How to solve 5 equations with 5 unknowns

Hi, I don't have any idea how to solve this system. I'm beginner so i don't knew how to write corect code. Pls help me. Equations:
(A-B-C)*5=C*6 ////
(A-B-C)*2+B*3=(A-B)*D ////
(A-B)*(2-D)=B*3 ////
(A-B)*2+B*4=B*E ////
A*2+(A-B)*3+(A-B-C)*2=50

7 Kommentare

use this http://math.stackexchange.com/questions/255750/5-linear-equations-in-5-unknowns and to solve 5 equations with 5 unknown variables use this http://math.cowpi.com/systemsolver/5x5.html hope this is fine...
Roger Stafford
Roger Stafford am 29 Mai 2015
Not all these equations are linear.
@Roger check the equations again..
John D'Errico
John D'Errico am 29 Mai 2015
Bearbeitet: John D'Errico am 29 Mai 2015
Oh, gosh. I learn something new every day. Apparently (A-B)*D, (A-B)*(2-D), and B*E are all linear in the unknowns! I guess I'll need to go back and review my notes for all of those linear algebra classes. Darn this new math.
Walter Roberson
Walter Roberson am 29 Mai 2015
Muhammad, the expressions involve A*D, B*D, and B*E, so they certainly are non-linear.
Hinko Fic
Hinko Fic am 30 Mai 2015
Bearbeitet: Hinko Fic am 30 Mai 2015
Yeah, i knew how to solve linear, but this isn't. That's reason why i need help. I need to solve these equations with different numbers (more complicated) for my project in school
Roger Stafford
Roger Stafford am 11 Mär. 2018
I don't remember this problem from almost two years ago, but apparently all of us (except perhaps Muhammad Saleem) overlooked the fact that it is actually a system of five linear equations in the five unknown quantities, A, B, C, d=(A-B)*D, and e=B*E, so it can be solved for these latter five variables using matrix division. From those solutions, it is easy to then find D and E: D = d/(A-B) and E = e/B.

Melden Sie sich an, um zu kommentieren.

 Akzeptierte Antwort

Roger Stafford
Roger Stafford am 29 Mai 2015
Bearbeitet: Roger Stafford am 29 Mai 2015

0 Stimmen

You can use 'solve' in the Symbolic Toolbox to solve them.
My alternate method was in error. Just use 'solve'. It does very well.

5 Kommentare

Hinko Fic
Hinko Fic am 30 Mai 2015
Bearbeitet: Walter Roberson am 30 Mai 2015
I used this code:
equ1='(a-b-c)*5-6*c' ////
equ2='(a-b-c)*2+b*3-(a-b)*d' ////
equ3='(a-b)*(2-d)-b*3' ////
equ4='(a-b)*2+b*4-b*e' ////
equ5='a*2+(a-b)*3+(a-b-c)*2-50' ////
sol=solve(equ1,equ2,equ3,equ4,equ5) ////
but now i don't knew how to transform my results into numerical because i got "a: [1x1 sym]...
solve by substitution and simultaneously methods to make this in 1*1 system
(1) firstly you have to make a function that take these 5 equations as input (2) make an algorithm by using methods of simultaneously or substitutions.. (3) set output of this function as system of 1*1 equation..
structfun(@double, sol, 'Uniform', 0)
will return back a structure whose fields are a, b, c, d, e. For any particular one of them you can use (e.g.) double(sol.b)
Hinko Fic
Hinko Fic am 30 Mai 2015
That works, ty a lot, you saved me a lot of time

Melden Sie sich an, um zu kommentieren.

Weitere Antworten (1)

Alex Sha
Alex Sha am 12 Okt. 2019

0 Stimmen

Numercial solution:
a: 9.00473933649289
b: 1.18483412322275
c: 3.55450236966825
d: 1.54545454545455
e: 17.2

Kategorien

Mehr zu Optimization finden Sie in Hilfe-Center und File Exchange

Produkte

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by