Real World Systems of Equations. Purpose Purpose Use systems of equations to determine exact...

Post on 17-Dec-2015

216 views 0 download

transcript

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?

AbsolutelyMichelle has 12 total pets in her home that are either fish or birds.- Are there multiple

possible solutions to this situation?

AbsolutelyMichelle 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:

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:

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

Let’s create our equations:

x + y = 12The # 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 = 12x = 2y – 3

Now we solve using, most likely, what method?

Substitution

Let’s create our equations:

x + y = 12x = 2y – 3

(2y – 3) + y = 12

Now we solve using, most likely, what method?

Substitution

Let’s create our equations:

x + y = 12x = 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 = 12x = 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 = 12x = 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 = 12x = 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 = 12x = 2y – 3

y = 5x = 7

Now we solve using, most likely, what method?

Substitution

Common problem

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 birds7 fish

As a result:

As a result:• We often use variables that

abbreviate the categories of items we have.

New example:

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 $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?

Setting up the system:

Setting up the system:What letters shall we use?

Setting up the system:a and bWhat letters

shall we use?

Setting up the system:a and bWhat 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 = 240

240

What letters shall we use?

How many total shoes were sold?

Setting up the system:a + b = 240

240

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: $66.95B: $84.95

What letters shall we use?

What does style A and style B cost?

Setting up the system:a + b = 240

66.95a + 84.95b = 17652

A: $66.95B: $84.95

What letters shall we use?

What does style A and style B cost?

If we want to use elimination:

a + b = 24066.95a + 84.95b =

17652

A: $66.95B: $84.95

What letters shall we use?

What does style A and style B cost?

If we want to use elimination:

a + b = 24066.95a + 84.95b =

17652

A: $66.95B: $84.95

What types of variable terms do we need?

If we want to use elimination:

a + b = 24066.95a + 84.95b =

17652

A: $66.95B: $84.95

What types of variable terms do we need?

Opposites

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

If we want to use elimination:

-66.95a – 66.95b = -16068

66.95a + 84.95b = 17652

Adding straight down:

If we want to use elimination:

-66.95a – 66.95b = -16068

66.95a + 84.95b = 17652

18b = 1584

Adding straight down:

If we want to use elimination:

-66.95a – 66.95b = -16068

66.95a + 84.95b = 17652

18b = 1584b = 88

If we want to use elimination:

-66.95a – 66.95b = -16068

66.95a + 84.95b = 17652

18b = 1584b = 88

Subtracting from 240:

If we want to use elimination:

-66.95a – 66.95b = -16068

66.95a + 84.95b = 17652

18b = 1584b = 88a = 152

Subtracting from 240:

Let’s get out your books:

Let’s get out your books:

• Page 398

Breaking things into categories…

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 = 9000.05f + .06s = 510

x + y = n

Ax + By =

C

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

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

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 = 510

450 – 0.05s + .06s = 510

x + y = n

Ax + By =

C

Model:

f = 9000 – s .05(9000 – s) + .06s = 510

450 – 0.05s + .06s = 510

s + 450 = 510

x + y = n

Ax + By =

C

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

Model:

f = 9000 – s .05(9000 – s) + .06s = 510

450 – 0.05s + .06s = 510

.01s + 450 = 510.01s = 60s = 6000

x + y = n

Ax + By =

C

Model:

f = 9000 – s .05(9000 – s) + .06s = 510

450 – 0.05s + .06s = 510

.01s + 450 = 510.01s = 60s = 6000

Mental

math does

the rest…

Model:

f = 9000 – s .05(9000 – s) + .06s = 510

450 – 0.05s + .06s = 510

.01s + 450 = 510.01s = 60s = 6000f = 3000

Mental

math does

the rest…

Thus, we conclude…

Thus, we conclude…6000 was invested in the 6% accounts and 3000 was invested in the 5% accounts.

Back to your books…

Back to your books…Page 410

Rate problemsp 391 in your book:

Rate problemsp 391 in your book:

y = mx + b

Rate problemsp 391 in your book:

y = 25x + 400

y = mx + b

Rate problemsp 391 in your book:

y = 25x + 400

y = mx + by = 50x +

200

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

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

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

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

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

Let’s verify algebraically…

Let’s verify algebraically…

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

Let’s verify algebraically…

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

25x + 400 = 50x + 200

Let’s verify algebraically…

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

25x + 400 = 50x + 200200 = 25x

Let’s verify algebraically…

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

25x + 400 = 50x + 200200 = 25x

8 = x

Let’s verify algebraically…

y = 25(8) + 400y = 50x + 200

25x + 400 = 50x + 200200 = 25x

8 = x

Let’s verify algebraically…

y = 25(8) + 400y = 600

25x + 400 = 50x + 200200 = 25x

8 = x

Let’s verify algebraically…

y = 25(8) + 400y = 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…

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)

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.