# Set of non linear equations

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}
• Thank you. Could you tell me how $(1-x)(1-y)(1-z)=(1-x)(1-y)-(1-x)(1-y)z$ – matt Aug 15 '17 at 4:09
• @matt That's by distributivity $\,a(1-z)=a-az\,$ for $\,a=(1-x)(1-y)\,$. – dxiv Aug 15 '17 at 4:10