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we define the following three dimensional system:

dyn3={b + r Subscript[\[Gamma], r] - s (b + Subscript[\[Gamma], s]) + 
  i s (-\[Beta] + \[Delta]), -b + s \[Beta] + 
  r Subscript[\[Beta], 
   r] - \[Gamma] + (-1 + i) \[Delta], -i r Subscript[\[Beta], r] + 
  i \[Gamma] - r (b + Subscript[\[Gamma], r]) + 
  s Subscript[\[Gamma], s] + i r \[Delta]}

which can be reduced to the following two dimensional system

dyn2={b - (-1 + i + s) Subscript[\[Gamma], r] - 
  s (b + Subscript[\[Gamma], s]) + i s (-\[Beta] + \[Delta]), -b + 
  s (\[Beta] - Subscript[\[Beta], r]) + Subscript[\[Beta], r] - 
  i Subscript[\[Beta], r] - \[Gamma] + (-1 + i) \[Delta]}

Now, solving the two systems yields a different number of solutions:

Solve[Thread[(dyn3)==0],{s,i,r}]
Solve[Thread[(dyn2)]==0,{s,i}]

I was expecting to get the same number of solutions. Could someone clarify what is the problem precisely in my code? Thank you!

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  • $\begingroup$ How do you know that dyn3 reduces to dyn2 ? Did you solve for r to get there ? If so how ? $\endgroup$ Nov 30, 2022 at 11:30
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    $\begingroup$ If you are sure that dyn3 is equivalent to dyn2 then you might want to use Reduce and include all assumptions you have about the variables $\endgroup$ Nov 30, 2022 at 11:36
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    $\begingroup$ I did not try but you might also want to use LinearSolve. The problem seems linear. $\endgroup$ Nov 30, 2022 at 11:47
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    $\begingroup$ The reason is that one of the three solutions does not satisfy the relation r+i+s==1. So when you impose that condition, you knock out a solution. $\endgroup$ Nov 30, 2022 at 15:21
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    $\begingroup$ found this thread while looking for something else. I should mention that perhaps my previous claim that there was a problem with Reduce and Subscript might have been linked to using both variables with and without subscripts in my notebook. I might have used Reduce with both but I do not know really. $\endgroup$ Dec 6, 2022 at 16:43

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