Question : If $\frac{xy}{x+y}=a$, $\frac{xz}{x+z}=b$ and $\frac{yz}{y+z}=c$, where $a,b,c$ are all non-zero numbers, $x$ equals to:
Option 1: $\frac{2abc}{ab+bc–ac}$
Option 2: $\frac{2abc}{ab+ac–bc}$
Option 3: $\frac{2abc}{ac+bc–ab}$
Option 4: $\frac{2abc}{ab+bc+ac}$
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Correct Answer: $\frac{2abc}{ac+bc–ab}$
Solution :
Given: $\frac{xy}{x+y}=a$, $\frac{xz}{x+z}=b$ and $\frac{yz}{y+z}=c$, where $a,b,c$ are all non-zero numbers.
We can write $\frac{xy}{x+y}=\frac{1}{\frac{1}{x}+\frac{1}{y}}$
⇒ $\frac{1}{a}=\frac{1}{x}+\frac{1}{y}$ -------------------------------(1)
Similarly, we can write,
$\frac{xz}{x+z}=\frac{1}{\frac{1}{x}+\frac{1}{z}}$
⇒ $\frac{1}{b}=\frac{1}{x}+\frac{1}{z}$ --------------------------------(2)
Similarly, $\frac{1}{c}=\frac{1}{y}+\frac{1}{z}$ ----------------------(3)
Adding equation (1) and equation (2),
⇒ $\frac{1}{a}+\frac{1}{b}=\frac{2}{x}+\frac{1}{y}+\frac{1}{z}$
⇒ $\frac{1}{a}+\frac{1}{b}=\frac{2}{x}+\frac{1}{c}$
⇒ $\frac{1}{a}+\frac{1}{b}–\frac{1}{c}=\frac{2}{x}$
⇒ $\frac{(bc+ac–ab)}{abc}=\frac{2}{x}$
$\therefore x=\frac{2abc}{(bc+ac–ab)}$
Hence, the correct answer is $\frac{2abc}{(bc+ac–ab)}$.
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