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| ==Problem==
| | #REDIRECT [[2000 AMC 12 Problems/Problem 11]] |
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| Two non-zero real numbers, <math>a</math> and <math>b</math>, satisfy <math>ab=a-b</math>. Find a possible value of <math>\frac{a}{b}+\frac{b}{a}-ab</math>.
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| <math>\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</math>
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| ==Solution==
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| <center>
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| <math>\begin{align*}
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| \frac{a}{b}+\frac{b}{a}-ab &= \frac{a^2+b^2}{ab}-ab \\
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| &= \frac{-a^2b^2+a^2+b^2}{ab}\\
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| \end(align*}</math>
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| </center>
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| Substituting <math>ab=a-b</math>, we get
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| <center>
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| <math>\begin{align*}
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| \frac{-(a-b)^2+a^2+b^2}{ab} &= \frac{-a^2+2ab-b^2+a^2+b^2}{ab} \\
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| &= \frac{2ab}{ab} \\
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| &= 2 \\
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| \end{align*}</math>
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| </center>
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| <math>\boxed{\text{E}}</math>
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| ==See Also==
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| {{AMC10 box|year=2000|num-b=14|num-a=16}}
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