Solve the following system of equations:

$$
\begin{aligned}
& 3 x^{2}-x y=1 \\
& 9 x y+y^{2}=22
\end{aligned}
$$
Solution. By adding three times the first equation to the second equation, we obtain the following quadratic equation:

$$
9 x^{2}+6 x y+y^{2}=25
$$

The left side of the equation is a perfect square:

$$
(3 x+y)^{2}=25
$$

Two cases are possible.

Case I: $3 x+y=5$, from which $y=5-3 x$. Substituting this into the first equation:

$$
3 x^{2}-x(5-3 x)=1, \quad 6 x^{2}-5 x-1=0
$$

from which $x_{1}=1, x_{2}=-\frac{1}{6} ; y_{1}=2, y_{2}=5.5$.

Case II: $3 x+y=-5$, from which $y=-5-3 x$. Substituting this into the first equation:

$$
3 x^{2}-x(-5-3 x)=1, \quad 6 x^{2}+5 x-1=0
$$

from which $x_{3}=-1, x_{4}=\frac{1}{6} ; y_{3}=-2, y_{4}=-5.5$.

The solutions to the system of equations are: $x_{1}=1, y_{1}=2 ; x_{3}=-1, y_{3}=-2$; and $x_{2}=-\frac{1}{6}, y_{2}=5.5 ; x_{4}=\frac{1}{6}, y_{4}=-5.5$. These are indeed solutions to the system of equations, as we only performed equivalent transformations.