#### bryanandrew

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It goes like this... -3y over 6(y-1) multiplied by 2y-2 over y squared. I think I have the rigth answer but what do you think the answer is ? Can you help me please ?!!!!!!!! :mrgreen:

- Thread starter bryanandrew
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It goes like this... -3y over 6(y-1) multiplied by 2y-2 over y squared. I think I have the rigth answer but what do you think the answer is ? Can you help me please ?!!!!!!!! :mrgreen:

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bryanandrew said:

It goes like this... -3y over 6(y-1) multiplied by 2y-2 over y squared. I think I have the rigth answer but what do you think the answer is ? Can you help me please ?!!!!!!!! :mrgreen:

I'd suggest that you begin by factoring 2y - 2...then, see if there are any common factors that can be divided out of the numerator and denominator. After you've "reduced" as much as you can, THEN do the multiplication.

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\(\displaystyle \frac{-3y}{6(y - 1)} \cdot \frac{2y - 2}{y^2} \neq \frac{-6y^2 - 2}{6(y^3 - 1)}\)bryanandrew said:

How did you evaluate \(\displaystyle 6(y - 1)y^2\) and end up with \(\displaystyle 6(y^3 - 1)\)?

You are forgetting to multiply into all terms contained within the ( )

Start by factoring the numerator of the second rational:

\(\displaystyle \frac{-3y}{6(y - 1)} \cdot \frac{2y - 2}{y^2} = \frac{-3y}{6(y - 1)} \cdot \frac{2(y - 1)}{y^2}\)

Now you can start canceling

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o.k. well thank you very much for helping me, and for clarifying the problem.