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Chapter 9 Systems of Equations and Inequalities

Matrices

9.1 Systems of Equations

- A set of equations is called a system of

equations. - The solutions must satisfy each equation in the

system. - A linear equation in n unknowns has the form
- where

the variables are of first-degree. - If all equations in a system are linear, the

system is a system of linear equations, or a

linear system.

9.1 Linear System in Two Variables

- Three possible solutions to a linear system in

two variables - One solution coordinates of a point,
- No solutions inconsistent case,
- Infinitely many solutions dependent case.

9.1

- Characteristics of a system of two linear

equations in two variables.

Graphs of solutions Classification

Nonparallel lines one consistent

Identical lines infinite dependent consistent

Parallel lines No solutions inconsistent

9.1 Substitution Method

- Example Solve the system.
- Solution

(1) (2)

Solve (2) for y.

Substitute y x 3 in (1).

Solve for x.

Substitute x 1 in y x 3.

Solution set (1, 4)

9.1 Solving a Linear System in Two Variables

Graphically

- Example Solve the system graphically.
- Solution Solve (1) and (2) for y.

(1) (2)

- Solve using the method of graphing.
- x2 y2 25
- x2 y 19

Solving a Nonlinear System of Equations

- Example Solve the system.
- Solution Choose the simpler equation, (2), and
- solve for y since x is squared in (1).
- Substitute for y into (1) .

(1) (2)

Solving a Nonlinear System of Equations

- Substitute these values for x into (3).
- The solution set is

9.2 Elimination Method

- Example Solve the system.
- Solution To eliminate x, multiply (1) by 2 and

(2) - by 3 and add the resulting equations.

(1) (2)

(3) (4)

9.2 Elimination Method

- Substitute 2 for y in (1) or (2).
- The solution set is (3, 2).
- Consistent System

Solving an Inconsistent System

- Example Solve the system.
- Solution Eliminate x by multiplying (1) by 2 and
- adding the result to (2).
- Solution set is ?.

(1) (2)

Inconsistent System

Solving a System with Dependent Equations

- Example Solve the system.
- Solution Eliminate x by multiplying (1) by 2 and

adding the - result to (2).
- Consistent

Dependent System - Solution set is all Rs.

(1) (2)

9.3 Graphing Systems of Inequalities

Look at the two graphs. Determine the

following A. The equation of each line. B. How

the graphs are similar. C. How the graphs are

different.

- The equation of each line is y x 3.
- The lines in each graph are the same and

represent all of the solutions to the equation y

x 3. - The graph on the right is shaded above the line

and this means that all of these points are

solutions as well.

Inequalities with Greater Than

Point (-4, 5)

Pick a point from the shaded region and test that

point in the equation y x 3.

This is incorrect. Five is greater than or equal

to negative 1.

If a solid line is used, then the equation would

be 5 -1. If a dashed line is used, then the

equation would be 5 gt -1. The area above the line

is shaded.

Inequalities with Less Than

Point (1, -3)

Pick a point from the shaded region and test that

point in the equation y -x 4.

This is incorrect. Negative three is less than

or equal to 3.

If a solid line is used, then the equation would

be -3 3. If a dashed line is used, then the

equation would be -3 lt 3. The area below the line

is shaded.

Graphing Linear Inequalities

- Write the inequality in slope-intercept form.
- Use the slope and y-intercept to plot two points.
- Draw in the line. Use a solid line for less than

or equal to (?) or greater than or equal to ().

Use a dashed line for less than (lt) or greater

than (gt). - Pick a point above the line or below the line.

Test that point in the inequality. If it makes

the inequality true, then shade the region that

contains that point. If the point does not make

the inequality true, shade the region on the

other side of the line. - Systems of inequalities Follow steps 1-4 for

each inequality. Find the region where the

solutions to the two inequalities would overlap

and this is the region that should be shaded.

Example

Graph the following linear system of inequalities.

Use the slope and y-intercept to plot two points

for the first inequality.

Draw in the line. For ? use a solid line.

Pick a point and test it in the inequality.

Shade the appropriate region.

Example

Graph the following linear system of inequalities.

The region above the line should be shaded.

Now do the same for the second inequality.

Example

Graph the following linear system of inequalities.

