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Problem 434

AMC 12 late, AIME early

Algebra Difficulty 4.8 Find the answer

6. Let $x, y, z > 0$ , satisfying $x + y = xy$ and $x + y + z = xyz$ . Then the range of $z$ is ( ).
(A) $(0, \sqrt{3}]$
(B) $(1, \sqrt{3}]$
(C) $\left(0, \frac{4}{3}\right]$
(D) $\left(1, \frac{4}{3}\right]$

Your answer

Official solution

6. D.

Given $y=\frac{x}{x-1}$ , we have
$\begin{aligned} z & =\frac{x+y}{x y-1}=\frac{x y}{x y-1}=\left(1-\frac{1}{x y}\right)^{-1} \\ & =\left(1-\frac{1}{x}+\frac{1}{x^{2}}\right)^{-1} . \end{aligned}$

Since $x>1$ , i.e., $0<\frac{1}{x}<1$ , we have
$\frac{3}{4} \leqslant 1-\frac{1}{x}+\frac{1}{x^{2}}<1 \text {. }$

Therefore, $1<z \leqslant \frac{4}{3}$ .

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.