Irish Mathematical Olympiad

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
$$2ax = c^2 + a^2 - b^2, \quad 2ay = a^2 + b^2 - c^2.$$

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