2000 AMC 10 Problems/Problem 15

Problem

Two non-zero real numbers, $a$ and $b$ , satisfy $ab=a-b$ . Find a possible value of $\frac{a}{b}+\frac{b}{a}-ab$ .

$\mathrm{(A)}\ -2 \qquad\mathrm{(B)}\ -\frac{1}{2} \qquad\mathrm{(C)}\ \frac{1}{3} \qquad\mathrm{(D)}\ \frac{1}{2} \qquad\mathrm{(E)}\ 2$

Solution

$\begin{align*} \frac{a}{b}+\frac{b}{a}-ab &= \frac{a^2+b^2}{ab}-ab \ &= \frac{-a^2b^2+a^2+b^2}{ab}\ \end(align*}$ (Error compiling LaTeX. Unknown error_msg)

Substituting $ab=a-b$ , we get

$ $\begin{aligned} \frac{- (a + b)^{2} + a^{2} + b^{2}}{a b} & = \frac{- a^{2} + 2 a b - b^{2} + a^{2} + b^{2}}{a b} \\ = \frac{2 a b}{a b} \\ = 2 \end{aligned}$ $ (Error compiling LaTeX. Unknown error_msg)

$\boxed{\text{E}}$

2000 AMC 10 (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Problem 16
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2000 AMC 10 Problems/Problem 15

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