Unable to find explicit solution

Hello all,

I am trying to solve the following system of equations:

So, my variables are R1, Q1, and b. My main concern is to find b and after that I can substitute it in the rest equations and solve. For now I want a solution including all the parameters (not algebraic solution) and after that I want to play around with the parameter values and compare the results. I am using the following code to do that:

syms A z w g u r o c b d y a1 a2 k m n t s v1 V1 K1 L1 R1 K2 L2 R2 Q1 Q2  %y for b in P, b for beta, n for z
v1 = k + m*b
V1 = v1/s
Pr1tilda = a1/(1-a1) * (c/a1 + t*(1-o*b) + v1 * b + w*b*o/j)
Pr1hat = (1-o*b*(1-n)*a1/(1-a1))*Pr1tilda

MC1 = 1/A * (z/w)^(g+u-1) *(r/g)^g * (Pr1hat/u)^u

Pr2tilda2 = 1/(1-a2) * (c+a2)
MC2 = 1/A * (z/w)^(g+u-1) *(r/g)^g * (Pr2tilda2/u)^u
 
Q1 = (d+MC2-2*MC1)/(3*b)
Q2 = (d+MC1-2*MC2)/(3*b)

L1 = Q1/A * (z/w)^(g+u) *(r/g)^g * (Pr1hat/u)^u
K1 = Q1/A * (z/w)^(g+u-1) *(r/g)^(g-1) * (Pr1hat/u)^u
R1 = Q1/A * (z/w)^(g+u-1) *(r/g)^g * (Pr1hat/u)^(u-1)
TC1 = Q1/A * (z/w)^(g+u-1) *(r/g)^g * (Pr1hat/u)^u

L2 = Q2/A * (z/w)^(g+u) *(r/g)^g * (Pr2tilda2/u)^u
K2 = Q2/A * (z/w)^(g+u-1) *(r/g)^(g-1) * (Pr2tilda2/u)^u
R2 = Q2/A * (z/w)^(g+u-1) *(r/g)^g * (Pr2tilda2/u)^(u-1)
TC2 = w*L2 + r*K2 + Pr2tilda2*R2 + V1*z*o*b*R1*a1/(1-a1)

P1 = (d-y*(Q1+Q2))*Q1 + z*V1*o*b*R1*a1/(1-a1) - TC1
P2 = (d-y*(Q1+Q2))*Q2 - TC2

eqn = z*m*o*b*a1*R1/(s*(1-a1)) + z*(k+m*b)*o*a1*R1/(1-a1) - Q1/A*(z/w)^(g+u-1)*(r/g)^g*(1/u)^u*Pr1hat^(u-1)*(((-(1-z)*o*a1/(1-a1))*a1/(1-a1)*(c/a1 + t*(1-o*b)+(k+m*b)*b+w*b*o/j)) + ((1-(1-z)*o*b*a1/(1-a1))*a1/(1-a1)*(-t*o+k+2*m*b+w*o/j))) == 0
B = solve(eqn,b)

However, when I am using the solve function for both B, I am getting back the following error

Warning: Unable to find explicit solution. For options, see help. 
> In solve (line 317) 
 
B =
 
Empty sym: 0-by-1
 

I would like to know if there is any other way to solve that equations, or a different methodology I could use.

I know very basic Matlab, so I am sorry if the question is simple or "stupid".

Thanks a lot in advance!!