## Task 1 - 221211

Determine all pairs $(x ; y)$ of real numbers with $x \neq 0$ and $y \neq 0$ that satisfy the following system of equations (1), (2):

$$
\begin{aligned}
& x+\frac{x}{y}=\frac{8}{3} \\
& y-\frac{1}{x}=\frac{5}{2}
\end{aligned}
$$
I. If a pair of numbers $(x ; y)$ with $x \neq 0, y \neq 0$ satisfies the system of equations (1), (2), then from (2) it follows that

$$
y=\frac{5 x+2}{2 x}
$$

Since $y \neq 0$, it follows that $5 x+2 \neq 0$, and from (1) by substituting (3) we get

$$
\begin{aligned}
x+\frac{2 x^{2}}{5 x+2} & =\frac{8}{3} \\
3 x(5 x+2)+6 x^{2} & =8(5 x+2) \\
x^{2}-\frac{34}{21} x-\frac{16}{21} & =0
\end{aligned}
$$

Since $\left(\frac{17}{21}\right)^{2}+\frac{16}{21}=\frac{625}{21^{2}}=\left(\frac{25}{21}\right)^{2}>0$, thus

$$
x=\frac{17}{21} \pm \sqrt{\left(\frac{17}{21}\right)^{2}+\frac{16}{21}}=\frac{17}{21} \pm \frac{25}{21}
$$

i.e., $x=2$ or $x=-\frac{8}{21}$, and thus from (3) $y=3$ or $y=-\frac{1}{8}$.

Therefore, only the pairs $(2 ; 3)$ and $\left(-\frac{8}{21} ;-\frac{1}{8}\right)$ can satisfy the system of equations. The verification confirms this.

Adapted from $[5]$