1. Slide 1- 1 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

2. Systems of Linear Equations and Problem Solving 8.1Systems of Equations in Two Variables 8.2Solving by Substitution or Elimination 8.3Solving Applications: Systems of Two Equations 8.4Systems of Equations in Three Variables 8.5Solving Applications: Systems of Three Equations 8.6Elimination Using Matrices 8.7 Determinants and Cramer’s Rule 8.8 Business and Economic Applications 8

3. Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Systems of Equations in Two Variables Translating Identifying Solutions Solving Systems Graphically 8.1

4. Slide 3- 4 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley System of Equations A system of equations is a set of two or more equations, in two or more variables, for which a common solution is sought.

5. Slide 3- 5 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution T-shirt Villa sold 52 shirts, one kind at $8.25 and another at $11.50 each. In all, $464.75 was taken in for the shirts. How many of each kind were sold? Set up the equations but do not solve. 1. Familiarize. To familiarize ourselves with this problem, guess that 26 of each kind of shirt was sold. The total money taken in would be The guess is incorrect, now turn to algebra. Example

6. Slide 3- 6 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 2. Translate. Let x = the number of $8.25 shirts and y = the number of $11.50 shirts. We have the following system of equations: x + y = 52 8.25x + 11.50y = 464.75 Kind of Shirt $8.25 shirt $11.50 shirt Total Number sold xy52 Price$8.25$11.50 Amount$8.25x$11.50y$464.75

7. Slide 3- 7 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Identifying Solutions A solution of a system of two equations in two variables is an ordered pair of numbers that makes both equations true.

8. Slide 3- 8 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution Determine whether (1, 5) is a solution of the system x – y = – 4 1 – 5 – 4 – 4 = – 4 TRUE 2x + y = 7 2(1) + 5 7 7 = 7 TRUE The pair (1, 5) makes both equations true, so it is a solution of the system. Example

9. Slide 3- 9 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solving Systems Graphically One way to solve a system of two equations is to graph both equations and identify any points of intersection. The coordinates of each point of intersection represent a solution of that system.

10. Slide 3- 10 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solve the system graphically. Solution It appears that (3, 2) is the solution. A check by substituting into both equations shows that (3, 2) is indeed the solution. Example Graph both equations. x – y = 1 x + y = 5 (3, 2)

11. Slide 3- 11 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solve the system graphically. Solution Example Graph both equations. The lines have the same slope and different y-intercepts, so they are parallel. The system has no solution.

12. Slide 3- 12 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solve the system graphically. Solution Example Graph both equations. The same line is drawn twice. Any solution of one equation is a solution of the other. There is an infinite number of solutions. The solution set is

13. Slide 3- 13 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley When we graph a system of two linear equations in two variables, one of the following three outcomes will occur. 1.The lines have one point in common, and that point is the only solution of the system. Any system that has at least one solution is said to be consistent. 2.The lines are parallel, with no point in common, and the system has no solution. This type of system is called inconsistent. 3.The lines coincide, sharing the same graph. This type of system has an infinite number of solutions and is also said to be consistent.

14. Slide 3- 14 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley When one equation in a system can be obtained by multiplying both sides of another equation by a constant, the two equations are said to be dependent. If two equations are not dependent, they are said to be independent.

15. Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solving by Substitution or Elimination The Substitution Method The Elimination Method 8.2

16. Slide 3- 16 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley The Substitution Method Algebraic (nongraphical) methods for solving systems are often superior to graphing, especially when fractions are involved. One algebraic method, the substitution method, relies on having a variable isolated.

17. Slide 3- 17 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution Solve the system(1) (2) The equations are numbered for reference. Equation (2) says that y and x – 1 name the same number. Thus we can substitute x – 1 for y in equation (1): x + y = 2 x + (x – 1) = 2 Equation (1) Substituting x – 1 for y We solve the last equation: Example

18. Slide 3- 18 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution (continued) x + (x – 1) = 2 Now return to the original pair of equations and substitute 3/2 for x in either equation so that we can solve for y 2x = 3 x =. Equation (2) Substituting 3/2 for x y = 3/2 – 1 y =. 2x – 1 = 2

19. Slide 3- 19 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution (continued) We obtain the ordered pair We can plug the ordered pair into the original pair of equations to check that it is the solution.

