Chapter 7: Systems of Equations and Inequalities; Matrices

16
Copyright © 2011 Pearson Education, Inc. Slide 7.1-1

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Chapter 7: Systems of Equations and Inequalities; Matrices. 7.1Systems of Equations 7.2Solution of Linear Systems in Three Variables 7.3Solution of Linear Systems by Row Transformations 7.4Matrix Properties and Operations 7.5Determinants and Cramer’s Rule - PowerPoint PPT Presentation

Transcript of Chapter 7: Systems of Equations and Inequalities; Matrices

Page 1: Chapter 7: Systems of Equations and          Inequalities; Matrices

Copyright © 2011 Pearson Education, Inc. Slide 7.1-1

Page 2: Chapter 7: Systems of Equations and          Inequalities; Matrices

Copyright © 2011 Pearson Education, Inc. Slide 7.1-2

Chapter 7: Systems of Equations and Inequalities; Matrices

7.1 Systems of Equations

7.2 Solution of Linear Systems in Three Variables

7.3 Solution of Linear Systems by Row Transformations

7.4 Matrix Properties and Operations

7.5 Determinants and Cramer’s Rule

7.6 Solution of Linear Systems by Matrix Inverses

7.7 Systems of Inequalities and Linear Programming

7.8 Partial Fractions

Page 3: Chapter 7: Systems of Equations and          Inequalities; Matrices

Copyright © 2011 Pearson Education, Inc. Slide 7.1-3

7.1 Systems of Equations

• A linear equation in n unknowns has the form where the variables are of first-degree.

• A set of equations is called a system of equations.

• The solutions of a system of equations must satisfy every equation in the system.

• If all equations in a system are linear, the system is a system of linear equations, or a linear system.

1 1 2 2 ,n na x a x a x b

Page 4: Chapter 7: Systems of Equations and          Inequalities; Matrices

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• Three possible solutions to a linear system in two variables:

1. One solution: coordinates of a point,

2. No solutions: inconsistent case,

3. Infinitely many solutions: dependent case.

7.1 Linear System in Two Variables

Page 5: Chapter 7: Systems of Equations and          Inequalities; Matrices

Copyright © 2011 Pearson Education, Inc. Slide 7.1-5

7.1 Substitution Method

Example Solve the system.

Solution

4

31

1

55

11623

11)3(23

3

y

y

x

x

xx

xx

xy Solve (2) for y.

Substitute y = x + 3 in (1).

Solve for x.

Substitute x = 1 in y = x + 3.

Solution set: {(1, 4)}

3

1123

yx

yx (1)

(2)

Page 6: Chapter 7: Systems of Equations and          Inequalities; Matrices

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7.1 Solving a Linear System in Two Variables Graphically

ExampleSolve the system graphically.

Solution Solve (1) and (2) for y.

3

1123

yx

yx (1)

(2)

Page 7: Chapter 7: Systems of Equations and          Inequalities; Matrices

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7.1 Elimination Method

Example Solve the system.

Solution To eliminate x, multiply (1) by –2 and (2)

by 3 and add the resulting equations.

3696

286

yx

yx (3)

(4)

2

3417

y

y

1232

143

yx

yx (1)

(2)

Page 8: Chapter 7: Systems of Equations and          Inequalities; Matrices

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7.1 Elimination Method

Substitute 2 for y in (1) or (2).

The solution set is {(3, 2)}.

• Check the solution set by substituting 3 in for x and 2 in for y in both of the original equations.

3

93

1)2(43

x

x

x

Page 9: Chapter 7: Systems of Equations and          Inequalities; Matrices

Copyright © 2011 Pearson Education, Inc. Slide 7.1-9

7.1 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 .

746

423

yx

yx (1)

(2)

746

846

yx

yx

150 Inconsistent System

Page 10: Chapter 7: Systems of Equations and          Inequalities; Matrices

Copyright © 2011 Pearson Education, Inc. Slide 7.1-10

7.1 Solving a System with Dependent Equations

Example Solve the system.

Solution Eliminate x by multiplying (1) by 2 and adding the result to (2).

Each equation is a solution of the other. Choose either equation and solve for x.

The solution set is e.g. y = –2:

42824

yxyx (1)

(2)

428

428

yx

yx

00

42

24

y

xyx

.,4

2

yy

)}2,1()}2,{( 422

Page 11: Chapter 7: Systems of Equations and          Inequalities; Matrices

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7.1 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) .

43

523 2

yx

yx (1)

(2)

(3)3

4

43

xy

yx

34 x

Page 12: Chapter 7: Systems of Equations and          Inequalities; Matrices

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7.1 Solving a Nonlinear System of Equations

Substitute these values for x into (3).

The solution set is

97

or 10)1)(79(

0729

15)4(29

53

423

2

2

2

xxxx

xx

xx

xx

1

314

or2743

34 9

7

yy

.1,1,, 2743

97

Page 13: Chapter 7: Systems of Equations and          Inequalities; Matrices

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7.1 Solving a Nonlinear System Graphically

Example Solve the system.

Solution (1) yields Y1 = 2x; (2) yields Y2 = |x + 2|.

The solution set is {(2, 4), (–2.22, .22), (–1.69, .31)}.

(2)

(1)

02

2

yx

y x

Page 14: Chapter 7: Systems of Equations and          Inequalities; Matrices

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7.1 Applications of Systems

• Solving Application Problems1. Read the problem carefully and

assign a variable to represent each quantity.

2. Write a system of equations involving these variables.

3. Solve the system of equations and determine the solution.

4. Look back and check your solution. Does it seem reasonable?

Page 15: Chapter 7: Systems of Equations and          Inequalities; Matrices

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7.1 Applications of Systems

Example In a recent year, the national average spent on two varsity athletes, one female and one male, was $6050 for Division I-A schools. However, average expenditures for a male athlete exceeded those for a female athlete by $3900. Determine how much was spent per varsity athlete for each gender.

Page 16: Chapter 7: Systems of Equations and          Inequalities; Matrices

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7.1 Applications of Systems

Solution Let x = average expenditures per male y = average expenditures per female

100,1260502

yxyxAverage spent on

one male and one female

(2)

(1)

3900

12100

yx

yx

160002 x8000x

Average Expenditure per male: $8000, andper female: from (2) y = 8000 – 3900 = $4100.