Among all solutions of the system

$$
\begin{aligned}
& x^{2}+y^{2}=4 \\
& z^{2}+t^{2}=9 \\
& x t+y z=6
\end{aligned}
$$

select those for which the quantity $x+z$ takes the maximum value.
Consider the vectors $\boldsymbol{a}=(x, y)$ and $\boldsymbol{b}=(t, z)$. By the condition $|\boldsymbol{a}|=2,|\boldsymbol{b}|=3,(\boldsymbol{a}, \boldsymbol{b})=6=|\boldsymbol{a}||\boldsymbol{b}|$. Therefore, the vectors $\boldsymbol{a}$ and $\boldsymbol{b}$ are directed in the same way, that is,

$\boldsymbol{a}=2(u, v), \boldsymbol{b}=3(u, v)$.

Let $u=\cos \varphi, v=\sin \varphi$, then $x+z=2 \cos \varphi+3 \sin \varphi=\sqrt{13} \sin (\varphi+\alpha) \leq \sqrt{13}$.

The maximum value $\sqrt{13}$ is achieved when $\cos \varphi=\frac{2}{\sqrt{13}}, \sin \varphi=\frac{3}{\sqrt{13}}$.

## Answer

$x=\frac{4}{\sqrt{13}}, y=\frac{6}{\sqrt{13}}, z=\frac{9}{\sqrt{13}}, t=\frac{6}{\sqrt{13}}$.