MATLAB: Matlab solve indeterminate polynomial equations numerically

MATLABnumerical solverpolynomial equations

I want to solve a problem like this:
syms u1 u2 u3 u4 u5 u6 a2 b2 b3;
eqs = [ u6*(3*u5 - 1) - u4 - u1 - 3*u2*u3 - 3*u5*u6 + 3*u3*(u2 - 1) + 1/24,...
u2^2*u3*(u2 - 1) - 2*u5^3*u6 - u2*u3*(u2 - 1)^2 - (2*u4)/27 - u5*u6*(3*u5 - 1)^2 + u5^2*u6*(3*u5 - 1) + 1/360,...
1/1260 - 2*u5^4*u6 - 2*u5^2*u6*(3*u5 - 1)^2 - 2*u2^2*u3*(u2 - 1)^2 - (2*u4)/81,...
u5^2*u6*(3*u5 - 1)^3 - 3*u5^5*u6 - u4/81 - 4*u5^3*u6*(3*u5 - 1)^2 + 4*u5^4*u6*(3*u5 - 1) + 1/2880,...
9*u5^4*u6*(3*u5 - 1) - 3*u5^3*u6*(3*u5 - 1)^2 + 1/6720,...
3*b2*(u5^4*u6*(3*u5 - 1)^3 - u5^5*u6*(3*u5 - 1)^2 + 2*u5^6*u6*(3*u5 - 1) + 7808145639665419/2361183241434822606848) + 3*b3*(u5^6*u6*(3*u5 - 1) - 3*u5^5*u6*(3*u5 - 1)^2 + 1301357606610903/590295810358705651712) + a2*(u5^3*u6*(3*u5 - 1)^3 + 3*u5^5*u6*(3*u5 - 1) + 1/60480) + 3*a2*(2*u5^5*u6*(3*u5 - 1) - 2*u5^4*u6*(3*u5 - 1)^2 + 1/90720),...
b2*(u5^4*u6*(3*u5 - 1)^3 - u5^5*u6*(3*u5 - 1)^2 + 2*u5^6*u6*(3*u5 - 1) + 7808145639665419/2361183241434822606848) + 2*b3*(u5^4*u6*(3*u5 - 1)^3 - u5^5*u6*(3*u5 - 1)^2 + 2*u5^6*u6*(3*u5 - 1) + 7808145639665419/2361183241434822606848) + 2*b2*(u5^6*u6*(3*u5 - 1) - 3*u5^5*u6*(3*u5 - 1)^2 + 1301357606610903/590295810358705651712) + b3*(u5^6*u6*(3*u5 - 1) - 3*u5^5*u6*(3*u5 - 1)^2 + 1301357606610903/590295810358705651712) + a2*(u5^3*u6*(3*u5 - 1)^3 + 3*u5^5*u6*(3*u5 - 1) + 1/60480) + 3*a2*(2*u5^5*u6*(3*u5 - 1) - 2*u5^4*u6*(3*u5 - 1)^2 + 1/90720)];
unknows = [u1,u2,u3,u4,u5,u6,a2,b2,b3];
guess = [0 Inf; 0 1; 0 Inf; 0 Inf; 0 0.3333; 0 Inf; -Inf Inf; -Inf Inf; -Inf Inf];
S = vpasolve(eqs, unknows, guess);
vpasolve can't give me an answer, However when I do this
a2 = 1; b2 = -1; b3 = 1;
neweqs = subs(eqs);
newunknows = unknows(1,1:6);
newguess = guess(1:6,:);
Y = vpasolve(neweqs, newunknows, newguess);
I get an answer, Y.u1 is not empty. So how can I make it work, I don't want to substitue a2 b2 b3, I just want get an answer directly, no need to get neweqs, newunknows, newguess. Please help me, thanks for any comments!

Best Answer

Finally, I find out the answer, it seems matlab will do some approximations, the true system is
eqstrue=[ u6*(3*u5 - 1) - u4 - u1 - 3*u2*u3 - 3*u5*u6 + 3*u3*(u2 - 1) + 1/24, u2^2*u3*(u2 - 1) - 2*u5^3*u6 - u2*u3*(u2 - 1)^2 - (2*u4)/27 - u5*u6*(3*u5 - 1)^2 + u5^2*u6*(3*u5 - 1) + 1/360, 1/1260 - 2*u5^4*u6 - 2*u5^2*u6*(3*u5 - 1)^2 - 2*u2^2*u3*(u2 - 1)^2 - (2*u4)/81, u5^2*u6*(3*u5 - 1)^3 - 3*u5^5*u6 - u4/81 - 4*u5^3*u6*(3*u5 - 1)^2 + 4*u5^4*u6*(3*u5 - 1) + 1/2880, 9*u5^4*u6*(3*u5 - 1) - 3*u5^3*u6*(3*u5 - 1)^2 + 1/6720, u5^3*u6*(3*u5 - 1)^3 - 6*u5^4*u6*(3*u5 - 1)^2 - 3*u5^4*u6*(3*u5 - 1)^3 - 6*u5^5*u6*(3*u5 - 1)^2 + 9*u5^5*u6*(3*u5 - 1) - 3*u5^6*u6*(3*u5 - 1) + 1/21600, u5^3*u6*(3*u5 - 1)^3 - 6*u5^4*u6*(3*u5 - 1)^2 + u5^4*u6*(3*u5 - 1)^3 + 2*u5^5*u6*(3*u5 - 1)^2 + 9*u5^5*u6*(3*u5 - 1) + u5^6*u6*(3*u5 - 1) + 11/226800]
after I make
a2 = 1; b2 = -1; b3 = 1;
matlab solves the approximatied system
eqs=[ u6*(3*u5 - 1) - u4 - u1 - 3*u2*u3 - 3*u5*u6 + 3*u3*(u2 - 1) + 1/24, u2^2*u3*(u2 - 1) - 2*u5^3*u6 - u2*u3*(u2 - 1)^2 - (2*u4)/27 - u5*u6*(3*u5 - 1)^2 + u5^2*u6*(3*u5 - 1) + 1/360, 1/1260 - 2*u5^4*u6 - 2*u5^2*u6*(3*u5 - 1)^2 - 2*u2^2*u3*(u2 - 1)^2 - (2*u4)/81, u5^2*u6*(3*u5 - 1)^3 - 3*u5^5*u6 - u4/81 - 4*u5^3*u6*(3*u5 - 1)^2 + 4*u5^4*u6*(3*u5 - 1) + 1/2880, 9*u5^4*u6*(3*u5 - 1) - 3*u5^3*u6*(3*u5 - 1)^2 + 1/6720, u5^3*u6*(3*u5 - 1)^3 - 6*u5^4*u6*(3*u5 - 1)^2 - 3*u5^4*u6*(3*u5 - 1)^3 - 6*u5^5*u6*(3*u5 - 1)^2 + 9*u5^5*u6*(3*u5 - 1) - 3*u5^6*u6*(3*u5 - 1) + 34433922270924495617/743772721051969121157120, u5^3*u6*(3*u5 - 1)^3 - 6*u5^4*u6*(3*u5 - 1)^2 + u5^4*u6*(3*u5 - 1)^3 + 2*u5^5*u6*(3*u5 - 1)^2 + 9*u5^5*u6*(3*u5 - 1) + u5^6*u6*(3*u5 - 1) + 37713343439583972437/743772721051969121157120];
so it gives me an answer, I don't know why it does this. In the last equation.
vpa(11/226800 - 37713343439583972437/743772721051969121157120)
ans = -0.0000022045855379188659798651850341144609
It's a bad approximation.