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

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...

• Come on, no need to close this immediately as "unclear what you're asking". Give the OP the time to correct it. Apr 13, 2015 at 14:38
• I am sorry. It has been edited now. Apr 13, 2015 at 14:39
• 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. Apr 13, 2015 at 14:44
• 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 $x + 1/y = z$. Apr 13, 2015 at 14:46
• 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) Apr 13, 2015 at 14:49

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 $abc=1$.

• Very neatly done. Apr 13, 2015 at 15:46
• Thankyou! Appreciate it a lot... Apr 13, 2015 at 18:27
• 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. Apr 14, 2015 at 5:21

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

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 $c$