20. Slide 3- 20 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution Solve the system(1) (2) First, select an equation to solve for one variable. To isolate y, subtract 3x from both sides of equation (1): (1) (3) Next, proceed as in the last example, by substituting: Example

21. Slide 3- 21 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution (continued) Equation (2) Substituting 5 – 3x for y We can substitute 2 for x in either equation (1), (2), or (3). It is easiest to use (3) because it has already been solved for y: 2x – 3y = 7 y = –1. 2x – 3(5 – 3x) = 7 2x – 15 + 9x = 7 11x = 22 x = 2. y = 5 – 3(2)

22. Slide 3- 22 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution (continued) We obtain the ordered pair (2, –1). We can plug the ordered pair into the original pair of equations to check that it is the solution.

23. Slide 3- 23 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution Solve the system (1) (2) We can substitute 7y + 2 for x in equation (1) and solve: 7y + 2 = 7y + 5 When the y terms drop out, the result is a contradiction. We state that the system has no solution. 2 = 5. Substituting 7y + 2 for x Subtracting 7y from both sides Example

24. Slide 3- 24 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley The Elimination Method The elimination method for solving systems of equations makes use of the addition principle: If a = b, then a + c = b + c.

25. Slide 3- 25 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 0x + 2y = 14. Solution Solve the system (1) (2) Note that according to equation (2), –x + y and 9 are the same number. Thus we can work vertically and add –x + y to the left side of equation (1) and 9 to the right side: x + y = 5(1) –x + y = 9(2) Adding Example

26. Slide 3- 26 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution (continued) This eliminates the variable x, and leaves an equation with just one variable, y for which we solve: 2y = 14 y = 7 x + 7 = 5 Substituting. We also could have used equation (2). Next, we substitute 7 for y in equation (1) and solve for x: x = –2

27. Slide 3- 27 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution (continued) We obtain the ordered pair (–2, 7). We can plug the ordered pair into the original pair of equations to check that it is the solution.

28. Slide 3- 28 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution Solve the system (1) (2) To clear the fractions in equation (2), we multiply both sides of equation (2) by 12 to get equation (3): (3) Example

29. Slide 3- 29 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution (continued) Now we solve the system (1) (3) Notice that if we add equations (1) and (3), we will not eliminate any variables. If the –2y in equation (1) were changed to –4y, we would. To accomplish this change, we multiply both sides of equation (1) by 2: 2x – 4y = –12 3x + 4y = 12 5x + 0y = 0 x = 0. Adding Solving for x (3) Multiply eqn. (1) by 2

30. Slide 3- 30 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution (continued) Then 3(0) + 4y = 12 Substituting 0 for x in equation (3) 4y = 12 y = 3 We obtain the ordered pair (0, 3). We can plug the ordered pair into the original pair of equations to check that it is the solution.

31. Slide 3- 31 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution Solve the system (1) (2) Multiply equation (1) by 2 4x – 6y = 4 – 4x + 6y = –4 Add 0 = 0 Note that what remains is an identity. Any pair that is a solution of equation (1) is also a solution of equation (2). The equations are dependent and the solution set is infinite: Example

32. Slide 3- 32 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Rules for Special Cases When solving a system of two linear equations in two variables: 1. If we obtain an identity such as 0 = 0, then the system has an infinite number of solutions. The equations are dependent and, since a solution exists, the system is consistent. 2. If we obtain a contradiction such as 0 = 7, then the system has no solution. The system is inconsistent.

33. Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solving Applications: Systems of Two Equations Total-Value and Mixture Problems Motion Problems 8.3

34. Slide 3- 34 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Total-Value Problems The next example involves two types of items, the quantity of each type bought, and the total value of the items. We refer to this type of problem as a total-value problem.

