Linear equations

ELECTRICAL-1

MATHS GROUP

ID NUMBER:-31,32,33,34,35,36

Systems of Linear Equations

How to: solve by graphing, substitution, linear combinations, and special types of linear systems

What is a Linear System, Anyways?

A linear system includes two, or more, equations, and each includes two or more variables.

When two equations are used to model a problem, it is called a linear system.

a1,a2,a3,…,an, b: real number

a1: leading coefficient

x1: leading variable

a linear equation in n variables:

a system of m linear equations in n variables:

5

mnmnmmm

nn

nn

nn

bxaxaxaxa

bxaxaxaxa

bxaxaxaxa

bxaxaxaxa

332211

33333232131

22323222121

11313212111

Consistent:

A system of linear equations has at least one solution.

Inconsistent:

A system of linear equations has no solution.

Elementary Linear Algebra: Section 1.1, p.4

Finding a Solution by Graphing

• Since our chances of guessing the right coordinates to try for a solution are not that high, we’ll be more successful if we try a different technique.

• Since a solution of a system of equations is a solution common to both equations, it would also be a point common to the graphs of both equations.

• So to find the solution of a system of 2 linear equations, graph the equations and see where the lines intersect.


Solve the following system of equations by graphing.

2x – y = 6 and

x + 3y = 10

x

y

First, graph 2x – y = 6.

(0, -6)

(3, 0)

(6, 6)

Second, graph x + 3y = 10

(1, 3)

(-2, 4)

(-5, 5)

The lines APPEAR to intersect at (4, 2).

(4, 2)

Example

Continued.


Although the solution to the system of equations appears to be (4, 2), you still need to check the answer by substituting x = 4 and y = 2 into the two equations.

First equation,

2(4) – 2 = 8 – 2 = 6 true

Second equation,

4 + 3(2) = 4 + 6 = 10 true

The point (4, 2) checks, so it is the solution of the system.

Example continued


Solve the following system of equations by graphing.

– x + 3y = 6 and

3x – 9y = 9

x

y

First, graph – x + 3y = 6

(-6, 0)

(0, 2)

(6, 4)

Second, graph 3x-9y=9

(0, -1)

(6, 1)

(3, 0)

The lines APPEAR to be parallel.

Example

Continued.


Although the lines appear to be parallel, you still need to check that they have the same slope. You can do this by solving for y.

First equation,

–x + 3y = 6

3y = x + 6 (add x to both sides)

3

1y = x + 2 (divide both sides by 3)

Second equation,

3x – 9y = 9

–9y = –3x + 9 (subtract 3x from both sides)

3

1y = x – 1 (divide both sides by –9)

Example continued

Both lines have a slope of , so they are parallel and do not intersect. Hence, there is no solution to the system

SLOPESlope is the ratio of the vertical rise to the horizontal run between any two points on a line. Usually referred to as the rise over run. Slope triangle between two

points. Notice that the slope triangle can be drawn two different ways.

Rise is -10 because we went down

Run is -6 because we went to the left

3

5

6

10

iscasethisinslopeThe

Rise is 10 because we went up

Run is 6 because we went to the right

3

5

6

10iscasethisinslopeThe

Another way to find slope

FORMULA FOR FINDING SLOPE

12

12

XX

YY

RUN

RISESLOPE

The formula is used when you know two points of a line.

),(),( 2211 YXBandYXAlikelookThey

EXAMPLE

Find the slope of the line between the two points (-4, 8) and (10, -4)

If it helps label the points. 1X 1Y2X 2Y

Then use the formula

12

12

XX

YY

)4()10(

)8()4(

FORMULAINTOSUBSTITUTE

7

6

14

12

)4()10(

)8()4(

SimplifyThen

Matrix equation

mn matrix:

14

(4) For a square matrix, the entries a11, a22, …, ann are called

the main diagonal entries.

mnmmm

n

n

n

aaaa

aaaa

aaaa

aaaa

321

3333231

2232221

1131211

rows m

columns n

(3) If , then the matrix is called square of order n.nm

Notes:

(1) Every entry aij in a matrix is a number.

(2) A matrix with m rows and n columns is said to be of size mn .

Elementary Linear Algebra: Section 1.2, p.14

mnmmm

n

n

n

aaaa

aaaa

aaaa

aaaa

A

321

3333231

2232221

1131211

mb

b

b

b2

1

nx

x

x

x2

1

bAx Matrix form:

Coefficient matrix:

A

aaaa

aaaa

aaaa

aaaa

mnmmm

n

n

n

321

3333231

2232221

1131211

The Substitution Method

Another method (beside getting lucky with trial and error or graphing the equations) that can be used to solve systems of equations is called the substitution method.

You solve one equation for one of the variables, then substitute the new form of the equation into the other equation for the solved variable.


Solve the following system using the substitution method.

