# Silly problem with joint variation.

So, I was recently revisiting some old competition math problems and came to the jarring (and embarrassing) realization that I apparently don't understand joint variation as well as I thought. By way of example, if we have $4 apples = 3 oranges$ and $3 apples = 2 pears$, we could say that apples and oranges have a direct relationship and that apples and pears have a direct relationship, so: $apples/oranges = 3/4$ and $apples/pears = 2/3$.

My problem is that, we could also say that apples are jointly proportional to pears and oranges, which would yield: $4 apples = 6 oranges \times pears$ according to the joint variation formula. My question is how do we go from two direct variation equations to 1 joint variation equation?

This all came about when I was working through the following problem:

Given that x is directly proportional to y and to z and is inversely proportional to w and that x = 4 when (w,y,z) = (6,8,5), what is x when (w,y,z) = (4,10,9).

I know how to solve this problem, when I was doing this sort of thing, I was just taught to multiply when I saw inverse and divide for direct variation, so: $xw/yz=...$ and solve. I'm a little confused as to why this works.

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To say that $x$ is directly proportional to $y$ and to $z$, and inversely proportional to $w$, is to say that there is a constant $k$ such that $$x=k\frac{yz}{w}.$$ (Here there is an implicit assumption that $y$, $z$, and $w$ are independent factors).
Since you know the value of $x$ when $(w,y,z)=(6,8,5)$, you can calculate $k$. And now that you know $k$, you can find $x$ when $(w,y,z)=(4,10,9)$.
We can short-circuit this process, at the cost of perhaps losing control over what's happening. To switch $y$ from $8$ to $10$, we multiply by $\frac{10}{8}$, that is, scale by that factor. To switch $z$ from $5$ to $9$, multiply by $\frac{9}{5}$. Finally, to switch $w$ from $6$ to $4$, we multiply by $\frac{6}{4}$ (note the changed order). So we multiply the original value of $x$ by $\frac{10}{8}\cdot\frac{9}{5}\cdot\frac{6}{4}$.
More unpleasantly still, let $(x_1,w_1,y_1,z_1)$ be the first values, and $(x_2,w_2,y_2,z_2)$ be the second values. Then $$x_1=k\frac{y_1z_1}{w_1}\qquad\text{and}\qquad x_2=k\frac{y_2z_2}{w_2}.$$ Divide. We get after a little algebra $$\frac{x_2}{x_1}=\frac{y_2z_2w_1}{y_1z_1w_2}.$$ To finish, substitute.