Given
$\begin {align} x&=(1-x)y\\x&=(1-x)(1-y)z\\1&=(1-x)(1-y)(1-z)+x+(1-x)y+(1-x)(1-y)z\end{align}$
I end up solving 1=1. I don't think there is anything wrong with my working. Is there something about this set of equations that makes it unsolvable?
The third "equation" is in fact an algebraic identity, valid for $\,\forall x,y,z\,$:
$$\require{cancel} \begin {align} 1 &=(1-x)(1-y)(1-z)+x+(1-x)y+(1-x)(1-y)z \\ &= (1-x)(1-y) - \cancel{(1-x)(1-y)z} + x + (1-x)y + \cancel{(1-x)(1-y)z} \\ &= (1- \cancel{x} - \bcancel{y} + \xcancel{xy}) + \cancel{x} + (\bcancel{y} - \xcancel{xy}) \\ &= 1 \end{align} $$
This leaves you with two actual equations in three unknowns.