3. (3 points) Solve the system of equations

$$
\left\{\begin{array}{c}
x + xy + y = 2 + 3 \sqrt{2} \\
x^{2} + y^{2} = 6
\end{array}\right.
$$
Solution: Substitution

$$
u=x+y, v=x \cdot y
$$

The original system will take the form

$$
\left\{\begin{array}{l}
u+v=2+3 \sqrt{2} \\
u^{2}-2 v=6
\end{array}\right.
$$

Multiplying equation (2) by 2 and adding to (3), we get $u^{2}+2 u=10+6 \sqrt{2} \Leftrightarrow(u+1)^{2}=11+6 \sqrt{2}=(3+\sqrt{2})^{2}$. $u_{1}=2+\sqrt{2}, u_{2}=-4-\sqrt{2}$, and from (2) we find the corresponding $v_{1}=2 \sqrt{2}, v_{2}=6+4 \sqrt{2}$. Considering (1), we obtain two systems

1) $\left\{\begin{array}{c}x+y=u_{1} \\ x \cdot y=v_{1}\end{array} \Leftrightarrow\left[\begin{array}{c}x_{1}=2, y_{1}=\sqrt{2} \\ x_{2}=\sqrt{2}, y_{2}=2 .\end{array}\right.\right.$
2) $\left\{\begin{array}{c}x+y=u_{2} \\ x \cdot y=v_{2}\end{array} \Rightarrow x, y\right.$-roots of the equation $t^{2}-u_{2} t+v_{2}=0$, which does not have real roots, since $D<0$.

Answer: $\left[\begin{array}{l}x_{1}=2, y_{1}=\sqrt{2} \\ x_{2}=\sqrt{2}, y_{2}=2\end{array}\right.$