anonymous
  • anonymous
Simplify:
Mathematics
jamiebookeater
  • jamiebookeater
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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anonymous
  • anonymous
simplify what?
lilg132
  • lilg132
simplified
anonymous
  • anonymous
\[2x-2y \over 3x+3y \] . \[6x+6y \over x ^{2}-y ^{2}\]

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

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