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 $$\begin{array}{rl}x+y+z+u& =\lambda \\ a+b+c+d& =t\end{array}$$ then we have $$x=\frac{\lambda }{4}+\frac{4a-t}{4\lambda },y=\frac{\lambda }{4}+\frac{4b-t}{4\lambda },z=\frac{\lambda }{4}+\frac{4c-t}{4\lambda },u=\frac{\lambda }{4}+\frac{4d-t}{4\lambda }.(1)$$ 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 $x,y,z,u$.