Problem 1342

1. Given the equations:
$$x + y + z = 5$$
$$xy + yz + xz = 3$$
we need to determine the largest real number $z$ such that $x$ and $y$ are also real.

2. Express $x + y$ and $xy$ in terms of $z$:
$$x + y = 5 - z$$
$$xy + yz + xz = 3 \implies xy + z(x + y) = 3 \implies xy + z(5 - z) = 3$$
Simplifying the second equation:
$$xy + 5z - z^2 = 3 \implies xy = z^2 - 5z + 3$$

3. For $x$ and $y$ to be real, the discriminant of the quadratic equation $t^2 - (x+y)t + xy = 0$ must be non-negative. The quadratic equation is:
$$t^2 - (5 - z)t + (z^2 - 5z + 3) = 0$$
The discriminant $\Delta$ of this quadratic equation is:
$$\Delta = (5 - z)^2 - 4(z^2 - 5z + 3)$$
Simplifying the discriminant:
$$\Delta = (5 - z)^2 - 4(z^2 - 5z + 3) = 25 - 10z + z^2 - 4z^2 + 20z - 12 = -3z^2 + 10z + 13$$

4. For $x$ and $y$ to be real, the discriminant must be non-negative:
$$-3z^2 + 10z + 13 \geq 0$$
Solving the inequality:
$$3z^2 - 10z - 13 \leq 0$$
The roots of the quadratic equation $3z^2 - 10z - 13 = 0$ are:
$$z = \frac{10 \pm \sqrt{100 + 156}}{6} = \frac{10 \pm \sqrt{256}}{6} = \frac{10 \pm 16}{6}$$
Thus, the roots are:
$$z = \frac{26}{6} = \frac{13}{3} \quad \text{and} \quad z = \frac{-6}{6} = -1$$
Therefore, the inequality $3z^2 - 10z - 13 \leq 0$ holds for:
$$-1 \leq z \leq \frac{13}{3}$$

5. To find the maximum value of $z$, we check the upper bound $z = \frac{13}{3}$:
$$x + y = 5 - \frac{13}{3} = \frac{15}{3} - \frac{13}{3} = \frac{2}{3}$$
$$xy = \left(\frac{13}{3}\right)^2 - 5 \left(\frac{13}{3}\right) + 3 = \frac{169}{9} - \frac{65}{3} + 3 = \frac{169}{9} - \frac{195}{9} + \frac{27}{9} = \frac{1}{9}$$
The quadratic equation for $x$ and $y$ is:
$$t^2 - \frac{2}{3}t + \frac{1}{9} = 0$$
Solving this quadratic equation:
$$t = \frac{\frac{2}{3} \pm \sqrt{\left(\frac{2}{3}\right)^2 - 4 \cdot \frac{1}{9}}}{2} = \frac{\frac{2}{3} \pm \sqrt{\frac{4}{9} - \frac{4}{9}}}{2} = \frac{\frac{2}{3}}{2} = \frac{1}{3}$$
Thus, $x = y = \frac{1}{3}$, and $z = \frac{13}{3}$ is valid.

Conclusion:
The largest value of $z$ is $\boxed{\frac{13}{3}}$.