Example 4 Given that $a$ and $b$ are the two real roots of the quadratic equation $t^{2}-t-1=0$. Solve the system of equations
$$
\left\{\begin{array}{l}
\frac{x}{a}+\frac{y}{b}=1+x, \\
\frac{x}{b}+\frac{y}{a}=1+\hat{y} .
\end{array}\right.
$$
(1000th Zu Chongzhi Cup Junior High School Mathematics Invitational Competition)
Solving (1)-(2), factorizing, we get
$$
\left(\frac{1}{a}-\frac{1}{b}-1\right)(x-y)=0 \text {. }
$$

From the given conditions, we have
$$
\begin{array}{l}
a+b=1, a b=-1 . \\
\therefore \frac{1}{a}-\frac{1}{b}-1=-b+a-(a+b) . \\
\quad=-2 b \neq 0 .
\end{array}
$$

Thus, from (3) we get $x=y$. Substituting into (1), simplifying, we get $(a+b-a b) x=a b$ or $2 x=-1$,
which gives $x=-\frac{1}{2}$.
Therefore, the solution to the original system of equations is $\left\{\begin{array}{l}x=-\frac{1}{2}, \\ y=-\frac{1}{2} .\end{array}\right.$