Section 7 4 applications of linear systems
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Section 7-4: Applications of Linear Systems. Objective: To model rate-time-distance problems using two variables To use systems of equations to solve problems graphically. Applications of Linear Systems. We have solved systems by graphing, substitution and elimination.

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Section 7-4: Applications of Linear Systems

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Section 7 4 applications of linear systems

Section 7-4: Applications of Linear Systems

Objective:

To model rate-time-distance problems using two variables

To use systems of equations to solve problems graphically


Applications of linear systems

Applications of Linear Systems

  • We have solved systems by graphing, substitution and elimination.

  • We are now going to apply these skills toward work problems (real world problems)

  • Let’s look at the following examples


Definitions

Definitions

  • Ingot

    • A metal melted down and cast into a bar or other shape

  • Alloy

    • Any combination of metals fused together

  • Kg

    • Abbreviation of Kilograms. 1000 grams.

    • A gram is about the weight of a paperclip

    • A kilogram is about 2.2 lbs.


Example 1

Example #1

A metalworker has some ingots of metal alloy that are 20% copper and others that are 60% copper. How many kilograms of each type of ingot should the metalworker combine to create 80kg of a 52% copper alloy?


Example 11

Example #1

  • Let x = the mass of the 20% alloy

  • Let y = the mass of the 60% alloy

  • What is the mass of the alloys?

    - x + y = 80

  • What is the mass of the copper?

    - 0.2x + 0.6y = 0.52(80)

  • What does that leave us with?


Example 12

Example #1

x + y = 80

0.2 x + 0.6 y = 0.52(80)

x + y = 80

- y -y

x = 80 – y

0.2(80-y) + 0.6 y =0.52(80)

  • A SYSTEM OF EQUATIONS!!! : )

  • Think of which method would be easiest…

  • Would it be easier to solve for x or y OR would be easier to multiply either or both equations to eliminate one of the variables?

  • Substitution is probably easier.


Example 13

Example #1

0.2(80-y) + 0.6y =0.52(80)

16 – 0.2y + 0.6y = 41.6

16 + 0.4y = 41.6

-16 -16

0.4y = 25.6

0.4 0.4

y = 64

  • Solve for y

  • Now plug in y in either equation to find x


Example 14

Example #1

x = 80 – y

x = 80 – (64)

x = 16

Mass of 20% alloy needed is 16kg.

Mass of 60% alloy needed is 64kg.

  • I chose this equation

  • Now, we have an answer for both x and y.

  • Remember what x and y stood for


Example 2 finding a break even point

Example #2: Finding a break even point

Suppose a gum chewing club publishes a newsletter. Expenses are $0.90 for printing and mailing each copy, plus $600 total for gum (both chewing and bubble), research, and writing. The price of the newsletter is $1.50 per copy. How many copies of the newsletter must the club sell to break even?


Example 2 finding a break even point1

Example #2: Finding a break even point

Let x = the number of copies

Let y = the amount of dollars of expenses or income.

Expenses =

y = 0.9x + 600

Income =

y = 1.5x

  • We need to create two equations.

  • One for expenses

  • One for income


Example 2 finding a break even point2

Example #2: Finding a break even point

y = 0.9x + 600

y = 1.5x

1.5 x = 0.9 x + 600

-0.9x -0.9x

0.6 x = 600

0.6 0.6

x = 1000

  • This gives us …

  • Another system

  • Since y = y always

  • The club must sell 1000 copies of just to break even


Example 3

Example #3

  • Suppose you fly from Miami to San Francisco. It takes 6.5 hours to fly 2600 mile against a head wind. At the same time, your friends flies from San Francisco to Miami. Her plane travels at the same average airspeed, but her flight only takes 5.2 hour. Find the average airspeed of the planes. Find the average wind speed.


Example 31

Example #3

Let A = Air speed

Let W = Wind speed

Remember:

Rate x Time = Distance

(A + W)(Time) = Distance

(A – W)(Time) = Distance

  • We need to create two equations

  • Which formula do we need?

    This is the formula with the tailwind

    This is the formula with the head wind

    Now, plug in values


Example 32

Example #3

(A + W)5.2 = 2600

(A – W)6.5 = 2600

(A + W)5.2 = 2600

5.2 5.2

A + W = 500

  • Plug in the values that we know

  • Take the 1st equation and divide by 5.2


Example 33

Example #3

(A – W)6.5 = 2600

6.5 6.5

A – W = 400

A + W = 500

A – W = 400

  • Now divide the 2nd equation by 6.5

  • This leaves us with…

    … an easier system to work with


Example 34

Example #3

A + W = 500

A – W = 400

2A = 900

2 2

A = 450

450 + W = 500

-450 -450

W = 50

  • Solve this by elimination

  • Airspeed = 450 mph

  • Now, solve for W

  • Wind speed = 50 mph


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