equation solving – A quadratic system with two solutions which may be solved with numeric coefficients, but not with symbolic ones

The system below may be solved using the observation that the second and third equation admit solution (,0,0,); alternatively, the determinant of these two equations must be 0. Mathematica 11.3 succeeds with a numeric instance

s1 = (CapitalLambda) - (Beta) s i + Subscript((Gamma), r) r - 
   Subscript((Gamma), s) s - (Mu) s;
e1 = (Beta) s i - e (Subscript((Gamma), e) + (Mu));
i1 =  e Subscript((Gamma), e) - (Gamma) i - ((Mu) + (Nu)) i;
r1 = (Gamma) i - Subscript((Gamma), r) r + 
   Subscript((Gamma), s) s - (Mu) r;
dyn = {s1, e1, i1, r1}
vz = {0, 0, 0, 0};
eq = Thread(dyn == vz)
cb = { (Beta) -> 5, (Gamma) -> 1/2, (Mu) -> (CapitalLambda), 
   Subscript((Gamma), r) -> 1/6,  Subscript((Gamma), s) -> 1/100, 
   Subscript((Gamma), e) -> 1/100, (CapitalLambda) -> 
    40/400, (Nu) -> 
    Subscript((Gamma), r) (1 + (Gamma)/(CapitalLambda)) - 1/10};

es = FullSimplify(Solve((eq //. cb), {s, e, i, r}))

but does not succeed with a symbolic instance, even after adding positivity assumptions

    cp = {(Beta) > 0, (CapitalLambda) > 0, (Nu) > 0, (Gamma) > 
    0, (Mu) > 0, Subscript((Gamma), e) > 0, 
   Subscript((Gamma), r) > 0};
es = FullSimplify(Solve(eq~Join~cp, {s, e, i, r}))

Is there a better trick to solve the symbolic case?