Solving systems of equations >2 simultaneous

In summary, the conversation is about solving a system of equations using nodal analysis. The problem is that the equations are not in a matrix-friendly form, making it difficult to use certain methods. Some suggestions are to use linear algebra or Gaussian elimination, but both require rearranging the equations. The matrix method is powerful once learned, but can be tedious by hand.
  • #1
pat666
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0

Homework Statement



Hi, I'm doing nodal analysis and I end up with 3-5 equations that need solving simultaneously. I have been sticking them in mathematica and letting it do its thing but I need to be able to do it on my 89 too.
The trouble is that there not in a matrix friendly form - - - something like this:
[tex] 0=(v_1-10)/4700+v_1/10000+(v_1-v_2)/1000 [/tex]
[tex] 0=(v_2-v_1)/1000+(v_2-v_4)/470+(v_2- v_4)/1000 [/tex]
[tex] 0=(v_4-v_2)/1000+(v_4-v_2)/470+v_4/10000+(v_4-5)/10000 [/tex]

Can someone help me find a simple way of solving these

Thanks

Homework Equations


The Attempt at a Solution

 
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  • #2
hi pat666! :smile:

(try using the X2 icon just above the Reply box :wink:)

for just 3 equations, use the first one to write v1 as a function of v2

then substitute that value of v1 into the second equation …

you second and third equations are now two linear euqations in v2 and v4 :smile:
 
  • #3
How about for systems of 4 or 5 eq's. From my experience matrices (can't remember the name of the mathematician or consequently the procedure's name) are the way to go. My question is, would that require rearrangement to or is there a shortcut I haven't seen yet. There must be simple ways to do this aside from mathematica, MATLAB etc??

Thanks
 
  • #4
U can use linear algebra to solve it.

A*x = b
x = (inv)A*b

where in your case:
A ={{(1/4700+1/10000+1/1000), -1/1000,0};{-1/1000,(1/1000+1/470+1/1000),(-1/470-1/1000)};{0,(-1/1000-1/470),(1/1000+1/470+1/10000+1/10000)}}

b = (10/4700,0,5/10000)'
x = (v1,v2,v4)'

(inv)A you can easily get in a calculator which can handle matrixes.
By hand its very tedious :)
 
  • #5
pat666 said:
From my experience matrices (can't remember the name of the mathematician or consequently the procedure's name) are the way to go.

you mean http://en.wikipedia.org/wiki/Gaussian_elimination" [Broken] ? :wink:
 
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  • #6
Gaussian elimination is a pain in the arse, what I meant gives the same result but is far easier. Not trying (GE) that for these, I wanted to shoot myself in the head when I was doing that in HS.
Jaynte has the method I meant but no name? and I still think it requires rearrangement so that you get; ax+bx+...fx for the three equations
then you stick the coefficients in a matrix - invert it then multiply by the rhs. Still painful and slow for my liking but I am probably used to using a computer to solve and don't want to go back to calculator and hand!
 
  • #7
Do you mean how I rearrange it?
 
  • #8
I agree, when you first learn the matrix method (still no name??) its not "very easy". But the reason its so popular is because once you learn it it becomes VERY easy and powerful (no match for Mathematica though) I havn't done it for years so I was't sure if the rearrangement was necessary or not? Thanks, Ill do some surfing on the subject until I remember again.

P.S. I'm interested in the name if you know it off the top of your head?
 
  • #9
Hm, I don't know the name. Just that its a part of linear algebra :)
Its very powerful when you know it, I've use it in many differenent areas. But not by hand
for a long time.
But rearrange it so you get A and b is nessesary to get x.
 

1. How do you solve a system of equations with more than two variables?

Solving a system of equations with more than two variables requires a method called elimination. This involves eliminating one variable at a time by combining equations and solving for the remaining variables until all variables have been eliminated.

2. What is the difference between a consistent and inconsistent system of equations?

A consistent system of equations has at least one solution that satisfies all equations. An inconsistent system of equations has no solutions that satisfy all equations.

3. Can a system of equations have more than one solution?

Yes, a system of equations can have infinitely many solutions. This occurs when there are more variables than equations, allowing for multiple combinations that satisfy the equations.

4. How do you use matrices to solve systems of equations?

Matrices can be used to organize and solve systems of equations, particularly when there are more than two variables. The coefficients of each variable are placed in a matrix, and the constants are placed in a separate column. The matrix can then be manipulated using elimination or other methods to solve for the variables.

5. Are there any shortcuts or tricks for solving systems of equations with more than two variables?

There are various methods and techniques that can be used to solve systems of equations with more than two variables, such as substitution, elimination, and matrices. However, there is no one-size-fits-all shortcut or trick for solving these systems. The best approach will depend on the specific equations and variables involved.

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