Annual Harvard-MIT Mathematics Tournament

Solution:
Squaring both sides of $x - y^{2} = 3$ gives $x^{2} + y^{4} - 2x y^{2} = 9$.

Subtract this equation from twice the second given to get
$$2(x^{2} + y^{4}) - (x^{2} + y^{4} - 2x y^{2}) = 2 \times 13 - 9$$
which simplifies to
$$x^{2} + 2x y^{2} + y^{4} = 17$$
So $(x + y^{2})^{2} = 17$, hence $x + y^{2} = \pm \sqrt{17}$.

But $x$ and $y$ are positive reals, so $x + y^{2} = \sqrt{17}$.

From $x - y^{2} = 3$ and $x + y^{2} = \sqrt{17}$, adding gives $2x = 3 + \sqrt{17}$, so
$$x = \frac{3 + \sqrt{17}}{2}$$
Since $x$ is a positive real, $x = \frac{3 + \sqrt{17}}{2}$.