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

$\begin{align*} \frac{-(a+b)^2+a^2+b^2}{ab} &= \frac{-a^2+2ab-b^2+a^2+b^2}{ab} \\ &= \frac{2ab}{ab} \\ &= 2 \\ \end{align*}$ (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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