jmc

Since $y=|x|,$ either $x=y$ (if $x\ge 0$) or $x=-y$ (if $x<0$). In the first case, substituting into the first equation, we get $y=y^2-6y+5,$ or $$0=y^2-7y+5.$$This equation has the roots $$y=\frac{7\pm \sqrt{7^2-4\cdot 5}}{2}=\frac{7\pm \sqrt{29}}{2},$$which are both positive and have a sum of $7.$ Since $x=y$ in this case, the sum of the possible values of $x$ is also $7.$

In the second case, substituting into the first equation, we get $-y=y^2-6y+5,$ or $$0=y^2-5y+5.$$This equation has the roots $$y=\frac{5\pm \sqrt{5^2-4\cdot 5}}{2}=\frac{5\pm \sqrt{5}}{2},$$which are both positive and have a sum of $5.$ Since $x=-y$ in this case, the sum of the possible values of $x$ is $-5.$

We conclude that the sum of all possible values for $x$ is $7+(-5)=\boxed{2}.$