solving three equations for 3 unknows

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.

Answer 1

[Star Strider]

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])

[Valerie Cala]

Thank you so much, it worked! However, I have other question, why the answer is like a vector? I mean, it is not supposed to be just one value per each unknown?

[Star Strider]

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]

Thank you! everything is clear now :3

[Star Strider]

As always, my pleasure!

Answer 2

[Jon Wieser]

try:

D=solve('(2*c*w1)+b2 = 0.3205','((2*c*b2)+ w1)*w1 = 214200','(b2*w1)=1.2260e+05;','c','w1','b2')