1. Copyright © 2011 Pearson Education, Inc. Systems of Linear Equations and Inequalities CHAPTER 9.1Solving Systems of Linear Equations Graphically 9.2Solving Systems of Linear Equations by Substitution; Applications 9.3Solving Systems of Linear Equations by Elimination; Applications 9.4Solving Systems of Linear Equations in Three Variables; Applications 9.5Solving Systems of Linear Equations Using Matrices 9.6 Solving Systems of Linear Equations Using Cramer’s Rule 9.7Solving Systems of Linear Inequalities 9

2. Copyright © 2011 Pearson Education, Inc. Solving Systems of Linear Equations Graphically 9.1 1.Determine whether an ordered pair is a solution for a system of equations. 2.Solve a system of linear equations graphically. 3.Classify systems of linear equations in two unknowns.

3. Slide 9- 3 Copyright © 2011 Pearson Education, Inc. System of equations: A group of two or more equations. Solution for a system of equations: An ordered set of numbers that makes all equations in the system.

4. Slide 9- 4 Copyright © 2011 Pearson Education, Inc. Checking a Solution to a System of Equations To verify or check a solution to a system of equations, 1. Replace each variable in each equation with its corresponding value. 2. Verify that each equation is true.

5. Slide 9- 5 Copyright © 2011 Pearson Education, Inc. Example 1 Determine whether each ordered pair is a solution to the system of equations. a. (  3, 2)b) (3, 7) Solution a. (  3, 2) x + y = 7y = 3x – 2  3 + 2 = 7 2 = 3(  3) – 2  1 = 72 =  11 FalseFalse Because (  3, 2) does not satisfy both equations, it is not a solution for the system.

6. Slide 9- 6 Copyright © 2011 Pearson Education, Inc. continued b. (3, 4) x + y = 7y = 3x – 2 3 + 4 = 7 4 = 3(3) – 2 7 = 74 = 7 TrueFalse Because (3, 4) does not satisfy both equations, it is not a solution to the system of equations.

7. Slide 9- 7 Copyright © 2011 Pearson Education, Inc. A system of two linear equations in two variables can have one solution, no solution, or an infinite number of solutions. The graphs intersect at a single point. There is one solution. The equations have the same slope, the graphs are parallel. There is no solution. The graphs are identical. There are an infinite number of solutions.

8. Slide 9- 8 Copyright © 2011 Pearson Education, Inc. Solving Systems of Equations Graphically To solve a system of linear equations graphically, 1. Graph each equation. a. If the lines intersect at a single point, then the coordinates of that point form the solution. b. If the lines are parallel, there is no solution. c. If the lines are identical, there are an infinite number of solutions, which are the coordinates of all the points on that line. 2. Check your solution.

9. Slide 9- 9 Copyright © 2011 Pearson Education, Inc. Example 2a Solve the system of equations graphically. Solution Graph each equation. The lines intersect at a single point, (  2, 4). We can check the point in each equation to verify and we will leave that to you. y = 2 – x 2x + 4y = 12 (  2, 4)

10. Slide 9- 10 Copyright © 2011 Pearson Education, Inc. Example 2b Solve the system of equations graphically. Solution Graph each equation. The lines appear to be parallel, which we can verify by comparing the slopes. The slopes are the same, so the lines are parallel. The system has no solution

11. Slide 9- 11 Copyright © 2011 Pearson Education, Inc. Example 2c Solve the system of equations graphically. Solution Graph each equation. The lines appear to be identical. The equations are identical. All ordered pairs along the line are solutions.

12. Slide 9- 12 Copyright © 2011 Pearson Education, Inc. Consistent system of equations: A system of equations that has at least one solution. Inconsistent system of equations: A system of equations that has no solution.

13. Slide 9- 13 Copyright © 2011 Pearson Education, Inc. Classifying Systems of Equations To classify a system of two linear equations in two unknowns, write the equations in slope-intercept form and compare the slopes and y-intercepts. Consistent system with independent equations: The system has a single solution at the point of intersection. The graphs are different. They have different slopes. Consistent system with dependent equations: The system has an infinite number of solutions. The graphs are identical. They have the same slope and same y-intercept. Inconsistent system: The system has no solution. The graphs are parallel lines. They have the same slope, but different y- intercepts.

14. Slide 9- 14 Copyright © 2011 Pearson Education, Inc. Example 3 For each of the systems of equations in Example 2, determine whether the system is consistent with independent equations, consistent with dependent equations, or inconsistent. How many solutions does the system have? 2a. The graphs intersected at a single point. The system is consistent. The equations are independent (different graphs), and the system has one solution: (  2, 4).

