## Task 2 - 241212

Determine all real numbers $x, y$ with $x \neq 0$ and $y \neq 0$ that satisfy the system of equations

$$
\begin{aligned}
x^{2} y-\frac{6}{x y} & =13 \\
x y+x^{2} y & =6 \quad \text { satisfy. }
\end{aligned}
$$
Let $x$ and $y$ be real numbers with $x \neq 0$ and $y \neq 0$, which satisfy the given system of equations. Then it follows:

The numbers $a = xy$ and $b = x^2 y$ satisfy the system of equations

$$
\begin{aligned}
b - \frac{6}{a} & = 13 \\
a + b & = 6
\end{aligned}
$$

Subtracting (1) from (2), we get $a + \frac{6}{a} = -7$, thus

$$
a^2 + 7a + 6 = 0
$$

This equation has only $a_1 = -1$ and $a_2 = -6$ as solutions. According to (2), the corresponding values are $b_1 = 7$ and $b_2 = 12$. From $a = xy$ and $b = x^2 y$ it follows that $b = ax$, so, since we can divide by $x (\neq 0)$ and by $a ( = xy \neq 0)$,

$$
x = \frac{b}{a} \quad ; \quad y = \frac{a}{x}
$$

Thus, the only solutions of the given system of equations are $x_1 = -7, y_1 = \frac{1}{7}$ and $x_2 = -2, y_2 = 3$, which are confirmed by the verification. The given $x_1, y_1$ and $x_2, y_2$ are the solutions of the system of equations.

Adapted from $[5]$