2023 AMC 12A Problem 10 — Solution | LIVE by Po-Shen Loh

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2023 AMC 12A Problem 10

Below is the professionally curated solution for Problem 10 of the 2023 AMC 12A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2023 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:system of equations substitution

Difficulty rating: 1560

10.

Positive real numbers $x$ and $y$ satisfy $y^3=x^2$ and $(y-x)^2=4y^2.$ What is $x+y?$

$12$

$18$

$24$

$36$

$42$

Solution:

From $(y-x)^2=4y^2$ we get $y-x=\pm 2y.$ The choice $y-x=2y$ gives $x=-y\lt 0,$ impossible, so $y-x=-2y,$ meaning $x=3y.$

Substituting into $y^3=x^2=9y^2$ gives $y=9,$ hence $x=27$ and $x+y=36.$

Thus, the correct answer is D.

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