# Can two unknowns of two *unrelated* linear equations be determined?

This question is based on the storm caused by this: https://twitter.com/ddmeyer/status/549965027948507136. Since a twitter discussion is not an objective question, I'll simply write out the key elements of the problem.

Movie earning estimates were reported as roughly \$15 million in total revenue across about 2 million transactions where the rental option was \$6 and the sale purchase option was \$15 dollars. Emphasizing that the given number of transactions and revenue are rough estimates, can this be solved algebraically? That is, can the two "rough" equations (in whatever form that maybe) simply be solved to determine the two unknowns r and s (rentals and sales)? To be clear, r and s represent the number of transactions. If solving these equations is valid, then what is the graphical explanation for why these two seemingly unrelated equations (one is # of transactions the other is revenue) can be solved? • Are you looking too deeply into the meaning of the words "roughly" and "about"? They mean the same thing in this context, "approximately". The author used different words to seem more creative, and used them in the first place to avoid taking too much responsibility for the numbers in question. – Anthony Jan 5 '15 at 5:13 • Also, these two equations are not considered unrelated in math. Both equations,$r+s=2\mathrm{M}$and$6r+15s=15\mathrm{M}$are functions of the same two variables,$r$and$s$. – Anthony Jan 5 '15 at 5:16 • In this case, maybe this question could be formulated better in the form "How can this scenario be setup algebraically?" – Anthony Jan 5 '15 at 5:17 ## 2 Answers If I'm understanding the scenario correctly, wouldn't it just be: $$\begin{cases} r + s = 2 \;\mathrm{million} \\ 6r + 15s = 15 \;\mathrm{million} \end{cases}$$ To take the roughness of the equations into account, we can edit our constants to incorporate an error term: $$\begin{cases} r + s = (2 \;\mathrm{million} \pm e_c) \\ 6r + 15s = (15 \;\mathrm{million} \pm e_r) \end{cases}$$ where$e_c$is the "error in the count" and$e_r$is the "error in the revenue." We can just solve these normally for$r$and$sand see what happens to the error terms. Scratching it out, I get something like: \begin{align} s = \frac{1}{3} \;\mathrm{million} \mp \frac{2}{3}e_c \pm \frac{1}{9}e_r \\ r = \frac{5}{3} \;\mathrm{million} \pm \frac{5}{3}e_c \mp \frac{1}{9}e_r \end{align} This quantifies how much the error in the original equations propagates through to the values fors$and$r$. We see that any error in the estimation for the total count has much more influence over the values of$r$and$s$than errors in the revenue. • That's the trivial part. The issue that I have is understanding the scenario. The meaning of the words "rough" and "about" as well as the fact that the known info is across revenue and transactions is what I am having issues wrapping my head around. – RecursiveCall Jan 4 '15 at 20:56 • @RecursiveCall I edited my answer to take the roughness of the original equations into account. – Mike Pierce Jan 4 '15 at 21:11 • What is the actual numerical answer, if any? What I am getting at is that the author insists that there is a numerical answer (which is the solution of those equations), but it looks like the "roughness" disclaimer justifies not posting any numerical data. – RecursiveCall Jan 6 '15 at 18:13 • Numerically, we could say the the number of rentals is about$\frac{5}{3} \mathrm{million} = 1,666,667$and the number sold is about$\frac{1}{3} \mathrm{million} = 333,333$. – Mike Pierce Jan 7 '15 at 1:59 Why do you think the total revenue and total number of transactions are unrelated? My understanding of the scenario is this:$15000000 was generated out of the sale of 2000000 tickets.

Tickets being something physical makes it easier to visualize the problem. And using 000000 instead of millions avoids mistakenly treating the word millions as a unit.

A ticket to own the movie costs \$15 and a ticket to rent the movie cost \$6.

So that's the gist of the problem. So the question is, what combination of \$15 and \$6 tickets generate \\$150000000 when the total number of tickets are 2000000?

This gives us two equations:

1. total number of sale tickets and rental tickets = 2000000

2. total price of sale tickets and rental tickets = 150000000