3x – y = 6 and – 4x + 2y = –8

Solving the first equation for y,3x – y = 6

–y = –3x + 6 (subtract 3x from both sides)

y = 3x – 6 (multiply both sides by – 1)

Substitute this value for y in the second equation.–4x + 2y = –8

–4x + 2(3x – 6) = –8 (replace y with result from first equation)

–4x + 6x – 12 = –8 (use the distributive property)

2x – 12 = –8 (simplify the left side)

2x = 4 (add 12 to both sides)

x = 2 (divide both sides by 2)

Example

Continued.


Substitute x = 2 into the first equation solved for y.

y = 3x – 6 = 3(2) – 6 = 6 – 6 = 0

Our computations have produced the point (2, 0).

Check the point in the original equations.

First equation,

3x – y = 6

3(2) – 0 = 6 true

Second equation,

–4x + 2y = –8

–4(2) + 2(0) = –8 true

The solution of the system is (2, 0).

Example continued


Solving a System of Linear Equations by the Substitution Method

1) Solve one of the equations for a variable.

2) Substitute the expression from step 1 into the other equation.

3) Solve the new equation.

4) Substitute the value found in step 3 into either equation containing both variables.

5) Check the proposed solution in the original equations.


Solve the following system of equations using the substitution method.

y = 2x – 5 and 8x – 4y = 20

Since the first equation is already solved for y, substitute this value into the second equation.

8x – 4y = 20

8x – 4(2x – 5) = 20 (replace y with result from first equation)

8x – 8x + 20 = 20 (use distributive property)

20 = 20 (simplify left side)

Example

Continued.


When you get a result, like the one on the previous slide, that is obviously true for any value of the replacements for the variables, this indicates that the two equations actually represent the same line.

There are an infinite number of solutions for this system. Any solution of one equation would automatically be a solution of the other equation.

This represents a consistent system and the linear equations are dependent equations.

Example continued


Solve the following system of equations using the substitution

method.3x – y = 4 and 6x – 2y = 4

Solve the first equation for y.

3x – y = 4

–y = –3x + 4 (subtract 3x from both sides)

y = 3x – 4 (multiply both sides by –1)

Substitute this value for y into the second equation.

6x – 2y = 4

6x – 2(3x – 4) = 4 (replace y with the result from the first equation)

6x – 6x + 8 = 4 (use distributive property)

8 = 4 (simplify the left side)

Example

Continued.


When you get a result, like the one on the previous slide, that is never true for any value of the replacements for the variables, this indicates that the two equations actually are parallel and never intersect.

There is no solution to this system.

This represents an inconsistent system, even though the linear equations are independent.

Example continued

Solving systems of linear equations by addition

The Elimination Method

Another method that can be used to solve systems of equations is called the addition or elimination method.

You multiply both equations by numbers that will allow you to combine the two equations and eliminate one of the variables.


Solve the following system of equations using the elimination method.

6x – 3y = –3 and 4x + 5y = –9

Multiply both sides of the first equation by 5 and the second equation by 3.

First equation,

5(6x – 3y) = 5(–3)

30x – 15y = –15 (use the distributive property)

Second equation,

3(4x + 5y) = 3(–9)

12x + 15y = –27 (use the distributive property)

Example

Continued.


Combine the two resulting equations (eliminating the variable y).

30x – 15y = –15

12x + 15y = –27

42x = –42

x = –1 (divide both sides by 42)

Example continued

Continued.


Substitute the value for x into one of the original equations.

6x – 3y = –3

6(–1) – 3y = –3 (replace the x value in the first equation)

–6 – 3y = –3 (simplify the left side)

–3y = –3 + 6 = 3 (add 6 to both sides and simplify)

y = –1 (divide both sides by –3)

Our computations have produced the point (–1, –1).

Example continued

Continued.


Check the point in the original equations.

First equation,

6x – 3y = –3

6(–1) – 3(–1) = –3 true

Second equation,

4x + 5y = –9

4(–1) + 5(–1) = –9 true

The solution of the system is (–1, –1).

Example continued

The cryptographic method

Use of matrix

1 2

0 3

To obtain the Hill cipher for the obain text message

I AM HIDING

Example

The Cryptographic method

Solution

If we group the plaintext into pairs and add the dummy letter G to fill out the last pair we obtain

IA MH ID IN GG

Example Continued


91 13 8 94 9 14 77

To cipher the paintext, we form the matrix product

1 2 9 11

=

0 3 1 3

Example Continued

or equivalently


Example Continued

Which from , yields ciphertext KCTo encipher the pair MH, we form the product

1 2 13 29=

0 3 8 24


Example Continued

Whenever an integer greater than 25 occurs, it will be by the remainder that results when this integer is divided by 26

1 2 9 17=

0 3 4 12

1 2 9 37 11= or

0 3 14 42 16

1 2 7 21


Example Continued

The entire ciphertext message is

KC CX QL KP UU

which would usually be transmitted as a single string Without spaces

KCCXQLKPUU

Engineering

Linear equations