Example 1.4.25. The Interview.
Once upon a time, the
New York Times published an article about the movie,
The Interview. It included the following quote:
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.
Note the article is suggesting that we do not know how many rentals there were and how many sales there were. A few days later, Joey Devilla cleverly pointed out in
his blog that there is enough information here to use algebra to figure out the number of sales and the number of rentals. We can write a system of equations and solve it to find the two quantities. (Since the numbers given in the article are only approximations, the solutions we find will also only be approximations.)
First, we will define variables. 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 measured in “transactions”. The article says that the total number of transactions is two million. So our first equation will add the total number of rental and sales transactions and set that equal to two 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 that would give us the total number of dollars generated from rentals. This is \(6r\text{.}\) 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, with the commas removed:
\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 a method called substitution. The idea is to use one equation to isolate \(r\text{.}\) Then substitute this for the “\(r\)” that’s in the other equation. This leaves you with one equation where the only variable is \(s\text{.}\) And we can handle that directly.
The first equation from the system is an easy one to isolate \(r\text{:}\)
\begin{align*}
r+s \amp=2\,000\,000\amp\text{(the system's first equation)}\\
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 that in place of \(r\) in the second equation:
\begin{align*}
6r+15s \amp=15\,000\,000\amp\text{(the system's second equation)}\\
6(\substitute{2\,000\,000-s})+15s \amp=15\,000\,000
\end{align*}
Now we have an equation with only one variable, \(s\text{.}\)
\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\\
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 isolated \(r\) previously 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*}
In summary, there were roughly \(333{,}333\) sales and roughly \(1{,}666{,}667\) rentals.