15. Slide 9- 15 Copyright © 2011 Pearson Education, Inc. Example 3 For each of the systems of equations in Example 2, determine whether the system is consistent with independent equations, consistent with dependent equations, or inconsistent. How many solutions does the system have? 2b. The graphs were parallel lines. The system is inconsistent and has no solutions.

16. Slide 9- 16 Copyright © 2011 Pearson Education, Inc. Example 3 For each of the systems of equations in Example 2, determine whether the system is consistent with independent equations, consistent with dependent equations, or inconsistent. How many solutions does the system have? 2c. The graphs coincide. The system is consistent (it has a solution) with dependent equations (same graph) and has an infinite number of solutions.

17. Slide 5- 17 Copyright © 2011 Pearson Education, Inc. Which set of points is a solution to the system? a) (–1, 1) b) (–1, –1) c) (0, 2) d) (–3, 7) 9.1

18. Slide 5- 18 Copyright © 2011 Pearson Education, Inc. Which set of points is a solution to the system? a) (–1, 1) b) (–1, –1) c) (0, 2) d) (–3, 7) 9.1

19. Copyright © 2011 Pearson Education, Inc. Solving Systems of Linear Equations by Substitution; Applications 9.2 1.Solve systems of linear equations using substitution. 2.Solve applications involving two unknowns using a system of equations.

20. Slide 9- 20 Copyright © 2011 Pearson Education, Inc. Solving Systems of Two Equations Using Substitution To find the solution of a system of two linear equations using the substitution method, 1. Isolate one of the variables in one of the equations. 2. In the other equation, substitute the expression you found in step 1 for that variable. 3. Solve this new equation. (It will now have only one variable.) 4. Using one of the equations containing both variables, substitute the value you found in step 3 for that variable and solve for the value of the other variable. 5. Check the solution in the original equations.

21. Slide 9- 21 Copyright © 2011 Pearson Education, Inc. Example 1 Solve the system of equations using substitution. Solution Step 1: Isolate a variable in one equation. The second equation is solved for x. Step 2: Substitute x = 1 – y for x in the first equation.

22. Slide 9- 22 Copyright © 2011 Pearson Education, Inc. continued Step 3: Solve for y. Step 4: Solve for x by substituting 4 for y in one of the original equations. The solution is (  3, 4).

23. Slide 9- 23 Copyright © 2011 Pearson Education, Inc. Example 2 Solve the system of equations using substitution. Solution Step 1: Isolate a variable in one equation. Use either equation. 2x + y = 7 y = 7 – 2x Step 2: Substitute y = 7 – 2x for y in the second equation.

24. Slide 9- 24 Copyright © 2011 Pearson Education, Inc. continued Step 3: Solve for x. Step 4: Solve for y by substituting 2 for x in one of the original equations. The solution is (2, 3).

25. Slide 9- 25 Copyright © 2011 Pearson Education, Inc. Inconsistent Systems of Equations The system has no solution because the graphs of the equations are parallel lines. You will get a false statement such as 3 = 4. Consistent Systems with Dependent Equations The system has an infinite number of solutions because the graphs of the equations are the same line. You will get a true statement such as 8 = 8.

26. Slide 9- 26 Copyright © 2011 Pearson Education, Inc. Example 4 Solve the system of equations using substitution. Solution Substitute y = 3x – 5 into the first equation. False statement. The system is inconsistent and has no solution.

27. Slide 9- 27 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example 5 Solve the system of equations using substitution. Solution Substitute y = 4 – 3x into the second equation. True statement. The number of solutions is infinite.

28. Slide 9- 28 Copyright © 2011 Pearson Education, Inc. Solving Applications Using a System of Equations To solve a problem with two unknowns using a system of equations, 1. Select a variable for each of the unknowns. 2. Translate each relationship to an equation. 3. Solve the system.

29. Slide 9- 29 Copyright © 2011 Pearson Education, Inc. Example 6 Terry is designing a table so that the length is twice the width. The perimeter is to be 216 inches. Find the length and width of the table. Understand We are given two relationships and are to find the length and width. Plan Select a variable for the length and another variable for the width, translate the relationships to a system of equations, and then solve the system.

30. Slide 9- 30 Copyright © 2011 Pearson Education, Inc. continued Execute Let l represent the length and w the width. Relationship 1:The length is twice the width. Translation: l = 2w Relationship 2:The perimeter is 216 inches. Translation: 2l + 2w = 216 System:

31. Slide 9- 31 Copyright © 2011 Pearson Education, Inc. continued Solve. Now find the value of l. l = 2w = 2(36) = 72 Answer The length should be 72 inches and the width 36 inches. Substitute 2w for l. Combine like terms. Divide both sides by 6 to isolate w.

