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Applications of Linear Systems

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Point of Intersection
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Learning Goal

By the end of this lesson you will be able to determine the point of intersection in context and state when to choose each scenario.


What if you were playing a game in which you collected houses and hotels. Three houses and one hotel are worth $1750. One house and three hotels are worth $3250. How could you find the value of each house and each hotel? After completing this Concept, you'll be able to solve real-world applications like this one that involve linear systems.

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CK-12 Foundation: 0709S Applications of Linear Systems

Guidance

In this section we will use Real Life Applications of Linear Relations to determine under what conditions we should choose each scenario.

Example A

The movie rental store CineStar offers customers two choices. Customers can pay a yearly membership of $45 and then rent each movie for $2 or they can choose not to pay the membership fee and rent each movie for $3.50. Determine under which conditions each option would be a better choice.

Solution

Let’s translate this problem into algebra. Since there are two different options to consider, we can write two different equations and form a system.

The choices are “membership” and “no membership.” We’ll call the number of movies you rent x and the total cost of renting movies for a year y .

flat fee rental fee total
membership $45 2x y = 45 + 2x
no membership $0 3.50x y = 3.5x

The flat fee is the dollar amount you pay per year and the rental fee is the dollar amount you pay when you rent a movie. For the membership option the rental fee is 2x , since you would pay $2 for each movie you rented; for the no membership option the rental fee is 3.50x , since you would pay $3.50 for each movie you rented.

Our two equations are:

y = 45 + 2x\!\\y = 3.50x

Here’s a graph of the system:

Now we need to find the exact intersection point.

We can use the graph to find the Point of Intersection.

The Point of Intersection occurs where the two lines cross each other. 

Looking at the graph you can see at 30 movies per year both options would cost $105 .

We need to report both numbers.

Now we need to decide when each option would be a better choice.

BEFORE the point of intersection:

If you are going to rent less than 30 movies than being a Non-Member would cost less and therefore be the better choice.

AFTER the point of intersection:

If you are going to rent more than 30 movies than being a Member would cost less and therefore be the better choice.

AT the point of intersection:

If you rent 30 movies then you can either buy the membership or not and it will cost the same - $105.

This example shows a real situation where a consistent system of equations is useful in finding a solution. Remember that for a consistent system, the lines that make up the system intersect at single point. In other words, the lines are not parallel or the slopes are different.

In this case, the slopes of the lines represent the price of a rental per movie. The lines cross because the price of rental per movie is different for the two options in the problem

Watch this video for help with the Examples above.

CK-12 Foundation: Applications of Linear Systems

Vocabulary

  • A linear system of equations is a set of equations that must be solved together to find the one solution that fits them both.
  • A consistent system will always give exactly one solution.

Practice

  1. Juan is considering two cell phone plans. The first company charges $120 for the phone and $30 per month for the calling plan that Juan wants. The second company charges $40 for the same phone but charges $45 per month for the calling plan that Juan wants. Determine under which conditions it is better to choose each plan.

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