Example5.2.2The Interview

In 2014, the New York Times 1  posted the following about the movie, “The Interview”:

“The Interview” generated roughly \(\$15\) million in online sales and rentals during its first four days of availability, Sony Pictures said on Sunday.

Sony did not say how much of that total represented \(\$6\) digital rentals versus \(\$15\) sales. The studio said there were about two million transactions overall.

A few days later, Joey Devilla cleverly pointed out in his blog 2 , that there is enough information given to find the amount of sales versus rentals. Using algebra, we can write a system of equations and solve it to find the two quantities. 3 

First, we will define variables. We need two variables, because there are two unknown quantities: how many sales there were and how many rentals there were. Let \(r\) be the number of rental transactions and let \(s\) be the number of sales transactions.

If you are unsure how to write an equation from the background information, use the units to help you. The units of each term in an equation must match because we can only add like quantities. Both \(r\) and \(s\) are in transactions. The article says that the total number of transactions is \(2\) million. So our first equation will add the total number of rental and sales transactions and set that equal to \(2\) million. Our equation is:

\begin{equation*} (r\,\text{transactions})+(s\,\text{transactions})=2{,}000{,}000\,\text{transactions} \end{equation*}

Without the units:

\begin{equation*} r+s=2{,}000{,}000 \end{equation*}

The price of each rental was \(\$6\text{.}\) That means the problem has given us a rate of \(6\,\frac{\text{dollars}}{\text{transaction}}\) to work with. The rate unit suggests this should be multiplied by something measured in transactions. It makes sense to multiply by \(r\text{,}\) and then the number of dollars generated from rentals was \(6r\text{.}\) Similarly, the price of each sale was \(\$15\text{,}\) so the revenue from sales was \(15s\text{.}\) The total revenue was \(\$15\) million, which we can represent with this equation:

\begin{equation*} \left(6\,\tfrac{\text{dollars}}{\text{transaction}}\right)(r\,\text{transactions})+\left(15\,\tfrac{\text{dollars}}{\text{transaction}}\right)(s\,\text{transactions})=\$15{,}000{,}000 \end{equation*}

Without the units:

\begin{equation*} 6r+15s=15{,}000{,}000 \end{equation*}

Here is our system of equations:

\begin{equation*} \left\{ \begin{alignedat}{4} r\amp+{}\amp s\amp={}\amp2{,}000{,}000 \\ 6r\amp+{}\amp 15s\amp={}\amp15{,}000{,}000 \end{alignedat} \right. \end{equation*}

To solve the system, we will use the substitution method. The idea is to use one equation to find an expression that is equal to \(r\) but, cleverly, does not use the variable “\(r\)”. Then, substitute this for \(r\) into the other equation. This leaves you with one equation that only has one variable.

The first equation from the system is an easy one to solve for \(r\text{:}\)

\begin{align*} r+s \amp=2{,}000{,}000 \\ r+s \subtractright{s} \amp=2{,}000{,}000\subtractright{s} \\ r \amp=2{,}000{,}000-s \end{align*}

This tells us that the expression \(2{,}000{,}000-s\) is equal to \(r\text{,}\) so we can substitute it for \(r\) in the second equation:

\begin{align*} 6r+15s \amp=15{,}000{,}000\\ 6(\substitute{2{,}000{,}000-s})+15s \amp=15{,}000{,}000\\ \end{align*}

Now we have an equation with only one variable, \(s\text{,}\) which we will solve for:

\begin{align*} 6(2{,}000{,}000-s)+15s \amp=15{,}000{,}000\\ 12{,}000{,}000-6s+15s \amp=15{,}000{,}000\\ 12{,}000{,}000+9s \amp= 15{,}000{,}000\\ 12{,}000{,}000+9s\subtractright{12{,}000{,}000} \amp= 15{,}000{,}000\subtractright{12{,}000{,}000}\\ 9s \amp= 3{,}000{,}000\\ \divideunder{9s}{9} \amp= \divideunder{3{,}000{,}000}{9}\\ s \amp= 333{,}333.\overline{3} \end{align*}

At this point, we know that \(s=333{,}333.\overline{3}\text{.}\) This tells us that out of the \(2\) million transactions, roughly \(333{,}333\) were from online sales. Recall that we solved the first equation for \(r\text{,}\) and found \(r=2{,}000{,}000-s\text{.}\)

\begin{align*} r \amp=2{,}000{,}000-s\\ r \amp=2{,}000{,}000-\substitute{333{,}333.\overline{3}}\\ r \amp=1{,}666{,}666.\overline{6} \end{align*}

To check our answer, we will see if \(s=333{,}333.\overline{3}\) and \(r=1{,}666{,}666.\overline{6}\) make the original equations true:

\begin{align*} r+s \amp=2{,}000{,}000\\ \substitute{1{,}666{,}666.\overline{6}}+\substitute{333{,}333.\overline{3}} \amp\stackrel{?}{=}2{,}000{,}000\\ 2{,}000{,}000\amp\stackrel{\checkmark}{=}2{,}000{,}000 \end{align*}
\begin{align*} 6r+15s \amp=15{,}000{,}000\\ 6\left(\substitute{1{,}666{,}666.\overline{6}}\right)+15\left(\substitute{333{,}333.\overline{3}}\right) \amp\stackrel{?}{=}15{,}000{,}000\\ 10{,}000{,}000+5{,}000{,}000 \amp\stackrel{\checkmark}{=}15{,}000{,}000 \end{align*}

In summary, there were roughly \(333{,}333\) copies sold and roughly \(1{,}666{,}667\) copies rented.

in-context