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}
 A: find sum of 2nd  and 3rd equation  :$$(xz+y-z-2)+(xy-2x-y+z+2)=0+0\\xz+xy-2x=0\\x(z+y-2)=0$$you can go on from there 
A: 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
A: Start by factorizing your equations:
$$z(y-2) + (x-1) = 0$$
$$z(x-1) + (y-2) = 0$$
$$(y-2)(x-1)  + z = 0$$
From this we see that we get a much simpler system if we define $\tilde{x} = x-1$, $\tilde{y} = y-2$ and $\tilde{z} = -z$ (this procedure backward is likely how the author of this problem constructed it):
$$\tilde{x} = \tilde{z}\tilde{y},~~~~~~~~~~~\tilde{y} = \tilde{z}\tilde{x},~~~~~~~~~~~\tilde{z} = \tilde{x}\tilde{y}$$
The system is now also invariant under a cyclic permutation of the variables so the solution set will also have this propery. All the combinations of any two of the three equaitions above gives us
$$\tilde{x}=\tilde{z}^2\tilde{x} = \tilde{y}^2\tilde{x}$$
$$\tilde{y}=\tilde{z}^2\tilde{y} = \tilde{x}^2\tilde{y}$$
$$\tilde{z}=\tilde{y}^2\tilde{z} = \tilde{x}^2\tilde{z}$$
From this it follows that $\tilde{x},\tilde{y},\tilde{z} \in\{-1,0,1\}$. Next we see that if one variable is $0$ then all variables must be $0$ and $(0,0,0)$ is indeed a solution so any other solution must have $\tilde{x},\tilde{y},z \in\{-1,1\}$ (which gives $8$ possible solutions). We can also rule out the possibillity that two of the variables being $1$ and the last being $-1$ or all three being $-1$ (since $\tilde{x}=\tilde{z}\tilde{y}$). This leaves us with $(\tilde{x},\tilde{y},\tilde{z})=(0,0,0)$, $(\tilde{x},\tilde{y},\tilde{z})=(1,1,1)$, $(\tilde{x},\tilde{y},\tilde{z})=(-1,-1,1)$, $(\tilde{x},\tilde{y},\tilde{z})=(1,-1,-1)$ and $(\tilde{x},\tilde{y},\tilde{z})=(-1,1,-1)$ which all satisfy the equation-set.
In terms of the original variables we get $(x,y,z)=(0,1,-1)$, $(x,y,z)=(0,3,1)$, $(x,y,z)=(1,2,0)$, $(x,y,z)=(2,1,1)$ and $(x,y,z)=(2,3,-1)$
