If $x$ and $y$ are rational numbers such that : $(x+y)+(x-2y)\sqrt {2}=2x-y+(x-y-1)\sqrt {6}$

Question

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?

Answer 1

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 $\boxed{B)}$

Answer 2

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.