Mediterranean Mathematics Competition

Subtracting from the first equation the other 3, we obtain
$$(x - y)(x + y + z + u) = a - b, \text{ etc}$$
If we add these 3 new equations, we get
$$[4x - (x + y + z + u)](x + y + z + u) = 3a - b - c - d,$$
and so, making the substitutions
$$\begin{align*} x + y + z + u &= \lambda \\ a + b + c + d &= t \end{align*}$$
then we have
$$x = \frac{\lambda}{4} + \frac{4a-t}{4\lambda}, \quad y = \frac{\lambda}{4} + \frac{4b-t}{4\lambda}, \quad z = \frac{\lambda}{4} + \frac{4c-t}{4\lambda}, \quad u = \frac{\lambda}{4} + \frac{4d-t}{4\lambda}. \quad (1)$$
By substitution of these values in the proposed equations we get
$$\lambda^4 + 2\lambda^2 (\sum a) + 2(\sum bc) - 3(\sum a^2) = 0,$$
biquadratic in $\lambda$, giving
$$\lambda^2 = -\sum a \pm 2\sqrt{\sum a^2},$$
and computing $\lambda$, by substitution in (1) we obtain $x, y, z, u$.