0
$\begingroup$

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

  • 1
    Notice $x^2-y^2=(x+y)(x-y)$.2012-07-09
  • 0
    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?2012-07-09
  • 0
    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 b2012-07-09
  • 0
    If by "pull $a$ out" you mean, "solve $2a(x^2+y^2)/(x+y)=b$ for $a$" then the answer is no.2012-07-09

1 Answers 1

1

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