8. (Canada National Training Team Practice Question) Solve the system of equations in the set of real numbers:
$$
\left\{\begin{array}{l}
x + xy + y = 2 + 3 \sqrt{2}, \\
x^2 + y^2 = 6.
\end{array}\right.
$$
8. The original system of equations can be transformed into $\left\{\begin{array}{l}(x+y)+xy=2+3\sqrt{2}, \\ (x+y)^{2}-2xy=6 .\end{array}\right.$ Thus, $(x+y)^{2}+2(x+y)-10-6\sqrt{2}=0$. This gives $x+y=2+\sqrt{2}$ or $x+y=-4-\sqrt{2}$.

When $x+y=2+\sqrt{2}$, $xy=2\sqrt{2}$, in this case, the solutions for $(x, y)$ are $(2, \sqrt{2})$ or $(\sqrt{2}, 2)$; when $x+y=-4-\sqrt{2}$, $xy=6+4\sqrt{2}$, there are no real solutions.