Use the slope and y-intercept to plot two points

for the second inequality.

Draw in the line. For lt use a dashed line.

Pick a point and test it in the inequality.

Shade the appropriate region.

Example

Graph the following linear system of inequalities.

The region below the line should be shaded.

Example

Graph the following linear system of inequalities.

The solution to this system of inequalities is

the region where the solutions to each inequality

overlap. This is the region above or to the left

of the green line and below or to the left of the

blue line. Shade in that region.

You Try It

Graph the following linear systems of

inequalities.

Problem 1

Use the slope and y-intercept to plot two points

for the first inequality.

Draw in the line.

Shade in the appropriate region.

Problem 1

Use the slope and y-intercept to plot two points

for the second inequality.

Draw in the line.

Shade in the appropriate region.

Problem 1

The final solution is the region where the two

shaded areas overlap (purple region).

Sect. 9.4 Linear Programming

Goal 1 Find Maximum and Minimum values of a

function Goal 2 Solve Real World Problems

with Linear Programming

When graphing a System of Linear Inequalities

Each linear inequality is called a Constraint

The intersection of the graphs is called the

Feasible Region

When the graphs of the constraints is a

polygonal region, we say the region is Bounded.

Sometimes it is necessary to find the Maximum or

Minimum values that a linear function has for the

points in a feasible region.

The Maximum or Minimum value of a related

function Always occurs at one of the Vertices of

the Feasible Region.

Graph the system of inequalities. Name the

coordinates of the vertices of the feasible

region. Find the Maximum and Minimum values of

the function f(x, y) 3x 2y for this

polygonal region.

The polygon formed is a quadrilateral with

vertices at (0, 4), (2, 4), (5, 1), and (- 1, -

2).

Use a table to find the maximum and minimum

values of the function.

(x, y) 3x 2y f(x, y)

(0, 4) 3(0) 2(4) 8

(2, 4) 3(2) 2(4) 14

(5, 1) 3(5) 2(1) 17

(- 1, - 2) 3(- 1) 2(- 2) - 7

The maximum value is 17 at (5, 1). The Minimum

value is 7 at (- 1, - 2).

Bounded Region

Graph the system of inequalities. Name the

coordinates of the vertices of the feasible

region. Find the Maximum and Minimum values of

the function f(x, y) 2x - 5y for this

polygonal region.

The polygon formed is a triangle with vertices at

(- 2, 12), (- 2, - 3), and (4, - 3)

(x, y) 2x 5y F(x, y)

(- 2, 12) 2(-2) 5(12) - 64

(- 2, - 3) 2(- 2) 5(- 3) 11

(4, - 3) 2(4) 5(- 3) 23

The Maximum Value is 23 at (4, - 3). The Minimum

Value is 64 at (- 2, 12).

Sometimes a system of Inequalities forms a region

that is not a closed polygon.

In this case, the region is said to be Unbounded.

Unbounded Region

Graph the system of inequalities. Name the

coordinates of the vertices of the feasible

region. Find the Maximum and Minimum values of

the function f(x, y) 4y 3x for this region.

y 3x ? 4 y ? - 3x 4 y ? 8 x

There are only two points of intersection (- 1,

7) and (- 3, 5). This is an Unbounded Region.

(x, y) 4y 3x F(x, y)

(- 1, 7) 4(7) 3(- 1) 31

(- 3, 5) 4(5) 3(- 3) 29

The Maximum Value is 31 at (- 1, 7). There is no

minimum value since there are other points in the

solution that produce lesser values. Since the

region is Unbounded, f(x, y) has no minimum value.

Linear Programming Procedures.

Step 1 Define the Variables

Step 2 Write a system of Inequalities

Step 3 Graph the System of Inequalities

Step 4 Find the coordinates of the vertices of

the feasible region.

Step 5 Write a function to be maximized or

minimized.

Step 6 Substitute the coordinates of the

vertices into the function.

Step 7 Select the greatest or least result.

Answer the problem.

Ingrid is planning to start a home-based

business. She will be baking decorated cakes and

specialty pies. She estimates that a decorated

cake will take 75 minutes to prepare and a

specialty pie will take 30 minutes to prepare.

