Solving the system $xy+x+y=-13$, $yz+y+z=-9$, $zx+z+x=5$

Question

Solve for $x,\,y$ and $z$ \begin{align*} xy + x + y &= -13\\ yz + y + z &= -\phantom{1}9\\ zx +z + x &= \phantom{-1}5 \end{align*}

Obviously one way would be to substitute the value of $x$, $y$ and $z$ but that would get really ugly. Can anybody suggest a cleaner way to solve this equation? I tried factoring, denoting the sum of the three variables as 's' and adding all the equations but couldn't make any progress

Answer 1

It's $$(x+1)(y+1)=-12,$$ $$(y+1)(z+1)=-8$$ and $$(z+1)(x+1)=6.$$

Thus, $(x+1)^2(y+1)^2(z+1)^2=576,$ which gives two cases:

1. $$(x+1)(y+1)(z+1)=24,$$ which gives $$x+1=-3,$$ $$y+1=4$$ and $$z+1=-2,$$ which gives $$(x,y,z)=(-4,3,-3).$$

2. $$(x+1)(y+1)(z+1)=-24,$$ which gives $$x+1=3,$$ $$y+1=-4$$ and $$z+1=2,$$ which gives $$(x,y,z)=(2,-5,1).$$