32. Slide 9- 32 Copyright © 2011 Pearson Education, Inc. Example 8 Anita and Ernesto are traveling north in separate cars on the same highway. Anita is traveling at 55 miles per hour and Ernesto is traveling at 70 miles per hour. Anita passes Exit 54 at 1:30 p.m. Ernesto passes the same exit at 1:45 p.m. At what time will Ernesto catch up with Anita? Understand To determine what time Ernesto will catch up with Anita, we need to calculate the amount of time it will take him to catch up to her. We can then add the amount to 1:45.

33. Slide 9- 33 Copyright © 2011 Pearson Education, Inc. continued Plan and Execute Let x = Anita’s travel time after passing the exit. Let y = Ernesto’s travel time after passing the exit. Relationship 1: Ernesto passes the exit 15 minutes after Anita; Anita will have traveled 15 minutes longer. Translation: x = y + ¼ (1/4 of an hour) Relationship 2: When Ernesto catches up, they will have traveled the same distance. Translation: 55x = 70y CategoryRateTimeDistance Anita55x55x Ernesto70y70y

34. Slide 9- 34 Copyright © 2011 Pearson Education, Inc. continued Our system: Use substitution: Substitute to find x.

35. Slide 9- 35 Copyright © 2011 Pearson Education, Inc. continued Answer Ernesto will catch up to Anita in a little over 1 hour (1.17, which is 1 hour 10 minutes). The time will be 1:45 p.m. + 1 hr 10 minutes = 2:55 p.m. Check Verify both given relationships.

36. Slide 5- 36 Copyright © 2011 Pearson Education, Inc. Solve the system. a) (10, 3) b) (1,  4) c) (10, 14) d) (3, 6) 9.2

37. Slide 5- 37 Copyright © 2011 Pearson Education, Inc. Solve the system. a) (10, 3) b) (1,  4) c) (10, 14) d) (3, 6) 9.2

38. Slide 5- 38 Copyright © 2011 Pearson Education, Inc. Solve the system. a) (2, 3) b) (  2, 6) c) (2,  6) d) (3, 2) 9.2

39. Slide 5- 39 Copyright © 2011 Pearson Education, Inc. Solve the system. a) (2, 3) b) (  2, 6) c) (2,  6) d) (3, 2) 9.2

40. Copyright © 2011 Pearson Education, Inc. Solving Systems of Linear Equations by Elimination; Applications 9.3 1.Solve systems of linear equations using elimination. 2.Solve applications using elimination.

41. Slide 9- 41 Copyright © 2011 Pearson Education, Inc. Example 1 Solve the system of equations using elimination. Solution We can add the equations. Notice that y is eliminated, so we can easily solve for the value of x. Divide both sides by 4 to isolate x. Now that we have the value of x, we can find y by substituting 4 for x in one of the original equations. x + y = 9 4 + y = 9 y = 5 The solution is (4, 5).

42. Slide 9- 42 Copyright © 2011 Pearson Education, Inc. Example 2 Solve the system of equations. Solution Because no variables are eliminated if we add, we rewrite one of the equations so that it has a term that is the additive inverse of one of the terms in the other equation. Multiply the first equation by 4. Solve for y. x + y = 8 3 + y = 8 y = 5 The solution is (3, 5).

43. Slide 9- 43 Copyright © 2011 Pearson Education, Inc. Example 3 Solve the system of equations. Solution Choose a variable to eliminate, y, then multiply both equations by numbers that make the y terms additive inverses. Multiply the first equation by 2. Multiply the second equation by  5. Multiply by 2. Multiply by  5. Add the equations to eliminate y.

44. Slide 9- 44 Copyright © 2011 Pearson Education, Inc. continued Substitute x = 1 into either of the original equations. 4x – 5y = 19 4(1) – 5y = 19 4 – 5y = 19 –5y = 15 y =  3 The solution is (1, –3).

45. Slide 9- 45 Copyright © 2011 Pearson Education, Inc. Example 4 Solve the system of equations. Solution To clear the decimals in Equation 1, multiply by 100. To clear the fractions in Equation 2, multiply by 5. Multiply by 100. Multiply by 5. Multiply equation 2 by  1 then combine the equations.

