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.

Recognize the squares of binomials in both the numerator and denominator to rewrite the equation as $$\begin{align}\frac{x^2-(y-1)^2}{y^2-(x-1)^2} &= 2, \end{align}$$ and thus, factoring, $$\require\cancel \begin{align}\frac{(x-y+1)\cancel{(x+y-1)}}{(y-x+1)\cancel{(y+x-1)}}&=2. \end{align}$$ Finally, let $t=x-y$ and multiply both sides by $1-t$ to have $$\begin{align} t+1&=2(1-t) \\ 3t&=1 \\ t&=\frac{1}{3}. \end{align}$$
