Real World Systems of Equations

Purpose

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

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

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

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?

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

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.

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.

Method to solve:

1)Establish the relationships between the 2 quantities.

Examples

Examples x + y = n

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

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

Examples y = ax ± b

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

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

Examples y = mx ± b

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

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

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

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

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.

Let’s create our equations:

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

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

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

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

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

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

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

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

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

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

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

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

Common problem

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

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

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

As a result:

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

New example:

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 $ 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?

Setting up the system:

What letters shall we use?

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

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

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

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

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

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

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?

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

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

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

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

If we want to use elimination: a – 66.95b = a b = What types of variable terms do we need? Opposites

If we want to use elimination: a – 66.95b = a b = Adding straight down:

If we want to use elimination: a – 66.95b = a b = b = 1584 Adding straight down:

If we want to use elimination: a – 66.95b = a b = b = 1584 b = 88

If we want to use elimination: a – 66.95b = a b = b = 1584 b = 88 Subtracting from 240:

If we want to use elimination: a – 66.95b = a b = b = 1584 b = 88 a = 152 Subtracting from 240:

Let’s get out your books:

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Breaking things into categories…

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?

Model: x + y = n

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

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

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

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

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

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

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

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

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

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

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

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

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

Thus, we conclude…

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

Back to your books…

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Rate problems p 391 in your book:

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

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

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

Solved by graphing y = 25x y = 50x + 200

Solved by graphing y = 25x y = 50x + 200

Solved by graphing y = 25x y = 50x + 200

Solved by graphing y = 25x y = 50x + 200

Appears to be approx. (8, 600) y = 25x y = 50x + 200

Let’s verify algebraically…

y = 25x y = 50x + 200

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

Let’s verify algebraically… y = 25x y = 50x x = 50x = 25x

Let’s verify algebraically… y = 25x y = 50x x = 50x = 25x 8 = x

Let’s verify algebraically… y = 25(8) y = 50x x = 50x = 25x 8 = x

Let’s verify algebraically… y = 25(8) y = x = 50x = 25x 8 = x

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

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.

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.

P 393 # 25 (Further interpretation)

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

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.

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.

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

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

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.