Solving Systems of Equations and Inequalities Section 3.1A-B Two variable linear equations Section...

Solving Systems of Equations and Inequalities

Section 3.1A-B Two variable linear equationsSection 3.1C Matrices Resolution of linear systemsSection 3.1D Three variable linear equationsSection 3.1D Determinants & Inverses of MatricesSection 3.2A Solving Systems of Linear InequalitiesSection 3.2B-D Linear Programming

Section 3.1 Two Variable Linear Equations

A set of two or more linear equations that each contain two variables.

These equations represent lines that will intersect, overlap, or be parallel to each other.

When these lines intersect or overlap each other they are said to be consistent. This means they have at least one point of intersection. If there is only one point of intersection, the lines are independent. If there is more than one point of intersection, the lines are dependent.

When two lines do not intersect, they are said to be inconsistent because the lines are parallel.



Check your understanding: Given the following two linear equations,

determine the consistency & dependence of the system. 2x + y = 5 & 4x + 2y = 8

a. Consistent, independentb. Consistent, dependentc. Inconsistent



determine the consistency & dependence of the system. 3x + 2y = 5 & 6x + 2y = 8

a. Consistent, independentb. Consistent, dependentc. Inconsistent

Section 3.1 Graphing Linear Equations

Systems of two linear equations that intersect at one point (consistent & independent) can be graphed separately by finding ordered pairs of the solution set for each and then locate the coordinates for the point of intersection.


In blue we have y = 3x - 2, and in red we have y = -x + 2.

y = 3x - 2 y = -x + 2

-1

0

1

2

-5

-2

1

4

-1

0

1

2

3

2

1

0


In blue we have y = 2x + 1, and in red we have y = x + 3.

y = 2x + 1 y = x + 3

-1

0

1

2

-1

1

3

5

-1

0

1

2

2

3

4

5


In blue we have y = 2x + 1, and in red we have y = 2x + 3.

y = 2x + 1 y = 2x + 3

-1

0

1

2

-1

1

3

5

-1

0

1

2

1

3

5

7

Homework

Section 3.1A pages 163-164 complete problems 1 – 17.


Systems of two linear equations are usually solved using either the elimination method or the substitution method.

The elimination method uses a process of removing one of the two variables simultaneously from both equations through addition of equal but opposite coefficients.




Blue 3x – y = 2

Red x + y = 2

4x = 4

With the variable y having equal but opposite values we can add the two equations together and eliminate the y variable. We get 4x = 4. This allows x = 1, and by inserting x = 1 back into the equation we can see that y = 1.


The substitution method uses a process of rewriting one of the two equations by isolating one of the variables and then substituting the equation into the other equation.


With both equations in the form of y equals, we can substitute the 2nd equation into the first and we get: -x + 2 = 3x – 2. Solving for x we get -4x = -4 and x = 1.

Practice Substitution Method2x + y = 10 y = -2x + 10

3x – 4y = -10 3x - 4(-2x + 10) = -10

3x + 8x – 40 = -1011x – 40 = -1011x = -10 + 40 = 30

x = 30/11 y = -2(30/11) + 10 y = -60/11 + 110/11 = 50/115x - y = 9 y = 5x - 9

2x + 4y = 42 2x + 4(5x - 9) = -10

2x + 20x – 45 = -1022x – 45 = -1022x = -10 + 45 = 35

x = 35/22 ; y = 5(35/22) – 9 = 175/22 – 198/22 = -23/22

Practice Elimination Method2x + y = 10 4(2x + y = 10) 8x + 4y = 403x – 4y = -10 3x – 4y = -10

11x = 30x = 30/11

2(30/11) + y = 10 y = 110/11 - 60/11 = 50/11 ; y = 50/11

5x - y = 9 4(5x - y = 9) 20x - 4y = 362x + 4y = 42 2x + 4y = 42

22x = 78x = 39/11

5(39/11) - y = 9 y = 195/11 - 99/11 = 96/11 ; y = 96/11



use the substitution method to start the solution of the system.

5x + 3y = 22 6x + 2y = 20

a. Y = 10 – 6xb. Y = 10 – 3xc. Y = 10 + 3x

Homework

Section 3.1B pages 173 – 174 problems 3 – 21 odd

Section 3.1 - Matrices

Matrix – A rectangle array of terms (elements) arranged in columns and rows. A matrix with m rows and n columns is called an m x n matrix, (read m by n matrix).

Matrices are also used to determine solutions for multiple variable linear equations. This technique can be used as an alternative to elimination or substitution methods.


a11 a12 a13

a21 a22 a23

a31 a32 a33

3 x 3 Matrix

The first number indicates the row (horizontal) and the second number indicates the column number (vertical).

Equal Matrices – Two matrices are equal if and only they have the same dimensions and are equal element by element.

=Y

X

2x – 6

2y

This expression states that

Y = 2x – 6 and x = 2y. Using the substitution method, we see that

Y = 2(2y) – 6 and so y = 2, x = 4.


Addition of Matrices – The sum of two m x n matrices is a m x n matrix in which the elements are the sum of the corresponding elements of the given matrices.

=A + B-2 + (-6) 0 + 7 1 + (-1)

0 + 4 5 + (-3) -8 + 10

A= -2 0 1

0 5 -8

= -6 7 -1

4 -3 10

B Solve for A + B.