46. Slide 9- 46 Copyright © 2011 Pearson Education, Inc. continued Substitute to find y. The solution is (  1,  3).

47. Slide 9- 47 Copyright © 2011 Pearson Education, Inc. Solving Systems of Two Linear Equations Using Elimination To solve a system of two linear equations using the elimination method, 1. Write the equations in standard form (Ax + By = C). 2. Use the multiplication principle to clear fractions or decimals (optional). 3. If necessary, multiply one or both equations by a number (or numbers) so that they have a pair of terms that are additive inverses. 4. Add the equations. The result should be an equation in terms of one variable. 5. Solve the equation from step 4 for the value of that variable. 6. Using an equation containing both variables, substitute the value you found in step 5 for the corresponding variable and solve for the value of the other variable. 7. Check your solution in the original equations.

48. Slide 9- 48 Copyright © 2011 Pearson Education, Inc. Inconsistent Systems and Dependent Equations When both variables have been eliminated and the resulting equation is false, such as 0 = 5, there is no solution. The system is inconsistent. When both variables have been eliminated and the resulting equation is true, such as 0 = 0, the equations are dependent. There are an infinite number of solutions.

49. Slide 9- 49 Copyright © 2011 Pearson Education, Inc. Example 5a Solve the system of equations. Solution Notice that the left side of the equations are additive inverses. Adding the equations will eliminate both variables. False statement. The system is inconsistent and has no solution.

50. Slide 9- 50 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example 5b Solve the system of equations. Solution To eliminate y multiply the first equation by 2. True statement. The equations are dependent. There are an infinite number of solutions. Multiply by 2.

51. Slide 9- 51 Copyright © 2011 Pearson Education, Inc. Example 7 Mia sells concessions at a movie theatre. In one hour, she sells 78 popcorns. The popcorn sizes are small, which sell for $4 each, and large, which sell for $6 each. If her sales totaled $420, then how many of each size popcorn did she sell? Understand The unknowns are the number of each size popcorn. One relationship involves the number of popcorns (78 total), and the other relationship involves the total sales in dollars ($420).

52. Slide 9- 52 Copyright © 2011 Pearson Education, Inc. continued Plan and Execute Let x represent the number of small popcorns. Let y represent the number of large popcorns. Relationship 1: The total number sold is 78. Translation: x + y = 78 Relationship 2: The total sales revenue is $420. Translation: 4x + 6y = 420 CategoryPriceNumberRevenue Small$4x4x4x Large$6y6y6y

53. Slide 9- 53 Copyright © 2011 Pearson Education, Inc. continued Our system: Use elimination; choose x. Multiply by  4.  Substitute to find x.

54. Slide 9- 54 Copyright © 2011 Pearson Education, Inc. continued Answer Mia sold 54 large popcorns and 24 small popcorns. Check Verify both given relationships.

55. Slide 9- 55 Copyright © 2011 Pearson Education, Inc. Example 8 How many milliliters of a 20% HCl solution and 50% HCl solution must be mixed together to make 500 milliliters of 35% HCl solution? Understand The unknowns are the volumes of 20% and 50% solution that are mixed. One relationship involves the concentrations of each solution in the mixture and the other relationship involves the total volume of the final mixture (500 ml).

56. Slide 9- 56 Copyright © 2011 Pearson Education, Inc. continued Plan and Execute Let x and y represent the two amounts to be mixed. Relationship 1: The total volume is 500 ml. Translation: x + y = 500 Relationship 2: The combined volumes of HCl in the two mixtures is to be 35% of the total mixture. Translation: 0.20x + 0.50y = 0.35(500) SolutionConcentrationVolumeAmount of HCl 20%0.20x0.20x 50%0.50y0.50y 35%0.35x + y0.35(500)

57. Slide 9- 57 Copyright © 2011 Pearson Education, Inc. continued Our system: Use elimination; choose x. Multiply by  0.20.  Substitute to find x.

58. Slide 9- 58 Copyright © 2011 Pearson Education, Inc. continued Answer Mixing 250 ml of 20% solution with 250 ml of 50% solution gives 500 ml of 35% solution. Check Verify both given relationships.

