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, then x equals to :

• A. $\frac{2abc}{ab+bc-ac}$
• B. $\frac{2abc}{ab+ac-bc}$
• C. $\frac{2abc}{ac+bc-ab}$
• D. $\frac{2abc}{ab+bc-ac}$

Answer 1

The correct option is C $\frac{2abc}{ac+bc-ab}$

$\frac{xy}{(x+y)}$ is the same as $\frac{1}{(\frac{1}{x}+\frac{1}{y}})$

so we can rewrite $\frac{xy}{(x+y)}=a$

as

$\frac{1}{x}+\frac{1}{y}=\frac{1}{a}-\left(1\right)$

also rewrite $\frac{xz}{(x+z)}=b$

as

$\frac{1}{x}+\frac{1}{z}=\frac{1}{b}-\left(2\right)$

add (1) and (2)

$⟹\frac{2}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{a}+\frac{1}{b}$

but we also know that $\frac{1}{z}+\frac{1}{y}=\frac{1}{c}$

$⟹\frac{2}{x}+\frac{1}{c}=\frac{1}{a}+\frac{1}{b}$

$⟹\frac{2}{x}=\frac{1}{a}+\frac{1}{b}-\frac{1}{c}=\frac{(bc+ac-ab)}{abc}$

$⟹x=\frac{2abc}{(bc+ac-ab)}$