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Real World Systems of Equations
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Purpose
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Purpose Use systems of equations to determine exact quantities that satisfy two requirements;
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Purpose Use systems of equations to determine exact quantities that satisfy two requirements; hence the need for a system.
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Example (Don’t copy) Michelle has 12 total pets in her home that are either fish or birds.
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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?
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Absolutely Michelle has 12 total pets in her home that are either fish or birds. -Are there multiple possible solutions to this situation?
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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.
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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.
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Method to solve:
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1)Establish the relationships between the 2 quantities.
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Examples
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Examples x + y = n
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Examples You will often be told the total number of objects involved in the problem. x + y = n
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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
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Examples y = ax ± b
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Examples You will be told that the number of one object is “b more than a times” the other. y = ax ± b
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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
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Examples y = mx ± b
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Examples m is some type of rate, while b is a starting point. y = mx ± b
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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
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Examples When dealing with measures of 2 quantities, we often use this method, ax + by = c
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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
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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.
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Let’s create our equations:
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Michelle has 12 total pets in her home that are either fish or birds.
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Let’s create our equations: x + y = 12 Michelle has 12 total pets in her home that are either fish or birds.
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Let’s create our equations: x + y = 12 The # of fish is 3 less than twice the number of birds.
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Let’s create our equations: x + y = 12 x = 2y – 3 The # of fish is 3 less than twice the number of birds.
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Let’s create our equations: x + y = 12 x = 2y – 3 Now we solve using, most likely, what method?
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Let’s create our equations: x + y = 12 x = 2y – 3 Now we solve using, most likely, what method? Substitution
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Let’s create our equations: x + y = 12 x = 2y – 3 (2y – 3) + y = 12 Now we solve using, most likely, what method? Substitution
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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
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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
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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
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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
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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
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Common problem
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Which one represents the birds and which one represents the fish?
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Common problem Which one represents the birds and which one represents the fish? 5 birds
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Common problem Which one represents the birds and which one represents the fish? 5 birds 7 fish
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As a result:
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We often use variables that abbreviate the categories of items we have.
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New example:
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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?
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Setting up the system:
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What letters shall we use?
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Setting up the system: a and b What letters shall we use?
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Setting up the system: a and b What letters shall we use? How many total shoes were sold?
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Setting up the system: a and b 240 What letters shall we use? How many total shoes were sold?
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Setting up the system: a + b = 240 240 What letters shall we use? How many total shoes were sold?
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Setting up the system: a + b = 240 240 What letters shall we use? What does style A and style B cost?
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Setting up the system: a + b = 240 A:$66.95 What letters shall we use? What does style A and style B cost?
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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?
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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?
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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?
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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?
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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
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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
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If we want to use elimination: -66.95a – 66.95b = -16068 66.95a + 84.95b = 17652 Adding straight down:
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If we want to use elimination: -66.95a – 66.95b = -16068 66.95a + 84.95b = 17652 18b = 1584 Adding straight down:
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If we want to use elimination: -66.95a – 66.95b = -16068 66.95a + 84.95b = 17652 18b = 1584 b = 88
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If we want to use elimination: -66.95a – 66.95b = -16068 66.95a + 84.95b = 17652 18b = 1584 b = 88 Subtracting from 240:
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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:
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Let’s get out your books:
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Page 398
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Breaking things into categories…
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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?
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Model: x + y = n
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Model: f + s = 9000 x + y = n
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Model: f + s = 9000 f represents amount invested in 5% and s represents amount invested in 6%. x + y = n
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Model: f + s = 9000.05f +.06s = 510 x + y = n Ax + By = C
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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
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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
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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
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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
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Model: f = 9000 – s.05(9000 – s) +.06s = 510 450 – 0.05s +.06s = 510 x + y = n Ax + By = C
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Model: f = 9000 – s.05(9000 – s) +.06s = 510 450 – 0.05s +.06s = 510 s + 450 = 510 x + y = n Ax + By = C
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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
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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
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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…
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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…
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Thus, we conclude…
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6000 was invested in the 6% accounts and 3000 was invested in the 5% accounts.
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Back to your books…
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Page 410
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Rate problems p 391 in your book:
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Rate problems p 391 in your book: y = mx + b
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Rate problems p 391 in your book: y = 25x + 400 y = mx + b
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Rate problems p 391 in your book: y = 25x + 400 y = mx + b y = 50x + 200
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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
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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
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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
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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
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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
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Let’s verify algebraically…
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y = 25x + 400 y = 50x + 200
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Let’s verify algebraically… y = 25x + 400 y = 50x + 200 25x + 400 = 50x + 200
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Let’s verify algebraically… y = 25x + 400 y = 50x + 200 25x + 400 = 50x + 200 200 = 25x
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Let’s verify algebraically… y = 25x + 400 y = 50x + 200 25x + 400 = 50x + 200 200 = 25x 8 = x
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Let’s verify algebraically… y = 25(8) + 400 y = 50x + 200 25x + 400 = 50x + 200 200 = 25x 8 = x
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Let’s verify algebraically… y = 25(8) + 400 y = 600 25x + 400 = 50x + 200 200 = 25x 8 = x
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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.
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How to interpret the solution to these…
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Comparing 2 rates, the solution is the point when both quantities will be the same & the amount both will be.
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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.
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P 393 # 25 (Further interpretation)
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When you are asked which one is a “better deal”:
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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.
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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.
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P 393 # 25 (Further interpretation) In this case:
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P 393 # 25 (Further interpretation) In this case: Before 125,000 miles, Car A is the better buy;
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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.
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