Equations

Linear, quadratic, module, parametric equations

Equations

Postby Guest » Fri Nov 16, 2018 3:36 pm

Hello.I need resolve this:
A+C+E=6
B+D+F=3
B+G+I=2
A+H+J=2
D+H+K=3
C+G+L=3
F+J+L=2
E+I+K=5
E=F
J=I
A+C+E+G+I+K=13
B+D+F+H+J+L=13
Guest
 

Re: Equations

Postby Guest » Thu Jun 06, 2019 4:29 pm

Well, to start with, you have 12 unknowns and 12 equations. That's a good thing! As long as those equations are all "independent" there will be a unique solution. The equations are:
A+C+E=6
B+D+F=3
B+G+I=2
A+H+J=2
D+H+K=3
C+G+L=3
F+J+L=2
E+I+K=5
E=F
J=I
A+C+E+G+I+K=13
B+D+F+H+J+L=13

My eye is caught by the two simplest equations, E= F and J= I. I would go through and replace every "F" by "E" and every "J" by "I". Then we have
A+C+E=6
B+D+E=3
B+G+I=2
A+H+I=2
D+H+K=3
C+G+L=3
E+I+L=2
E+I+K=5
A+C+E+G+I+K=13
B+D+E+H+I+L=13 10 equations in 10 unknowns.
Of those the simplest are E+I+L= 2 and E+I+K= 2. Subtracting the second of those from the first, L-K= 1, so L= K+ 1. Replacing every L by K+1 we have:
A+C+E=6
B+D+E=3
B+G+I=2
A+H+I=2
D+H+K=3
C+G+K=2
E+I+K=1
E+I+K=5
A+C+E+G+I+K=13
B+D+E+H+I+K=12

But now we have a serious problem! One equation says "E+I+K= 1" and another "E+I+K= 5". We can stop now. NO value of E, I, and K can satisfy both of those so there is no solution.
Guest
 

Re: Equations

Postby Guest » Fri Nov 15, 2019 2:02 pm

how you guys are doing Pythagorean theorem calculation
Guest
 


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