Browse · harp

smc

algebra senior

Problem

If

\frac{x y}{x + y} = a, \frac{x z}{x + z} = b, \frac{y z}{y + z} = c

, where

a, b, c

are other than zero, then

x

equals:

(A)

\frac{ab c}{ab + a c + b c}

(B)

\frac{2 ab c}{ab + b c + a c}

(C)

\frac{2 ab c}{ab + a c - b c}

(D)

\frac{2 ab c}{ab + b c - a c}

(E)

\frac{2 ab c}{a c + b c - ab}

Solution

Note that

\frac{1}{a} = \frac{1}{x} + \frac{1}{y}

,

\frac{1}{b} = \frac{1}{x} + \frac{1}{z}

, and

\frac{1}{c} = \frac{1}{y} + \frac{1}{z}

. Therefore

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

Therefore

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

A little algebraic manipulation yields that

x = \frac{2 ab c}{a c + b c - ab}

Final answer

E