3. Solve the system of equations:

$$
\left\{\begin{array}{l}
x^{2}-6 \sqrt{3-2 x}-y+11=0 \\
y^{2}-4 \sqrt{3 y-2}+4 x+16=0
\end{array}\right.
$$

(P. Alishev)
Answer: $x=-3, y=2$.

Solution. Add these equations and gather complete squares: $x^{2}+4 x-6 \sqrt{3-2 x}+y^{2}-y-4 \sqrt{3 y-2}+27=x^{2}+$ $6 x+9+3-2 x-6 \sqrt{3-2 x}+9+y^{2}-4 y+4+3 y-2-4 \sqrt{3 y-2}+4=$ $(x+3)^{2}+(\sqrt{3-2 x}-3)^{2}+(y-2)^{2}+(\sqrt{3 y-2}-2)^{2}=0$. It is clear that only when $x=-3$ and $y=2$ do the squares take their minimum values, which are zero. By substituting the found values of the variables into the original system, we verify that they satisfy both equations.