## Task A-2.1.

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

$$
x^{2}+y^{2}=5, \quad x z+y=7, \quad y z-x=1
$$
## First Solution.

From the second equation, we have $y=7-x z$.

By substituting into the third equation, we get $x=y z-1=(7-x z) z-1=7 z-x z^{2}-1$, which simplifies to $x\left(1+z^{2}\right)=7 z-1$.

Since for real $z$, the value $1+z^{2}$ is never zero, rearranging and substituting gives us

$$
x=\frac{7 z-1}{1+z^{2}}, \quad y=7-x z=\frac{z+7}{1+z^{2}}
$$

By substituting into the first equation, we get

$$
\begin{aligned}
5 & =x^{2}+y^{2}=\left(\frac{7 z-1}{1+z^{2}}\right)^{2}+\left(\frac{z+7}{1+z^{2}}\right)^{2} \\
& =\frac{1}{\left(1+z^{2}\right)^{2}}\left(49 z^{2}-14 z+1+z^{2}+14 z+49\right) \\
& =\frac{50\left(z^{2}+1\right)}{\left(1+z^{2}\right)^{2}}=\frac{50}{1+z^{2}}
\end{aligned}
$$

It follows that $1+z^{2}=10$, or $z^{2}=9$, i.e., $z= \pm 3$.

Now $x$ and $y$ can be calculated from the first equations where we expressed these variables in terms of $z$.

For $z=3$, we get $x=\frac{20}{10}=2, y=\frac{10}{10}=1$.

For $z=-3$, we get $x=-\frac{22}{10}=-\frac{11}{5}, y=\frac{4}{10}=\frac{2}{5}$.

The second and third equations are satisfied because we obtained $x$ and $y$ as solutions to a system of two linear equations, and the first equation is satisfied because we determined $z$ in this way. Therefore, the solutions are $(2,1,3)$ and $\left(-\frac{11}{5}, \frac{2}{5},-3\right)$.

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