If $x$ and $y$ are rational numbers such that : $(x+y)+(x-2y)\sqrt {2}=2x-y+(x-y-1)\sqrt {6}$ 
If $x$ and $y$ are rational numbers such that : $$(x+y)+(x-2y)\sqrt {2}=2x-y+(x-y-1)\sqrt {6}$$ then which of the following is true?

A) $\;x=1$, $y=1$
B) $\;x=2$, $y=1$
C) $\;x=5$, $y=1$
D) $\;x$ and $y$ can take infinitely many values.
Given:
$$(x+y)+(x-2y)\sqrt {2} = (2x-y)+(x-y-1)\sqrt {6}$$
$$x+y-2x+y+(x-2y) \sqrt {2}=(x-y-1)\sqrt {6}$$
$$(x-2y)\sqrt {2} - (x-2y)=(x-y-1)\sqrt {6}$$
$$(x-2y)(\sqrt {2} -1)=(x-y-1)\sqrt {6}$$
Only the values at option B satisfies the equation. I just did that by hit and trial. Is there any elaboration on that?

And why is this down voted?
$$2y-x = \sqrt{2}\Big(2y-x+\sqrt{3}(x-y-1)\Big)\;\;\;\;\;\;/^2$$
$${(2y-x)^2\over 2}+x-2y = \sqrt{3}(x-y-1)$$
If $x-y-1\ne 0$ we have
$$\sqrt {3}= \underbrace{{(2y-x)^2\over 2}+x-2y\over x-y-1}_{\in\mathbb{Q}}$$ 
which is impossible. So $y=x-1$ and $${(2y-x)^2\over 2}+x-2y=0$$
Solve this system and you are done.
Edit: So $(x-2)^2+2x-4x+4=0$ so $x^2-6x+8=0$ So $x=2$ (and $y=1$) or $x=4$ and ($y=3$). So the answer is $\boxed{B)}$
