2000 AMC 12 Problems/Problem 11

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Problem

Two non-zero real numbers, $a$ and $b,$ satisfy $ab = a - b$ . Which of the following is a possible value of $\frac {a}{b} + \frac {b}{a} - ab$ ?

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

Solution

$\frac {a}{b} + \frac {b}{a} - ab = \frac{a^2 + b^2}{ab} - (a - b) = \frac{a^2 + b^2}{a-b} - \frac{(a-b)^2}{(a-b)} = \frac{2ab}{a-b} = 2 \Rightarrow \text{(E)}$ .

Alternatively, we could test simple values, like $(a,b)=\left(1, \frac{1}{2}\right)$ , which would yield $\frac {a}{b} + \frac {b}{a} - ab=2$ .

2000 AMC 12 (Problems)
Preceded by Problem 10	Followed by Problem 12
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2000 AMC 12 Problems/Problem 11

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Problem

Solution

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