system of equations $\sqrt{x}+y = 11$ and $x+\sqrt{y} = 7$. If $x,y\in \mathbb{R}$ and $\sqrt{x}+y = 11\;$ and $x+\sqrt{y} = 7$. Then $(x,y) = $
$\underline{\bf{My\;\; Try::}}$ Let $x=a^2$ and $y=b^2$, Then equation is $a+b^2 = 11$ and $a^2+b = 7$.
$(a+b)+(a+b)^2-2ab = 18$ and Now Let $a+b=S$ and $ab=P$, we get $S+S^2-P=18$
Now I did not understand how can I solved it.
Help required
Thanks 

HINT:
Assuming $x,y$ are real we have $x=7-\sqrt y\le 7\implies a^2\le 7$
Putting $b=7-a^2,$
$$a+(7-a^2)^2=11\implies a^4-14a^2+a+38=0$$
Observe that $a=2$ is a  solution
For the general case, we need to use the formula described here or here
