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?

Since
$\sqrt{2}$
and
$\sqrt{6}$
are linearly independent, their coefficients must be zero.
Therefore
$0 = x-2y = x-y-1$
so $x=2, y=1$.
They threw in
$x+y=2x-y$
for the fun of it.
