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 March 4th, 2012, 08:31 PM #1 Member   Joined: Feb 2012 Posts: 97 Thanks: 0 Add and Subtract Indicated $\frac{5}{2r+12} - \frac{1}{r}$ Okay, I know that I'm suppose to multiply the top and bottom so that the denominator is alike, but it's not. Ugh, I broke the bottom down to be $2(r+6)$ in the first column and this is where I'm lost, the "r" . . . I've tried making both denominatiors 2r(r+6) by multiplying 2(r+6) just lost
March 4th, 2012, 08:39 PM   #2
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Quote:
 Originally Posted by HellBunny ... I've tried making both denominators 2r(r+6) by multiplying [color=#000080]*** 1/r by ****[/color]2(r+6) just lost
That's good.
You'll also need to multiply the first term by r/r

 March 4th, 2012, 08:43 PM #3 Member   Joined: Feb 2012 Posts: 97 Thanks: 0 Re: Add and Subtract Indicated It's still not coming out the right answer,
 March 4th, 2012, 08:55 PM #4 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1038 Re: Add and Subtract Indicated What's the "right answer"? I get: 3(r - 4) --------- 2r(r+6) ??
 March 4th, 2012, 08:57 PM #5 Member   Joined: Feb 2012 Posts: 97 Thanks: 0 Re: Add and Subtract Indicated That's the answer. But what do you have on the second denominator when you're multiplying?
 March 4th, 2012, 09:12 PM #6 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1038 Re: Add and Subtract Indicated Not sure what you're asking, HB, but here's my steps: 5 / (2r + 12) - 1 / r = [5r - 1(2r + 12)] / [r(2r + 12)] = (5r - 2r - 12) / (2r^2 + 12r) = (3r - 12) / [2r(r + 6)] = [3(r - 4)] / [2r(r + 6)] Like: a/b - c/d = (ad - bc) / bd Hope that helps instead of confuses Being 7/10 bald-headed, I musy say I'm jealous of your Afro
 March 4th, 2012, 11:25 PM #7 Member   Joined: Feb 2012 Posts: 97 Thanks: 0 Re: Add and Subtract Indicated Thanks, I'll give you some hair.

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