7.9: Applications of Linear Systems
Applications of Linear Systems
In this section, we’ll see how consistent, inconsistent and dependent systems might arise in applications.
Real-World Application: Yearly Membership
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. How many movies would you have to rent before the membership becomes the cheaper option?
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 \begin{align*}x\end{align*} and the total cost of renting movies for a year \begin{align*}y\end{align*}.
flat fee | rental fee | total | |
---|---|---|---|
membership | $45 | \begin{align*}2x\end{align*} | \begin{align*}y = 45 + 2x\end{align*} |
no membership | $0 | \begin{align*}3.50x\end{align*} | \begin{align*}y = 3.5x\end{align*} |
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 \begin{align*}2x\end{align*}, since you would pay $2 for each movie you rented; for the no membership option the rental fee is \begin{align*}3.50x\end{align*}, since you would pay $3.50 for each movie you rented.
Our system of equations is:
\begin{align*}y = 45 + 2x\!\\ y = 3.50x\end{align*}
Here’s a graph of the system:
Now we need to find the exact intersection point. Since each equation is already solved for \begin{align*}y\end{align*}, we can easily solve the system with substitution. Substitute the second equation into the first one:
\begin{align*}y = 45 + 2x\!\\ {\;} \qquad \qquad \qquad \ \Rightarrow 3.50x = 45 + 2x \Rightarrow 1.50x = 45 \Rightarrow x = 30 \ \text{movies}\!\\ y = 3.50x\end{align*}
You would have to rent 30 movies per year before the membership becomes the better option.
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
Now let’s look at a situation where the system is inconsistent. From the previous explanation, we can conclude that the lines will not intersect if the slopes are the same (and the \begin{align*}y-\end{align*}intercept is different). Let’s change the previous problem so that this is the case.
Real-World Application: Comparing Options
Two movie rental stores are in competition. Movie House charges an annual membership of $30 and charges $3 per movie rental. Flicks for Cheap charges an annual membership of $15 and charges $3 per movie rental. After how many movie rentals would Movie House become the better option?
It should already be clear to see that Movie House will never become the better option, since its membership is more expensive and it charges the same amount per movie as Flicks for Cheap.
The lines on a graph that describe each option have different \begin{align*}y-\end{align*}intercepts—namely 30 for Movie House and 15 for Flicks for Cheap—but the same slope: 3 dollars per movie. This means that the lines are parallel and so the system is inconsistent.
Now let’s see how this works algebraically. Once again, we’ll call the number of movies you rent \begin{align*}x\end{align*} and the total cost of renting movies for a year \begin{align*}y\end{align*}.
flat fee | rental fee | total | |
---|---|---|---|
Movie House | $30 | \begin{align*}3x\end{align*} | \begin{align*}y = 30 + 3x\end{align*} |
Flicks for Cheap | $15 | \begin{align*}3x\end{align*} | \begin{align*}y = 15 + 3x\end{align*} |
The system of equations that describes this problem is:
\begin{align*}y &= 30 + 3x\!\\ y &= 15 + 3x\end{align*}
Let’s solve this system by substituting the second equation into the first equation:
\begin{align*}y &= 30 + 3x\!\\ & \Rightarrow 15 + 3x = 30 + 3x \Rightarrow 15 = 30 \qquad \text{This statement is always false.}\!\\ y &= 15 + 3x\end{align*}
This means that the system is inconsistent.
Real-World Application: Price of Fruit
Peter buys two apples and three bananas for $4. Nadia buys four apples and six bananas for $8 from the same store. How much does one banana and one apple costs?
We must write two equations: one for Peter’s purchase and one for Nadia’s purchase.
Let’s say \begin{align*}a\end{align*} is the cost of one apple and \begin{align*}b\end{align*} is the cost of one banana.
cost of apples | cost of bananas | total cost | |
---|---|---|---|
Peter | \begin{align*}2a\end{align*} | \begin{align*}3b\end{align*} | \begin{align*}2a + 3b = 4\end{align*} |
Nadia | \begin{align*}4a\end{align*} | \begin{align*}6b\end{align*} | \begin{align*}4a + 6b = 8\end{align*} |
The system of equations that describes this problem is:
\begin{align*}2a + 3b &= 4 \\ 4a + 6b &= 8\end{align*}
Let’s solve this system by multiplying the first equation by -2 and adding the two equations:
\begin{align*}-2(2a + 3b = 4) \qquad \qquad \quad -4a - 6b = -8\!\\ {\;} \qquad \qquad \qquad \qquad \ \Rightarrow\!\\ \ 4a + 6b = 8 \qquad \qquad \qquad \qquad \underline{\;\;4a + 6b = 8\;\;}\!\\ {\;}\qquad \qquad \qquad \qquad \qquad \qquad \quad \ \ \ 0 + 0 = 0\end{align*}
This statement is always true. This means that the system is dependent.