59. Slide 5- 59 Copyright © 2011 Pearson Education, Inc. Solve the system. a) (2, 3) b) (7, 0) c) (–2,  3) d) (5, 5) 9.3

60. Slide 5- 60 Copyright © 2011 Pearson Education, Inc. Solve the system. a) (2, 3) b) (7, 0) c) (–2,  3) d) (5, 5) 9.3

61. Slide 5- 61 Copyright © 2011 Pearson Education, Inc. Solve the system. a) (0, 10) b) (1, 5) c) (–2,  3) d) no solution 9.3

62. Slide 5- 62 Copyright © 2011 Pearson Education, Inc. Solve the system. a) (0, 10) b) (1, 5) c) (–2,  3) d) no solution 9.3

63. Slide 5- 63 Copyright © 2011 Pearson Education, Inc. Solve the system. a) (1, 1) b) (1, 5) c) (–1, 1) d) no solution 9.3

64. Slide 5- 64 Copyright © 2011 Pearson Education, Inc. Solve the system. a) (1, 1) b) (1, 5) c) (–1, 1) d) no solution 9.3

65. Copyright © 2011 Pearson Education, Inc. Solving Systems of Linear Equations in Three Variables; Applications 9.4 1.Determine whether an ordered triple is a solution for a system of equations. 2.Understand the types of solution sets for systems of three equations. 3.Solve a system of three linear equations using the elimination method. 4.Solve application problems that translate to a system of three linear equations.

66. Slide 9- 66 Copyright © 2011 Pearson Education, Inc. Example 1 Determine whether (2, –1, 3) is a solution of the system Solution In all three equations, replace x with 2, y with –1, and z with 3. x + y + z = 4 2 + (–1) + 3 = 4 4 = 4 TRUE 3 = 3 TRUE 2x – 2y – z = 3 2(2) – 2(–1) – 3 = 3 – 4x + y + 2z = –3 – 4(2) + (–1) + 2(3) = –3 –3 = –3 TRUE Because (2,  1, 3) satisfies all three equations in the system, it is a solution for the system.

67. Slide 9- 67 Copyright © 2011 Pearson Education, Inc. Types of Solution Sets A Single Solution: If the planes intersect at a single point, that ordered triple is the solution to the system.

68. Slide 9- 68 Copyright © 2011 Pearson Education, Inc. Types of Solution Sets Infinite Number of Solutions: If the three planes intersect along a line, the system has an infinite number of solutions, which are the coordinates of any point along that line. Infinite Number of Solutions: If all three graphs are the same plane, the system has an infinite number of solutions, which are the coordinates of any point in the plane.

69. Slide 9- 69 Copyright © 2011 Pearson Education, Inc. Types of Solution Sets No Solution: If all of the planes are parallel, the system has no solution. No Solution: Pairs of planes also can intersect, as shown. However, because all three planes do not have a common intersection, the system has no solution.

70. Slide 9- 70 Copyright © 2011 Pearson Education, Inc. Example 2a Solution Solve the system using elimination. We select any two of the three equations and work to get one equation in two variables. Let’s add equations (1) and (2): (1) (2) (3) (1) (2) (4) 2x + 3y = 8 Adding to eliminate z

71. Slide 9- 71 Copyright © 2011 Pearson Education, Inc. Next, we select a different pair of equations and eliminate the same variable. Let’s use (2) and (3) to again eliminate z. (5) x – y + 3z = 8 Multiplying equation (2) by 3 4x + 5y = 14. 3x + 6y – 3z = 6 Now we solve the resulting system of equations (4) and (5). That will give us two of the numbers in the solution of the original system, 4x + 5y = 14 2x + 3y = 8 (5) (4)

72. Slide 9- 72 Copyright © 2011 Pearson Education, Inc. We multiply both sides of equation (4) by –2 and then add to equation (5): Substituting into either equation (4) or (5) we find that x = 1. 4x + 5y = 14 –4x – 6y = –16, –y = –2 y = 2 Now we have x = 1 and y = 2. To find the value for z, we use any of the three original equations and substitute to find the third number z.

73. Slide 9- 73 Copyright © 2011 Pearson Education, Inc. Let’s use equation (1) and substitute our two numbers in it: We have obtained the ordered triple (1, 2, 3). It should check in all three equations. x + y + z = 6 1 + 2 + z = 6 z = 3. The solution is (1, 2, 3).

74. Slide 9- 74 Copyright © 2011 Pearson Education, Inc. Example 2b Solution Solve the system using elimination. The equations are in standard form. (1) (2) (3) (2) (3) (4) 3x + 2y = 4 Adding Eliminate z from equations (2) and (3).

