4-108 Find the real solutions of the system of equations
$$
\left\{\begin{array}{l}
x+y=3 \\
x y-y^{2}=1 .
\end{array}\right.
$$
[Solution] From (1) $x=3-y$,
substitute into (2) $2 y^{2}-3 y+1=0$, i.e., $(2 y-1)(y-1)=0$, hence
$$
\begin{array}{l}
y_{1}=\frac{1}{2}, x_{1}=\frac{5}{2}, \\
y_{2}=1, x_{2}=2 .
\end{array}
$$

We get two sets of solutions
$$
\left\{\begin{array} { l } 
{ x _ { 1 } = \frac { 5 } { 2 } , } \\
{ y _ { 1 } = \frac { 1 } { 2 } ; }
\end{array} \quad \left\{\begin{array}{l}
x_{2}=2, \\
y_{2}=1 .
\end{array}\right.\right.
$$