Question

If $\dfrac{xy}{x+y}=a,$ $\dfrac{xz}{x+z}=b,$ and $\dfrac{yz}{y+z}=c,$ where $a,$ $b,$ $c$ are other than zero, then $x$ equals:

$\dfrac{abc}{ab+ac+bc}$

$\dfrac{2abc}{ab+bc+ac}$

$\dfrac{2abc}{ab+ac-bc}$

$\dfrac{2abc}{ab+bc-ac}$

$\dfrac{2abc}{ac+bc-ab}$

Answer: E

Concepts:system of equations algebraic manipulation

Difficulty rating: 1830

Small Hint:

Invert all three given equations to make them linear in $\frac{1}{x},\frac{1}{y},\frac{1}{z}$

Big Hint:

Compute $\dfrac1b+\dfrac1a-\dfrac1c=\dfrac2x$

Solution:

Inverting gives $\begin{gathered} \frac1x+\frac1y=\frac1a,\\ \frac1x+\frac1z=\frac1b,\\ \frac1y+\frac1z=\frac1c. \end{gathered}$ Adding the first two and subtracting the third yields $\begin{aligned} \frac2x &=\frac1a+\frac1b-\frac1c\\ &=\frac{ac+bc-ab}{abc}. \end{aligned}$ Hence $x=\dfrac{2abc}{ac+bc-ab}.$

Thus, the correct answer is E.

1951 AMC 12 Problem 44

Answer: E

Difficulty rating: 1830

Small Hint:

Big Hint:

Solution:

Problem 44 in Other Years