1. Real World Systems of Equations

2. Purpose

3. Purpose Use systems of equations to determine exact quantities that satisfy two requirements;

4. Purpose Use systems of equations to determine exact quantities that satisfy two requirements; hence the need for a system.

5. Example (Don’t copy) Michelle has 12 total pets in her home that are either fish or birds.

6. Example (Don’t copy) Michelle has 12 total pets in her home that are either fish or birds. -Are there multiple possible solutions to this situation?

7. Absolutely Michelle has 12 total pets in her home that are either fish or birds. -Are there multiple possible solutions to this situation?

8. Absolutely Michelle has 12 total pets in her home that are either fish or birds. The # of fish is 3 less than twice the number of birds.

9. Now we have one situation that represents this… Michelle has 12 total pets in her home that are either fish or birds. The # of fish is 3 less than twice the number of birds.

10. Method to solve:

11. 1)Establish the relationships between the 2 quantities.

12. Examples

13. Examples x + y = n

14. Examples You will often be told the total number of objects involved in the problem. x + y = n

15. Examples You will often be told the total number of objects involved in the problem. This is what we use to represent this, where n is the total number of objects. x + y = n

16. Examples y = ax ± b

17. Examples You will be told that the number of one object is “b more than a times” the other. y = ax ± b

18. Examples You will be told that the number of one object is “b more than a times” the other. a can also be a fraction, and it can also be “b less than”. y = ax ± b

19. Examples y = mx ± b

20. Examples m is some type of rate, while b is a starting point. y = mx ± b

21. Examples m is some type of rate, while b is a starting point. In this type of problem, the solution is usually a break –even point or where two of these quantities are equal. y = mx ± b

22. Examples When dealing with measures of 2 quantities, we often use this method, ax + by = c

23. Examples When dealing with measures of 2 quantities, we often use this method, where a and b represent the “weight” of each individual object and c is the “total”. ax + by = c

24. Let’s return to our intro problem. Michelle has 12 total pets in her home that are either fish or birds. The # of fish is 3 less than twice the number of birds.

25. Let’s create our equations:

26. Michelle has 12 total pets in her home that are either fish or birds.

27. Let’s create our equations: x + y = 12 Michelle has 12 total pets in her home that are either fish or birds.

28. Let’s create our equations: x + y = 12 The # of fish is 3 less than twice the number of birds.

29. Let’s create our equations: x + y = 12 x = 2y – 3 The # of fish is 3 less than twice the number of birds.

30. Let’s create our equations: x + y = 12 x = 2y – 3 Now we solve using, most likely, what method?

31. Let’s create our equations: x + y = 12 x = 2y – 3 Now we solve using, most likely, what method? Substitution

32. Let’s create our equations: x + y = 12 x = 2y – 3 (2y – 3) + y = 12 Now we solve using, most likely, what method? Substitution

33. Let’s create our equations: x + y = 12 x = 2y – 3 (2y – 3) + y = 12 3y – 3 = 12 Now we solve using, most likely, what method? Substitution

34. Let’s create our equations: x + y = 12 x = 2y – 3 (2y – 3) + y = 12 3y – 3 = 12 3y = 15  Now we solve using, most likely, what method? Substitution

35. Let’s create our equations: x + y = 12 x = 2y – 3 (2y – 3) + y = 12 3y – 3 = 12 3y = 15  y = 5 Now we solve using, most likely, what method? Substitution

36. Mental math should be used to find y: x + y = 12 x = 2y – 3 (2y – 3) + y = 12 3y – 3 = 12 3y = 15  y = 5 Now we solve using, most likely, what method? Substitution

37. Mental math should be used to find y: x + y = 12 x = 2y – 3 y = 5 x = 7 Now we solve using, most likely, what method? Substitution

38. Common problem

39. Which one represents the birds and which one represents the fish?

40. Common problem Which one represents the birds and which one represents the fish? 5 birds

41. Common problem Which one represents the birds and which one represents the fish? 5 birds 7 fish

42. As a result:

43. We often use variables that abbreviate the categories of items we have.

44. New example:

45. You are the manager of a shoe store. On Sunday morning, you are going over the sales receipts for the past week. They show that 240 pairs of walking shoes were sold. Style A sells for $66.95, and Style B sells for $84.95. The total receipts for the two types were $17,652. The cash register was supposed to keep track of the number of each type sold. It malfunctioned. Can you find out how many of each type were sold?

