Solve the system of equations

$$
\begin{aligned}
x-x y+y & =1 \\
x^{2}+y^{2} & =17
\end{aligned}
$$
Solution. Let's transform the first equation into a product, we get

$$
(1-y)(x-1)=0
$$

A product is zero if and only if one of its factors is zero. From the first factor, we get $y=1$. Substituting this into the second equation, we get $x^{2}=16$, which means $x=4$ or -4. From the symmetry of the equations, it follows that if $x=1$, then $y=4$ or -4.

The solutions are: $x_{1}=1, y_{1}=4 ; x_{2}=1, y_{2}=-4 ; x_{3}=4, y_{3}=1 ; x_{4}=-4, y_{4}=1$. It is easy to verify that all four pairs of numbers satisfy the equations.