Your tolerance is too high so you are not as close to the roots as you would like to be. Also, whenever there is a denominator, it is better to get rid of it as this helps the numerical process.
>>> nsolve([numer(i.normal()).expand() for i in (eqn1,eqn2,eqn3)],(p,q,r),(.01, .2,18)) matrix( [['0.00959710286420681'], ['0.25'], ['18.9205683344602']]) >>> reps = dict(zip([p,q,r],[S(i) for i in _])) >>> eqn1.subs(reps).n(3) -3.55e-15 >>> eqn2.subs(reps).n(3) 0.0906 >>> eqn3.subs(reps).n(3) 2.22e-16 It looks like that particular solution sets the denominator of eqn2 to zero: >>> denom(eqn2.normal()).subs(reps) -1.33226762955019e-15 Another thing to try when solving a nasty set of equations is to introduce a parameter `a` that can be varied from 0 to 1: when 0 it shuts off a nonlinear portion of your equations making them easier to solve. From that solution you increase `a` a little and use the last guess as your next guess, increasing `a` and updating the guess until you get to `a`=1. Hope that gets you a little further down the road. /chris -- You received this message because you are subscribed to the Google Groups "sympy" group. To unsubscribe from this group and stop receiving emails from it, send an email to sympy+unsubscr...@googlegroups.com. To post to this group, send email to sympy@googlegroups.com. Visit this group at http://groups.google.com/group/sympy?hl=en-US. For more options, visit https://groups.google.com/groups/opt_out.
