Can two unknowns of two *unrelated* linear equations be determined? This question is based on the storm caused by this twitter post. 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 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 number of transactions the other is revenue) can be solved?
 A: 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:


*

*total number of sale tickets and rental tickets = 2000000

*total price of sale tickets and rental tickets = 150000000
A: 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 $s$ and 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 for $s$ 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.
