
$$

From this, either $\sqrt{x}-\sqrt{y}=0$ or $\sqrt{x}+\sqrt{y}-2=0$. In the first case, substituting $\sqrt{y}=\sqrt{x}$ into (1), we get the quadratic equation in $\sqrt{x}$:

$$
x+2 \sqrt{x}=2
$$

Since $\sqrt{x}$ is non-negative, the only solution is:

$$
\sqrt{x_{1}}=-1+\sqrt{3}, \quad x_{1}=(-1+\sqrt{3})^{2}=4-2 \sqrt{3}
$$

In the second case,

$$
\sqrt{x}=2-\sqrt{y}
$$

Substituting this into (2),

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

After rearranging, we get the quadratic equation in $\sqrt{y}$:

$$
\sqrt{y}^{2}-2 \sqrt{y}+2=0
$$

The discriminant of this equation is $D=4-8<0$, so this equation has no real solutions.

Therefore, the only solution to the system of equations is

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

(We can verify by substitution that this is indeed a solution.)