Multiplying rational expressions

bryanandrew

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Feb 1, 2008
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I think I know how to do this kind of work but I just want to make sure that I'm doing the math right. The 1st subject today is multiplying rational expressions. The problem is... well it's kinda hard to list fractions so.....

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:
 

o_O

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

Why don't you show us what you have done and we'll tell you whether you made any errors or not?
 

bryanandrew

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o.k. well what I did is I multiplied and got -6y squared minus 2 over 6(y cubed -1) then I simplified the equation and got -3y squared minus 1 over 3(y cubed - 1), is there anything wrong that I did in my steps?
 

Mrspi

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Dec 17, 2005
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bryanandrew said:
I think I know how to do this kind of work but I just want to make sure that I'm doing the math right. The 1st subject today is multiplying rational expressions. The problem is... well it's kinda hard to list fractions so.....

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.
 

bryanandrew

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Feb 1, 2008
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I see where you're going, but the steps I showed is how I was taught. I'm not trying to say you're wrong, because I know that you know more than me in math, so do you know if you can solve it the way she taught me because the answer that you'll give me and the answer that is in the book, they might be different, and I'll get it wrong. Can you still help me, please?
 

jwpaine

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Mar 10, 2007
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bryanandrew said:
o.k. well what I did is I multiplied and got -6y squared minus 2 over 6(y cubed -1) then I simplified the equation and got -3y squared minus 1 over 3(y cubed - 1), is there anything wrong that I did in my steps?
\(\displaystyle \frac{-3y}{6(y - 1)} \cdot \frac{2y - 2}{y^2} \neq \frac{-6y^2 - 2}{6(y^3 - 1)}\)

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 very easily
 

bryanandrew

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Feb 1, 2008
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o.k. well thank you very much for helping me, and for clarifying the problem.
 
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