75. Slide 9- 75 Copyright © 2011 Pearson Education, Inc. Eliminate z from equations (1) and (2). Multiplying equation (2) by 6 (1) (2) (3) Adding 15x + 15y = 15 Eliminate x from equations (4) and (5). 3x + 2y = 4 15x + 15y = 15 Multiplying top by  5  15x – 10y =  20 15x + 15y = 15 Adding 5y =  5 y =  1 Using y =  1, find x from equation 4 by substituting. 3x + 2y = 4 3x + 2(  1) = 4 x = 2 continued

76. Slide 9- 76 Copyright © 2011 Pearson Education, Inc. continued Substitute x = 2 and y =  1 to find z. x + y + z = 2 2 – 1 + z = 2 1 + z = 2 z = 1 The solution is the ordered triple (2,  1, 1). (1) (2) (3)

77. Slide 9- 77 Copyright © 2011 Pearson Education, Inc. Solving Systems of Three Linear Equations Using Elimination To solve a system of three linear equations with three unknowns using elimination. 1. Write each equation in the form Ax + By+ Cz = D. 2. Eliminate one variable from one pair of equations using the elimination method. 3. If necessary, eliminate the same variable from another pair of equations.

78. Slide 9- 78 Copyright © 2011 Pearson Education, Inc. continued 4. Steps 2 and 3 result in two equations with the same two variables. Solve these equations using the elimination method. 5. To find the third variable, substitute the values of the variables found in step 4 into any of the three original equations that contain the third variable. 6. Check the ordered triple in all three original equations.

79. Slide 9- 79 Copyright © 2011 Pearson Education, Inc. Example 3 Solution Solve the system using elimination. The equations are in standard form. (1) (2) (3) (1) (2) (4) 5y  4z = 0 Adding Eliminate x from equations (1) and (2).

80. Slide 9- 80 Copyright © 2011 Pearson Education, Inc. Eliminate x from equations (1) and (3). (1) (2) (3)  5y + 4z = 2 (5) Eliminate y from equations (4) and (5). 5y  4z = 0  5y + 4z = 2 0 = 2 All variables are eliminated and the resulting equation is false, which means that this system has no solution; it is inconsistent. continued

81. Slide 9- 81 Copyright © 2011 Pearson Education, Inc. Example 4 At a movie theatre, Kara buys one popcorn, two drinks and 2 candy bars, all for $12. Rebecca buys two popcorns, three drinks, and one candy bar for $17. Leah buys one popcorn, one drink and three candy bars for $11. Find the individual cost of one popcorn, one drink and one candy bar. Understand We have three unknowns and three relationships, and we are to find the cost of each. Plan Select a variable for each unknown, translate the relationship to a system of three equations, and then solve the system.

82. Slide 9- 82 Copyright © 2011 Pearson Education, Inc. continued Execute: p = popcorn, d = drink, and c = candy Relationship 1: one popcorn, two drinks and two candy bars, cost $12 Translation: p + 2d + 2c = 12 Relationship 2: two popcorns, three drinks, and one candy bar cost $17 Translation: 2p + 3d + c = 17 Relationship 3: one popcorn, one drink and three candy bars cost $11 Translation: p + d + 3c = 11

83. Slide 9- 83 Copyright © 2011 Pearson Education, Inc. continued Our system: Choose to eliminate p: Start with equations 1 and 3. Choose to eliminate p: Start with equations 1 and 2.

84. Slide 9- 84 Copyright © 2011 Pearson Education, Inc. continued Use equations 4 and 5 to eliminate d. Substitute for c in d – c = 1 Substitute for c and d in p + d + 3c = 11 p + 2.5 + 3(1.5) = 11 p + 7 = 11 p = 4 The cost of one candy bar is $1.50. The cost of one drink is $2.50. The cost of one popcorn is $4.00.

85. Slide 4- 85 Copyright © 2011 Pearson Education, Inc. Determine if (2, –5, 3) is a solution to the given system. a) Yes b) No 9.4

86. Slide 4- 86 Copyright © 2011 Pearson Education, Inc. Determine if (2, –5, 3) is a solution to the given system. a) Yes b) No 9.4

87. Slide 4- 87 Copyright © 2011 Pearson Education, Inc. Solve the system. a) (–2, 2, –5) b) (–5, 2, –2) c) (–2,  5, 2) d) no solution 9.4

88. Slide 4- 88 Copyright © 2011 Pearson Education, Inc. Solve the system. a) (–2, 2, –5) b) (–5, 2, –2) c) (–2,  5, 2) d) no solution 9.4

89. Copyright © 2011 Pearson Education, Inc. Solving Systems of Linear Equations Using Matrices 9.5 1.Write a system of equations as an augmented matrix. 2.Solve a system of linear equations by transforming its augmented matrix into row echelon form.

90. Slide 9- 90 Copyright © 2011 Pearson Education, Inc. Matrix: A rectangular array of numbers. The following are examples of matrices: The individual numbers are called elements.