She plans to work no more than 40 hours per week

and does not want to make more than 60 pies in

any one week. If she plans to charge 34 for a

cake and 16 for a pie, find a combination of

cakes and pies that will maximize her income for

a week.

Step 1 Define the Variables

C number of cakes P number of pies

Step 2 Write a system of Inequalities

Since number of baked items cant be negative, c

and p must be nonnegative c ? 0 p ? 0

A cake takes 75 minutes and a pie 30 minutes.

There are 40 hours per week available. 75c 30p

? 2400 40 hours 2400 min.

She does not want to make more than 60 pies each

week p ? 60

Step 3 Graph the system of Inequalities

cakes

Pies

Step 4 Find the Coordinates of the vertices of

the feasible region.

The vertices of the feasible region are (0, 0),

(0, 32), (60, 8), and (60, 0).

Step 5 Write a function to be maximized or

minimized.

The function that describes the income is f(p,

c) 16p 34c

Step 6 Substitute the coordinates of the

vertices into function.

(p, c) 16p 34c f(p, c)

(0, 0) 16(0) 34(0) 0

(0, 32) 16(0) 34(32) 1088

(60, 8) 16(60) 34(8) 1232

(60, 0) 16(60) 34(0) 960

Step 7 Select the Greatest or Least result.

Answer the Problem.

The maximum value of the function is 1232 at (60,

8). This means that the maximum income is 1232

when Ingrid makes 60 pies and 8 cakes per week.

Section 9.5Solving Systems using Matrices

- What is a Matrix?
- Augmented Matrices
- Solving Systems of 3 Equations
- Inconsistent Dependent Systems

ConceptA Matrix

- Any rectangular array of numbers arranged in rows

and columns, written within brackets - Examples

- Matrix Row Transformations
- Streamlined use of echelon methods

9.5 Solution of Linear Systems by Row

Transformations

- This is called an augmented matrix where each

member of the array is called an element or

entry. - The rows of an augmented matrix can be treated

just like the equations of a linear system.

ConceptAugmented Matrices

- Are used to solve systems of linear equations
- Arrange equations in simplified standard form
- Put all coefficients into a 2x3 or 3x4 augmented

matrix

Reduced Row Echelon Method with the Graphing

Calculator

- Example Solve the system.
- Solution The augmented matrix of the system is
- shown below.

9.5 Reduced Row Echelon Method with the

Graphing Calculator

- Using the rref command we obtain the row reduced
- echelon form.

9.5 Solving a System with No Solutions

- Example Show that the following system is
- inconsistent.
- Solution The augmented matrix of the system is

9.5 Solving a System with No Solutions

- The final row indicates that the system is
- inconsistent and has solution set ?.

9.5 Solving a System with Dependent Equations

- Example Show that the system has dependent
- equations. Express the general solution using an
- arbitrary variable.
- Solution The augmented matrix is

9.5 Solving a System with Dependent Equations

- The final row of 0s indicates that the system has

- dependent equations. The first two rows represent

- the system

9.5 Solving a System with Dependent Equations

- Solving for y we get
- Substitute this result into the expression to

find x. - Solution set written with z arbitrary

Matrix Algebra Basics

- 9.6

Algebra

Matrix

A matrix is any doubly subscripted array of

elements arranged in rows and columns.

Row Matrix

1 x n matrix

Column Matrix

m x 1 matrix

Square Matrix

Same number of rows and columns

The Identity

Identity Matrix

Square matrix with ones on the diagonal and zeros

elsewhere.

Matrix Addition and Subtraction

A new matrix C may be defined as the additive

combination of matrices A and B where C

A B is defined by

Note all three matrices are of the same dimension

Addition

If

and

then

Matrix Addition Example

Matrix Subtraction

C A - B Is defined by

Matrix Multiplication

Matrices A and B have these dimensions

r x c and s x d

Matrix Multiplication

Matrices A and B can be multiplied if

r x c and s x d

c s

Matrix Multiplication

The resulting matrix will have the dimensions

r x c and s x d

r x d

Computation A x B C

2 x 2

2 x 3

2 x 3

Computation A x B C

3 x 2

2 x 3

A and B can be multiplied

3 x 3

Computation A x B C

3 x 2

2 x 3

Result is 3 x 3

3 x 3