Difference between revisions of "2000 AMC 10 Problems/Problem 15"

Latest revision as of 23:54, 26 November 2011

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-==Problem==
+#REDIRECT [[2000 AMC 12 Problems/Problem 11]]
-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>.
-<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>
-==Solution==
-<math>ab=a-b</math>
-<math>\frac{a}{b}+\frac{b}{a}-ab=\frac{a^2+b^2}{ab}-ab=\frac{-a^2b^2+a^2+b^2}{ab}</math>
-<math>\frac{-a^2+2ab-b^2+a^2+b^2}{ab}=2</math>.
-E.
-==See Also==
-{{AMC10 box|year=2000|num-b=14|num-a=16}}