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Algebra Difficulty 4.8 AIME Prove it Slovenia

Find all real numbers xx, yy and zz which are solutions of the system of equations
x+y−2z=0,xy−z2=0,y2+5z+6=0. x + y - 2z = 0, \quad xy - z^2 = 0, \quad y^2 + 5z + 6 = 0.

Solution

From the first equation we get x=−2z−yx = -2z - y. Inserting this into the second equation we get (−2z−y)y−z2=0(-2z - y)y - z^2 = 0 or, equivalently, −(z+y)2=0-(z + y)^2 = 0. This implies that z=−yz = -y. Finally, we use this together with the third equation to obtain y2−5y+6=0y^2 - 5y + 6 = 0 or (y−2)(y−3)=0(y - 2)(y - 3) = 0. Hence, y=2y = 2 or y=3y = 3.
There are two solutions: x=2x = 2, y=2y = 2, z=−2z = -2 and x=3x = 3, y=3y = 3, z=−3z = -3.

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