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$$0 = \frac2{r-1}-\frac3{r+4}+\frac1{r+5}$$

So to my understanding I could give them all the same denominator by multiplying their denominators with each others denominators and numerators. Or could I just flip them all around? But that would mean that 0 would be a denominator and I don't think I'm allowed to do that :p. Can someone please help me out?

Thanks!

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Is that $0 = \frac{2}{r-1}-\frac{3}{r+4}+\frac{1}{r+5}$? –  Peter Taylor Aug 21 '11 at 12:22
    
Yes, sorry im really bad with those math characters :( –  John Aug 21 '11 at 12:23
    
peter has it right –  John Aug 21 '11 at 12:24
    
Yes, bring them together; your denominator would be $(r-1)(r+4)(r+5)$, and the first term of your numerator will look like $2(r+4)(r+5)$... –  J. M. Aug 21 '11 at 12:29
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up vote 3 down vote accepted

As clarified in Peter's comment, the problem is

$$0 = \frac{2}{r-1}-\frac{3}{r+4}+\frac{1}{r+5}\;.$$

Excluding the poles $r=1$, $r=-4$ and $r=-5$, multiply through by the product of the denominators:

$$0=2(r+4)(r+5)-3(r-1)(r+5)+1(r-1)(r+4)\;.$$

The quadratic terms cancel, which leaves

$$ \begin{eqnarray} 0 &=& 18r+40-12r+15+3r-4 \\ &=& 9r+51\;, \end{eqnarray} $$

and so $r=-51/9=-17/3$.

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Thanks a lot for your reply! However im a bit confused how you canceled the quadratic terms. Could you please elaborate on the second step you have done? Thanks again! –  John Aug 21 '11 at 12:37
    
@John: The quadratic terms are $2r^2-3r^2+1r^2=(2-3+1)r^2=0r^2=0$. –  joriki Aug 21 '11 at 12:47
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