I have a system of 3 linear equations with 2 variables. I can solve it by substitution but not with linsolve.

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3*x/5+y/5=x
2*x/5+0.8*y=y
x+y=1

Réponse acceptée

Oddur Bjarnason
Oddur Bjarnason le 14 Oct 2016
I found the answer to my question:
3*x/5+y/5=x 2*x/5+0.8*y=y x+y=1
3*x/5-5/5*x+y/5=0 2*x/5+4/5*y-5/5*y=0 x+y=1
-2/5*x+y/5=0 2/5*x-1/5*y=0 x+y=1
A=[-2/5,1/5;2/5,-1/5;1,1] B=[0;0;1] linsolve(A,B)
ans = 0.3333 0.6667

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