Question about a solution of a system of three non linear equations in three unknowns

Question

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...

Answer 1

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 $$b=\frac54,a=\frac{11}5,c=\frac4{11}.$$

Answer 2

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 $c$

Answer 3

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 $abc=1$.