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Let $a$, $b$ and $c$ be positive real numbers such that $$ a + \frac{1}{b} = 3$$ $$b + \frac{1}{c} = 4$$ $$ c + \frac{1}{a} = \frac{9}{11} $$ then $$ a \times b \times c =?$$

I tried doing this problem but I was unsuccessful. Tried a lot but couldn't get the answer! The answer is a numerical value...

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    $\begingroup$ Come on, no need to close this immediately as "unclear what you're asking". Give the OP the time to correct it. $\endgroup$ Apr 13, 2015 at 14:38
  • $\begingroup$ I am sorry. It has been edited now. $\endgroup$
    – Aditya Pai
    Apr 13, 2015 at 14:39
  • $\begingroup$ Could you give an outline of your approach? I'm not asking to include everything you've tried, but you may make it easier for 'us' if you show some possible approaches. $\endgroup$ Apr 13, 2015 at 14:44
  • $\begingroup$ But sir, I guess it is not the same case here in this question. Here, it isn't $ x + 1/x$, here it is sort of $ x + 1/y = z $. $\endgroup$
    – Aditya Pai
    Apr 13, 2015 at 14:46
  • $\begingroup$ Okay, so basically there are 3 variables and 3 equations, so it is possible to find the value of a, b and c. However, I was unsuccessful in doing so. (I tried substituting) $\endgroup$
    – Aditya Pai
    Apr 13, 2015 at 14:49

3 Answers 3

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Let $a$, $b$ and $c$ be positive real numbers such that

\begin{align} a + \frac{1}{b} &= 3 \tag{1}\label{1} \\ b + \frac{1}{c} &= 4 \tag{2}\label{2} \\ c + \frac{1}{a} &= \frac{9}{11} \tag{3}\label{3} \end{align}

\eqref{1}$\times$\eqref{2}$\times$\eqref{3} $-$\eqref{1}$-$\eqref{2}$-$\eqref{3} gives: \begin{align} abc+\frac{1}{abc}&=2, \end{align}

hence $abc=1$.

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  • $\begingroup$ Very neatly done. $\endgroup$ Apr 13, 2015 at 15:46
  • $\begingroup$ Thankyou! Appreciate it a lot... $\endgroup$
    – Aditya Pai
    Apr 13, 2015 at 18:27
  • $\begingroup$ While the deduction is in itself complete, I think an argument to show that a solution actually exists would be appropriate; if this were false then any other outcome could also be obtained. $\endgroup$ Apr 14, 2015 at 5:21
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Eliminate $a$ and $c$: $$a=3-\frac1b$$ $$c=\frac1{4-b}$$

Then

$$c+\frac1a=\frac1{4-b}+\frac b{3b-1}=\frac9{11},$$ can be rewritten $$16b^2-40b+25=(4b-5)^2=0.$$ Hence $$b=\frac54,a=\frac{11}5,c=\frac4{11}.$$

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Here is a way we have : $$b=\dfrac{1}{3-a}\ \ \ c=\dfrac{1}{4-b}\tag1$$ hence : $$\dfrac{1}{4-\dfrac{1}{3-a}}+\frac{1}{a}=\frac{3-a}{11-4a}+\frac{1}{a}=\frac{9}{11}$$

so $11((3-a)a-11+4a)=9a(11-4a)$ or $25a^2-110a+121=0$ so $a=\frac{11}{5}$ and here you can find $a$ and repalce in $(1)$ to find $b$ and $c$

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