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Find the value of r in the following expression.

$(J - p - r - s)/q + (J - p - q - s)/r + (J - q - r - s)/p + (J - p - q - r)/s = 4$

$J, p, q, r, s$ are real and $p + q + s = 3$.

Find the value of $r$.

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    Probably because it is rude to phrase your post as a command, rather than a request for help. **Also, to get the best possible answers, you should explain what your thoughts on the problem are so far**. That way, people won't tell you things you already know, and they can write answers at an appropriate level; also, people are much more willing to help you if you show that you've tried the problem yourself.2012-07-14

2 Answers 2

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Note $(J - p - r - s)/q =(J+q-3-r)/q$ $(J - p - q - s)/r =(J-3)/r$$(J - q - r - s)/p=(J+p-3-r)/p$$ (J - p - q - r)/s = (J+s-3-r)/s$

Summing them up we get $ (J-3)(\frac{1}{p}+\frac{1}{q}+\frac{1}{s}+\frac{1}{r})-r(\frac{1}{p}+\frac{1}{q}+\frac{1}{s})+3=4$ Assuming $\frac{1}{p}+\frac{1}{q}+\frac{1}{s}=A$And$(J-3)=B$ We get $ B(A+\frac{1}{r})-rA-1=0$ Rearranging we get the following quadratic$Ar^2-rAB-B+r=0$$(rA+1)(r-B)=0$ $r=-\frac{1}{A}=\frac{-pqs}{ps+pq+qs}$$r=B=(J-3)$

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    I am not sure if under special circumstances $r=1$ but if you replace $r=J-3 $ or $r=-\frac{pqs}{ps+pq+qs}$ in your equation then you get $LHS=RHS$2012-07-14
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Also $r=\frac{-qp(-3+p+q)}{(qp-3p+p^2-3q+q^2)}$ (In addition to $J-3$)

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    Did you pay a lot of money for the book?2012-07-18