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simplify what?
simplified
\[2x-2y \over 3x+3y \] . \[6x+6y \over x ^{2}-y ^{2}\]

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Other answers:

Answer choices are: A. 4 B. \[2 \over 3(x-y)\] C. \[4 \over x+y\]
I put the . there to show its times because I don't know how to put the dot in between the problems
Yes that is the problem
\[\frac{2x-2y}{3x+3y}\times \frac{6x+6y}{x^2-y^2}\]
\[\frac{4}{x+y}\]
lot of factoring and canceling for this one you need the steps?
Yes please! (:
actually factors as this \[\frac{2(x-y)6(x+y)}{3(x+y)(x+y)(x-y)}\]
What do I have to do to factor it? I'm confused.
\[2x-2y=2(x-y)\]
\[6x+6y=6(x+y)\]
\[3x+3y=3(x+y)\]
\[x^2-y^2=(x+y)(x-y)\]
so you factor as much as you can to see what factors you might have in common in the numerator and denominator. then you cancel each factor you have in common to reduce the fraction
Alright thanks! (:
when you get this \[\frac{2(x-y)6(x+y)}{3(x+y)(x+y)(x-y)}\] then you get rid of all common factors
\[\frac{2\cancel{(x-y)}6(x+y)}{3(x+y)(x+y)\cancel{(x-y)}}\]
\[\frac{12(x+y)}{3(x+y)(x+y)}\]
\[\frac{12\cancel{(x+y)}}{3\cancel{(x+y)}(x+y)}\]
so finally you get \[\frac{4}{x+y}\]
Exactly what I was thinking!! (: Thank you very much!!
yw

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