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Slovenia — National Math Olympiad

Algebra Difficulty 4.8 AIME Prove it Slovenia

Find all real numbers $x$ , $y$ and $z$ which are solutions of the system of equations
$x + y - 2z = 0, \quad xy - z^2 = 0, \quad y^2 + 5z + 6 = 0.$

Solution

From the first equation we get $x = -2z - y$ . Inserting this into the second equation we get $(-2z - y)y - z^2 = 0$ or, equivalently, $-(z + y)^2 = 0$ . This implies that $z = -y$ . Finally, we use this together with the third equation to obtain $y^2 - 5y + 6 = 0$ or $(y - 2)(y - 3) = 0$ . Hence, $y = 2$ or $y = 3$ .
There are two solutions: $x = 2$ , $y = 2$ , $z = -2$ and $x = 3$ , $y = 3$ , $z = -3$ .

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