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

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

Answer 1

[Star Strider]

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. .

[Michael Garvin]

I promise you, this is not homework.

[Star Strider]

O.K.

f = @(p) p(1)*B + p(2)*D - E;
p0 = [1;1];
P = fsolve(f, p0);

with:

a = P(1)
c = P(2)

Different values for ‘p0’ may be necessary. That may require a bit of experimentation.

[John D'Errico]

Except you can't use one equation to compute a solution for two variables. Yes. You will get A solution. And depending on the initial values posed, you will get completely different solutions for every set of initial values.

[Star Strider]

Definitely true. However if we had vectors for ‘B’. ‘D’ and ‘E’, this becomes a simple linear regression problem, solved with the mldivide,\ operator.

We only know the information we have been given, and thus far, that indicates that the constants are scalars.

[Michael Garvin]

And ‘B’. ‘D’ and ‘E’ are all vectors so this would be the next step. Thank you!

I was looking at fsolve and @ but had never used them before and was constantly getting errors. Thank you fro clearing it up!

[Star Strider]

As always, my pleasure!

[Walter Roberson]

Is each B D E tuple to be solved independently, or are you needing to find a single a, c that together are "best fits" over all of the B D E together?

[Walter Roberson]

If you have more than one B D E and they are considered to be related, then you can find both a and c simultenously as best-fit using techniques similar to what Ivo Houtzager shows, or using the \ operator.

[Michael Garvin]

I'm needing to find a single ‘A’ & ‘C’ that best fits ‘B’, ‘D’, and ‘E’. I think the ‘\’will work, as described above by Star Strider, but I will definitely look at Ivo Houtzagar's link. Thank you.

[Star Strider]

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.

Answer 2

[Walter Roberson]

((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.

Answer 3

[Ivo Houtzager]

A = E*pinv([B; D]);
a = A(1);
c = A(2);

Answer 4

[Steven Lord]

This is a generalization of Cleve's simplest impossible problem. Cleve's has B = 1/2, D = 1/2, E = 3.