I haven't looked at your equations all that carefully, but maybe the following is relevant:
Pasting an approximate solution into an equation yields the so-called residual. A "small" residual does not imply an accurate solution: for
example,
(%i1) eq : (x-1)^10000 = 0;
(%o1) (x-1)^10000=0
(%i2) subst(x=1.99,eq);
(%o2) 2.248774849816094*10^-44=0
The residual (2.248774849816094*10^-44) is small, but 1.99 is not close to any solution of (x-1)^10000 = 0.
A large residual does not imply an inaccurate answer; for example,
(%i3) eq : (x-1)^(1/10000)=0;
(%o3) (x-1)^(1/10000)=0
(%i4) subst(x=1.000000000000001,eq);
(%o4) 0.996562500229301=0
Here 1.000000000000001 is close to a solution of (x-1)^(1/10000)=0 but the residual 0.9965... isn't small.
Of course, "large" and "small" have no definite meaning. The standard advice is to consult a good numerical analysis text for details.
Specifically for this, I don't have a good book in mind.
--bw