Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 4.7 AIME Prove it United States

Problem:

Find all ordered pairs of real numbers $(x, y)$ such that $x^{2} y = 3$ and $x + x y = 4$ .

Solution

Solution:

Answer: $(1, 3), \left(3, \frac{1}{3}\right)$

Multiplying the second equation by $x$ gives
$x^{2} + x^{2} y = 4x$
and substituting our known value of $x^{2} y$ gives the quadratic
$x^{2} - 4x + 3 = 0$
so $x = 1$ or $x = 3$ . Hence, we obtain the solutions $(x, y) = (1, 3), (3, 1/3)$ .

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