# Question:How do I solve a non-linear equation system

## Question:How do I solve a non-linear equation system

Maple

Hello.I could use some help to solve the folowing system, please.

Basically I have the 4 unknowns (A,P,C2,alfa)and I want to know their values by changing values to "a2" (eg. varying a2 from [0;80]).

With another software I solved matricially the EQ2;EQ3 and EQ4 for P,A and C2 with different values of a2 and then I replaced the P,C2 solutions (for each a2 value) on the EQ1 and tooked the value of "alfa" for each corresponding "a2". At the final I had the 4 unknows solved varying the "a2" value.

The problem is that I have a bigger system than this example (and non linear this time).thus I can not solve it matricially and have to do it solving the non linear equations of the system. For instance, if I use "fsolve" or "solve" for the previous example I got strange results for "alfa" (enourms results comparing with the real ones), A and C2....P appears to be correct..(?)..., but I have no confidence on the solution I got from Maple.

Can you suggest me some way of solving a non-linear system of equations? or for the example below How could I solve it? I used "for a from 0 to 80 do solutions := solve(sys) end do;" but as I wrote before the results are not correct.

EQ1: 0.9868421053+74479.54250*a^4-1.*alfa-33391.41112*P*a^3-.5000000000*C2*a^2

EQ2: 2.9791817*10^5*a^3-0.4044127197e-1-1.001742334*10^5*P*a^2-1.*C2*a+0.4098048893e-1*A

EQ3: -2.003484667*10^5*P*a-0.3358800946e-2*A-1.*C2+8.937545100*10^5*a^2

EQ4: -2.003484667*10^5*P+0.1376453050e-3*A+0.1113310242e-4*a+0.1358341826e-3