35. Slide 3- 35 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution T-shirt Villa sold 52 shirts, one kind at $8.25 and another at $11.50 each. In all, $464.75 was taken in for the shirts. How many of each kind were sold? 1. Familiarize. To familiarize ourselves with this problem, guess that 26 of each kind of shirt was sold. The total money taken in would be The guess is incorrect, now turn to algebra. Example

36. Slide 3- 36 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 2. Translate. Let x = the number of $8.25 shirts and y = the number of $11.50 shirts. We have the following system of equations: x + y = 52 8.25x + 11.50y = 464.75 Kind of Shirt $8.25 shirt $11.50 shirt Total Number sold xy52 Price$8.25$11.50 Amount$8.25x$11.50y$464.75

37. Slide 3- 37 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 3. Carry out. We are to solve the system of equations Where x is the number of $8.25 shirts and y is the number of $11.50 shirts. We can eliminate x by multiplying both sides of equation (1) by –8.25 and adding them to the corresponding sides of equation (2): (1) (2) 3.25y = 35.75 8.25x + 11.50y = 464.75 –8.25x – 8.25y = – 429 y = 11.

38. Slide 3- 38 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley To find x, we substitute 11 for y in equation (1) and then solve for x: x + 11 = 52 x = 41 We obtain (41, 11), or x = 41 and y = 11. 4. Check. Recall x is the number of $8.25 shirts and y is the number of $11.50 shirts. Number of shirts: x + y = 41 + 11 = 52 Money taken in: $8.25x + $11.50y = $8.25(41) + $11.50(11) = 464.75 The numbers check.

39. Slide 3- 39 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 5. State. T-shirt Villa sold 41 $8.25 shirts and 11 $11.50 shirts.

40. Slide 3- 40 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Mixture Problems The next example is similar to the last example. Note that in each case, one of the equations in the system is a simple sum while the other equation represents a sum of products. We refer to this type of problem as a mixture problem.

41. Slide 3- 41 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Problem-Solving Tip When solving a problem, see if it is patterned or modeled after a problem that you have already solved.

42. Slide 3- 42 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution An employee at a small cleaning company wishes to mix a cleaner that is 30% acid and another cleaner that is 50% acid. How many liters of each should be mixed to get 20 L of a solution that is 35% acid? 1. Familiarize. We make a drawing and then make a guess to gain familiarity with the problem. + 30% acid 50% acid 35% acid = t liters f liters 20 liters Example

43. Slide 3- 43 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley To familiarize ourselves with this problem, guess that 10 liters of each are mixed. The resulting mixture will be the right size but we need to check the strength: The guess is incorrect, now turn to algebra. That is 8 L of acid for our guess but we want 0.35(20) = 7 L.

44. Slide 3- 44 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 2. Translate. Let t = the number of liters of the 30% solution and f = the number of liters of the 50% solution. We have the following system of equations: t + f = 20 0.30t + 0.50f = 7 First Solution Second Solution Mixture Number of Liters tf20 Percentage of Acid 30%50%35% Amount of Acid 0.30t0.50f7

45. Slide 3- 45 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 3. Carry out. We are to solve the system of equations: We can eliminate f by multiplying both sides of equation (1) by –0.5 and adding them to the corresponding sides of equation (2): (1) (2) –0.20t = –3 0.30t + 0.50f = 7 –0.50t – 0.50f = –10 t = 15.

46. Slide 3- 46 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley To find f, we substitute 15 for t in equation (1) and then solve for f: 15 + f = 20 f = 5 We obtain (15, 5), or t = 15 and f = 5. 4. Check. Recall t is the number of liters of 30% solution and f is the number of liters of 50% solution. Number of liters: t + f = 15 + 5 = 20 Amount of Acid: 0.30t + 0.50f = 0.30(15) + 0.50(5) = 7 The numbers check.

47. Slide 3- 47 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 5. State. The employee should mix 15 L of the 30% solution with 5 L of the 50% solution to get 20 L of a 35% solution.

