Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 5.2 AIME, harder Prove it Philippines

Problem:
Find all real solutions of the following nonlinear system:
$\begin{aligned} x + 4y + 6z &= 16 \\ x + 6y + 12z &= 24 \\ x^{2} + 4y^{2} + 36z^{2} &= 76 \end{aligned}$

Solution

Solution:
Using the first two equations, we can obtain $y = 4 - 3z$ (by eliminating $x$ ) and $x = 6z$ (by eliminating $y$ ).

Substituting to the third equation gives $(6z)^2 + 4(4 - 3z)^2 + 36z^2 = 76$ , which simplifies to $9z^2 - 8z - 1 = 0$ .

This gives $z = -1/9$ or $z = 1$ .

Substituting back to the expressions for $x$ and $y$ gives the required answers.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.