Looking at the problem again, we can see that we were given exactly the same information in both statements. If Peter buys two apples and three bananas for $4, it makes sense that if Nadia buys twice as many apples (four apples) and twice as many bananas (six bananas) she will pay twice the price ($8). Since the second equation doesn’t give us any new information, it doesn’t make it possible to find out the price of each fruit.
Example
Example 1
A baker sells plain cakes for $7 and decorated cakes for $11. On a busy Saturday the baker started with 120 cakes, and sold all but three. His takings for the day were $991. How many plain cakes did he sell that day, and how many were decorated before they were sold?
plain cakes | decorated cakes | total | |
---|---|---|---|
Cakes sold | p | d | \begin{align*} 120-3=117\end{align*} |
Cost of cakes | 7p | 11d | $991 |
The system of equations that describes this problem is:
\begin{align*}p+d=117\!\\ 7p+11d=991\end{align*}
Let’s solve this system by substituting the second equation into the first equation:
\begin{align*}p+d=117 \Rightarrow p=117-d\end{align*}
\begin{align*}7p+11d=991 & \Rightarrow 7(117-d)+11d=991\\ & \Rightarrow 819-7d+11d=991\\ & \Rightarrow 819+4d=991 \\ & \Rightarrow 4d=172 \\ & \Rightarrow d=43\end{align*}
We can substitute \begin{align*}d\end{align*} into the first equation to solve for \begin{align*}p\end{align*}.
\begin{align*} p=117-d=117-(43)=74\end{align*}
The baker sold 74 plain cakes and 43 decorated cakes.
Review
- Twice John’s age plus five times Claire’s age is 204. Nine times John’s age minus three times Claire’s age is also 204. How old are John and Claire?
- 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. After how many months would the total cost of the two plans be the same?
- Jamal placed two orders with an internet clothing store. The first order was for 13 ties and 4 pairs of suspenders, and totaled $487. The second order was for 6 ties and 2 pairs of suspenders, and totaled $232. The bill does not list the per-item price, but all ties have the same price and all suspenders have the same price. What is the cost of one tie and of one pair of suspenders?
- An airplane took four hours to fly 2400 miles in the direction of the jet-stream. The return trip against the jet-stream took five hours. What were the airplane’s speed in still air and the jet-stream's speed?
For questions 5-7, a movie theater charges $4.50 for children and $8.00 for adults.
- On a certain day, 1200 people enter the theater and $8375 is collected. How many children and how many adults attended?
- The next day, the manager announces that she wants to see them take in $10000 in tickets. If there are 240 seats in the house and only five movie showings planned that day, is it possible to meet that goal?
- At the same theater, a 16-ounce soda costs $3 and a 32-ounce soda costs $5. If the theater sells 12,480 ounces of soda for $2100, how many people bought soda? (Note: Be careful in setting up this problem!)
For questions 8-10, consider the situation: Nadia told Peter that she went to the farmer’s market and bought two apples and one banana, and that it cost her $2.50. She thought that Peter might like some fruit, so she went back to the seller and bought four more apples and two more bananas. Peter thanked Nadia, but told her that he did not like bananas, so he would only pay her for four apples. Nadia told him that the second time she paid $6.00 for the fruit.
- What did Peter find when he tried to figure out the price of four apples?
- Nadia then told Peter she had made a mistake, and she actually paid $5.00 on her second trip. Now what answer did Peter get when he tried to figure out how much to pay her?
- Alicia then showed up and told them she had just bought 3 apples and 2 bananas from the same seller for $4.25. Now how much should Peter pay Nadia for four apples?
Review (Answers)
To view the answers, open this PDF file and look for section 7.9.
Notes/Highlights Having trouble? Report an issue.
Color | Highlighted Text | Notes | |
---|---|---|---|
Show More |
Term | Definition |
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elimination | The elimination method for solving a system of two equations involves combining the two equations in order to produce one equation in one variable. |
linear equation | A linear equation is an equation between two variables that produces a straight line when graphed. |
Linear System of Equation(s) | A linear system of equations is a set of equations that must be solved together to find the one solution that fits them both. |
system of equations | A system of equations is a set of two or more equations. |
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Here you'll learn how consistent, inconsistent, and dependent systems arise in real-world applications and you'll solve such problems.
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