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If $$y \propto x$$ and $$y \propto \frac1z$$ does this imply $$y \propto \frac{x}{z}$$ is true? I've tried a few cases and it has worked every time, however I can't find any information on the internet and I'm not sure how to derive a proof.

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  • $\begingroup$ Sometime, they called joint variation. $\endgroup$ – Ng Chung Tak Apr 25 at 19:40
  • $\begingroup$ Thank you, just what i was looking for. I have no idea why i couldn't find this. $\endgroup$ – user668526 Apr 25 at 19:42
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$y\propto x$ means $y=k_1x$, for some $k_1$ when $z$ is constant. $y\propto\frac 1z$ when $x$ is constant. From second proportionality, we can rewrite $k_1x\propto 1/z$ when $x$ is const. As, $x$ is const. we have $k_1\propto k_2/z$, we can write it like $k_1=k_3/z$ or, $y=\frac{k_3x}{z}$ which means $y\propto\frac xz$

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No, it's not true in general, because the variables $x$ and $z$ are often not independent. For example, suppose $x$ and $z$ are constrained so that $xz = 1$. To be even more explicit, suppose the set of attainable triples $(x,y,z)$ is $\{(t, t, 1/t) : t > 0\}$. Then certainly $y \propto x$ and $y \propto 1/z$, but $x/z = y^2$, so $y \propto \sqrt{x/z}$.

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