In fact the unknown "A" and "a" are different things. Since I was using a simplified case, probably I was not clear describing it. Sorry for making you loose time thinking on it....!

So, is better if I put here the complete non linear system I am interested to solve(already with a2 replaced on it).

Please consider only these case and ignore the previous one: (12 EQs vs 12 unknowns{A1,B1,A2,B2,C2,D2,A3,B3,D3,P,alfa,a3}

EQ1:=-16.75+(37.45*10^(-3))*A2+(961.07*10^(-3))*B2+(40.48*10^(-3))*C2+1.04*D2-1.00*B1=0;

EQ2:=(7.19*10^(-3))*A2-(7.78*10^(-3))*B2+(8.41*10^(-3))*C2+(7.78*10^(-3))*D2-(8.12*10^(-3))*A1+(8.12*10^(-3))*B1=0;

EQ3:=(2.27*10^(-6))*A2+(5.83*10^(-5))*B2-(2.46*10^(-6))*D2+(6.30*10^(-5))*C2-(6.59*10^(-5))*B1=0;

EQ4:=(4.36*10^(-7))*A2-(4.72*10^(-7))*B2+(4.72*10^(-7))*C2-(5.10*10^(-7))*D2-(5.35*10^(-7))*A1+(5.35*10^(-7))*B1=0;

EQ5:=B1-1.53*10^0=0;

EQ6:=-(1.55*10^(-8))*a3^(4.00*10^0)+(3.33*10^(-1))*A3*a3^(3.00*10^0)+(5.00*10^(-1))*B3*a3^(2.00*10^0)+D3+1.67*10-(1.00*10^0)*B2-(1.00*10^0)*D2=0;

EQ7:=-(6.21*10^(-8))*a3^(3.00*10^0)+A3*a3^(2.00*10^0)+B3*a3-(7.79*10^(-3))*A2+(7.79*10^(-3))*B2-(7.79*10^(-3))*C2-(7.79*10^(-3))*D2=0;

EQ8:=-(1.86*10^(-7))*a3^(2.00*10^0)+(2.00*10^0)*A3*a3+B3-(1.21*10^(-4))*B2-(1.21*10^(-4))*C2=0;

EQ9:=-(3.73*10^(-7))*a3+(2.00*10^0)*A3-(9.45*10^(-7))*A2+(9.45*10^(-7))*B2-(9.45*10^(-7))*C2+(9.45*10^(-7))*D2=0;

EQ10:=-2.52*10+B2+D2=0;

EQ11:=D3-1.*alfa=0;

EQ12:=A3-(6.24*10^(-7))*P=0;

thus, solving the system:

> solve(sys, {A1, A2, A3, B1, B2, B3, C2, D2, D3, P, a3, alfa});

 I got multiple solutions: {A1 = -164.3765226, A2 = -166.1535085, A3 = 0.0001929630612, B1 = 1.533000000, B2 = 17.74633654, B3 = -0.1507247679, C2 = -8.532505619, D2 = 7.499026172, D3 = 31202.14335, P = 309.2788275, a3 = 1541.895107, alfa = 31202.14335}, { A1 = -164.3765226, A2 = -166.1535085, A3 = -0.00009648153046 + 0.00002277073987 I, B1 = 1.533000000, B2 = 17.74633654, B3 = 0.001854207963 + 0.02356930294 I, C2 = -8.532505619, D2 = 7.499026172, D3 = -3.444079163 + 117.2116020 I, P = -154.6394136 + 36.49666255 I, a3 = -10.70944769 + 122.1441183 I, alfa = -3.444079163 + 117.2116020 I}, {A1 = -164.3765226, A2 = -166.1535085, A3 = -0.00009648153046 - 0.00002277073987 I, B1 = 1.533000000, B2 = 17.74633654, B3 = 0.001854207963 - 0.02356930294 I, C2 = -8.532505619, D2 = 7.499026172, D3 = -3.444079163 - 117.2116020 I, P = -154.6394136 - 36.49666255 I, a3 = -10.70944769 - 122.1441183 I, alfa = -3.444079163 - 117.2116020 I}

using the command "fsolve" I have no solution.

My doubt is if I was doing the right approach using "solve" for non-linear equations and if yes, which solution por P and alfa is reasonable to choose? For instance, physically I can not expect an alfa=31202. ..

Thanks again.