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2000 AMC 10 Problems/Problem 15

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Problem

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

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

Solution

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

Substituting $ab=a-b$ , we get

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

$\boxed{\text{E}}$

See Also

2000 AMC 10 (Problems • Resources)
Preceded by Problem 14	Followed by Problem 16
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

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