Cannot find ALL solutions to simultaneous equation
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Xinzhe Chai
on 8 Jan 2020
Commented: Xinzhe Chai
on 8 Jan 2020
Hi everyone, I would like to find ALL the solutions to this series of equations. Here is my code.
My problem is that it prompts me to use vpasolve, which doesn't give ALL the solutions. I tried "ReturnCondition", but to no avail.
My code is here:
clear all
syms x y z
eq1 = exp(x^2 + y^2 + z^2) == 9;
eq2 = x + x*y - z == 1;
eq3 = x + y + z == 1;
[solx,soly,solz] = solve([eq1,eq2,eq3],[x y z])
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Accepted Answer
David Goodmanson
on 8 Jan 2020
Hi CZ
If
exp(x^2+y^2+z^2) = 9
then
x^2+y^2+z^2 = log(9)
which is a lot less challenging for symbolic evaluation.
clear all
syms x y z C
eq1 = x^2 + y^2 + z^2 == C;
eq2 = x + x*y - z == 1;
eq3 = x + y + z == 1;
[solx,soly,solz] = solve([eq1,eq2,eq3],[x y z])
solz =
root(z1^4 - 5*z1^3 - z1^2*(C - 7) + z1*((5*C)/2 - 1/2) - C/2 + C^2/4 + 1/4, z1, 1)
root(z1^4 - 5*z1^3 - z1^2*(C - 7) + z1*((5*C)/2 - 1/2) - C/2 + C^2/4 + 1/4, z1, 2)
root(z1^4 - 5*z1^3 - z1^2*(C - 7) + z1*((5*C)/2 - 1/2) - C/2 + C^2/4 + 1/4, z1, 3)
root(z1^4 - 5*z1^3 - z1^2*(C - 7) + z1*((5*C)/2 - 1/2) - C/2 + C^2/4 + 1/4, z1, 4)
There are expressions that are twice as long for x and y. You can get numerical results by plugging in log(9) for C and using the roots function to get four solutions. I wish I knew how to get the coefficients in solx, soly and solz into the roots function in a convenient way, but I don't.
More Answers (1)
Walter Roberson
on 8 Jan 2020
Ookkay... The 55 kilobyte solution is attached.
Now what are you going to do with it?
5 Comments
Walter Roberson
on 8 Jan 2020
solve eq2 for x (in terms of y and z) . Substitute that x into eq3 and solve for z. substitute for x and then z into eq1, and solve for y asking solve for MaxDegree 4. The result will be four exact roots for y. simplify(). Now do back substitution.
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