Maybe substitution?

Solving $x+y=15$ for $x$ and using it in the secound equation we get $y=15-y+\sqrt y$ . When substituting $\sqrt y$ with $a$ we get an quadratic equation $2a^2-a-15=0$ with the soultions $a=3$ and $a=-\frac{5}{2}$ . Since $a$ is the square root of a natural number it cant be negative, therefore $a=3\Rightarrow y=9\Rightarrow x=6$