How to solve the following system : $ begin{cases} 2x^2 +x + z = 0 \ (1-z) x^2 + 3x - y = 0 \ (2+y) x^2 + 4x - 2 = 0 end{cases} $?

$$ begin{cases} 2x^2 +x + z = 0 \ (1-z) x^2 + 3x - y = 0 \ (2+y) x^2 + 4x - 2 = 0 end{cases} $$

subtract equation 1 from 3 :

$$yx^2+3x-z=0$$

Make a system with equation 2:

$$ begin{cases} yx^2+3x-z=0 \ (1-z) x^2 + 3x - y = 0 end{cases} $$

ُThe value of x from both equations must be equal, so we must have:

$x=frac {3 ± sqrt{9+4yz}}y=frac{3 ±sqrt{9+4y(1-z)}}{1-z}$

Now we may assume that:

$3 ±sqrt{9+4yz}= 3 ±sqrt{9+4y(1-z)}$ ⇒ $yz=y(1-z)$

$y=1-z$

Now solve following system and find y and z. Plugging z in first equation will give you two roots for x.

$$ begin{cases} y(1-z)=yz\ y=1-z end{cases} $$

Hijacking the concept behind Michael Rozenberg's comment, the first thing to do is check that you copied the problem correctly.

Assuming so,
let $f(x) = -(2x^2 + x)$
and let $g(x) = [1 - f(x)]x^2 + 3x.$

From the first equation, $z = f(x)$, so
the 2nd equation gives $y = g(x).$

Thus, the 3rd equation becomes
$[2 + g(x)]x^2 + 4x - 2 = 0.$