6.12. (GDR, 78). Solve the system

$$
\left\{\begin{array}{l}
x + xy + y = 2 + 3 \sqrt{2} \\
x^{2} + y^{2} = 6
\end{array}\right.
$$
6.12. Multiplying the first equation by 2 and adding it to the second, we get for any solution of the system

$$
x^{2}+2 x y+y^{2}+2(x+y)=(x+y)^{2}+2(x+y)=10+6 \sqrt{2},
$$

i.e., $(x+y+1)^{2}=(3+\sqrt{2})^{2}$, hence $x+y+1= \pm(3+\sqrt{2})$. If $x+y=-4-\sqrt{2}$, then $x y=6+4 \sqrt{2}$ and $(x-y)^{2}=(x+y)^{2}-4 x y=$ $=(4+\sqrt{2})^{2}-4(6+4 \sqrt{2})=-6-8 \sqrt{2}<0$, which is impossible. Therefore, $x+y=2+\sqrt{2}, x y=2 \sqrt{2}$, from which we obtain two pairs $x_{1}=2, y_{1}=\sqrt{2}$ and $x_{2}=\sqrt{2}, y_{2}=2$, each of which (as verification shows) satisfies the original system.