Ireland

Label the equations 1, 2, 3, 4, 5, 6 in the order of their appearance. Using equations 3, 6 eliminate $w$ from the system, thereby adding to equations 1, 2, 4, 5 the equation 7: $ax+cv=c^2$. Next, use this and 5 to eliminate $v$ to produce equation 8: $ax+bu=c^2$. Eliminate $z$ from 2, 4 to produce equation 9: $ay+bu=b^2$. Now eliminate $u$ from equations 8, 9 giving $ax-ay=c^2-b^2$. Finally, using this in conjunction with 1, we see that $$2ax=c^2+a^2-b^2,2ay=a^2+b^2-c^2.$$

And so, using 6 and 4, too, $$2cw=c^2+a^2-b^2,2bz=a^2+b^2-c^2.$$ Hence, using equations 2, 5, we get that $$2cv=2bu=2b^2-2bz=b^2+c^2-a^2.$$ Thus, letting $$\alpha =\frac{b^2+c^2-a^2}{2bc},\beta =\frac{c^2+a^2-b^2}{2ca},\gamma =\frac{a^2+b^2-c^2}{2ab},$$ we see that $$x=c\beta ,y=b\gamma ,z=a\gamma ,u=c\alpha ,v=b\alpha ,w=a\beta .$$