4・105 Solve the system of equations $\left\{\begin{array}{l}x+x y+y=2+3 \sqrt{2}, \\ x^{2}+y^{2}=6 .\end{array}\right.$
[Solution] 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, $(x+y)^{2}+2(x+y)-10-6 \sqrt{2}=0$,

which gives
$$
x+y=2+\sqrt{2}, \quad x+y=-4-\sqrt{2} \text{.}
$$

When $x+y=2+\sqrt{2}$, $xy=2 \sqrt{2}$, in this case
$$
\left\{\begin{array}{l}
x=2, \\
y=\sqrt{2}.
\end{array} \quad \left\{\begin{array}{l}
x=\sqrt{2}, \\
y=2.
\end{array}\right.\right.
$$

When $x+y=-4-\sqrt{2}$, $xy=6+4 \sqrt{2}$, there are no real solutions.
Therefore, the solutions to the system of equations are $\left\{\begin{array}{l}x_{1}=2, \\ y_{1}=\sqrt{2} ;\end{array} \quad\left\{\begin{array}{l}x_{2}=\sqrt{2}, \\ y_{2}=2 .\end{array}\right.\right.$