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Subtracting the equations $y+xz=10$ and $z+xy=10,$ we get $$y+xz-z-xy=0.$$Then $y-z+x(z-y)=0,$ so $(y-z)(1-x)=0.$ Hence, $y=z$ or $x=1.$

If $x=1,$ then $yz=6$ and $y+z=10.$ Then by Vieta's formulas, $y$ and $z$ are the roots of $t^2-10t+6=0.$ Thus, $x=1$ for two ordered triples $(x,y,z).$

If $y=z,$ then $$\begin{array}{rl}x+y^2& =7,\\ y+xy& =10.\end{array}$$Squaring the second equation, we get $(x+1{)}^2{y}^2=100.$ Then $(x+1{)}^2(7-x)=100,$ which simplifies to $x^3-5x^2-13x+93=0.$ By Vieta's formulas, the sum of the roots is 5, so the sum of all the $x_i$ is $2+5=\boxed{7}.$