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 $$x=\frac{37+\sqrt{13}}{2},y=\frac{5+\sqrt{13}}{2}.$$