Solving Word Problems with Linear Systems

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 $\begin{align*}y\end{align*}$.

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.

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 $\begin{align*}y\end{align*}$.

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.

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.

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 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?

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.

1. Claire is 28 and John is 32.
2. The cost is the same after $5 \frac{1}{3}$ months.
3. Ties are $23, and suspenders are $47.
4. The plane flies at 540mph in still air, and the jet stream is 60mph.
5. 350 children and 850 adults
6. Not possible, the maximum take is $9,600.
7. 300 16-oz sodas, and 240 32-oz sodas
8. Peter discovered that Nadia would have had to pay a different price for each apple or banana (or both) each time for the second purchase to have been $6.
9. Apples cost less than $1.25 each, and bananas cost less than $2.50 each.
10. Four apples cost $3.00. Apples cost $0.75 each and bananas $1.00 each.