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

Combine

\dfrac{a}{b} + \dfrac{b}{a}

over the common denominator

ab

Replace

ab

with

a - b

inside

(ab)^2

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.

Problem 15 in other years