10. Real numbers $x, y, z$ satisfy $x+y+z=5, xy+yz+zx=$ 3. Then the maximum value of $z$ is $\qquad$.
10. $\frac{13}{3}$.

Since $x+y=5-z$, $xy=3-z(x+y)=3-z(5-z)=z^{2}-5z+3$, therefore, $x, y$ are the two real roots of the quadratic equation in $t$
$$
t^{2}-(5-z)t+z^{2}-5z+3=0
$$

Also, 
$$
\begin{array}{l}
\text { and } \Delta=(5-z)^{2}-4\left(z^{2}-5z+3\right) \geqslant 0, \text { then } \\
3z^{2}-10z-13 \leqslant 0, (3z-13)(z+1) \leqslant 0 .
\end{array}
$$

Therefore, $z \leqslant \frac{13}{3}$.
When $x=y=\frac{1}{3}$, $z=\frac{13}{3}$. Hence, the maximum value of $z$ is $\frac{13}{3}$.