Solve (a*B) + (c*D) = E without the Symbolic Toolbox

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Michael Garvin on 25 Sep 2020
Commented: Star Strider on 28 Sep 2020
Solve (a*B) + (c*D) = E without the Symbolic Toolbox
where, B, D, & E are all known.
If the Symbolic Toolbox was available it would looke like this:
syms a c
eqn = ((a*B) + (c*D)) / E == 1;
x = solve( eqn );
Any help would be greatly appreciated.
(Available toolboxes include: Image Processing, Signal Processing, & Statistical and Machine Learning

Star Strider on 25 Sep 2020
This would seem to be homework, and for homework we only give guidance and hints.
I would set it up as an implicit equation (so it equals 0), and use fsolve. To do this, ‘a’ and ‘c’ would have to be parameterized as ‘p(1)’ and ‘p(2)’, and you would have to code it as an anonymous function. .
Star Strider on 28 Sep 2020
Experiment with something like this:
p = [B(:) D(:)] \ E(:);
a = p(1)
c = p(2)
If I understand correctly what you are doing, that should work.
To also get statistics with the parameter estimates, use the regress or fitlm functions, depending on what you want to do.

Walter Roberson on 25 Sep 2020
((a*B) + (c*D)) / E == 1
((a*B) + (c*D)) == 1 * E
a*B + c*D == E
a*B == E - c*D
a == (E-c*D) / B
a == E/B - D/B * c
a == (-D/B) * c + (E/B)
Parameterized:
c = t
a = (-D/B) * t + (E/B)
You have one equation in two variables; you are not going to be able to solve for both variables simultaneously.

Ivo Houtzager on 25 Sep 2020
Edited: Ivo Houtzager on 25 Sep 2020
A = E*pinv([B; D]);
a = A(1);
c = A(2);

Steven Lord on 26 Sep 2020
This is a generalization of Cleve's simplest impossible problem. Cleve's has B = 1/2, D = 1/2, E = 3.

R2020a

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