Find the maximum value of $z+x$, if $(x, y, z, t)$ satisfies the conditions:

$$
\left\{\begin{array}{l}
x^{2}+y^{2}=4 \\
z^{2}+t^{2}=9 \\
x t+y z \geq 6
\end{array}\right.
$$
I: From the conditions we have

$$
36=\left(x^{2}+y^{2}\right)\left(z^{2}+t^{2}\right)=(x t+y z)^{2}+(x z-y t)^{2} \geq 36+(x z-y t)^{2}
$$

and this implies $x z-y t=0$.

Now it is clear that

$$
x^{2}+z^{2}+y^{2}+t^{2}=(x+z)^{2}+(y-t)^{2}=13
$$

and the maximum value of $z+x$ is $\sqrt{13}$. It is achieved for $x=\frac{4}{\sqrt{13}}, y=t=\frac{6}{\sqrt{13}}$ and $z=\frac{9}{\sqrt{13}}$.

Solution II: From inequality $x t+y z \geq 6$ and problem conditions we have:

$$
\begin{gathered}
(x t+y z)^{2}-36 \geq 0 \Leftrightarrow \\
(x t+y z)^{2}-\left(x^{2}+y^{2}\right)\left(z^{2}+t^{2}\right) \geq 0 \Leftrightarrow \\
2 x y z t-x^{2} y^{2}-y^{2} t^{2} \geq 0 \Leftrightarrow \\
-(x z-y t)^{2} \geq 0
\end{gathered}
$$

From here we have $x z=y t$.

Furthermore,

$$
x^{2}+y^{2}+z^{2}+t^{2}=(x+z)^{2}+(y-t)^{2}=13
$$

and it follows that

$$
(x+z)^{2} \leq 13
$$

Thus,

$$
x+z \leq \sqrt{13}
$$

Equality $x+z=\sqrt{13}$ holds if we have $y=t$ and $z^{2}-x^{2}=5$, which leads to $z-x=\frac{5}{\sqrt{13}}$. Therefore, $x=\frac{4}{\sqrt{13}}, y=t=\frac{6}{\sqrt{13}}, z=\frac{9}{\sqrt{13}}$.