solving three equations for 3 unknows
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Hi Guys, I have the three following equations:
(2*c*w1)+b2 == 0.3205;
((2*c*b2)+ w1)*w1 == 214200;
(b2*w1)==1.2260e+05;
And I need to find the values of: c, w1 and b2.... is there a function for me to do it?
Thanks for your time.
采纳的回答
Star Strider
2019-6-10
For your system, use vpasolve (link) to get numeric results. You may also need to use the double function if you want to use the results in other calculations.
syms c w1 b2
Eqs = [(2*c*w1)+b2 == 0.3205;
((2*c*b2)+ w1)*w1 == 214200;
(b2*w1)==1.2260e+05];
[c,w1,b2] = vpasolve(Eqs, [c,w1,b2])
4 个评论
Valerie Cala
2019-6-10
Star Strider
2019-6-10
As always, my pleasure.
Each equaton is defined by the product of two of the variables, and in the second equation ‘w1’ is also squared.
It is a vector because all the variables are then defined as 
degree polynomials (the Symbolic Math Toolbox defines them as such with respect to the parameter ‘z’):
degree polynomials (the Symbolic Math Toolbox defines them as such with respect to the parameter ‘z’): c =
(641*root(z^4 - (641*z^3)/2000 + 214200*z^2 - 15030760000, z, 1))/490400000 - root(z^4 - (641*z^3)/2000 + 214200*z^2 - 15030760000, z, 1)^2/245200
(641*root(z^4 - (641*z^3)/2000 + 214200*z^2 - 15030760000, z, 2))/490400000 - root(z^4 - (641*z^3)/2000 + 214200*z^2 - 15030760000, z, 2)^2/245200
(641*root(z^4 - (641*z^3)/2000 + 214200*z^2 - 15030760000, z, 3))/490400000 - root(z^4 - (641*z^3)/2000 + 214200*z^2 - 15030760000, z, 3)^2/245200
(641*root(z^4 - (641*z^3)/2000 + 214200*z^2 - 15030760000, z, 4))/490400000 - root(z^4 - (641*z^3)/2000 + 214200*z^2 - 15030760000, z, 4)^2/245200
w1 =
root(z^4 - (641*z^3)/2000 + 214200*z^2 - 15030760000, z, 1)^3/122600 - (641*root(z^4 - (641*z^3)/2000 + 214200*z^2 - 15030760000, z, 1)^2)/245200000 + (1071*root(z^4 - (641*z^3)/2000 + 214200*z^2 - 15030760000, z, 1))/613
root(z^4 - (641*z^3)/2000 + 214200*z^2 - 15030760000, z, 2)^3/122600 - (641*root(z^4 - (641*z^3)/2000 + 214200*z^2 - 15030760000, z, 2)^2)/245200000 + (1071*root(z^4 - (641*z^3)/2000 + 214200*z^2 - 15030760000, z, 2))/613
root(z^4 - (641*z^3)/2000 + 214200*z^2 - 15030760000, z, 3)^3/122600 - (641*root(z^4 - (641*z^3)/2000 + 214200*z^2 - 15030760000, z, 3)^2)/245200000 + (1071*root(z^4 - (641*z^3)/2000 + 214200*z^2 - 15030760000, z, 3))/613
root(z^4 - (641*z^3)/2000 + 214200*z^2 - 15030760000, z, 4)^3/122600 - (641*root(z^4 - (641*z^3)/2000 + 214200*z^2 - 15030760000, z, 4)^2)/245200000 + (1071*root(z^4 - (641*z^3)/2000 + 214200*z^2 - 15030760000, z, 4))/613
b2 =
root(z^4 - (641*z^3)/2000 + 214200*z^2 - 15030760000, z, 1)
root(z^4 - (641*z^3)/2000 + 214200*z^2 - 15030760000, z, 2)
root(z^4 - (641*z^3)/2000 + 214200*z^2 - 15030760000, z, 3)
root(z^4 - (641*z^3)/2000 + 214200*z^2 - 15030760000, z, 4)
So there are 4 roots of each variable.
Valerie Cala
2019-6-10
Star Strider
2019-6-10
As always, my pleasure!
更多回答（1 个）
Jon Wieser
2019-6-10
0 个投票
try:
D=solve('(2*c*w1)+b2 = 0.3205','((2*c*b2)+ w1)*w1 = 214200','(b2*w1)=1.2260e+05;','c','w1','b2')
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