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I have a system with 5 equations and several parameters and Mma 11.3 finds a solution after a few minutes. But if I make a harmless modification, by inserting (kfus*wtss/kfis+1) in front of the two Log[2], solve finds no solution (aborted after 90 minutes).

But why? The inserted term is basically a constant, since it only contains parameters. So, it should not effect finding a solution for the variables !?

This works:

    eqSSL = {0 == 
    f*(wtf[t] + wts[t]) - v1*ATP[t] - v2*(mtf[t] + mts[t] + wtf[t] + wts[t]), 
   0 == (sw*(wtf[t] + wts[t]))/(ATP[t]/c + 1) - kfis*wtf[t] + 
     kfus*wts[t]*(wts[t] + wtf[t]) + sf*kfus*wts[t]*(mtf[t] + mts[t]),
   0 == kfis*wtf[t] - kfus*wts[t]*(wts[t] + wtf[t]) - 
     sf*kfus*wts[t]*(mtf[t] + mts[t]) - wts[t]*Log[2]/halfL,
   0 == (sm*(mtf[t] + mts[t]))/(ATP[t]/c + 1) - kfis*mtf[t] + 
     sf*kfus*mts[t]*(wts[t] + wtf[t] + mts[t] + mtf[t]),
   0 == kfis*mtf[t] - sf*kfus*mts[t]*(wts[t] + wtf[t] + mts[t] + mtf[t]) - 
     mts[t]*sd*Log[2]/halfL};
Simplify[Solve[eqSSL, {ATP[t], wtf[t], wts[t], mtf[t], mts[t]}]]

This does not work (aborted after 90 minutes):

    eqSSL = {0 == 
    f*(wtf[t] + wts[t]) - v1*ATP[t] - v2*(mtf[t] + mts[t] + wtf[t] + wts[t]), 
   0 == (sw*(wtf[t] + wts[t]))/(ATP[t]/c + 1) - kfis*wtf[t] + 
     kfus*wts[t]*(wts[t] + wtf[t]) + sf*kfus*wts[t]*(mtf[t] + mts[t]),
   0 == kfis*wtf[t] - kfus*wts[t]*(wts[t] + wtf[t]) - 
     sf*kfus*wts[t]*(mtf[t] + mts[t]) - wts[t]*(kfus*wtss/kfis+1)*Log[2]/halfL,
   0 == (sm*(mtf[t] + mts[t]))/(ATP[t]/c + 1) - kfis*mtf[t] + 
     sf*kfus*mts[t]*(wts[t] + wtf[t] + mts[t] + mtf[t]),
   0 == kfis*mtf[t] - sf*kfus*mts[t]*(wts[t] + wtf[t] + mts[t] + mtf[t]) - 
     mts[t]*sd*(kfus*wtss/kfis+1)*Log[2]/halfL};
Simplify[Solve[eqSSL, {ATP[t], wtf[t], wts[t], mtf[t], mts[t]}]]
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  • $\begingroup$ Is it a constant? Can that just be replaced with a single random variable ? Is it solvable then? $\endgroup$ – morbo May 15 '19 at 9:49
  • $\begingroup$ Yes it is a constant. It can be replaced with a single parameter (e.g. xxx) and then it is solvable. $\endgroup$ – Axel May 15 '19 at 15:30

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