solving three equations for 3 unknows

Question

Valerie Cala 2019-6-10

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/466392-solving-three-equations-for-3-unknows

评论： Star Strider 2019-6-10

采纳的回答： Star Strider

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.

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Answer 1

Star Strider 2019-6-10

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https://ww2.mathworks.cn/matlabcentral/answers/466392-solving-three-equations-for-3-unknows#answer_378601

在 MATLAB Online 中打开

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])

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Star Strider 2019-6-10

在 MATLAB Online 中打开

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’):

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

Thank you! everything is clear now :3

Star Strider 2019-6-10

As always, my pleasure!

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Answer 2

Jon Wieser 2019-6-10

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/466392-solving-three-equations-for-3-unknows#answer_378599

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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solving three equations for 3 unknows

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