# How is the joint proportionality (joint variation) equation (i.e. $y = kxz$) logically correct?

If $$y$$ is proportional to $$x$$, and $$y$$ is also proportional to $$z$$, then how are we able to arrive at the equation: $$y = kxz$$ ?

My understanding so far of proportionality, which comes from this Wikipedia article, is that two variables are directly proportional if their ratio yields a constant.

Thus, if $$y$$ is proportional to $$x$$, then $$\frac yx = k$$, where $$k$$ is a constant.

So, if $$\frac yx = k*z$$, then how are $$y$$ and $$x$$ still proportional if $$z$$ is not a constant?

Two variables $$\alpha$$ and $$\beta$$ are directly proportional if their ratio is constant, relative to $$\alpha$$ and $$\beta$$. That last phrase is missing from the Wiki article, because it is not relevant to the two-variable models discussed there. But it is crucial here.
With that correction, direct proportionality tells us that $$\frac{y}{x}=\lambda_1(z)$$ where $$\lambda_1(z)$$ is independent of $$x$$ and $$y$$, but might depend on $$z$$. Similarly, $$\frac{y}{z}=\lambda_2(x)$$ where $$\lambda_2(x)$$ is independent of $$y$$ and $$z$$, but might depend on $$x$$. Dividing the two equations, $$\frac{z}{x}=\frac{\lambda_1(z)}{\lambda_2(x)}$$ or, rearranging, $$\frac{\lambda_1(z)}{z}=\frac{\lambda_2(x)}{x}$$
Each side of the latter equation is constant with respect to the other side, so let their common value be $$k$$. Then $$\lambda_1(z)=kz$$, whence the claim.
• @johnnyodonnell No, that definition suffers the same problem. Let $x$, $y$, and $z$ be three mutually-related variables, with $x$ directly proportional to $y$. Holding $y$ constant means $y$ changes with ratio $1$, so $x$ does too, so $x$ is constant. But changing $z$ might induce a change in $x$. Commented Feb 14, 2019 at 19:12