Chapter 7 Systems of Linear Equations and Inequalities

1

• Chapter 7 Systems of Linear Equations and
  Inequalities
• 7.1 Systems of Linear Equations
• 7.2 Solving Systems of Equations by
  Substitution and Addition Methods

2

• 7.1 Systems of Linear Equations
• A system of linear equations is a group of two or
  more linear equations in two variables. It is
  often necessary to find the common solution to
  all of these equations.
• This chapter presents different methods of
  solving systems of linear equations, and
  highlights how modeling problems can be solved
  using these methods.

3

• Systems of Linear Equations
• A system of linear equations is also called
  simultaneous linear equations.
• A solution to a system of equations is an ordered
  pair or pairs that satisfy all the equations in
  the system.
• A system of linear equations can have
• 1. Exactly one solution
• 2. No solutions
• 3. Infinitely many solutions

4

• Systems of Linear Equations
• To check if an ordered pair is a solution to a
  given system of equations, you simply substitute
  the ordered pair into all of the equations in the
  system. If each time we substitute, a true
  statement is found, then the ordered pair is a
  solution to the system.
• Look at Example 1 on page 346 of the text. This
  example shows how to check if an ordered pair is
  a solution to a given system of equations.

5

• Systems of Linear Equations
• There are four ways to solve systems of linear
  equations
• 1. By graphing
• 2. By substitution
• 3. By addition (also called elimination)
• 4. By matrices
• This section of the text covers solving systems
  of linear equations by graphing.

6

• Solving Systems by Graphing
• When solving a system by graphing
• Find ordered pairs that satisfy each of the
  equations.
• Plot the ordered pairs and sketch the graphs of
  both equations on the same axis.
• The coordinates of the point or points of
  intersection of the graphs are the solution or
  solutions to the system of equations.

7

• Solving Systems by Graphing
• Remember there were three possible cases for
  systems of linear equations
• 1. Exactly one solution
• 2. No solution
• 3. Infinitely many solutions

8

• Exactly One Solution
• When solving by graphing, if there is exactly one
  solution, then the lines intersect, and the
  solution is an ordered pair.
• This type of system, where there is exactly one
  solution, is called a consistent system.
• Look at Example 2 on page 347 in the text. It
  illustrates a system with one solution.

9

• No Solution
• When solving by graphing, if there is no
  solution, then the lines are parallel.
• Since the lines never intersect, there is no
  solution to be found.
• This type of system, where there is no solution,
  is called an inconsistent system.
• Look at Example 3 on page 347 in the text. It
  illustrates a system with no solution.

10

• Infinitely Many Solutions
• When solving by graphing, if there are infinitely
  many solutions, then the lines coincide (they are
  the same line).
• This type of system, where there are infinitely
  many solutions, is called a dependent system.
• Look at Example 4 on page 348 in the text. It
  illustrates this type of system.

11

• Summary of All Three

12

• Modeling Examples
• A mathematical model is an equation or system of
  equations that represent a real life situation.
  One main reason to solve systems of linear
  equations is to solve real world problems. Study
  Examples 5 and 6 on pages 348 and 349 in the
  text. Most of the modeling problems in the
  homework are similar to these two examples.
• Notice the accounting equations in Example 6, and
  how they are set up using the information in the
  question. Make sure to remember
• Profit Revenue Cost or P R - C

13

• Modeling Examples
• 1. Read problem 48 on page 352 in the text
  Security Systems. Write the system of equations
  to represent the cost of each system.

14

• Modeling Examples
• 1. Read problem 48 on page 352 in the text
  Security Systems. Write the system of equations
  to represent the cost of each system.
• let x of months of service
• Write the 1st equation

15

• Modeling Examples
• 1. Read problem 48 on page 352 in the text
  Security Systems. Write the system of equations
  to represent the cost of each system.
• let x of months of service
• Write the 1st equation
• A 18x 3380
• This is the equation for ABC Security. It is 18
  a month (18x) and a one time installation fee of
  3380.

16

• Modeling Examples
• 1. Continued
• Write the 2nd equation

17

• Modeling Examples
• 1. Continued
• Write the 2nd equation
• S 29x 2302
• This is the equation for Safe Homes Security. It
  is 29 a month (29x) and a one time installation
  fee of 2302.

18

• Modeling Examples
• 1. Continued
• The system of equations
• A 18x 3380
• S 29x 2302

19

• Modeling Examples
• 1. Continued
• The system of equations
• A 18x 3380
• S 29x 2302
• To complete the problem, graph the equations on
  the same axis, and determine where the two lines
  intersect. You will find that they intersect at
  98 months. So, at 98 months, the two systems
  cost the same amount.

