Determine whether there exist real numbers x, y and z satisfying the following system of equations:

x²+4yz+2z = 0; x+2xy+2z² = 0; 2xz+y²+y+1 = 0

Well, let's see.

Eq III:
2xz + y² + y + 1 = 0
x = - (y² + y + 1)/2z  ,  z = - (y² + y + 1)/2x
Eq I:
x² + 4yz + 2z = 0
x² - 4y * (y² + y + 1)/2x - (y² + y + 1)/x = 0
2x³ = (4y + 2)(y² + y + 1)
Eq II:
x + 2xy + 2z² = 0
x + 2xy + (y² + y + 1)²/2x² = 0
2x³(1 + 2y) = - (y² + y + 1)²
2x³ = - (y² + y + 1)²/(1 + 2y)

We can now equate our findings:
(4y + 2)(y² + y + 1) = - (y² + y + 1)²/(1 + 2y)
(4y + 2)(1 + 2y) = - (y² + y + 1)
8y² + 8y + 2 = -y² - y - 1
9y² + 9y + 3 = 0
3y² + 3y + 1  = 0
This equation has no real solution, and therefore the answer is no.

Posted by TamTam on 2007-02-07 17:14:44