Problem for system of equations Find the solution of the following set of equations
\begin{equation}
\begin{cases}
yz-2z+x-1=0 \nonumber \\
zx+y-z-2=0 \nonumber \\
xy-2x-y+z+2=0
\end{cases}
\end{equation}

By summing (2) and (3) you end up with the equation:
$$
x(z+y-2)=0.
$$
Which has two solutions, either $x=0$ or $z+y-2=0$.
For the first case you end up with the following system:
\begin{align}
z(y-2)&=1\\
y-z&=2,
\end{align}
which is equivalent to $(y-2)^2=1$ and whose solutions are $(0,1,-1)$ and $(0,3,1)$. 
For the other case you end up with the following system, given that $z=2-y$:
 \begin{align}
y(2-y)-2(2-y) +x = -y^2 +x+4y -4 &=1\\
x(2-y)+y-(2-y)=-xy+2x +2y -2 &=2\\
xy-2x-2y+4&=0.
\end{align}
Observe that the second and third equation are the same. 
Geomtrically the first equation is the horizontal parabola with vertex $(2,1)$:
$$
(y-2)^2 = (x-1).
$$
And the second is either a vertical or an horizontal line:
$$
x(y-2)-2(y-2) = (x-2)(y-2)=0.
$$
So the solutions are the possible intersections of them! To find them explicitly consider each possible case, $x=2$ or $y=2$, from which we obtain:
If $x=2$, then $y=1$ or $y=3$, with solutions $(2,1,1)$ and $(2,3,-1)$, respectively.
If $y=2$, then $x=1$ and $z=0$, with solution $(1,2,0)$.
These are all the possible solutions.
Plot: http://www.wolframalpha.com/input/?i=plot++%28y-2%29%5E2-%28x-1%29%3D0+and+x%3D2+and+y%3D2