20

• Modeling Examples
• 1. Continued
• What if Tamika plans to use the system for 10
  years, which plan is cheaper?

21

• Modeling Examples
• 1. Continued
• What if Tamika plans to use the system for 10
  years, which plan is cheaper? Ten years is 120
  months. Locate 120 months on your graph. Which
  line is lower at 120 months?

22

• Modeling Examples
• 1. Continued
• What if Tamika plans to use the system for 10
  years, which plan is cheaper? Ten years is 120
  months. Locate 120 months on your graph. Which
  line is lower at 120 months? ABC Security.
  Thus, ABC Security is cheaper if Tamika plans to
  use the system for 10 years.

23

• Modeling Examples
• 1. Continued
• What if Tamika plans to use the system for 10
  years, which plan is cheaper? Ten years is 120
  months. Locate 120 months on your graph. Which
  line is lower at 120 months? ABC Security.
  Thus, ABC Security is cheaper if Tamika plans to
  use the system for 10 years.
• Lets set up (determine the system of equations)
  another modeling problem.

24

• Modeling Examples
• 2. Read problem 50 on page 352 Selling Picture
  Frames. Write the system of equations.

25

• Modeling Examples
• 2. Read problem 50 on page 352 Selling Picture
  Frames. Write the system of equations.
• let x of picture frames sold
• Write the Revenue equation

26

• Modeling Examples
• 2. Read problem 50 on page 352 Selling Picture
  Frames. Write the system of equations.
• let x of picture frames sold
• Write the Revenue equation
• R 25x
• The revenue is 25 a picture frame.

27

• Modeling Examples
• 2. Read problem 50 on page 352 Selling Picture
  Frames. Write the system of equations.
• let x of picture frames sold
• Write the Revenue equation
• R 25x
• The revenue is 25 a picture frame.
• Write the Cost equation

28

• Modeling Examples
• 2. Read problem 50 on page 352 Selling Picture
  Frames. Write the system of equations.
• let x of picture frames sold
• Write the Revenue equation
• R 25x
• The revenue is 25 a picture frame.
• Write the Cost equation
• C 15x 400
• The cost is 15 per frame (15x) and a fixed cost
  of 400.

29

• Modeling Examples
• 2. Continued
• The system of equations
• R 25x
• C 15x 400

30

• Modeling Examples
• 2. Continued
• The system of equations
• R 25x
• C 15x 400
• To complete the problem, graph the equations on
  the same axis, and determine where the two lines
  intersect.

31

• Modeling Examples
• 2. Continued
• The system of equations
• R 25x
• C 15x 400
• To complete the problem, graph the equations on
  the same axis, and determine where the two lines
  intersect.
• You will find that they intersect at 40 picture
  frames. So, 40 picture frames is the break-even
  point, they make no money and they lose no money.

32

• Modeling Examples
• 2. Continued
• Write the profit formula

33

• Modeling Examples
• 2. Continued
• Write the profit formula
• Profit Revenue Cost
• P R C

34

• Modeling Examples
• 2. Continued
• Write the profit formula
• Profit Revenue Cost
• P R C
• P 25x ( 15x 400 )
• P 25x 15x 400
• P 10x 400

35

• Modeling Examples
• 2. Continued
• Write the profit formula
• Profit Revenue Cost
• P R C
• P 25x ( 15x 400 )
• P 25x 15x 400
• P 10x 400
• This equation is useful in answering questions e
  and f.

36

• Modeling Examples
• 2. Continued
• e) Determine whether the company makes a profit
  or loss if it sells 10 frames. What is the
  profit or loss?

37

• Modeling Examples
• 2. Continued
• e) Determine whether the company makes a profit
  or loss if it sells 10 frames. What is the
  profit or loss?
• P 10x 400

38

• Modeling Examples
• 2. Continued
• e) Determine whether the company makes a profit
  or loss if it sells 10 frames. What is the
  profit or loss?
• P 10x 400
• Substitute x 10 into profit equation.

39

• Modeling Examples
• 2. Continued
• e) Determine whether the company makes a profit
  or loss if it sells 10 frames. What is the
  profit or loss?
• P 10x 400
• Substitute x 10 into profit equation.
• P 10(10) 400
• P 100 400
• P -300

40

• Modeling Examples
• 2. Continued
• e) Determine whether the company makes a profit
  or loss if it sells 10 frames. What is the
  profit or loss?
• P 10x 400
• Substitute x 10 into profit equation.
• P 10(10) 400
• P 100 400
• P -300
• So, at 10 frames, there is a loss of 300.

