Number theory with binary quadratic I found this questions from past year maths competition in my country, I've tried any possible way to find it, but it is just way too hard.
Given $$ \frac {x^2-y^2+2y-1}{y^2-x^2+2x-1} = 2$$ find $x-y$
I'm not sure if given choices is right... (A)2 (B)3 (C)4 (D)5 (E)6
I've tried to move them $$x^2-y^2+2y-1 = 2y^2-2x^2+4x-2$$ 
$$x^2-y^2+2y-1 - 2y^2+2x^2-4x+2 = 0$$
$$3x^2-3y^2+2y-4x+1=0$$
$$(3x-1)(x-1)-(3y-1)(y+1)+1=0$$
I've stuck in here, not sure if I've found x and y, or not...
EDIT: I've move other questions to other posts, thanks for helping me identifying the questions category.

\begin{align} 2&=  \frac {x^2-y^2+2y-1}{y^2-x^2+2x-1} \\
 &=\frac {x^2-(y^2-2y+1)}{y^2-(x^2-2x+1)} \\
 &= \frac {x^2-(y-1)^2}{y^2-(x-1)^2} \\
 &= \frac {(x-y+1)(x+y-1)}{(y+x-1)(y-x+1)}\end{align}
$$\implies  {x-y+1}=2y-2x+2\\
3(x-y)=1 $$
Thus, $x-y=\dfrac13$
