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meow18Best ResponseYou've already chosen the best response.0
\[2x2y \over 3x+3y \] . \[6x+6y \over x ^{2}y ^{2}\]
 2 years ago

meow18Best ResponseYou've already chosen the best response.0
Answer choices are: A. 4 B. \[2 \over 3(xy)\] C. \[4 \over x+y\]
 2 years ago

meow18Best ResponseYou've already chosen the best response.0
I put the . there to show its times because I don't know how to put the dot in between the problems
 2 years ago

satellite73Best ResponseYou've already chosen the best response.2
\[\frac{2x2y}{3x+3y}\times \frac{6x+6y}{x^2y^2}\]
 2 years ago

satellite73Best ResponseYou've already chosen the best response.2
lot of factoring and canceling for this one you need the steps?
 2 years ago

satellite73Best ResponseYou've already chosen the best response.2
actually factors as this \[\frac{2(xy)6(x+y)}{3(x+y)(x+y)(xy)}\]
 2 years ago

meow18Best ResponseYou've already chosen the best response.0
What do I have to do to factor it? I'm confused.
 2 years ago

satellite73Best ResponseYou've already chosen the best response.2
\[x^2y^2=(x+y)(xy)\]
 2 years ago

satellite73Best ResponseYou've already chosen the best response.2
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
 2 years ago

satellite73Best ResponseYou've already chosen the best response.2
when you get this \[\frac{2(xy)6(x+y)}{3(x+y)(x+y)(xy)}\] then you get rid of all common factors
 2 years ago

satellite73Best ResponseYou've already chosen the best response.2
\[\frac{2\cancel{(xy)}6(x+y)}{3(x+y)(x+y)\cancel{(xy)}}\]
 2 years ago

satellite73Best ResponseYou've already chosen the best response.2
\[\frac{12(x+y)}{3(x+y)(x+y)}\]
 2 years ago

satellite73Best ResponseYou've already chosen the best response.2
\[\frac{12\cancel{(x+y)}}{3\cancel{(x+y)}(x+y)}\]
 2 years ago

satellite73Best ResponseYou've already chosen the best response.2
so finally you get \[\frac{4}{x+y}\]
 2 years ago

meow18Best ResponseYou've already chosen the best response.0
Exactly what I was thinking!! (: Thank you very much!!
 2 years ago
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