41

• Modeling Examples
• 2. Continued
• f) How many frames must the company sell to
  realize a profit of 1000?

42

• Modeling Examples
• 2. Continued
• f) How many frames must the company sell to
  realize a profit of 1000?
• P 10x 400

43

• Modeling Examples
• 2. Continued
• f) How many frames must the company sell to
  realize a profit of 1000?
• P 10x 400
• Substitute P 1000 into the profit equation.

44

• Modeling Examples
• 2. Continued
• f) How many frames must the company sell to
  realize a profit of 1000?
• P 10x 400
• Substitute P 1000 into the profit equation.
• 1000 10x 400
• 1400 10x
• 140 x

45

• Modeling Examples
• 2. Continued
• f) How many frames must the company sell to
  realize a profit of 1000?
• P 10x 400
• Substitute P 1000 into the profit equation.
• 1000 10x 400
• 1400 10x
• 140 x
• So, a profit of 1000 is realized at the sale of
  140 picture frames.

46

• Modeling Examples
• Graphing is not necessarily the most efficient
  way to solve systems of linear equations, but it
  is interesting to compare this type of solution
  method to what will follow in the next section of
  the text.

47

• Determine Without Graphing
• There is a somewhat shortened way to determine
  what type (one solution, no solutions, infinitely
  many solutions) of solution exists within a
  system. Notice we are not finding the solution,
  just what type of solution.
• To answer this type of question, write the
  equations in slope-intercept form y mx b.
• (i.e., solve the equations for y, remember that
  m slope, b y - intercept).

48

• Determine Without Graphing
• Once the equations are in slope-intercept form,
  compare the slopes and intercepts.
• One solution the lines will have different
  slopes.
• No solution the lines will have the same slope,
  but different y intercepts.
• Infinitely many solutions the lines will have
  the same slope and the same y intercept.

49

• Determine Without Graphing
• Given the following lines, determine what type of
  solution exists, without graphing.
• Equation 1 3x 6y 5
• Equation 2 y (1/2)x 3

50

• Determine Without Graphing
• Given the following lines, determine what type of
  solution exists, without graphing.
• Equation 1 3x 6y 5
• Equation 2 y (1/2)x 3
• Writing each in slope-intercept form (solve for y)

51

• Determine Without Graphing
• Given the following lines, determine what type of
  solution exists, without graphing.
• Equation 1 3x 6y 5
• Equation 2 y (1/2)x 3
• Writing each in slope-intercept form (solve for
  y)
• Equation 1 y (1/2)x 5/6
• Equation 2 y (1/2)x 3

52

• Determine Without Graphing
• Given the following lines, determine what type of
  solution exists, without graphing.
• Equation 1 3x 6y 5
• Equation 2 y (1/2)x 3
• Writing each in slope-intercept form (solve for
  y)
• Equation 1 y (1/2)x 5/6
• Equation 2 y (1/2)x 3
• Since the lines have the same slope but different
  y-intercepts, there is no solution to the system
  of equations. The lines are parallel.

53

• Determine Without Graphing
• This type of problem is covered in the homework,
  problems 31 37 odd on page 352. Be sure you
  can solve this type of problem.

54

• 7.2 Solving Systems of Equations by
  Substitution and Addition Methods
• This section covers two other methods to solve
  systems of linear equations the substitution
  method and the addition method (which is also
  called the elimination method). These methods
  are often more useful than the graphing method
  since the solution to a system of equations may
  not be an integer value.

55

• Substitution Method
• Procedure for Substitution Method (two equations,
  two unknowns)
• Solve one of the equations for one of the
  variables.
• Substitute the expression found in step 1 into
  the other equation. Now solve for the remaining
  variable.
• Substitute the value from step 2 into the
  equation written in step 1, and solve for the
  remaining variable.

56

• Substitution Method
• The procedure for substitution found on the
  previous slide is slightly different from the
  found in the text on page 354. The procedure in
  basically the text is the same, but it is broken
  down into four steps.
• Lets try an example and see how the procedure
  works.

57

• Substitution Method
• 1. Solve the following system of equations by
  substitution.

Step 1 is already completed.
Step 2Substitute x3 into 2nd equation and solve.
Step 3 Substitute 4 into 1st equation and solve.
The answer ( -4 , -1)
58

• Substitution Method
• The previous example had one solution, an ordered
  pair (dont forget to write it as an ordered
  pair!)
• Example 2 on page 355 in the text illustrates a
  system with no solution using the substitution
  method. Notice at the end of this problem, you
  find a false statement (i.e. 8 0).
• Example 3 on page 356 in the text illustrates a
  system with an infinite number of solutions.
  Notice at the end of this problem, you find a
  true statement (i.e. 2 2).

