2000 AMC 10 Problem 15 — Solution | LIVE by Po-Shen Loh

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2000 AMC 10 Problem 15

Attempt Problem 15 of the 2000 AMC 10 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 10 solutions, or check the answer key.

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15.

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

$-2$

$-\dfrac12$

$\dfrac13$

$\dfrac12$

$2$

Answer: E

Concepts:algebraic manipulation substitution

Difficulty rating: 1420

Solution:

Over the common denominator $ab,$ $\dfrac{a}{b} + \dfrac{b}{a} - ab = \dfrac{a^2 + b^2 - (ab)^2}{ab}.$

Substituting $ab = a - b$ gives $\dfrac{a^2 + b^2 - (a - b)^2}{ab} = \dfrac{2ab}{ab} = 2.$

Thus, the correct answer is E.

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