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I have the following input and try to do solve the equations for in total four unknowns. When I run the input only "solutions" for two of the four unknowns are given in the output. I am not quite sure, if my equations a linear independent, I think this is the problem why there is no solution. Is there a way of checking this? The input should be runable. The code is only that long as I need to substitute many varibles with long expressions, which would make the needed equations to be much longer and I decided that mathematica can do that for me.

RAB34 = RAB34m*K34;
RA34 = RA34m*K34;
RB34 = RB34m*K34;
RA2 = RA2m*K2;
RB2 = RB2m*K2;
RAB34 = RAB34m*K34;
RA34 = RA34m*K34;
RB34 = RB34m*K34;
RA2 = RA2m*K2;
RAB2 = RAB2m*K2;
RB2 = RB2m*K2;
RC2 = RC2m*K2;
RD2 = RD2m*K2;
RC34 = RC34m*K34;
RD34 = RD34m*K34;
RAC34 = RAC34m*K34;
RAD34 = RAD34m*K34;
RBC34 = RBC34m*K34;
RCD2 = RCD2m*K2;
RBD34 = RBD34m*K34;
RCD34 = RCD34m*K34;
RBC2 = RBC2m*K2;
RAC2 = RAC2m*K2;
RA342 = RA34/RA2;
RB342 = RB34/RB2;
RC342 = RC34/RC2;
RD342 = RD34/RD2;
RAD2 = K2*RAD2m;

RA3 = a*RA34;
RA4 = (1 - a)*RA34;

RB3 = b*RB34;
RB4 = (1 - b)*RB34;

RC3 = c*RC34;
RC4 = (1 - c)*RC34;

RD3 = d*RD34;
RD4 = (1 - d)*RD34;

RAB342 = RAB34/RAB2;
RAC342 = RAC34/RAC2;
RAD342 = RAD34/RAD2;
RA12 = 1/RA2;
RB12 = 1/RB2;
RC12 = 1/RC2;
RD12 = 1/RD2;

\[CapitalSigma]RAi2 = RA12 + 1 + RA342;
\[CapitalSigma]RBi2 = RB12 + 1 + RB342;
\[CapitalSigma]RCi2 = RC12 + 1 + RC342;
\[CapitalSigma]RDi2 = RD12 + 1 + RD342;

\[CapitalSigma]RAi = 1 + RA2 + RA3 + RA4;
\[CapitalSigma]RBi = 1 + RB2 + RB3 + RB4;
\[CapitalSigma]RCi = 1 + RC2 + RC3 + RC4;
\[CapitalSigma]RDi = 1 + RD2 + RD3 + RD4;

MA = 1/\[CapitalSigma]RAi*(M1 + M2*RA2 + M3*a*RA34 + M4*(1 - a)*RA34);
MB = 1/\[CapitalSigma]RBi*(M1 + M2*RB2 + M3*b*RB34 + M4*(1 - b)*RB34);
MC = 1/\[CapitalSigma]RCi*(M1 + M2*RC2 + M3*c*RC34 + M4*(1 - c)*RC34);
MD = 1/\[CapitalSigma]RDi*(M1 + M2*RD2 + M3*d*RD34 + M4*(1 - d)*RD34);


MA2 = 1/\[CapitalSigma]RAi2*(M1*RA12 + M2 + M3*a*RA342 + 
     M4*(1 - a)*RA342);
MB2 = 1/\[CapitalSigma]RBi2*(M1*RB12 + M2 + M3*b*RB342 + 
     M4*(1 - b)*RB342);
MC2 = 1/\[CapitalSigma]RCi2*(M1*RC12 + M2 + M3*c*RC342 + 
     M4*(1 - c)*RC342);
MD2 = 1/\[CapitalSigma]RDi2*(M1*RD12 + M2 + M3*d*RD342 + 
     M4*(1 - d)*RD342);

nAB = mAB/MA;
nAC = mAC/MA; 
nAD = mAD/MA;
nBA = mBA/MB;
nCA = mCA/MC;
nDA = mDA/MD;


nAB2 = mAB/MA2;
nAC2 = mAC/MA2; 
nAD2 = mAD/MA2;
nBA2 = mBA/MB2;
nCA2 = mCA/MC2;
nDA2 = mDA/MD2;



RAB2 == (RA2/\[CapitalSigma]RAi*nAB + 
     RB2/\[CapitalSigma]RBi*nBA)/(1/\[CapitalSigma]RAi*nAB + 
     1/\[CapitalSigma]RBi*nBA);
eq1 = %
RAB342 == (RA342/\[CapitalSigma]RAi2*nAB2 + 
     nBA2*RB342/\[CapitalSigma]RBi2)/(1/\[CapitalSigma]RAi2*nAB2 + 
     nBA2*1/\[CapitalSigma]RBi2);
eq2 = %

RAC34 == (RA34/\[CapitalSigma]RAi*nAC + 
     RC34/\[CapitalSigma]RCi*nCA)/(1/\[CapitalSigma]RAi*nAC + 
     1/\[CapitalSigma]RCi*nCA);
eq3 = %

RAC342 == (RA342/\[CapitalSigma]RAi2*nAC2 + 
     RC342/\[CapitalSigma]RCi2*nCA2)/(1/\[CapitalSigma]RAi2*nAC2 + 
     1/\[CapitalSigma]RCi2*nCA2);
eq4 = %


RAD34 == (RA34/\[CapitalSigma]RAi*nAD + 
     RD34/\[CapitalSigma]RDi*nDA)/(1/\[CapitalSigma]RAi*nAD + 
     1/\[CapitalSigma]RDi*nDA);
eq5 = %

RAD342 == (RA342/\[CapitalSigma]RAi2*nAD2 + 
     RD342/\[CapitalSigma]RDi2*nDA2)/(1/\[CapitalSigma]RAi2*nAD2 + 
     1/\[CapitalSigma]RDi2*nDA2);
eq6 = %


Solve[{eq1 && eq2 && eq3 && eq4 && eq5 && eq6}, {a, b, c, d, K2, K34}]
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  • $\begingroup$ The parameters you try to solve for are predefinded {a, b, c, d, K2, K34} (*{{a1, a2}, b, c, d, K2, K34}*), that's why Solve doesn't work. $\endgroup$ – Ulrich Neumann Jun 7 at 9:49
  • $\begingroup$ Thanks, but i am afraid i don not understand what this means. $\endgroup$ – Zorg Jun 7 at 9:52
  • $\begingroup$ With a refereshed kernel everthing is ok. $\endgroup$ – Ulrich Neumann Jun 7 at 10:34
  • $\begingroup$ Back to your question: MMA gives/finds one solution K2->, K34-> without showing the other parameters a,b,c,d. That means, the solution of your nonlinear eqations only depends on K2,K34 , parameters a,b,c,dare arbitrary! $\endgroup$ – Ulrich Neumann Jun 7 at 11:48
  • $\begingroup$ Okay, that is surprising and not the way I have designed it. But one of the solutions is'K34 -> 0' which is also not what I have expected. So maybe I have to reconsider how I can calculate my unknowns. $\endgroup$ – Zorg Jun 7 at 12:06

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