Find all real pairs $(a,b)$ that solve the system of equation \begin{align*} a^2+b^2 &= 25, \\ 3(a+b)-ab &= 15. \end{align*} [i](German MO 2016 - Problem 1)[/i]
1. We start with the given system of equations:
   \[
   a^2 + b^2 = 25 \quad \text{(1)}
   \]
   \[
   3(a + b) - ab = 15 \quad \text{(2)}
   \]

2. From equation (1), we can express \(a^2 + b^2\) in terms of \(a + b\) and \(ab\):
   \[
   a^2 + b^2 = (a + b)^2 - 2ab = 25
   \]
   Let \(s = a + b\) and \(p = ab\). Then:
   \[
   s^2 - 2p = 25 \quad \text{(3)}
   \]

3. Substitute \(s\) and \(p\) into equation (2):
   \[
   3s - p = 15 \quad \text{(4)}
   \]

4. We now have the system of equations:
   \[
   s^2 - 2p = 25 \quad \text{(3)}
   \]
   \[
   3s - p = 15 \quad \text{(4)}
   \]

5. Solve equation (4) for \(p\):
   \[
   p = 3s - 15
   \]

6. Substitute \(p\) into equation (3):
   \[
   s^2 - 2(3s - 15) = 25
   \]
   Simplify:
   \[
   s^2 - 6s + 30 = 25
   \]
   \[
   s^2 - 6s + 5 = 0
   \]

7. Solve the quadratic equation:
   \[
   s^2 - 6s + 5 = 0
   \]
   \[
   (s - 5)(s - 1) = 0
   \]
   \[
   s = 5 \quad \text{or} \quad s = 1
   \]

8. Case 1: \(s = 5\)
   \[
   p = 3s - 15 = 3(5) - 15 = 0
   \]
   \[
   ab = 0
   \]
   This implies \(a = 0\) or \(b = 0\). Since \(a + b = 5\):
   \[
   (a, b) = (0, 5) \quad \text{or} \quad (5, 0)
   \]

9. Case 2: \(s = 1\)
   \[
   p = 3s - 15 = 3(1) - 15 = -12
   \]
   \[
   ab = -12
   \]
   Since \(a + b = 1\), we solve the quadratic equation:
   \[
   t^2 - t - 12 = 0
   \]
   \[
   (t - 4)(t + 3) = 0
   \]
   \[
   t = 4 \quad \text{or} \quad t = -3
   \]
   Thus, the pairs \((a, b)\) are:
   \[
   (a, b) = (4, -3) \quad \text{or} \quad (-3, 4)
   \]

10. Therefore, the solutions to the system of equations are:
    \[
    (0, 5), (5, 0), (4, -3), (-3, 4)
    \]

The final answer is \(\boxed{(0, 5), (5, 0), (4, -3), (-3, 4)}\)