Determine the largest real number $z$ such that
\begin{align*}
x + y + z = 5 \\
xy + yz + xz = 3
\end{align*}
and $x$, $y$ are also real.
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 \(\boxed{\frac{13}{3}}\).