Spring Mathematical Tournament

Writing the second equation in the form
$x - y + 13\sqrt[4]{x - y} - 42 = 0$
and setting $u = \sqrt[4]{x-y}$, $u \ge 0$, we get that $u^4 + 13u - 42 = 0$. Note that this equation has only one positive root $u = 2$ and hence $x = y + 16$. Then the first equation easily gives
$y^2 - 5y + 3 = 0.$
The last equation has roots $y_{1,2} = \frac{5 \pm \sqrt{13}}{2}$. Since $y - 2 \ge 0$ and $\frac{5 - \sqrt{13}}{2} < 2$, it follows that
$$x = \frac{37 + \sqrt{13}}{2}, \quad y = \frac{5 + \sqrt{13}}{2}.$$