59

• Addition Method
• The ultimate goal of the addition method is to
  rewrite the two equations so when you add them
  together, one of the variables drops out and you
  are left with an equation containing only one
  variable.
• For example, if one equation had a 5x, rewrite
  the other equation so that the x term will be
  5x. To obtain the desired equation, it might be
  necessary to multiply one or both of the original
  equations by a number.

60

• Addition Method
• Study the procedure for the addition method found
  at the top of page 357 in the text. The
  procedure seems long, but the actual work
  involved in solving a system with the addition
  method is minimal. An example follows.

61

• Addition Method
• 2. Solve the following system of equations by
  the addition method.
• x y -3
• -x y 5

62

• Addition Method
• 2. Solve the following system of equations by
  the addition method.
• x y -3
• -x y 5

Skip steps 1 2
63

• Addition Method
• 2. Solve the following system of equations by
  the addition method.
• x y -3
• -x y 5
• - 2y 2
• y -1

Skip steps 1 2
Add down, and solve for y
64

• Addition Method
• 2. Solve the following system of equations by
  the addition method.
• x y -3
• -x y 5
• - 2y 2
• y -1
• x y -3
• x- (-1) -3
• x 1 -3
• x -4

Skip steps 1 2
Add down, and solve for y
Substitute y -1 into the 1st equation and solve
for x. The answer ( -4 , -1)
65

• Addition Method
• 3. Solve using the addition method
• 4x y 6
• -8x 2y 13

66

• Addition Method
• 3. Solve using the addition method
• 4x y 6
• -8x 2y 13

Neither variable will be eliminated if we add the
equations as written. We must choose a variable
to be eliminated. Lets eliminate x.
67

• Addition Method
• 3. Solve using the addition method
• 2( 4x y 6 )
• -8x 2y 13

Multiply the entire first equation by 2.
68

• Addition Method
• 3. Solve using the addition method
• 2( 4x y 6 )
• -8x 2y 13
• 8x 2y 12
• -8x 2y 13

Multiply the entire first equation by 2.
69

• Addition Method
• 3. Solve using the addition method
• 2( 4x y 6 )
• -8x 2y 13
• 8x 2y 12
• -8x 2y 13
• 0 25

Add down to eliminate x. But look what happens,
y is eliminated too. We now have a false
statement, thus the system has no solution, it is
inconsistent.
70

• Addition Method
• Study Example 5 on page 357 in the text. In this
  example, one of the equations must be multiplied
  by -1 for a variable to be eliminated.
• Study Example 7 on page 359 in the text. In this
  example, both equations must be multiplied by
  different values for one of the variables to be
  eliminated. This example also has a fractional
  answer. Yes, sometimes the answer is a fraction.

71

• Modeling Examples
• The reason to learn about systems of equations is
  to learn how to solve real world problems.
• Study Example 8 on page 360 in the text. Notice
  how the original equations are set up based on
  the data in the question.
• Also note that we are trying to determine when
  the total cost at each garage will be the same.
  To do this, set the two cost equations equal to
  each other and solve. You will see this type of
  problem often.

72

• Modeling Examples
• Study Example 9 on page 361 in the text. This is
  a mixture problem. Notice how the original
  equations are set up based on the data in the
  question.
• Once the equations are set up, the 2nd equation
  is multiplied by 100 to remove the decimal. This
  is a common occurrence, so make sure you know how
  to do this.
• Note The example is solved using the addition
  method. It can also be solved by substitution.

73

• Modeling Examples
• 4. Read problem 40 on page 362 of the text
  basketball game.

74

• Modeling Examples
• 4. Read problem 40 on page 362 of the text
  basketball game.
• First assign the variables
• let x of 2 point shots
• let y of 3 point shots

75

• Modeling Examples
• 4. Read problem 40 on page 362 of the text
  basketball game.
• First assign the variables
• let x of 2 point shots
• let y of 3 point shots
• Writing the 1st equation
• They made 45 goals in a recent game
• x y 45

76

• Modeling Examples
• 4 continued.
• Writing the 2nd equation

77

• Modeling Examples
• 4 continued.
• Writing the 2nd equation
• Some 2 pointers, some 3 pointers, for a total
  score of 101 points
• 2x 3y 101

78

• Modeling Examples
• 4 continued.
• Writing the 2nd equation
• Some 2 pointers, some 3 pointers, for a total
  score of 101 points
• 2x 3y 101
• In words, the equation says 2 times the number of
  2 point shots plus 3 times the number of 3 point
  shots totals 101 points.

