Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 4.9 AIME Find the answer

Let $x, y, z$ be real numbers satisfying $\begin{aligned} 2 x+y+4 x y+6 x z & =-6 \\ y+2 z+2 x y+6 y z & =4 \\ x-z+2 x z-4 y z & =-3 \end{aligned}$ Find $x^{2}+y^{2}+z^{2}$ .

Your answer

A number or a short expression. Spacing and $ signs are ignored.

Solution

We multiply the first, second, and third equations by $\frac{1}{2},-\frac{1}{2}$ , and -1 , respectively, then add the three resulting equations. This gives $x y+x z+y z=-2$ . Doing the same with the coefficients $-1,2$ , and 3 gives $x+y+z=5$ , from which $(x+y+z)^{2}=25$ . So $x^{2}+y^{2}+z^{2}=25-2 \cdot-2=29$ .

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