# Simultaneous equations with 3 variables

Discussion in 'Mathematics and Physics' started by Overclocked, Oct 18, 2006.

1. ### OverclockedMember

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We are beginning Mesh analysis in Electric Circuits 1 (Im a graduate from high school with a major in Electronics Engineering, this is college), I remember how to use Simultaneous equations with 2 variables, but I cant seem to remember how to use it with 3 variables.

But Things dont always go the way as they show it. So do you multiply by the "other number" and add (With 2 variables thats What I do, for example:

3x+5y=2
6x+10y=4

To get rid of x you would multiply the top by 6 and the bottom by -3, and add. Can this be used to get rid of a third variable?)

2. ### PapabravoWell-Known Member

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Any of the three equations can be replaced by a linear combination of those equations and the solution set remains the same.

What you are trying for is to have one equation with a single nonzero coefficient, one equation with two nonzero coefficients, and one equation with three nonzero coefficients. This is called a triangular form. For example
Code (text):

3x + 5y + 6z = 13
4y + 2z = 7
9z = 18

for which the unique solution is { -11/12, 3/4, 2}

3. ### nyeNew Member

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yeah..from one equation, find an expression for say for example x..then substitute for x so tht u get 2 equations in z and y..and then solve as for a usual 2 variable simultaneous equation..

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5. ### GeoMXNew Member

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There are several methods, the one I like is sum/rest (well, sorry for my poor English , it means sum one equation to another so that you can get rid of one of the variables, then get the equation for one of the two remaining ones and finally substitute this in one of the others).

Also you can always use the simplest one: determinants (really good for 3 equations/variables).

Regards,
JJ Enríquez.

6. ### ljcoxWell-Known Member

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These 2 Eq are the same so there is no intersection.

Consider

3x + 5y = 2 -----1 and
6x + 2y = 1 ------2

Multiplty 1 by 2. this gives 6x+10y=4 ----3

Subtract the Eq 2 from Eq 3:-

0x + 8y = 3

Thus y = 3/8

7. ### HayatoMember

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3x+5y=2
6x+10y=4

The both equations are equal.
You have 1 equation, and 2 variables. You have infinites solutions.

So, you have to "parametrize".

y = r;

3x+5r = 2

x = (2 - 5r)/3

When r = 0 , x = 0 and y = 0

When r = 1 , x = -1 and y = 1

You have to choose aleatory values, but they need to obey the conditions:
y = r;
and
x = (2 - 5r)/3;

8. ### OverclockedMember

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So now, What if you have 4 Equations, and 4 Variables? How can that be solved? Is it even possible to solve something like that? Luckly, she just taught us another method-Super Position and Thevanins theorem. I know why she did it, to show us that there are easier ways to solve a circuit, along with the limitations (if any) that exsist in the circuit.

9. ### PapabravoWell-Known Member

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You can take n linear equations in n unknowns and:

a) find a unique solution if it exists
b) find a linear dependance to eliminate an equation
c) find that there are an infinite number of solutions

surprisingly the method remains the same regardless of the value of n.

10. ### HayatoMember

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Or you can solve the equation by the method:
Ax = b then x = [A^(-1)].b

Where 'A' is the coeficient matrix, 'x' is the variables matrix and 'b' is the solution matrix;
The determinant of 'A' != 0;

Or you can use the Gauss Elimination.

11. ### sci-3dMember

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I agree with this. By this method, you solve the n variables system linear equations. It just find the inverts matrix you can also use Matlab to do that.

12. ### HayatoMember

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Yes, it's a nice method to solve n variables linear eq. systems.

13. ### klmNew Member

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Its easier if you use an advanced calculator e.g Ti-83 etc, create a 3X4 matrix and enter the co-efficients, then use the rref function prebuilt into your calculator.

14. ### PommieWell-Known MemberMost Helpful Member

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It might be easier but you don't actually learn anything.

Mike.

15. ### ljcoxWell-Known Member

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I agree with Mike.

You should be able to do it by the basic method of eliminating variables.

Once you understand that procedure, then use either determinates or the Matrix method.

16. ### ljcoxWell-Known Member

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I find McMillman's Theorm very useful also.

17. ### vortechs2kNew Member

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You can always do it with a matrix, then work down to row-echelon form (I think that's the correct term). I know most people hate matrices, but that's what calculators are for.

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19. ### klmNew Member

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Speaking of learning, for just two equations elimination and substitution are easier. Find one and replace it in the other equation.

For 3 equations, use Gauss elimination or Cramer's rule, for more than 3 variables, Gauss elimination still applies however cramer's rule might be difficult. Or you could just start grouping equations and solving groups at a time.

Once learnt, learnt! as for applying it could be a tremendous time waster if you plan to do manually in the exams.

20. ### lcmvsvNew Member

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Dunno if this is already mentions:
where it's
Ax+By=C
Dx+Ey=F .........................for 3 equations
for 2 equations, variables
(F-C*(D/A))/(E-(D/A)*B)->Y
(C-BY)/A->X

Ax+By+Cy=D.........for the rest
for 3 equations,vars:
((DFK+BGL+CHJ-LFC-JGD-KHB)/M)->X
((AHK+DGI+CEL-IHC-LGA-KED)/M)->Y
((AFL+BHI+DEJ-IFD-JHA-LEB)/M)->Z

And plz help me on this:
I got these formula from looking at a program once, but I still have no clue how these formulas are derived from this :
Ax+By=C
Dx+Ey=F

21. ### littletransistorNew Member

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Solving simutaneous equations with three variables can be done with the row-echelon matrix way, or the Cramer's Rule if the matrix satisfy certain conditions.