Simplifying and solving surds with two variables

[miniradman]

Homework Statement

Find values for x, and y in the following statement:

$2x-y+\sqrt{4x-y}=x-2y+3+\sqrt{x+5}$

The Attempt at a Solution

I've managed to rearrange the equation, but I cannot eliminate or equate the square roots on the LHS:

$\sqrt{4x-y}-\sqrt{x+5}=-x-y+3$

 

[micromass]

Square both sides. This will give you one square root instead of two. Isolated the square root on one side. Square both sides again.

Be careful, however. It is of course true that if $x=y$, then $x^2 = y^2$. So if you square both sides of an equality, then the equality is preserved. But the converse is not true. If $x^2 = y^2$, then you can not deduce $x=y$.

So let's say you want to solve $\sqrt{x}=x-2$. Then you can square both sides and get $x = x^2 - 4x + 4$. If we solve this then we get $x=4$ and $x=1$.
So what we have shown is that if $\sqrt{x}=x-2$, then $x=4$ or $x=1$. We do not have the converse. Solutions of $x=x^2 - 4x + 4$ are not necessarily solutions of $\sqrt{x}=x-2$.

So any solution of $\sqrt{4x - y} - \sqrt{x+5} = -x - y +3$ is also a solutions of $(\sqrt{4x - y} - \sqrt{x+5})^2 = (-x - y +3)^2$. But not conversely. So even if you manage to solve $(\sqrt{4x - y} - \sqrt{x+5})^2 = (-x - y +3)^2$, then its solutions will not necessarily be solutions of the original equation. In fact, you will have to check which solutions are also solutions of the original equation.

 

[miniradman]

As it turned out... I could have equated both sides. I think I must have made an arithmetic errors when I tried it the first 5 times or so :p

Thank you micromass for the intuition