Simultaneous equations with 3 variables - Electronic Circuits Projects Diagrams Free

Overclocked · 19th October 2006, 12:20 AM

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.

I have found this page: http://www.themathpage.com/alg/simul...quations-3.htm

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?)

Papabravo · 19th October 2006, 12:56 AM

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:

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

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

nye · 19th October 2006, 05:19 AM

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..

GeoMX · 25th October 2006, 02:32 AM

There are several methods, the one I like is sum/rest (well, sorry for my poor English :P, 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.

ljcox · 4th November 2006, 10:05 PM

Quote:

Originally Posted by Overclocked

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.

I have found this page: http://www.themathpage.com/alg/simul...quations-3.htm

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?)

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

Hayato · 7th November 2006, 06:56 PM

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;

Overclocked · 8th November 2006, 01:32 AM

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.

Papabravo · 8th November 2006, 02:17 AM

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.

Hayato · 8th November 2006, 12:22 PM

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.

sci-3d · 11th November 2006, 07:01 AM

Quote:

Originally Posted by Hayato

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.

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.

Hayato · 11th November 2006, 02:36 PM

Quote:

Originally Posted by sci-3d

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.

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

klm · 12th November 2006, 01:51 PM

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.

Pommie · 14th November 2006, 02:25 PM

Quote:

Originally Posted by klm

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.

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

Mike.

ljcox · 14th November 2006, 10:20 PM

Quote:

Originally Posted by Pommie

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

Mike.

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.

ljcox · 14th November 2006, 10:23 PM

Quote:

Originally Posted by Overclocked

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.

I find McMillman's Theorm very useful also.

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19th October 2006, 12:20 AM	#1
Overclocked	Simultaneous equations with 3 variables 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. I have found this page: http://www.themathpage.com/alg/simul...quations-3.htm 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?)

19th October 2006, 12:56 AM	#2
Papabravo	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: 3x + 5y + 6z = 13 4y + 2z = 7 9z = 18 for which the unique solution is { -11/12, 3/4, 2}

19th October 2006, 05:19 AM	#3
nye	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.. __________________ \"electricity is a good slave, but a bad master\"

25th October 2006, 02:32 AM	#4
GeoMX	There are several methods, the one I like is sum/rest (well, sorry for my poor English :P, 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. __________________ La imaginación es el límite.

7th November 2006, 06:56 PM	#6
Hayato	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;

8th November 2006, 01:32 AM	#7
Overclocked	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.

8th November 2006, 02:17 AM	#8
Papabravo	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.

8th November 2006, 12:22 PM	#9
Hayato	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.

12th November 2006, 01:51 PM	#12
klm	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.