=A + B-8 7 0

4 2 2


Subtraction of Matrices – The difference of two m x n matrices is equal to the sum A + (-B) where (-B) is the additive inverse of B.

=A - B-2 - (-6) 0 - 7 1 - (-1)

0 - 4 5 - (-3) -8 - 10

A= -2 0 1

0 5 -8

= -6 7 -1

4 -3 10

B Solve for A - B.

=A - B4 -7 2

-4 8 -18


Scalar Product – The product of a scalar k and an m x n matrix A is an m x n matrix denoted by kA. Each element of kA equals k times the corresponding element of A.

=kA5(-2) 5(0 ) 5(1)

5(0) 5(5) 5(-8)

A= -2 0 1

0 5 -8

= 5k Solve for kA.

=kA-10 0 5

0 25 -40

Section 3.1 Determinants and Inverses A determinant is a square array of numbers (written within a

pair of vertical lines) which represents a certain sum of products.

Calculating a 2 × 2 Determinant In general, we find the value of a 2 × 2 determinant with elements a, b,

c, d as follows:

We multiply the diagonals (top left × bottom right first), (bottom left x top right) then subtract the first product minus the second.

a b

c ddet =

a b

c d= ad - cb

Section 3.1 Determinants and Inverses

3 4(3 2) (4 1) 6 4 2

1 2x x

2 6( 2 2) ( 1 6) 4 ( 6) 2

1 2x x

a b

c ddet =

a b

c d= ad - cb

Ex:

4 6(4 2) ( 1 6) 8 ( 6) 2

1 2x x

3.1 Determinants & InversesSolve for the inverse by crisscrossing the first and fourth

terms and reversing the sign of the second and third term. Multiply this with the reciprocal of the determinant for the matrix ( ).

4 6 2 (6) 2 6

1 2 ( 1) 4 1 4

Examples of inverses for a 2 x 2 matrix.

This is the inverse of the given matrix.

Determinant = {(4 -2) – (-16)} = -8-(-6) = -2

1

2

2 6

1 4

=

2 61 3

2 21

1 4 22

2 2

1

det

4 6

1 2

3.1 Determinants & Inverses

For 2 x 2 matrices, the product of a determinant and the inverse of a matrix can be used to solve systems of two linear equations.

3 1det (3 4) (( 2) (1)) 12 ( 2) 14

2 4

4 14 11 14 142 3 2 3

14 1

14

4

x x

inverse

Determinants & InversesIf we have a two variable system, we

can use the determinant and inverse of a matrix to find the solution to the system.

3 2 5

4 5

x y

x y

We solved for the value of x and y using matrices and found that x = -1 and y = 1.

3 2 5

1 4 5

det( 3 4) (1 2) 12 2

4 24 21 10 101 3 1 3

10 10

4 2 20 105 10 1010 10 10 10( , ) ( , ) (

1 3 5 5 15 10 10

10 1

10

0 1 10

10

0

x

y

inverse

x y

1,1)

Determinants & InversesIf we have a two variable

system, we can use the determinant and inverse of a matrix to find the solution to the system.

We solved for the value of x and y using matrices and found that x = -1 and y = 1.

2 3 3

5 7 9

x y

x y

2 3 3

5 7 9

det(2 7) (5 3) 14 15

7 3 7 31

5 2 5 2

7 3 3 21 27( , ) (6,3)

5 2 9 15 18

1

1

x

y

inverse

x y

Homework

Section 3.1D pages 190-191 problems 2 - 6

Cramer’s Rule

Cramer’s Rule begins with the solving of the determinant for the system followed by the determinants for each of the variables within the system.

The determinant for each of the variables is calculated by first substituting the solution column values for the variable column values and then worked on as a 2 x 2 matrix.

Cramer’s Rule (continued)

Cramer’s Rule (conclusion)

Section 3.1 Three variable linear systems

Three linear equation systems are solved using the elimination and substitution methods.

The technique requires the isolation of one of the variables amongst the three equations.

This gets repeated for a 2nd variable and the remaining variable is then determined.

Afterwards, the other variables are determined.


SampleX + 2y + 2z = 102x – y + 2z = 6X – 3y + 2z = 1

Solve for x, y,& z.

Isolate the z value first. Combine line 1 & 2 and then combine line 3 & 2.

X + 2y + 2z = 10 X + 2y + 2z = 102x – y + 2z = 6 -2x +y – 2z = -6 -x + 3y = 4

X – 3y + 2z = 1 X – 3y + 2z = 12x – y + 2z = 6 -2x +y – 2z = -6 -x - 2y = -5


SampleX + 2y + 2z = 102x – y + 2z = 6X – 3y + 2z = 1

Solve for x, y,& z.

Isolate the x value next. Combine line 1 & 2 answer and the line 3 & 2 answer.

X + 2y + 2z = 11 X + 2y + 2z = 112x – y + 2z = 6 -2x +y – 2z = -6 -x + 3y = 5

X – 3y + 2z = 1 X – 3y + 2z = 12x – y + 2z = 6 -2x +y – 2z = -6 -x - 2y = -5

-x + 3y = 5 -x + 3y = 5

-x -2y = -5 x + 2y = 5

5y = 10

Y = 2, x = 1

Solve for z by reinserting y & x values. 1 + 2(2) + 2z = 10

Z = 2.5

Solving Systems of Equations and Inequalities Section 3.1A-B Two variable linear equations Section...

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