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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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The answer given in book is 1. –  Hyperbola Jul 14 '12 at 5:04
    
Why are people downvoting my question due to no reason? –  Hyperbola Jul 14 '12 at 5:15
    
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. –  Zev Chonoles Jul 14 '12 at 8:14

2 Answers 2

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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The answer given in book is 1. –  Hyperbola Jul 14 '12 at 5:14
    
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 $LHS=RHS$ –  Barun Dasgupta Jul 14 '12 at 21:46

Also $r=\frac{-qp(-3+p+q)}{(qp-3p+p^2-3q+q^2)}$ (In addition to $J-3$)

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The answer given in book is 1. –  Hyperbola Jul 14 '12 at 5:14
    
My answer agrees with Barun. I used Maple which does not make too many mistakes! –  PAD Jul 14 '12 at 7:54
1  
Did you pay a lot of money for the book? –  PAD Jul 18 '12 at 12:21

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