Is this solution correct? $$\sqrt{x}+y=4\tag{A}$$
$$x+\sqrt{y}=6\tag{B}$$
Subtracting A from B, we have
$$x-y -\sqrt{x}+\sqrt{y}=2$$
$$(\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y}-1)=2$$
$$(\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y}-1)=(\sqrt{2})(\sqrt{2})$$
Now we have,
$$\sqrt{x}-\sqrt{y}=\sqrt{2}\tag{AA}$$
$$\sqrt{x}+\sqrt{y}-1=\sqrt{2}\tag{BB}$$
This gives,
$$\sqrt{x}=\sqrt{2}+\frac{1}{2}$$
$$\sqrt{y}=\frac{1}{2}$$
 A: That answer doesn't work, which you can see if you plug your values into your original equations. You made the mistake when you said
$\sqrt{x}-\sqrt{y}=\sqrt{2}$ -- AA
$\sqrt{x}+\sqrt{y}-1=\sqrt{2}$ -- BB
this doesn't follow from the previous equation because they don't both have to equal the squareroot of 2. They just have to multiply to 2. You are trying to solve 2 variable with one equation at this point. You lose information from the previous equations when you do this, so your final answer didn't work. What you might want to try instead is a substitution method, although this ends up with some very messy equations.
edit: so far i end up with 
$0 = y^2 - 8y + \sqrt y + 10$
A: Do a substitution $x= X^2$ and $y=Y^2$, to get 2 quadratic equations. We have the conditions that $x , y \geq 0$, which also imply that $y\leq 4, x \leq 6$.
Substitute one into the other and you a get $ (6-x^2)^2 = 4-x$.
Solve this quartic (which doesn't seem to have nice roots), subject to $ 0 \leq x \leq 4$. You will get 2 solutions. Verify if they work.
A: The solution as given by wolfram is as follows,