48. Slide 3- 48 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Motion Problems The next example deals with distance, speed (rate), and time. We refer to this type of problem as a motion problem.

49. Slide 3- 49 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Distance, Rate, and Time Equations If r represents rate, t represents time, and d represents distance, then

50. Slide 3- 50 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution Alex’s motorboat took 4 hr to make a trip downstream with a 5-mph current. The return trip against the same current took 6 hr. Find the speed of the boat in still water. 1. Familiarize. Note that the current speeds up the boat when going downstream, but slows down the boat when going upstream. For our guess, suppose that the speed of the boat with no current is 20 mph. The boat would then travel 20 + 5 = 25 mph downstream and 20 – 5 = 15 mph upstream. Example

51. Slide 3- 51 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley In 4 hr downstream the boat would travel 4(25) = 100 mi. In 6 hr upstream the boat would travel 6(15) = 90 mi. So our guess of 20 mph is incorrect. Let r = the rate of the boat in still water. Then r + 5 = the boat’s speed downstream, and r – 5 = the boat’s speed upstream. 2. Translate.

52. Slide 3- 52 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Use Distance = (Rate)(Time). We have the following system of equations: d = (r + 5)4 d = (r – 5)6 DistanceRateTime Downstreamdr + 54 Upstreamdr – 56

53. Slide 3- 53 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 3. Carry out. We solve the system by substitution: (r + 5)4 = (r – 5)6 4r + 20 = 6r – 30 50 = 2r 25 = r 4. Check. When r = 25 mph, the speed downstream is 30 mph and the speed upstream is 20 mph. The distance downstream is 30(4) = 120 mi and the distance upstream is 20(6) = 120 mi, so we have a check. 5. State. The speed of the boat in still water is 25 mph.

54. Slide 3- 54 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Tips for Solving Motion Problems 1.Draw a diagram using an arrow or arrows to represent distance and the direction of each object in motion. 2.Organize the information in a chart. 3.Look for times, distances, or rates that are the same. These often can lead to an equation. 4.Translating to a system of equations allows for the use of two variables. 5.Always make sure that you have answered the question asked.

55. Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Systems of Equations in Three Variables Identifying Solutions Solving Systems in Three Variables Dependency, Inconsistency, and Geometric Considerations 8.4

56. Slide 3- 56 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Identifying Solutions A solution of a system of three equations in three variables is an ordered triple (x, y, z) that makes all three equations true. A linear equation in three variables is an equation equivalent to one in the form Ax + By + Cz = D, where A, B, C, and D are real numbers. We refer to the form Ax + By + Cz = D as standard form for a linear equation in three variables.

57. Slide 3- 57 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Determine whether (2, –1, 3) is a solution of the system Example

58. Slide 3- 58 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution We substitute (2, –1, 3) into the three equations, using alphabetical order: x + y + z = 4 2 + (–1) + 3 4 4 = 43 = 3 TRUE The triple makes all three equations true, so it is a solution of the system. 2x – 2y – z = 3 2(2) – 2(–1) – 3 3 – 4x + y + 2z = –3 – 4(2) + (–1) + 2(3) –3 –3 = –3 TRUE

59. Slide 3- 59 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solving Systems in Three Variables The elimination method allows us to manipulate a system of three equations in three variables so that a simpler system of two equations in two variables is formed. Once that simpler system has been solved, we can substitute into one of the three original equations and solve for the third variable.

60. Slide 3- 60 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution Solve the following system of equations: 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 Example

61. Slide 3- 61 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 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)

62. Slide 3- 62 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 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.

63. Slide 3- 63 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 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).

64. Slide 3- 64 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solving Systems of Three Linear Equations To use the elimination method to solve systems of three linear equations: 1. Write all the equations in standard form Ax + By+ Cz = D. 2. Clear any decimals or fractions. 3.Choose a variable to eliminate. Then select two of the three equations and work to get one equation in which the selected variable is eliminated.

