5. Solve the system of equations:

$$
\begin{aligned}
x^{2}+y^{2} & =9 \\
y^{2}+z^{2} & =16 \\
y^{2} & =x z
\end{aligned}
$$

## Third Grade - B Category
5. Adding the first two equations from the problem statement gives $x^{2}+2 y^{2}+z^{2}=25$, and using the third equation, we get $x^{2}+2 x z+z^{2}=25$, i.e., $(x+z)^{2}=25$, from which it follows that $x+z= \pm 5$, i.e., $z= \pm 5-x$. Now, from the first and third equations, using the conclusion just made, we get $9=x^{2}+y^{2}=x^{2}+x z=x^{2}+x( \pm 5-x)= \pm 5 x$, i.e., $x= \pm \frac{9}{5}$, and from this it follows that $z= \pm 5-\left( \pm \frac{9}{5}\right)= \pm \frac{16}{5}$ and $y^{2}=x z=\left( \pm \frac{9}{5}\right) \cdot\left( \pm \frac{16}{5}\right)=\frac{144}{25}$. Therefore, the system under consideration has four solutions:

$$
(x, y, z) \in\left\{\left(\frac{9}{5}, \frac{12}{5}, \frac{16}{5}\right),\left(\frac{9}{5},-\frac{12}{5}, \frac{16}{5}\right),\left(-\frac{9}{5}, \frac{12}{5},-\frac{16}{5}\right),\left(-\frac{9}{5},-\frac{12}{5},-\frac{16}{5}\right)\right\}
$$

## Third Grade - B Category