2013 AMC 12A Problem 8 — Solution | LIVE by Po-Shen Loh

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2013 AMC 12A Problem 8

Below is the professionally curated solution for Problem 8 of the 2013 AMC 12A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2013 AMC 12A 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 factoring

Difficulty rating: 1400

8.

Given that $x$ and $y$ are distinct nonzero real numbers such that $x + \dfrac{2}{x} = y + \dfrac{2}{y},$ what is $xy?$

$\dfrac{1}{4}$

$\dfrac{1}{2}$

$1$

$2$

$4$

Solution:

Multiplying by $xy$ gives $x^2 y + 2y = xy^2 + 2x,$ so $x^2 y - xy^2 - 2x + 2y = (x - y)(xy - 2) = 0.$

Since $x \ne y,$ it follows that $xy = 2.$

Thus, the correct answer is D.

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