$\sqrt{x} +y = 4$, $x+ \sqrt{y}= 6$, find the solution $(x,y)$ $\sqrt{x} +y = 4$, $\sqrt{y} +x= 6$, find the solution (x,y). $NOTE$ : $\sqrt{4}+1= 4-1$, $\sqrt{1} +4 =1+4$ 
 A: 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
A: Let, $\sqrt{x}=s$, $\sqrt{y}=t$
we have, $s^4 -12s^2+s+32=0$, which is a 'biquadratic' equation of the form,
$$(s^2+ks+l)(s^2-ks+m)=0$$
i.e. $$s^4 -12s^2+s+32=(s^2+ks+l)(s^2-ks+m)$$
now by equating coefficients, we have
$$l+m-k^2 = -12, k(m-l) = 1, lm = 32$$
from the first two of these equations, we obtain
$$2m=k^2-12+(1/k), 2l=k^2-12-(1/k)$$
hence substituting in the third equation, the values of l,m,
$$(4)(32)=(k^2-12-1/k)(k^2-12+1/k)$$
$$128=(k^2-12)^2-1/k^2$$
$$128=k^4+144-24k^2-1/k^2$$
$$k^6-24k^4+16k^2-1=0$$
this is a cubic in $k^2$ which always has one real positive solution and we can find $k^{2}$,$l$,$m$
