$\sqrt{x} +y = 4$, $x+ \sqrt{y}= 6$, find the solution $(x,y)$ $\sqrt{x} +y = 4$, $\sqrt{y} +x= 6$, find the solution (x,y). $NOTE$ : $\sqrt{4}+1= 4-1$, $\sqrt{1} +4 =1+4$ 

This is basically the method which was suggested in the comments above - turning this into a quartic equation. We will see whether someone suggest a substantially more elegant solution.
$$\sqrt{x}+y=4\\
x+\sqrt{y}=6$$
Using the substitution
$\sqrt{x}=s$ and $\sqrt{y}=t$
we get:
$$s+t^2=4\\
t+s^2=6$$
Which gives
$$s=4-t^2=4-(6-s^2)^2\\
(s^2-6)^2+s-4=0\\
s^4-12s^2+s+32=0$$
It should be possible to solve this as a quartic equation, although it would be quite laborious. You can check what WolframAlpha is able to find out 
here
and
here
