Suppose I have a business that sells the same product in three different countries (countries $x$, $y$ and $z$).Product sales is given by price times quantity, so sales of the product in country $x$ is: $$ s_x = p_x*q_x $$ where $p_x$ represents the price of the product in country $x$ and $q_x$ represents the quantity sold. We also know that sales for the entire company is the sum of sales in all three countries, so: $$ s_t = s_x + s_y + s_z $$ which can be rewritten as: $$ p_tq_t =p_xq_x + p_yq_y + p_zq_z $$ I am provided the:
- total sales ($s_t$),
- sales from each country ($s_x,s_y,s_z$),
- total quantity sold for the company ($q_t$), and
- the relative pricing between each country (i.e. $p_y=0.75p_x$ and $p_z=0.5p_x$).
Do I have enough information to solve for the price and quantity sold in each country (assuming $p$ and $q$ must be positive)? If so, how would I do that?
Yes, you do.
There's a bit of redundancy in what you know: once you know $s_x, s_y, s_z$, you can figure out $s_t$ by summing them up, for instance.
There's also something a little mysterious: you write $p_t q_t$, and while $q_t$ makes a lot of sense, $p_t$ is ... what? It's not necessarily the price that any unit sold for. In fact, it's an "average price per unit," but seems oddly irrelevant here.
Anyhow, on with the show:
$$ \newcommand{\red}[1]{\color{red}{#1}} $$
For relative pricing, I'm going to write $$ p_x = \red{f_x} p_x \\ p_y = \red{f_y} p_x \\ p_z = \red{f_z} p_x \tag{1} \label{1} $$ so that the $f$ values are your "relative price" factors. Obviously $f_x = 1$, but in your example, $f_y = 0.75$ and $f_z = 0.5$. I'm using "f" so that I get a more generic answer. And I'm coloring them red because we know them.
We also know sales for each region, and we know that \begin{align} \red{s_x} &= p_x q_x \\ \red{s_y} &= p_y q_y \\ \red{s_z} &= p_z q_z \\ \end{align} which we can rewrite as \begin{align} \frac{\red{s_x}}{p_x} &= q_x \\ \frac{\red{s_y}}{p_y} &= q_y \\ \frac{\red{s_z}}{p_z} &= q_z \tag{2}\label{2} \end{align}
Substituting in the Equation \eqref{1} forms for $p_x, p_y$ and $p_z$ in Equation \eqref{2}, we get \begin{align} \frac{\red{s_x}}{\red{f_x}p_x} &= q_x \\ \frac{\red{s_y}}{\red{f_y}p_x} &= q_y \\ \frac{\red{s_z}}{\red{f_z}p_x} &= q_z \tag{3}\label{3} \end{align} and putting everything over a common denominator gives us \begin{align} \frac{\red{f_y f_z}\red{s_x}}{\red{f_x f_y f_z}p_x} &= q_x \\ \frac{\red{f_x f_z}\red{s_y}}{\red{f_x f_y f_z}p_x} &= q_y \\ \frac{\red{f_x f_y}\red{s_z}}{\red{f_x f_y f_z}p_x} &= q_z \tag{4}\label{4} \end{align} We can sum these up to get the known total quantity $q_t$: \begin{align} \red{q_t} &= \frac{\red{f_y f_z}\red{s_x}}{\red{f_x f_y f_z}p_x} + \frac{\red{f_x f_z}\red{s_y}}{\red{f_x f_y f_z}p_x} + \frac{\red{f_x f_y}\red{s_z}}{\red{f_x f_y f_z}p_x} \\ &= \frac{\red{f_y f_z}\red{s_x} + \red{f_x f_z}\red{s_y} + \red{f_x f_y}\red{s_z}}{\red{f_x f_y f_z}p_x} \tag{5}\label{5} \end{align} Multiplying both sides by $p_x$, and dividing both by $q_t$, we arrive at \begin{align} p_x &= \frac{\red{f_y f_z}\red{s_x} + \red{f_x f_z}\red{s_y} + \red{f_x f_y}\red{s_z}}{\red{f_x f_y f_z}\red{q_t}} \tag{6}\label{6} \end{align} which gives us $p_x$ in terms of things we know. We can then use Equation \eqref{1} to compute $p_y$ and $p_z$ as well, and then use Equation \eqref{2} to compute $q_x, q_y,$ and $q_z$, and we're done!
Note that by cancelling, Equation \eqref{6} can be simplified to \begin{align} p_x &= \frac{1}{q_t}\left( \frac{\red{s_x}}{\red{f_x}} + \frac{\red{s_y}}{\red{f_y}} + \frac{\red{s_z}}{\red{f_z}}\right) \tag{6a}\label{6a} \end{align} which may suggest something useful to you, although I don't immediately see what that would be. :)