4・114 Try to solve the system of equations
$$
\left\{\begin{array}{l}
\lg x+\lg y=1 \\
x^{2}+y^{2}-3 x-3 y=8
\end{array}\right.
$$
[Solution] From (1) we get $x y=10$.
(2) $+2 \times$ (3), and rearranging terms, we get
$$
(x+y)^{2}-3(x+y)-28=0 \text {, }
$$

So $x+y=7$ and $x+y=-4$.
Combining (4) with (3), we get the following system of equations:
$$
\begin{array}{l}
\left\{\begin{array}{l}
x+y=7 \\
x y=10
\end{array}\right. \\
\left\{\begin{array}{l}
x+y=-4 \\
x y=10
\end{array}\right.
\end{array}
$$

From the system of equations (5), we solve to get $\left\{\begin{array}{l}x_{1}=2, \\ y_{1}=5 ;\end{array}\left\{\begin{array}{l}x_{2}=5, \\ y_{2}=2 .\end{array}\right.\right.$ Substituting the values of $x$ and $y$ into the original system of equations, they all satisfy, so they are all solutions to the original system of equations.

From equation (1) in the original system, we know that $x, y$ must be positive numbers, so the solutions of system (6) cannot be solutions to the original system of equations.