Solving systems of equations >2 simultaneous

[pat666]

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:
$$0=(v_1-10)/4700+v_1/10000+(v_1-v_2)/1000$$
$$0=(v_2-v_1)/1000+(v_2-v_4)/470+(v_2- v_4)/1000$$
$$0=(v_4-v_2)/1000+(v_4-v_2)/470+v_4/10000+(v_4-5)/10000$$

Can someone help me find a simple way of solving these

Thanks

Homework Equations

The Attempt at a Solution

 

[tiny-tim]

hi pat666!

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

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

 

[pat666]

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

 

[Jaynte]

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 :)

 

[tiny-tim]

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" ?

 

[pat666]

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!

 

[Jaynte]

Do you mean how I rearrange it?

 

[pat666]

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?

 

[Jaynte]

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.