I have these two equations:

$$x=\frac{ab(1+k)}{b+ka}\\ y=\frac{ab(1+k)}{a+kb}$$

where $a,b$ are constants and $k$ is a parameter to be eliminated.

A relation between $x,y$ is to be found. What is the best way to do it? Cross multiplying and solving is a bit too hectic. Is there a way we can maybe exploit the symmetry of the situation? Thanks!!

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    $\begingroup$ Don't know if this helps but $\frac{1}{x} + \frac{1}{y} = \frac{1}{a} + \frac{1}{b}$ $\endgroup$ – iamwhoiam Jul 16 '18 at 3:37
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    $\begingroup$ @iamwhoiam. Is that easy to see? Oh yeah. That's easy. I think you should make that an answer. $\endgroup$ – Mason Jul 16 '18 at 3:40
  • $\begingroup$ I think it is, since the the numerator of $x$ and $y$ is the same. So it kinda makes sense that you might take a closer look on $x^{-1}$ and $y^{-1}$. $\endgroup$ – Cornman Jul 16 '18 at 3:42
  • $\begingroup$ @iamwhoiam This was what I was looking for!! Thanks a lot! $\endgroup$ – tatan Jul 16 '18 at 3:43
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    $\begingroup$ @iamwhoiam I think you should post it as an answer! $\endgroup$ – tatan Jul 16 '18 at 3:48

Notice that the numerators of the two fractions are equal. It might thus be helpful to consider $\frac{1}{x}$ and $\frac{1}{y}$. With this approach, we observe that $$\frac{1}{x} + \frac{1}{y} = \frac{1}{a} + \frac{1}{b}$$

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    $\begingroup$ Nicely done (+1). $\endgroup$ – dxiv Jul 16 '18 at 4:00
  • $\begingroup$ What step am I missing to get to $\frac{1}{a} + \frac{1}{b} $?$\frac{1}{x} + \frac{1}{y} = \frac{a + ak + b + kb}{ab(1 + k)} \iff \frac{a (1 + k) + b (1 + k)}{ab(1 + k)} \iff \frac{(a + b) (1 + k)}{ab(1 + k)} \iff \frac{(a + b)}{ab} \iff ???$ $\endgroup$ – Phil Patterson Jul 16 '18 at 20:43
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    $\begingroup$ @PhilPatterson $\frac{(a+b)}{ab}$ <-> $\frac{a}{ab} + \frac{b}{ab}$ <-> $\frac{1}{b} + \frac{1}{a}$ <-> $\frac1a+\frac1b$ $\endgroup$ – pizzapants184 Jul 16 '18 at 21:18
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    $\begingroup$ @pizzapants184 somehow I forgot that you could split the terms in that way ... amazing how much you can forget 15 years after university ... Thanks for spelling it out for me! $\endgroup$ – Phil Patterson Jul 17 '18 at 4:28
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    $\begingroup$ This was literally the first time in my life using MathJax to create equations ... now that you've said something about it @Mason, I have to assume those double arrows have another meaning and I just used them erroneously. Reflecting on it I picked them because I liked the way they looked ... FWIW $\endgroup$ – Phil Patterson Jul 19 '18 at 17:37

Direct elimination doesn't look so hectic in this case:

$$(b+ka)x=ab(1+k) \iff ka(x-b)=b(a-x)\iff k = - \frac{b(x-a)}{a(x-b)}$$

Doing the same for the second equation then equating eliminates $\,k\,$.

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    $\begingroup$ Thanks! But I think that adding $x^-1$ and $y^-1$ and adding them (as mentioned in the comments) is a nice trick here ;-) $\endgroup$ – tatan Jul 16 '18 at 3:44
  • $\begingroup$ @tatan I agree, and will upvote that once posted as an answer ;-) $\endgroup$ – dxiv Jul 16 '18 at 3:47

Alternatively, note that $$\frac xy=\frac{a+kb}{b+ka}\implies (bx-ay)=k(by-ax)\implies k=\frac{bx-ay}{by-ax}$$ and equate with @dxiv's answer.


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