Question

If $x,y$ are rational number such that $x+y+(x-2y)\sqrt{2}=2x-y+(x-y-1)\sqrt{6}$, then

• A. $x$ and $y$ cannot be determined
• B. $x=2,y=1$
• C. $x=5,y=1$
• D. none of these

Answer 1

Answer: (B)

$x=2,y=1$

Given, $x+y+(x-2y)\sqrt{2}=2x-y+(x-y-1)\sqrt{6}$

$\Rightarrow x+y+\sqrt{2}x-2\sqrt{2}y=2x-y+\sqrt{6}x-\sqrt{6}y-\sqrt{6}$

$\Rightarrow y-2\sqrt{2}+y+\sqrt{6}y+\sqrt{6}=2x+\sqrt{6}x-x-\sqrt{2}x$

$\Rightarrow x(1+\sqrt{6}-\sqrt{2})=y(2-2\sqrt{2}+\sqrt{6})+\sqrt{6}$

$x=2,y=1$ satisfies this.