Solve (a*B) + (c*D) = E without the Symbolic Toolbox
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Michael Garvin
le 25 Sep 2020
Commenté : Star Strider
le 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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Star Strider
le 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. .
Plus de réponses (3)
Walter Roberson
le 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
le 25 Sep 2020
Modifié(e) : Ivo Houtzager
le 25 Sep 2020
A = E*pinv([B; D]);
a = A(1);
c = A(2);
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Steven Lord
le 26 Sep 2020
This is a generalization of Cleve's simplest impossible problem. Cleve's has B = 1/2, D = 1/2, E = 3.
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