Decide for which $x,y,z$ the following equation system is met: ${1+x+y=xy}$ $2+y+z=yz$ and $5+z+x=zx$

Question

I need to decide for which $x,y,z$ the following equation system is met:

$$ 1+x+y=xy $$ $$ 2+y+z=yz $$ $$5+z+x=zx$$

I can see that $x=0, y=0, z=0$ is not a solution.

I tried to divide first equation with $xy$, second with $yz$, third with $zx$ and then substitute $\frac 1x=a$, $\frac 1y=b$ and $\frac 1z=c$ but it didn't work out.

Answer 1

Hint: rewrite the system as follow.

\begin{align} (x-1)(y-1)&=2 \\ (y-1)(z-1)&=3 \\ (z-1)(x-1)&=6 \end{align}

Answer 2

Rearranging we get $$xy-x-y+1=2$$ so : $$ (x-1)(y-1) = 2$$ and $$(y-1)(z-1)= 3$$ and $$(z-1)(x-1)= 6$$