How to solve simultaneous equations?

2021 2 14
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Hello,
How can I solve a non linear system of three equations with three variables? There is a variable raised to the fourth power that complicates the system. That is why I tried to get a only true solution using, 'PrincipalValue'.
Thanks for the help.
syms x y z
eqn1 = 1500*(20-x)+1.5309E-05*(293^4-(x+273)^4) == z;
eqn2 = 20000*(x-y) == z;
eqn3 = 3600*(y-10)+1.5309E-05*((y+273)^4-100^4) == z;
eqns = [eqn1,eqn2,eqn3];
vars = [x y z];
[solx, soly, solz] = solve(eqns, vars, 'PrincipalValue',true)

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Walter Roberson
Walter Roberson 2021-2-14
Ran in:

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Ran in:
syms x y z
eqn1 = 1500*(20-x)+1.5309E-05*(293^4-(x+273)^4) == z;
eqn2 = 20000*(x-y) == z;
eqn3 = 3600*(y-10)+1.5309E-05*((y+273)^4-100^4) == z;
eqns = [eqn1,eqn2,eqn3];
vars = [x y z];
sol = solve(eqns, vars, 'returnconditions',true)
sol = struct with fields:
x: [16×1 sym] y: [16×1 sym] z: [16×1 sym] parameters: [1×0 sym] conditions: [16×1 sym]
16 solutions.
sol.conditions
ans = 
and they are unconditional.
So there are 16 "true" solutions.
Perhaps you only want real-valued ones?
vx = vpa(sol.x); vy = vpa(sol.y); vz = vpa(sol.z);
mask = imag(vx) == 0 & imag(vy) == 0 & imag(vz) == 0;
realx = vx(mask);
realy = vy(mask);
realz = vz(mask);
([realx, realy, realz])
ans = 
So there are two "true" solutions that also happen to be real-valued. This was predictable: degree 16 polynomials in real roots always have an even number of real roots.
If you wanted to ignore some of the valid solutions even further, you could further test realx > 0

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Walter Roberson
Walter Roberson 2021-2-14
I would, by the way, point out that asking for exact solutions to systems with floating point coefficients is like asking for exactly how many molecules are in a bottle of beer that is "about" half-full. (How big is the bottle, exactly ? How close to "half full" is "about" half full, exactly ?). It is is a category error to use solve() with any equations with floating point coefficients.
eqn1 = 1500*(20-x)+1.5309E-05*(293^4-(x+273)^4) == z;
273 ? really?? You give 5 significant digits to 1.5309, so why did you only give 3 significant digits to 273 and 293? 0 C is 273.1600(1) K https://en.wikipedia.org/wiki/Kelvin -- 293.16 exactly until 2019.
What is the point of asking for exact solutions to equations that are wrong?
Valter Silva Nava
Valter Silva Nava 2021-2-14
Thanks for the answer, I will analyze it and understand all the comments that you made. In a second try I used "vpasolve" and showed me all the solutions and with the last part that you answered I could get only the real solutions. Thanks again.
syms x y z
eqn1 = 1500*(20-x)+1.5309E-05*(293.1600^4-(x+273.1600)^4) == z;
eqn2 = 20000*(x-y) == z;
eqn3 = 3600*(y-10)+1.5309E-05*((y+273.1600)^4-100^4) == z;
eqns = [eqn1 eqn2 eqn3];
vars = [x y z];
solution = vpasolve(eqns, vars);
solution.x
solution.y
solution.z
vx = vpa(solution.x); vy = vpa(solution.y); vz = vpa(solution.z);
mask = imag(vx) == 0 & imag(vy) == 0 & imag(vz) == 0;
realx = vx(mask); realy = vy(mask); realz = vz(mask);
([realx, realy, realz])

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