# Can we solve $2a(x^2-y^2)/(x-y)=b$ for $a$ without multiplying $b$ by $x-y$?

I would like to know if its possible to pull $a$ out of the following equation without multiplying $b$ by $(x-y)$

$$\frac{ 2a(x^2 - y^2)}{x - y} = b$$

Its part of a more complex problem I'm stuck on.

Cheers

-
Notice $x^2-y^2=(x+y)(x-y)$. – Jacqueline Pauwels Jul 9 '12 at 2:19
I'm not really sure what you're asking for. Perhaps $$\frac{2(x^2-y^2)}{(x-y)}=\frac{b}{a}$$ is what you want? – Alex Becker Jul 9 '12 at 2:19
Awesome, thanks Jakucha. What if the - was a + $$\frac{ 2a(x^2 + y^2) }{ (x + y)}$$ Is it possible to pull a out without multiplying or dividing b – user346443 Jul 9 '12 at 2:34
If by "pull $a$ out" you mean, "solve $2a(x^2+y^2)/(x+y)=b$ for $a$" then the answer is no. – Gerry Myerson Jul 9 '12 at 6:13

Yes indeed, you have the identity $$x^2 - y^2 = (x-y)(x+y)$$ So, $$\dfrac{2a (x^2-y^2)}{x-y}=b \Leftrightarrow 2a(x+y)=b$$