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 March 4th, 2012, 06:02 PM #1 Member   Joined: Feb 2012 Posts: 97 Thanks: 0 Add and Subtract Indicated $\frac{4p}{(p+1)(p+5)}-\frac{3p}{(p+1)(p+4)}$ I constantly get the wrong answer for this problem. I did multiply the denominator so that they both have equal sides a the bottom and multiplied the top by the denominator that I placed there. But the answer is still wrong.
 March 4th, 2012, 06:14 PM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Add and Subtract Indicated You want to multiply the first term by $\frac{p+4}{p+4}$ and the second term by $\frac{p+5}{p+5}$: $\frac{4p}{(p+1)(p+5)}\cdot\frac{p+4}{p+4}+\frac{3p }{(p+1)(p+4)}\cdot\frac{p+5}{p+5}=$ $\frac{4p(p+4)+3p(p+5)}{(p+1)(p+4)(p+5)}=\frac{4p^2 +16p+3p^2+15p}{(p+1)(p+4)(p+5)}=$ $\frac{7p^2+31p}{(p+1)(p+4)(p+5)}=\frac{p(7p+31)}{( p+1)(p+4)(p+5)}$
 March 4th, 2012, 06:21 PM #3 Member   Joined: Feb 2012 Posts: 97 Thanks: 0 Re: Add and Subtract Indicated For some reason the anw.outcome is$P/(p+4)(p+5)$
 March 4th, 2012, 06:30 PM #4 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Add and Subtract Indicated That's because I added instead of subtracting! You want to multiply the first term by $\frac{p+4}{p+4}$ and the second term by $\frac{p+5}{p+5}$: $\frac{4p}{(p+1)(p+5)}\cdot\frac{p+4}{p+4}-\frac{3p}{(p+1)(p+4)}\cdot\frac{p+5}{p+5}=$ $\frac{4p(p+4)-3p(p+5)}{(p+1)(p+4)(p+5)}=\frac{4p^2+16p-3p^2-15p}{(p+1)(p+4)(p+5)}=$ $\frac{p^2+p}{(p+1)(p+4)(p+5)}=\frac{p(p+1)}{(p+1)( p+4)(p+5)}=\frac{p}{(p+4)(p+5)}$
 March 4th, 2012, 06:54 PM #5 Member   Joined: Feb 2012 Posts: 97 Thanks: 0 Re: Add and Subtract Indicated Hahaha. It's okay, lol. Thanks. I
March 4th, 2012, 09:18 PM   #6
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Re: Add and Subtract Indicated

Quote:
 Originally Posted by MarkFL That's because I added instead of subtracting!
Go stand in the corner...

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