2000 AMC 12 Problems/Problem 11

The following problem is from both the 2000 AMC 12 #11 and 2000 AMC 10 #15, so both problems redirect to this page.

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 1

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

Another way is to solve the equation for $b,$ giving $b = \frac{a}{a+1};$ then substituting this into the expression and simplifying gives the answer of $2.$

Solution 2

This simplifies to $ab+b-a=0 \Rightarrow (a+1)(b-1) = -1$ . The two integer solutions to this are $(-2,2)$ and $(0,0)$ . The problem states than $a$ and $b$ are non-zero, so we consider the case of $(-2,2)$ . So, we end up with $\frac{-2}{2} + \frac{2}{-2} - 2 \cdot -2 = 4 - 1 - 1 = \boxed{2}~ \textbf{E}$

2000 AMC 12 (Problems • Answer Key • Resources)
Preceded by Problem 10	Followed by Problem 12
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All AMC 12 Problems and Solutions

2000 AMC 10 (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Problem 16
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All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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2000 AMC 12 Problems/Problem 11

Contents

Problem

Solution 1

Solution 2

See also