2000 AMC 12 Problem 11

Attempt Problem 11 of the 2000 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2000 AMC 12 solutions, or check the answer key.

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11.

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

$-2$

$-\dfrac{1}{2}$

$\dfrac{1}{3}$

$\dfrac{1}{2}$

$2$

Answer: E

Concepts:algebraic manipulation substitution

Difficulty rating: 1530

Solution:

Combining over a common denominator, $\frac{a}{b} + \frac{b}{a} - ab = \frac{a^2 + b^2 - (ab)^2}{ab}.$

Replacing $ab$ with $a - b$ in the numerator, $\begin{gathered} a^2 + b^2 - (a - b)^2 \\ = a^2 + b^2 \\ {}- (a^2 - 2ab + b^2) \\ = 2ab. \end{gathered}$

Therefore the expression equals $\dfrac{2ab}{ab} = 2.$

Thus, the correct answer is E.

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