79

• Modeling Examples
• 4 continued.
• The two equations are
• x y 45
• 2x 3y 101

80

• Modeling Examples
• 4 continued.
• The two equations are
• -2( x y 45 )
• 2x 3y 101

Lets eliminate x, multiply the entire 1st
equation by 2.
81

• Modeling Examples
• 4 continued.
• The two equations are
• -2( x y 45 )
• 2x 3y 101
• -2x -2y -90
• 2x 3y 101

Lets eliminate x, multiply the entire 1st
equation by 2.
82

• Modeling Examples
• 4 continued.
• The two equations are
• -2( x y 45 )
• 2x 3y 101
• -2x -2y -90
• 2x 3y 101
• y 11

• Add down to eliminate x. Substitute y into the
  1st equation. x 11 45, so x 34.
• - 2 point shots and 11 - 3 point shots.
  

83

• Modeling Examples
• 5. Read problem 44 on page 363 in the text
• A Milk Mixture.

84

• Modeling Examples
• 5. Read problem 44 on page 363 in the text
• A Milk Mixture.
• First assign the variables
• let x gallons of 5 milk
• let y gallons of skim (0) milk

85

• Modeling Examples
• 5. Read problem 44 on page 363 in the text
• A Milk Mixture.
• First assign the variables
• let x gallons of 5 milk
• let y gallons of skim (0) milk
• Writing the 1st equation
• x y 100
• This is because they want to make a mixture
  totaling 100 gallons of milk.

86

• Modeling Examples
• 5. Continued
• Writing the 2nd equation

87

• Modeling Examples
• 5. Continued
• Writing the 2nd equation
• 0.05x 0.0y 0.035(100)
• Basically, we are multiplying the 1st equation by
  the percent butterfat of the milk. Our final
  mixture should be 3.5, so we multiply
  0.035(100), since we want 100 total gallons.

88

• Modeling Examples
• 5. Continued
• The two equations are
• x y 100
• 0.05x 0.0y 0.035(100)

89

• Modeling Examples
• 5. Continued
• The two equations are
• x y 100
• 0.05x 0.0y 0.035(100)
• Next, multiply the 2nd equation by 1000 to remove
  the decimal. This gives us the following system
  of equations x y 100
• 50x 0y 35(100)

90

• Modeling Examples
• 5. Continued
• The two equations are
• x y 100
• 0.05x 0.0y 0.035(100)
• Next, multiply the 2nd equation by 1000 to remove
  the decimal. This gives us the following system
  of equations x y 100
• 50x 0y 35(100)
• Solve the system (use substitution since the 2nd
  equation has only one variable). The answer
  follows on the next slide.

91

• Modeling Examples
• 5. Continued
• The answer is 70 gallons of 5 milk and 30
  gallons of skim (0) milk.

92

• Modeling Examples
• 6. Read problem 48 on page 363 in the text
  School Play Tickets.

93

• Modeling Examples
• 6. Read problem 48 on page 363 in the text
  School Play Tickets.
• First assign the variables
• let x of adult tickets sold (5 per ticket)
• let y of student tickets sold (2 per ticket)

94

• Modeling Examples
• 6. Read problem 48 on page 363 in the text
  School Play Tickets.
• First assign the variables
• let x of adult tickets sold (5 per ticket)
• let y of student tickets sold (2 per ticket)
• Writing the 1st equation
• x y 250
• Since a total of 250 tickets were sold.

95

• Modeling Examples
• 6. Continued
• Writing the 2nd equation

96

• Modeling Examples
• 6. Continued
• Writing the 2nd equation
• 5x 2y 950
• Basically, we multiplied the 1st equation by the
  price of the tickets, and set it equal to the
  amount of money collected.

97

• Modeling Examples
• 6. Continued
• Writing the 2nd equation
• 5x 2y 950
• Basically, we multiplied the 1st equation by the
  price of the tickets, and set it equal to the
  amount of money collected.
• Do you see how this is similar to example 4?
  The 2 and 3 point shots?

98

• Modeling Examples
• 6. Continued
• The two equations are
• x y 250
• 5x 2y 950

99

• Modeling Examples
• 6. Continued
• The two equations are
• x y 250
• 5x 2y 950
• Can you solve the system using either
  substitution or addition? The answer follows on
  the next slide.

100

• Modeling Examples
• 6. Continued
• The answer is 150 adult tickets were sold, and
  100 student tickets were sold.

101

• Congratulations!
• You have finished the PowerPoint slides for
  Chapter 7!

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