Question

If $\dfrac{xy}{x+y}=a$ , $\dfrac{xz}{x+z}=b$ and $\dfrac{yz}{y+z}=c$ where $a,b,c$ are all non-zero numbers, then $x$ equals to:
(a) $\dfrac{2abc}{ab+bc-ac}$
(b) $\dfrac{2abc}{ab+ac-bc}$
(c) $\dfrac{2abc}{ac+bc-ab}$
(d) $\dfrac{2abc}{ab+bc-ac}$

Answer 1

Hint: For solving this question as it is given that $a,b,c$ are all non-zero numbers so, we will first take their reciprocals and find the expression of $\dfrac{1}{a},\dfrac{1}{b},\dfrac{1}{c}$ . After that, we will find a suitable linear combination between $\dfrac{1}{a},\dfrac{1}{b},\dfrac{1}{c}$ to get the value of $x$ easily.

Complete step-by-step solution -
Given:
It is given that $\dfrac{xy}{x+y}=a$ , $\dfrac{xz}{x+z}=b$ and $\dfrac{yz}{y+z}=c$ where $a,b,c$ are all non-zero numbers and we have to find the expression of $x$ in terms of $a,b$ and $c$ .
Now, we have the following equations:
$\begin{align}
  & \dfrac{xy}{x+y}=a...................\left( 1 \right) \\
 & \dfrac{xz}{x+z}=b....................\left( 2 \right) \\
 & \dfrac{yz}{y+z}=c....................\left( 3 \right) \\
\end{align}$
Now, as it given that $a,b,c$ are all non-zero numbers so, their reciprocal will de defined. Then,
Now, we will take reciprocals of the terms on the left-hand side and right-hand side in equation (1). Then,
$\begin{align}
  & \dfrac{xy}{x+y}=a \\
 & \Rightarrow \dfrac{x+y}{xy}=\dfrac{1}{a} \\
 & \Rightarrow \dfrac{1}{y}+\dfrac{1}{x}=\dfrac{1}{a}....................\left( 4 \right) \\
\end{align}$
Now, we will take reciprocals of the terms on the left-hand side and right-hand side in equation (2). Then,
$\begin{align}
  & \dfrac{xz}{x+z}=b \\
 & \Rightarrow \dfrac{x+z}{xz}=\dfrac{1}{b} \\
 & \Rightarrow \dfrac{1}{z}+\dfrac{1}{x}=\dfrac{1}{b}....................\left( 5 \right) \\
\end{align}$
Now, we will take reciprocals of the terms on the left-hand side and right-hand side in equation (3). Then,
$\begin{align}
  & \dfrac{yz}{y+z}=c \\
 & \Rightarrow \dfrac{y+z}{yz}=\dfrac{1}{c} \\
 & \Rightarrow \dfrac{1}{z}+\dfrac{1}{y}=\dfrac{1}{c}....................\left( 6 \right) \\
\end{align}$
Now, add the equation (4) and equation (5). Then,
$\begin{align}
  & \dfrac{1}{y}+\dfrac{1}{x}+\dfrac{1}{z}+\dfrac{1}{x}=\dfrac{1}{a}+\dfrac{1}{b} \\
 & \Rightarrow \left( \dfrac{1}{z}+\dfrac{1}{y} \right)+\dfrac{2}{x}=\dfrac{1}{a}+\dfrac{1}{b} \\
\end{align}$
Now, substitute the value of $\dfrac{1}{z}+\dfrac{1}{y}=\dfrac{1}{c}$ from equation (6) in the above equation. Then,
$\begin{align}
  & \left( \dfrac{1}{z}+\dfrac{1}{y} \right)+\dfrac{2}{x}=\dfrac{1}{a}+\dfrac{1}{b} \\
 & \Rightarrow \dfrac{1}{c}+\dfrac{2}{x}=\dfrac{1}{a}+\dfrac{1}{b} \\
 & \Rightarrow \dfrac{2}{x}=\dfrac{1}{a}+\dfrac{1}{b}-\dfrac{1}{c} \\
 & \Rightarrow \dfrac{2}{x}=\dfrac{bc+ac-ab}{abc} \\
 & \Rightarrow \dfrac{x}{2}=\dfrac{abc}{ac+bc-ab} \\
 & \Rightarrow x=\dfrac{2abc}{ac+bc-ab} \\
\end{align}$
Now, from the above result, we conclude that $x=\dfrac{2abc}{ac+bc-ab}$ .
Hence, option (c) will be the correct option.

Note: Here, the student should first understand what is asked in the question and then proceed in the right direction to get the correct answer quickly. Moreover, for solving this question quickly if we analyse the options like in option (a) it is given that $\dfrac{2abc}{ab+bc-ac}$ which can be written as $\dfrac{2}{\left( \dfrac{1}{c}+\dfrac{1}{a}-\dfrac{1}{b} \right)}$ and similar type of expression in other options also so, from here we come to know that for $x$ we have to first solve for $\dfrac{1}{a},\dfrac{1}{b},\dfrac{1}{c}$ . After that, we can easily solve for the $x$ .