91. Slide 9- 91 Copyright © 2011 Pearson Education, Inc. The rows of a matrix are horizontal, and the columns are vertical. column 1column 2column 3 row 1 row 2 row 3

92. Slide 9- 92 Copyright © 2011 Pearson Education, Inc. Augmented Matrix: A matrix made up of the coefficients and the constant terms of a system. The constant terms are separated from the coefficients by a dashed vertical line. Let’s write this system as an augmented matrix:

93. Slide 9- 93 Copyright © 2011 Pearson Education, Inc. Example 1 Solution Write the augmented matrix. Write as an augmented matrix.

94. Slide 9- 94 Copyright © 2011 Pearson Education, Inc. Row Operations The solution of a system is not affected by the following row operations in its augmented matrix. 1. Any two rows may be interchanged. 2. The elements of any row may be multiplied (or divided) by any nonzero real number. 3. Any row may be replaced by a row resulting from adding the elements of that row (or multiples of that row) to a multiple of the elements of any other row.

95. Slide 9- 95 Copyright © 2011 Pearson Education, Inc. Row echelon form: An augmented matrix whose coefficient portion has 1’s on the diagonal from upper left to lower right and 0s below the 1’s.

96. Slide 9- 96 Copyright © 2011 Pearson Education, Inc. Solution Example 2 Solve the following linear system by transforming its augmented matrix into row echelon form. We write the augmented matrix. We perform row operations to transform the matrix into echelon form. 3R 2 + R 1

97. Slide 9- 97 Copyright © 2011 Pearson Education, Inc. The resulting matrix represents the system: R 2  10 continued Since y = 1, we can solve for x using substitution. The solution is (2, 1). 3x + 1 = 7 3x + y = 7 3x = 6 x = 2

98. Slide 9- 98 Copyright © 2011 Pearson Education, Inc. Example 3 Solution Write the augmented matrix: Use the row echelon method to solve

99. Slide 9- 99 Copyright © 2011 Pearson Education, Inc. Our goal is to transform into the form Interchange Row 1 and Row 2 –2R 1 + R 2 continued

100. Slide 9- 100 Copyright © 2011 Pearson Education, Inc. –R 2 + R 3 (1/3)R 2 R 1 + R 3 (1/3)R 3 continued

101. Slide 9- 101 Copyright © 2011 Pearson Education, Inc. z = –1. Substitute z into y – z = 2 y –(  1) = 2 y + 1 = 2 y = 1 continued Substitute y and z into x – y + z = 1 x – 1 – 1 = 1 x – 2 = 1 x = 3 The solution is (3, 1, –1).

102. Slide 9- 102 Copyright © 2011 Pearson Education, Inc. Replace R 2 in with R 1 + R 2. a)b) c)d) 9.5

103. Slide 9- 103 Copyright © 2011 Pearson Education, Inc. Replace R 2 in with R 1 + R 2. a)b) c)d) 9.5

104. Slide 4- 104 Copyright © 2011 Pearson Education, Inc. Solve by transforming the augmented matrix into row echelon form. a) (2, 1,  3) b) (  3, 1, 2) c) (  3, 2, 1) d) No solution 9.5

105. Slide 4- 105 Copyright © 2011 Pearson Education, Inc. Solve by transforming the augmented matrix into row echelon form. a) (2, 1,  3) b) (  3, 1, 2) c) (  3, 2, 1) d) No solution 9.5

106. Copyright © 2011 Pearson Education, Inc. Solving Systems of Linear Equations Using Cramer’s Rule 9.6 1.Evaluate determinants of 2  2 matrices. 2.Evaluate determinants of 3  3 matrices. 3.Solve systems of equations using Cramer’s Rule.

107. Slide 9- 107 Copyright © 2011 Pearson Education, Inc. Square matrix: A matrix that has the same number of rows and columns. Every square matrix has a determinant. Determinant of a 2  2 Matrix If then det(A) =

108. Slide 9- 108 Copyright © 2011 Pearson Education, Inc. Example 1a Find the determinant of the following matrix. Solution

109. Slide 9- 109 Copyright © 2011 Pearson Education, Inc. Example 1b Find the determinant of the following matrix. Solution

110. Slide 9- 110 Copyright © 2011 Pearson Education, Inc. Example 1c Find the determinant of the following matrix. Solution

111. Slide 9- 111 Copyright © 2011 Pearson Education, Inc. Minor of an element of a matrix: The determinant of the remaining matrix when the row and column in which the element is located are ignored.

112. Slide 9- 112 Copyright © 2011 Pearson Education, Inc. Example 2 Find the minor of  4 in Solution To find the minor of  4, we ignore its row and column (shown in blue) and evaluate the determinant of the remaining matrix (shown in red).

