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How to Solve a System of 5 Equations

How to Solve a System of 5 Equations

Postby ealfano » Wed Feb 29, 2012 1:49 am
Hello,

I would like to solve this system of equations:

0=-0.0000011*L*R + 0.31*C – 0.08*R + 0.06*0.8*A + 0.000000000021;
0=0.0000011*L*R – 0.41*C + 0.042*0.45*B;
0=0.08*R - 0.06*A;
0=0.1*C - 0.042*B;
0=-0.0000011*L*R + 0.31*C

Where my variables are A, B, C, L, R

When I enter the command line below
0=-0.0000011*L*R + 0.31*C – 0.08*R + 0.06*0.8*A + 0.000000000021; 0=0.0000011*L*R – 0.41*C + 0.042*0.45*B; 0=0.08*R - 0.06*A; 0=0.1*C - 0.042*B; 0=-0.0000011*L*R + 0.31*C

into WolframAlpha, I get the following error message:
Using closest Wolfram|Alpha interpretation: 0.08*R + 0.06*0.8*A + 0.000000000021

Can someone please suggest a solution? Thank you.

Ed
ealfano
 
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Joined: Wed Feb 29, 2012 1:44 am

Re: How to Solve a System of 5 Equations

Postby jyellott » Thu Mar 01, 2012 7:58 pm
I haven't signed up as a power user (yet). As a free W|A user one sometimes runs out of computational power when attacking larger problems. The symptoms of this are a bit vague... This problem does seem to fit the diagnosis of insufficient computation power applied.

I first observe that each of the equations in your list works fine individually. And that some of the equations work together as listed. All of the equations together though, at present, result in W|A "refusing the gate" so to speak.

A convenent approach to this sort of problem with a problem, when practical, is to feed the equations in in smaller groups and pass the condensing results in too:

W|A (last 3): 0=0.08*R - 0.06*A; 0=0.1*C - 0.042*B; 0=-0.0000011*L*R + 0.31*C
Results:
C = (21 B)/50, A!=0, L = (1736000 B)/(11 A), R = (3 A)/4
A = 0, B = 0, C = 0, R = 0

Next the 2nd, assuming 1st solution above, to W|A:
0=0.0000011*L*R – 0.41*C + 0.042*0.45*B, C = (21 B)/50, A!=0, L = (1736000 B)/(11 A), R = (3 A)/4
Results:
B = 0, C = 0, L = 0, R = (3 A)/4, A!=0

Finally the 1st equation:
0=-0.0000011*L*R + 0.31*C – 0.08*R + 0.06*0.8*A + 0.000000000021, B = 0, C = 0, L = 0, R = (3 A)/4, A!=0
Resulting in a solution:
A~~1.75x10^-9, B~~0, C~~0, L~~0, R~~1.3125x10^-9

A quick inspection tells me none of the last two equations/solutions are inconsistent with the other 1st solution, of all zeros.

As some of the terms were of the same magnitude of the nonzero result I believe we are looking at a valid solution.
jyellott
 
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