46. Setting up the system:

47. What letters shall we use?

48. Setting up the system: a and b What letters shall we use?

49. Setting up the system: a and b What letters shall we use? How many total shoes were sold?

50. Setting up the system: a and b 240 What letters shall we use? How many total shoes were sold?

51. Setting up the system: a + b = 240 240 What letters shall we use? How many total shoes were sold?

52. Setting up the system: a + b = 240 240 What letters shall we use? What does style A and style B cost?

53. Setting up the system: a + b = 240 A:$66.95 What letters shall we use? What does style A and style B cost?

54. Setting up the system: a + b = 240 A:$66.95 B:$84.95 What letters shall we use? What does style A and style B cost?

55. Setting up the system: a + b = 240 66.95a + 84.95b = 17652 A:$66.95 B:$84.95 What letters shall we use? What does style A and style B cost?

56. If we want to use elimination: a + b = 240 66.95a + 84.95b = 17652 A:$66.95 B:$84.95 What letters shall we use? What does style A and style B cost?

57. If we want to use elimination: a + b = 240 66.95a + 84.95b = 17652 A:$66.95 B:$84.95 What types of variable terms do we need?

58. If we want to use elimination: a + b = 240 66.95a + 84.95b = 17652 A:$66.95 B:$84.95 What types of variable terms do we need? Opposites

59. If we want to use elimination: -66.95a – 66.95b = -16068 66.95a + 84.95b = 17652 What types of variable terms do we need? Opposites

60. If we want to use elimination: -66.95a – 66.95b = -16068 66.95a + 84.95b = 17652 Adding straight down:

61. If we want to use elimination: -66.95a – 66.95b = -16068 66.95a + 84.95b = 17652 18b = 1584 Adding straight down:

62. If we want to use elimination: -66.95a – 66.95b = -16068 66.95a + 84.95b = 17652 18b = 1584 b = 88

63. If we want to use elimination: -66.95a – 66.95b = -16068 66.95a + 84.95b = 17652 18b = 1584 b = 88 Subtracting from 240:

64. If we want to use elimination: -66.95a – 66.95b = -16068 66.95a + 84.95b = 17652 18b = 1584 b = 88 a = 152 Subtracting from 240:

65. Let’s get out your books:

66. Page 398

67. Breaking things into categories…

68. You invest a total of $9000 in two funds paying 5% and 6% annual interest. The combined annual interest is $510. How much of the $9000 is invested in each fund?

69. Model: x + y = n

70. Model: f + s = 9000 x + y = n

71. Model: f + s = 9000 f represents amount invested in 5% and s represents amount invested in 6%. x + y = n

72. Model: f + s = 9000.05f +.06s = 510 x + y = n Ax + By = C

73. Model: f + s = 9000.05f +.06s = 510 When dealing with percents, put the percent as a decimal in front of the variable. x + y = n Ax + By = C

74. Model: f + s = 9000.05f +.06s = 510 Now a little substitution after a bit of solving for a variable. x + y = n Ax + By = C

75. Model: f = 9000 – s.05f +.06s = 510 Now a little substitution after a bit of solving for a variable. x + y = n Ax + By = C

76. Model: f = 9000 – s.05(9000 – s) +.06s = 510 Now a little substitution after a bit of solving for a variable. x + y = n Ax + By = C