113. Slide 9- 113 Copyright © 2011 Pearson Education, Inc. Evaluating the Determinant of a 3  3 Matrix

114. Slide 9- 114 Copyright © 2011 Pearson Education, Inc. Example 3 Find the determinant of Solution Use the rule for expanding by minors along the first column, we have the following:

115. Slide 9- 115 Copyright © 2011 Pearson Education, Inc. Cramer’s Rule The solution to the system of linear equations

116. Slide 9- 116 Copyright © 2011 Pearson Education, Inc. The solution to the system of linear equations

117. Slide 9- 117 Copyright © 2011 Pearson Education, Inc. Solution Example 4 Use Cramer’s Rule to solve First, we find D, D x, and D y. The solution is (1/5, 4/5).

118. Slide 9- 118 Copyright © 2011 Pearson Education, Inc. Solution Example 5 Use Cramer’s Rule to solve We need to find D, D x, D y, and D z.

119. Slide 9- 119 Copyright © 2011 Pearson Education, Inc. continued

120. Slide 9- 120 Copyright © 2011 Pearson Education, Inc. continued The solution is (3, 1, –1). The check is left to the viewer.

121. Slide 4- 121 Copyright © 2011 Pearson Education, Inc. Find the determinant. a)  75 b)  72 c)  27 d) 72 9.6

122. Slide 4- 122 Copyright © 2011 Pearson Education, Inc. Find the determinant. a)  75 b)  72 c)  27 d) 72 9.6

123. Slide 4- 123 Copyright © 2011 Pearson Education, Inc. Solve using Cramer’s Rule. a) (  1,  4) b) (1,  3) c) (  2,  3) d) No solution 9.6

124. Slide 4- 124 Copyright © 2011 Pearson Education, Inc. Solve using Cramer’s Rule. a) (  1,  4) b) (1,  3) c) (  2,  3) d) No solution 9.6

125. Copyright © 2011 Pearson Education, Inc. Solving Systems of Linear Inequalities 9.7 1.Graph the solution set of a system of linear inequalities. 2.Solve applications involving a system of linear inequalities.

126. Slide 9- 126 Copyright © 2011 Pearson Education, Inc. Solving a System of Linear Inequalities in Two Variables To solve a system of linear inequalities in two variables, graph all of the inequalities on the same grid. The solution set for the system contains all ordered pairs in the region where the inequalities’ solution sets overlap along with ordered pairs on the portion of any solid line that touches the region of overlap.

127. Slide 9- 127 Copyright © 2011 Pearson Education, Inc. Example 1a Graph the solution set for the system of inequalities. Solution Graph the inequalities on the same grid. Both lines will be dashed. Solution

128. Slide 9- 128 Copyright © 2011 Pearson Education, Inc. Example 1b Graph the solution set for the system of inequalities. Solution Graph the inequalities on the same grid. Solution

129. Slide 9- 129 Copyright © 2011 Pearson Education, Inc. Example 2—Inconsistent System Graph the solution set for the system of inequalities. Solution Graph the inequalities on the same grid. The slopes are equal, so the lines are parallel. Since the shaded regions do not overlap there is no solution for the system.

130. Slide 9- 130 Copyright © 2011 Pearson Education, Inc. Example 3 Mr. Reynolds is landscaping his yard with some trees and bushes. He would like to purchase at least 3 plants. The trees cost $40 and the bushes cost $20. He cannot spend more than $300 for the plants. Write a system of inequalities that describes what Mr. Reynolds could purchase, then solve the system by graphing. Understand We must translate to a system of inequalities, and then solve the system.

131. Slide 9- 131 Copyright © 2011 Pearson Education, Inc. continued Plan and Execute Let x represent the trees and y represent the bushes. Relationship 1: Mr. Reynolds would like to purchase at least 3 plants. x + y  3 Relationship 2: Mr. Reynolds cannot spend more than $300. 40x + 20y  300

132. Slide 9- 132 Copyright © 2011 Pearson Education, Inc. continued Answer Since Mr. Reynolds cannot purchase negative plants, the solution set is confined to Quadrant 1. Any ordered pair in the overlapping region is a solution. Assuming that only whole trees and bushes can be purchased, only whole numbers would be in the solution set. For example: (4, 2); (3, 4)

133. Slide 5- 133 Copyright © 2011 Pearson Education, Inc. Graph. a)b) c)d) 9.7

134. Slide 5- 134 Copyright © 2011 Pearson Education, Inc. Graph. a)b) c)d) 9.7