#### Answer

x=5

#### Work Step by Step

Substitue $\sqrt x$ for y in the second equation, and solve for x
$x^{2}+(\sqrt x)^2=30$
$x^{2}+x=30$
$x^{2}+x-30=0$
$(x-5)(x+6)=0$
x=5 or x=-6
The solution -6 us discarded because we have noted that x must be nonnegative. To see this, we let x=-6 in the first equation to find the corresponding y-values.
Let x=-6,
$y=\sqrt x$
$y=\sqrt -6$ is not a real number
so the answer is x=5