65. Slide 3- 65 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solving Systems of Three Linear Equations (continued) 4. Next, use a different pair of equations and eliminate the same variable that you did in step (3). 5. Solve the system of equations that resulted from steps (3) and (4). 6.Substitute the solution from step (5) into one of the original three equations and solve for the third variable. Then check.

66. Slide 3- 66 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Dependency, Inconsistency, and Geometric Considerations The graph of a linear equation in three variables is a plane. Solutions are points common to the planes of each system. Since three planes can have an infinite number of points in common or no points at all in common, we need to generalize the concept of consistency in three dimensions.

67. Slide 3- 67 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

68. Slide 3- 68 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Consistency A system of equations that has at least one solution is said to be consistent. A system of equations that has no solution is said to be inconsistent.

69. Slide 3- 69 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution Solve the following system of equations: The variable x is missing in equation (1). Let’s add equations (2) and (3) to get another equation with x missing: (1) (2) (3) (2) (3) (4) y + 2z = 6 Adding Example

70. Slide 3- 70 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Equations (1) and (4) form a system in y and z. Solve as before: Multiplying equation (1) by –1 0 = 4. Since we ended up with a false equation, or contradiction, we know that the system has no solution. It is inconsistent. y + 2z = 6 y + 2z = 2–y – 2z = –2 y + 2z = 6

71. Slide 3- 71 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Recall that when dependent equations appeared in Section 8.1, the solution sets were always infinite in size and were written in set-builder notation. There, all systems of dependent equations were consistent. This is not always the case for systems of three or more equations. The following figures illustrate some possibilities geometrically.

72. Slide 3- 72 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

73. Solving Applications: Systems of Three Equations Applications of Three Equations in Three Unknowns 8.5

74. Slide 3- 74 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution The sum of three numbers is 6. The first number plus twice the second, minus the third is 2. The first minus the second, plus three times the third is 8. 1. Familiarize. There are three statements involving the same three numbers. Let’s label these numbers x, y, and z. Example

75. Slide 3- 75 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 2. Translate. We can translate directly as follows. The sum of three numbers is 6. x + y + z = 6 The first number plus twice the second, minus the third is 2. x + 2y – z = 2 The first minus the second, plus three times the third is 8. x – y + 3z = 8

76. Slide 3- 76 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley We now have the system of three equations: 3. Carry out. We need to solve the system of equations. Note that we found the solution, (1, 2, 3) in an example in Section 3.4. 5. State. The three numbers are 1, 2, and 3. 4. Check. The check is left to the student.

77. Slide 3- 77 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution In triangle ABC, the measure of angle B is three times the measure of angle A. The measure of angle C is 60 o greater than twice the measure of angle A. Find the measure of each angle. 1. Familiarize. We can make a sketch and label the angles A, B, and C. Recall that the measures of the angles in any triangle add to 180 o. A B C Example

78. Slide 3- 78 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 2. Translate. This geometric fact about triangles gives us one equation: A + B + C = 180. B = 3A C = 60 + 2A Angle C is 60 o greater than twice the measure of A. Angle B is three times the measure of angle A.

79. Slide 3- 79 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley We now have the system of three equations: 3. Carry out. We need to solve the system of equations. Through the techniques of Section 3.4 we find the solution to be (20, 60, 100). 5. State. The angles in the triangle measure 20 o, 60 o, and 100 o. 4. Check. The check is left to the student.

80. Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Elimination Using Matrices Matrices and Systems Row-Equivalent Operations 8.6

81. Slide 3- 81 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley In solving systems of equations, we perform computations with the constants. The variables play no important role until the end. simplifies to 3 1 5 2 –3 7 if we do not write the variables, the operation of addition, and the equals signs. For example, the system

82. Slide 3- 82 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Matrices and Systems In the previous slide, we have written a rectangular array of numbers. Such an array is called a matrix (plural, matrices). We ordinarily write brackets around matrices. The following are examples of matrices: The individual numbers are called elements or entries.

