Describing the number of real solutions $(x,y,z)$

Question

Let $x,y,z$ be real numbers satisfying the following equations $$x+yz=1\\ y+xz=1\\ z+xy=1$$ How can we best describe the number of real solutions $(x,y,z)$?

Answer 1

Subtracting the first two equations gives $(x-y)(1-z)=0$, so either $x=y$ or $z=1$.

If $z=1$, then $x+y=1$, $xy=0$, so $x=0, y=1$ or $x=1, y=0$. Thus, we have solutions $(1,0,1), (0,1,1)$.

If $x=y$, then $xz+x=1$ and $z+x^2=1$. Subtracting these two gives $(x-z)(1-x)=0$, so either $x=z$ or $x=1$.

$x=y=1$ implies $z=0$, so we get solution $(1,1,0)$.

$x=y=z$ implies $x+x^2=1$, so $x=y=z=\dfrac{-1\pm\sqrt{5}}{2}$, so we get solutions $\left (\dfrac{-1+\sqrt{5}}{2}, \dfrac{-1+\sqrt{5}}{2}, \dfrac{-1+\sqrt{5}}{2} \right ), \left (\dfrac{-1-\sqrt{5}}{2}, \dfrac{-1-\sqrt{5}}{2}, \dfrac{-1-\sqrt{5}}{2} \right )$.

Thus, there are $5$ solutions altogether.

Answer 2

Hint:   subtracting the first two equations gives $\,(x-y)(1-z) = 0\,$.