## Task 1 - 281221

Determine all triples $(x, y, z)$ of non-zero real numbers $x, y$, and $z$ that satisfy the following system of equations (1), (2), (3):

$$
\begin{aligned}
& \frac{1}{x}+\frac{1}{y}=z \\
& \frac{1}{y}+\frac{1}{z}=-x \\
& \frac{1}{x}+\frac{1}{x}=y
\end{aligned}
$$
Assume there is a triple $(x, y, z)$ of non-zero real numbers for which (1), (2), and (3) hold. From (1), (2), and (3) it follows that

\[
\begin{aligned}
& x+y=x y z \\
& y+z=-x y z \\
& x+z=x y z
\end{aligned}
\]

From (1') and (3') it follows that $y=z$ (4). Thus, from (1') and (2') by addition, we get $x+3 z=0$, so $x=-3 z \quad$ (5). With (4) and (5), (2') yields $2 z=3 z^{3}$.

Since $z \neq 0$, it follows that $z= \pm \frac{1}{3} \sqrt{6}$.

From (4) and (5), it follows that $y= \pm \frac{1}{3} \sqrt{6}$ and $x=\mp \sqrt{6}$.

Thus, the only solutions to the system of equations (1), (2), (3) are the triples

\[
\left(-\sqrt{6} ; \frac{1}{3} \sqrt{6} ; \frac{1}{3} \sqrt{6}\right) \quad \text { and } \quad\left(\sqrt{6} ;-\frac{1}{3} \sqrt{6} ;-\frac{1}{3} \sqrt{6}\right)
\]

Indeed, these triples satisfy the verification with (1), (2), and (3).

## Adopted from $[5]$