## Task B-2.5.

Determine all ordered pairs of real numbers $(x, y)$ for which

$$
\left\{\begin{array}{l}
x^{2}-3 x y-18 y^{2}=0 \\
x y-3 y^{2}+x-5 y=0
\end{array}\right.
$$
## First Solution.

One solution is obviously the ordered pair $(0,0)$.

To find other solutions, assume that $y \neq 0$ (if $y=0$, then $x=0$ must also be true). Divide the first equation by $y^{2}$. We get

$$
\left(\frac{x}{y}\right)^{2}-3 \frac{x}{y}-18=0
$$

The solutions to this quadratic equation are $\frac{x}{y}=6$ and $\frac{x}{y}=-3$.

If $x=6 y$, from the second equation of the initial system we get $3 y^{2}+y=0$. Due to the assumption $y \neq 0$, it follows that $y=-\frac{1}{3}$.

For $y=-\frac{1}{3}$, we get $x=-2$, so the ordered pair $\left(-2,-\frac{1}{3}\right)$ is the second solution to the system.