# 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

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### Accepted Answer

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. .

### More Answers (3)

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.

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Ivo Houtzager
on 25 Sep 2020

Edited: Ivo Houtzager
on 25 Sep 2020

A = E*pinv([B; D]);

a = A(1);

c = A(2);

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Steven Lord
on 26 Sep 2020

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