Systems of Equations
In Electronics and with Matrices


This site explores the solution of two simultaneous equations for two unknowns.

There are three different manners in which you may want to explore this site depending on your interest and needs. They can be interchanged or used by themselves.

Electronics Application of Simultaneous Linear Equations

Linear Simultaneous Equations are frequently encountered in electronics and technology. For example, when using Kirchhoff's Laws (KCL & KVL) for analyzing a circuit you eventually arrive at a system of loop equations that must be solved for CURRENTS I1 and I2. These two unknowns are solution to the network that is established by Kirchhoff's Laws.

A System of Equations has one solution. This solution is a set of ordered pairs of the form (x,y) that satisfies two or more equations. A System of Equations is generally of the (standard) form. It is helpful to arrange (rewrite the equations of the system) to look like the standard form:

AX + BY = C
DX + EY = F

and one ordered pair (x,y) satisfies these two equations.

Or in electronics

AI1 + BI2 = C
D
I1 + EI2 = F

and one ordered pair (I1,I2) satisfies these two equations.

Example:

Eq 1. ) 3I1 I2 = 14
Eq 2. ) 2I1 + 3I2 = 7

The next step is to perform operations on one equation in the system to eliminate one variable and solve for the other.

Using the example, we want to find a way to make the coefficient of either I1 or I2 equal so that we can isolate the other variable and solve for it. Try multiplying the entire Eq 1. by "3" so that we have

Eq 1. ) (3)3I1+ (3)I2 = (3)14
Eq 2. ) 2I1 + 3I2 = 7

Now we have

Eq 1. ) 9I1+ 3I2 = 42
Eq 2. ) 2I1 + 3I2 = 7

We can then subtract Eq. 2 from Eq. 1 and solve for I1.

7I1 = 35

and so,

I1 = 5

Insert this I1 = 5 back into either Eq. 1 or Eq. 2 and then you will be able to solve for X.
Let's Insert it into Eq. 2.

Eq 2. ) 2I1 + 3I2 = 7
Eq 2. ) 2(5) + 3I2 = 7
Eq 2. ) 10 + 3I2 = 7
Eq 2. ) 3I2 = 7 - 10
Eq 2. ) 3I2 = -3
Eq 2. ) I2 = -1

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Interactive Solver for I1 and I2

Eq 1. ) 3I1 I2 = 14
Eq 2. ) 2I1 + 3I2 = 7

Try these Equations in the Solver below and hit the "Solve" Button.

I1+ I2 =

I1+ I2 =

I1 =

I2 =

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Systems of Equations in terms of x and y and a SOLVER

A System of Equations has one solution. This solution is a set of ordered pairs of the form (x,y) that satisfies two or more equations. A System of Equations is generally of the (standard) form. It is helpful to arrange (rewrite the equations of the system) to look like the standard form:

AX + BY = C
DX + EY = F

and one ordered pair (x,y) satisfies these two equations.

One way to solve this system of equations is to graph both equations (make sure to set y = equation) and observe the intersection of the lines that are plotted. This intersection point is the ordered pair that represents the solution.

You can try this by hand on paper or by using a graphing utility like a graphing calculator, or even a tool available on the web.

Try this example by graphing it:

Eq 1. ) X+ 2Y = 10
Eq 2. ) 2X - Y = 10

Another method is to perform operations on one equation in the system to eliminate one variable and solve for the other.

Using the same example:

Eq 1. ) X+ 2Y = 10
Eq 2. ) 2X - Y = 10

We want to find a way to make the coefficient of either x or y equal so that we can isolate the other variable and solve for it.

Try multiplying the entire Eq 1. by "2" so that we have

Eq 1. ) (2)X+ (2)2Y = (2)10
=
Eq 1. ) 2X+ 4Y = 20

Now we have

Eq 1. ) 2X+ 4Y = 20
Eq 2. ) 2X - Y = 10

We can then subtract them and solve for Y.

5Y = 10
and so,
Y = 2

Insert this Y = 2 into either Equation and then you will be able to solve for X.

Eq 2. ) 2X - Y = 10
Eq 2. ) 2X - (2) = 10
Eq 2. ) 2X = 12
Eq 2. ) X = 6

Does this match with what you have graphed?
Try it out numerically by entering values here (remember to put "-" for a negative sign.) and then hit the "Solve" button.

X + Y =

X + Y =

X =

Y =

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 Solver: Just Enter your Coefficient Values and hit solve

X + Y =

X + Y =

X =

Y =


Matrices and Cramer's Rule for Solving a System of Equations

Solutions to systems of equations of the form

AX + BY = C
DX + EY = F

Can be found using the following:

(CE - BF)

=


(AE - BD)

(AF - CD)

Y =


(AE - BD)

This method is derived from the determinants of matrices that can be arranged from a system of equations

Recall that a matrix A is of the form

[A] =

| a

b |

| c

d |

The value of the determinant of the 2 X 2 matrix is the difference of the diagonal products.

a*d - c * b = determinant A

To solve a system of the form

AX + BY = C
DX + EY = F

1.) Find the determinants of the following matrices.

[A] =

| a

b |

| d

e |

a*e - d * b = determinant A

[Ax] =

| c

b |

| f

e |

c * e - f * b = determinant Ax

[Ay] =

| a

c |

| d

f |

a*f - d * c = determinant Ay

2.) Use the determinants to find the solution if determinant A is not zero.

X =

determinant Ax
determinant A

Y =

determinant Ay
determinant A

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Email Jim Dildine
Date Last Modified: 11/11/99
James P. Dildine