# help finding the common denominator with this rational expression

• Nov 18th 2010, 05:44 PM
NecroWinter
help finding the common denominator with this rational expression
4xy/x^2-y^2 + x-y/x+y

im completely lost on this particular problem.

if someone can point me in the correct direction here that would be great, somehow i got 4+x-y/(x-y)^2
• Nov 18th 2010, 05:54 PM
skeeter
Quote:

Originally Posted by NecroWinter
4xy/x^2-y^2 + x-y/x+y

im completely lost on this particular problem.

if someone can point me in the correct direction here that would be great, somehow i got 4+x-y/(x-y)^2

note that the common denominator is $x^2-y^2 = (x+y)(x-y)$
• Nov 18th 2010, 05:58 PM
harish21
$\dfrac{4xy}{x^2-y^2} + \dfrac{x-y}{x+y}$

$= \dfrac{4xy}{(x+y)(x-y)}+ \dfrac{x-y}{x+y}$

$=\dfrac{4xy+(x-y)(x-y)}{(x+y)(x-y)}$

$=\dfrac{4xy+(x-y)^2}{(x+y)(x-y)}$

simplify the numerator..
• Nov 18th 2010, 06:27 PM
NecroWinter
Quote:

Originally Posted by harish21
$\dfrac{4xy}{x^2-y^2} + \dfrac{x-y}{x+y}$

$= \dfrac{4xy}{(x+y)(x-y)}+ \dfrac{x-y}{x+y}$

$=\dfrac{4xy+(x-y)(x-y)}{(x+y)(x-y)}$

$=\dfrac{4xy+(x-y)^2}{(x+y)(x-y)}$

simplify the numerator..

im a little confused

the only thing i would know to do is to start cancelling things out

when i see (x-y)^2 i assume (x+y)(x-y) and i can cancel out the denom and be left with 4xy.
• Nov 18th 2010, 06:29 PM
skeeter
Quote:

Originally Posted by NecroWinter
im a little confused

the only thing i would know to do is to start cancelling things out

when i see (x-y)^2 i assume (x+y)(x-y) and i can cancel out the denom and be left with 4xy.

numerator ... $4xy + (x-y)^2 = 4xy + (x^2 - 2xy + y^2) = x^2 + 2xy + y^2 = (x+y)^2$
• Nov 18th 2010, 06:34 PM
NecroWinter
Quote:

Originally Posted by skeeter
numerator ... $4xy + (x-y)^2 = 4xy + (x^2 - 2xy + y^2) = x^2 + 2xy + y^2 = (x+y)^2$

thanks i get it now :)

but i have one more small question

when you have (x-y)^2 how do you know that it isnt a difference of two squares, but rather a perfect square trinomial?
• Nov 18th 2010, 06:38 PM
skeeter
Quote:

Originally Posted by NecroWinter
thanks i get it now :)

but i have one more small question

when you have (x-y)^2 how do you know that it isnt a difference of two squares, but rather a perfect square trinomial?

a difference of two squares is $x^2 - y^2$ , not $(x-y)^2$ , which is the square of a difference.