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I have the following system of 27 equations:

{pb λ10 + λ12 + 0.99 (-0.995 (0.04 + 0.96 pb) λ10 - 
  pb (0.0025 + 0.972 Rb - 0.972 Rd) λ3) + pb λ7 -    1/2 λ12 Erfc[
 2.11216 (0.021975 + 2 (0.150823 + Log[B]))] == 0,  1/c - 3.409 L^0.5 λ1 - λ13 - λ2/c^2 - (   0.99 (-Rd + Rk) λ5 (1 + 
  0.972 (-1 + (0.3863 νz)/(0.3863 - νk))))/   c + ((1 + Rd) λ2 + 
c (-Rd + Rk) λ5 (1 + 
   0.972 (-1 + (0.3863 νz)/(0.3863 - νk))) + 
c λ6 (1 + 
   0.972 (-1 + (0.3863 νz)/(0.3863 - νk))) (1 + 
   Rd - τz))/c^2 - (   0.99 λ6 (1 + 
  0.972 (-1 + (0.3863 νz)/(0.3863 - νk))) (1 + 
  Rd - τz))/c == 0, -1.728 inv pk λ8 == 0, (1/(  K^2 L^0.33))(-K^2 L^0.33 (0.058906 λ13 + 
   pk (0.002475 λ3 - 0.96228 Rd λ3 + 
      0.96228 Rk λ3 - λ7)) + 
0.218889 K^0.33 L λ9 + 
0.3267 K^1.33 (-0.67 λ1 (-1 + τh) + 
   L (λ13 + 0.67 λ10 τh))) == 0, (1/(  K L^1.33))(K (-3.409 L^0.83 (L + 0.5 c λ1) + 
   1. K^0.33 L (λ13 + λ10 τh)) + 
0.1089 K^0.33 (L λ9 + 
   K (λ1 - λ1 τh + L λ10 τh)) - 
0.33 K^0.33 (L λ9 + 
   K (λ1 + L λ13 - λ1 τh + 
      2 L λ10 τh))) == 0, (  0.995 (0.0396 - 0.0096 pb) λ11 +    B pb^2 (0.0448 λ10 - 0.002475 λ3 - 
  0.96228 Rb λ3 + 0.96228 Rd λ3 + λ7))/ pb^2 == 0,  K (-0.002475 λ3 + 0.96228 Rd λ3 - 
  0.96228 Rk λ3 + λ7) + λ8 + (0.0394 + 
  0.99 Rk) λ9 == 0, λ11 - 0.972 B pb λ3 ==   0, -(λ2/c) + 0.972 B pb λ3 +    0.972 K pk λ3 - 0.972 Z λ3 + 0.028 λ5 -    0.028 λ6 + 0.972 λ5 νz -    0.972 λ6 νz + (0.972 λ5 νk νz)/( 0.3863 - νk) - (0.972 λ6 νk νz)/(   0.3863 - νk) ==   0, -0.972 K pk λ3 + pk λ9 + λ5 (-0.028 - (0.375484 νz)/(
  0.3863 - νk)) ==   0, λ3 - (λ7 νz)/(0.3863 - νk) +    0.99 (-0.972 (1 +          Rd) λ3 + (λ10 + λ3) τz) ==   0, (-3.409 c L^0.83 - 0.67 K^0.33 (-1 + τh))/L^0.33 ==   0, -0.2 K^0.33 L^0.67 - 0.995 B (0.04 + 0.96 pb) + B pb +    0.67 K^0.33 L^0.67 τh + Z τz ==   0, -((0.995 (0.04 + 0.96 pb))/pb) + Rb ==   0, -0.86 + B + 0.431301 Erfc[4.22432 (0.133791 + Log[B])] -    1/2 B Erfc[4.22432 (0.16181 + Log[B])] ==   0, -c - 0.0494 K + 0.8 K^0.33 L^0.67 == 0, (1 - 0.99 (1 + Rd))/c ==   0, -B pb (0.0025 + 0.972 Rb - 0.972 Rd) +    K pk (-0.0025 + 0.972 Rd - 0.972 Rk) +    Z (1 - 0.972 (1 + Rd) + τz) ==   0, νk -    0.99 (-Rd + Rk) (1 + 
  0.972 (-1 + (0.3863 νz)/(0.3863 - νk))) ==   0, νz - 0.99 (1 + 0.972 (-1 + (0.3863 νz)/(0.3863 - νk))) (1 +       Rd - τz) == 0,  B pb + K pk - (Z νz)/(0.3863 - νk) == 0, -1 + pk == 0, -((0.33 L^0.67)/K^0.67) + pk (0.0494 + Rk) == 0, (  0.149228 λ5 + λ5 νk^2 - Z λ7 νz -    0.3863 (λ5 (2 νk + 0.972 (-Rd + Rk) νz) +       0.972 λ6 νz (1 + 
     Rd - τz)))/(0.3863 - νk)^2 ==   0, (-0.375484 Rk λ5 +    0.375484 Rd (λ5 - λ6) + 0.0108164 λ6 -    Z λ7 - λ6 νk + 0.375484 λ6 τz)/(0.3863 - νk) ==   0, -((0.67 K^0.33 (λ1 - L λ10))/L^0.33) == 0, Z (λ10 + λ3) + λ6 (1 +       0.972 (-1 + (0.3863 νz)/(0.3863 - νk))) == 0}

There are 27 Real variables : B, c, inv, K, L, pb, pk, Rb, Rd, Rk, Z, λ1, λ10, λ11, λ12, λ13, λ2, λ3, λ5, λ6, λ7, λ8, λ9, νk, νz, τh, τz.

Can anyone please give me some hint on how to find a solution of this system in Mathematica? I have tried the command NSolve and FindRoot but don't manage to calculate it. Note that Greek letter variables can be positive or negative, while the others can only be positive.

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  • $\begingroup$ Please also show the exact command you tried. Note that K is a system symbol and shouldn't be used as a variable name. $\endgroup$ – Szabolcs Dec 1 '16 at 11:43
  • $\begingroup$ I don't think NSolve will work unless you can bound the variables (because of Erfc and perhaps the inexact 0.33 powers). FindRoot might find one root at a time, if it has a good starting point for the variables. $\endgroup$ – Michael E2 Dec 1 '16 at 11:54
  • $\begingroup$ Thanks. How can I select only equations in which variables are not raised to power (except for 1 or -1)? $\endgroup$ – C. Basile Dec 1 '16 at 12:38
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    $\begingroup$ You are going to have to invest some thought into this. Naively throwing a large nonlinear system at Mathematica (or any other computational system, for that matter) is bound to fail. Things to consider are: Do you have any estimates for the solution values of (some of) the variables? Such estimates could come from the physics or other background of the problem. Can the equations be structured in such a way that the solution would become easier, perhaps by eliminating some of the variables in a smart way? Such simplifications usually still require human intelligence. $\endgroup$ – Pirx Dec 1 '16 at 15:10
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    $\begingroup$ It is sometimes useful to take the sum of squares of all your expressions and let FindMinimum work at it. However I have little hope you will solve that unless you can make a very good initial guess at all 27 unknowns. $\endgroup$ – george2079 Dec 1 '16 at 16:09

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