How would one solve this system $$12x^2=6z\\2y=-z\\6x-y=7$$
It's been many years since I've dealt with system equations, and now find myself in need to solve them. I am not quite sure what to do; I am interested in finding $x$ and $y$, so assuming that $z \neq 0$ (can I do this?), I isolate it and get
$$2x^2 = z = -2y \\ -x^2 = y \\ 6x - (-x^2) = 7  \\ 6x + x^2 = 7 $$ which gives $$x^2 + 6x - 7 = 0 \\ (x+7)(x-1)=0$$
so $$ x = -7 \\or \\x = 1$$
which gives $$y = -(-7)^2 = -49 \\or \\ y = -(1)^2= -1 $$
But the answer is supposed to be $(1,-1)$ and $(7,35)$ ??  Where's my mistake? 

eliminating $y,z$ we get
$x^2+6x-7=0$
with $x-1=1,x_2=-7$ therefore $6-y=7$ thus $y_1=-1$ or $-42-y=7$ thus $y_2=-49$ now you can calculate $z$