77. Model: f = 9000 – s.05(9000 – s) +.06s = 510 450 – 0.05s +.06s = 510 x + y = n Ax + By = C

78. Model: f = 9000 – s.05(9000 – s) +.06s = 510 450 – 0.05s +.06s = 510 s + 450 = 510 x + y = n Ax + By = C

79. Model: f = 9000 – s.05(9000 – s) +.06s = 510 450 – 0.05s +.06s = 510.01s + 450 = 510.01s = 60 x + y = n Ax + By = C

80. Model: f = 9000 – s.05(9000 – s) +.06s = 510 450 – 0.05s +.06s = 510.01s + 450 = 510.01s = 60 s = 6000 x + y = n Ax + By = C

81. Model: f = 9000 – s.05(9000 – s) +.06s = 510 450 – 0.05s +.06s = 510.01s + 450 = 510.01s = 60 s = 6000 Mental math does the rest…

82. Model: f = 9000 – s.05(9000 – s) +.06s = 510 450 – 0.05s +.06s = 510.01s + 450 = 510.01s = 60 s = 6000 f = 3000 Mental math does the rest…

83. Thus, we conclude…

84. 6000 was invested in the 6% accounts and 3000 was invested in the 5% accounts.

85. Back to your books…

86. Page 410

87. Rate problems p 391 in your book:

88. Rate problems p 391 in your book: y = mx + b

89. Rate problems p 391 in your book: y = 25x + 400 y = mx + b

90. Rate problems p 391 in your book: y = 25x + 400 y = mx + b y = 50x + 200

91. Solved by graphing 1000 900 800 700 600 500 400 300 200 0 2 4 6 8 10 12 y = 25x + 400 y = 50x + 200

92. Solved by graphing 1000 900 800 700 600 500 400 300 200 0 2 4 6 8 10 12 y = 25x + 400 y = 50x + 200

93. Solved by graphing 1000 900 800 700 600 500 400 300 200 0 2 4 6 8 10 12 y = 25x + 400 y = 50x + 200

94. Solved by graphing 1000 900 800 700 600 500 400 300 200 0 2 4 6 8 10 12 y = 25x + 400 y = 50x + 200

95. Appears to be approx. (8, 600) 1000 900 800 700 600 500 400 300 200 0 2 4 6 8 10 12 y = 25x + 400 y = 50x + 200

96. Let’s verify algebraically…

97. y = 25x + 400 y = 50x + 200

98. Let’s verify algebraically… y = 25x + 400 y = 50x + 200 25x + 400 = 50x + 200

99. Let’s verify algebraically… y = 25x + 400 y = 50x + 200 25x + 400 = 50x + 200 200 = 25x

100. Let’s verify algebraically… y = 25x + 400 y = 50x + 200 25x + 400 = 50x + 200 200 = 25x 8 = x

101. Let’s verify algebraically… y = 25(8) + 400 y = 50x + 200 25x + 400 = 50x + 200 200 = 25x 8 = x

102. Let’s verify algebraically… y = 25(8) + 400 y = 600 25x + 400 = 50x + 200 200 = 25x 8 = x

103. Let’s verify algebraically… y = 25(8) + 400 y = 600 Meaning after 8 months, they both will have 600 current visits, if they both continue at the same rate.

104. How to interpret the solution to these…

105. Comparing 2 rates, the solution is the point when both quantities will be the same & the amount both will be.

106. How to interpret the solution to these… Comparing 2 rates, the solution is the point when both quantities will be the same & the amount both will be. When you’re talking business (comparing sale price with cost), this is the break-even point.

107. P 393 # 25 (Further interpretation)

108. When you are asked which one is a “better deal”:

109. P 393 # 25 (Further interpretation) When you are asked which one is a “better deal”: The one with the smaller b is the better buy before the solution.

110. P 393 # 25 (Further interpretation) When you are asked which one is a “better deal”: The one with the smaller b is the better buy before the solution. The one with the smaller m is the better buy after the solution.

111. P 393 # 25 (Further interpretation) In this case:

112. P 393 # 25 (Further interpretation) In this case: Before 125,000 miles, Car A is the better buy;

113. P 393 # 25 (Further interpretation) In this case: Before 125,000 miles, Car A is the better buy; after this point, Car B is the better buy.