83. Slide 3- 83 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley The rows of a matrix are horizontal, and the columns are vertical. column 1column 2column 3 row 1 row 2 row 3

84. Slide 3- 84 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution Use matrices to solve the system. We write a matrix using only coefficients and constants, listing x-coefficients in the first column and y-coefficients in the second. Note that in each matrix a dashed line separates the coefficients from the constants: Example

85. Slide 3- 85 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Our goal is to transform The variables x and y can then be reinserted to form equations from which we can complete the solution. We do calculations that are similar to those that we would do if we wrote the entire equations. New Row 2 = 3(Row 2 from above) into the form

86. Slide 3- 86 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley If we now reinsert the variables, we have Now proceed as before, solving equation (2) for y: y = 1. (1) (2) 10y = 10, New Row 2 = Row 1 + 3(Row 2)

87. Slide 3- 87 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Next, we substitute 1 in for y in equation (1): The solution is (2, 1). The check is left to the student. 3x + 1 = 7 3x + y = 7 3x = 6 x = 2.

88. Slide 3- 88 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Use matrices to solve the system. Solution We write a matrix using only coefficients and constants. Example

89. Slide 3- 89 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Our goal is to transform into the form Interchange Row 1 and Row 2 New Row 2 = –2(Row 1) + Row 2

90. Slide 3- 90 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley If we now reinsert the variables, we have New Row 3 = –(Row 2) + Row 3 New Row 2 = (1/3)(Row 2) New Row 3 = Row 1 + Row 3

91. Slide 3- 91 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Now proceed as before solving equation (3) for z: z = –1. Substitute the z value into equation (2) to find that y = 1. Then substitute the z and y values into equation (1) to find x = 3. The solution is (3, 1, –1). The check is left to the student.

92. Slide 3- 92 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Row-Equivalent Operations Each of the following row-equivalent operations produces a row-equivalent matrix: a) Interchanging any two rows. b) Multiplying all elements of a row by a nonzero constant. c) Replacing a row with the sum of that row and a multiple of another row.

93. Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Determinants and Cramer’s Rule Determinants of 2 x 2 Matrices Cramer’s Rule: 2 x 2 Systems Cramer’s Rule: 3 x 3 Systems 8.7

94. Slide 3- 94 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Determinants of 2 x 2 Matrices When a matrix has m rows and n columns, it is called an “m by n” matrix. Thus its dimensions are denoted by m x n. If a matrix has the same number of rows and columns, it is called a square matrix. Associated with every square matrix is a number called its determinant, defined as follows for 2 x 2 matrices.

95. Slide 3- 95 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 2 x 2 Determinants The determinant of a two-by-two matrix and is defined as follows:

96. Slide 3- 96 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Evaluate: Solution = 3(1) – 7(–2) = 17. Multiply and subtract as follows: Example

97. Slide 3- 97 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Cramer’s Rule: 2 x 2 Matrices Using the elimination method, we can show that the solution to the system is These fractions can be rewritten using determinants.

98. Slide 3- 98 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Cramer’s Rule: 2 x 2 Systems The solution of the system if it is unique, is given by

99. Slide 3- 99 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Cramer’s Rule: 2 x 2 Systems (continued) These formulas apply only if the denominator is not 0. If the denominator is 0, then one of two things happens: 1.If the denominator is 0 and the numerators are also 0, then the equations in the system are dependent. 2.If the denominator is 0 and at least one numerator is not 0, then the system is inconsistent.

100. Slide 3- 100 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution Solve using Cramer’s rule: We have Example

101. Slide 3- 101 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution (continued) and The solution is (1/5, 4/5).

102. Slide 3- 102 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 3 x 3 Determinants The determinant of a three-by-three matrix is defined as follows: Subtract.Add.

103. Slide 3- 103 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution Evaluate: = 1(34) – 2(–9) + (–1)(–21) = 73. Example

104. Slide 3- 104 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Cramer’s Rule: 3 x 3 Systems The solution of the system can be found by using the following determinants:

105. Slide 3- 105 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Cramer’s Rule: 3 x 3 Systems (continued) If a unique solution exists, it is given by

106. Slide 3- 106 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution Solve using Cramer’s rule: Compute the determinants: Example

107. Slide 3- 107 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution (continued) Then The solution is (3, 1, –1). The check is left to the student.

108. Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Business and Economic Applications Break-Even Analysis Supply and Demand 8.8

109. Slide 3- 109 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Break-Even Analysis When a company manufactures x units of a product, it spends money. This is total cost and can be thought of as a function C, where C(x) is the total cost of producing x units. When a company sells x units of the product, it takes in money. This is total revenue and can be thought of as a function R, where R(x) is the total revenue from the sale of x units.

110. Slide 3- 110 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Break-Even Analysis Total profit is the money taken in less the money spent, or total revenue minus total cost. Total profit from the production and sale of x units is a function P given by Profit = Revenue – Cost, or P(x) = R(x) – C(x). (continued)

111. Slide 3- 111 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Cost There are two types of costs. Costs which must be paid whether a product is produced or not, are called fixed costs. Costs that vary according to the amount being produced are called variable costs. The sum of the fixed cost and variable cost gives the total cost.

112. Slide 3- 112 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley A specialty wallet company has fixed costs that are $2,400. Each wallet will cost $2 to produce (variable costs) and will sell for $10. a)Find the total cost C(x) of producing x wallets. b)Find the total revenue R(x) from the sale of x wallets. c)Find the total profit P(x) from the production and sale of x wallets. d)What profit will the company realize from the production and sale of 500 wallets? e)Graph the total-cost, total-revenue, and total-profit functions. Determine the break-even point. Example

113. Slide 3- 113 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution a) Total cost is given by C(x) = (Fixed costs) plus (Variable costs) C(x) = 2,400 + 2x. where x is the number of wallets produced. b) Total revenue is given by R(x) = 10x $10 times the number of wallets sold. c) Total profit is given by P(x) = R(x) – C(x) = 10x – (2,400 + 2x) = 8x – 2,400. A specialty wallet company has fixed costs that are $2,400. Each wallet will cost $2 to produce (variable costs) and will sell for $10.

114. Slide 3- 114 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution d) Total profit will be P(500) = 8(500) – 2,400 = 4,000 – 2,400 = $1,600. e) The graphs of each of the three functions are shown on the next slide. R(x), C(x), and P(x) are all in dollars. A specialty wallet company has fixed costs that are $2,400. Each wallet will cost $2 to produce (variable costs) and will sell for $10.

115. Slide 3- 115 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley e) 0 50 100 150 200 250 300 350 400 450 500 550 500 1,500 1,000 2,500 2,000 3,000 4,000 3,500 R(x) = 10x C(x) = 2400 + 2x P(x) = 8x – 2400 -2500 Break-even point Loss Gain Wallets sold

116. Slide 3- 116 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Gains occur where the revenue is greater than the cost. Losses occur where the revenue is less than the cost. The break-even point occurs where the graphs of R and C cross. Thus to find the break-even point, we solve the system: Using substitution we find that x = 300. The company will break even if it produces and sells 300 wallets and takes in a total of R(300) = $3,000 in revenue. Note that the break-even point can also be found by solving P(x) = 0.

117. Slide 3- 117 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Supply and Demand As the price of a product varies, the amount sold varies. Consumers will demand less as price goes up. Sellers will supply more as the price goes up.

118. Slide 3- 118 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Supply and Demand Supply Demand Equilibrium point Price Quantity The point of intersection is called the equilibrium point. At that price, the amount that the seller will supply is the same amount that the consumer will buy.

119. Slide 3- 119 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution Find the equilibrium point for the demand and supply functions given. Since both demand and supply are quantities and they are equal at the equilibrium point, we rewrite the system as (1) (2) (1) (2) Example

120. Slide 3- 120 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solving using substitution we find the equilibrium price is $32. To find the quantity, we substitute $32 into either equation D(p) or S(p). We use S(p): Thus, the equilibrium quantity is 440 units, and the equilibrium price is $32. Solution (continued)