0
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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?

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3
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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.

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

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