# 2000 AMC 12 Problems/Problem 20

## Problem

If $x,y,$ and $z$ are positive numbers satisfying

$$x + \frac{1}{y} = 4,\qquad y + \frac{1}{z} = 1, \qquad \text{and} \qquad z + \frac{1}{x} = \frac{7}{3}$$

Then what is the value of $xyz$ ?

$\text {(A)}\ 2/3 \qquad \text {(B)}\ 1 \qquad \text {(C)}\ 4/3 \qquad \text {(D)}\ 2 \qquad \text {(E)}\ 7/3$

## Solution

### Solution 1

We multiply all given expressions to get: $$(1)xyz + x + y + z + \frac{1}{x} + \frac{1}{y} + \frac{1}{z} + \frac{1}{xyz} = \frac{28}{3}$$ Adding all the given expressions gives that $$(2) x + y + z + \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 4 + \frac{7}{3} + 1 = \frac{22}{3}$$ We subtract $(2)$ from $(1)$ to get that $xyz + \frac{1}{xyz} = 2$. Hence, by inspection, $\boxed{xyz = 1 \rightarrow B}$. $$ ~AopsUser101

### Solution 2

We have a system of three equations and three variables, so we can apply repeated substitution.

$$4 = x + \frac{1}{y} = x + \frac{1}{1 - \frac{1}{z}} = x + \frac{1}{1-\frac{1}{7/3-1/x}} = x + \frac{7x-3}{4x-3}$$

Multiplying out the denominator and simplification yields $4(4x-3) = x(4x-3) + 7x - 3 \Longrightarrow (2x-3)^2 = 0$, so $x = \frac{3}{2}$. Substituting leads to $y = \frac{2}{5}, z = \frac{5}{3}$, and the product of these three variables is $1$.

## Also see

 2000 AMC 12 (Problems • Answer Key • Resources) Preceded byProblem 19 Followed byProblem 21 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.