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

Revision as of 23:15, 10 January 2009

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

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

$ab=a-b$

$\frac{a}{b}+\frac{b}{a}-ab=\frac{a^2+b^2}{ab}-ab=\frac{-a^2b^2+a^2+b^2}{ab}$

$\frac{-a^2+2ab-b^2+a^2+b^2}{ab}=2$ .

E.

2000 AMC 10 (Problems • Answer Key • 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
All AMC 10 Problems and Solutions

@@ Line 1: / Line 1: @@
 ==Problem==
+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==