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

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2000 AMC 10 Problem 15

Below is the professionally curated solution for Problem 15 of the 2000 AMC 10, 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.

All of the real AMC 8, AMC 10, AMC 12, and AIME problems in our complete solution collection are used with official legal permission of the Mathematical Association of America (MAA).

Concepts:algebraic manipulation substitution

Difficulty rating: 1420

15.

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

$-2$

$-\dfrac12$

$\dfrac13$

$\dfrac12$

$2$

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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