# Easiest way to solve this system of equations

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!!

• Don't know if this helps but $\frac{1}{x} + \frac{1}{y} = \frac{1}{a} + \frac{1}{b}$ Jul 16 '18 at 3:37
• @iamwhoiam. Is that easy to see? Oh yeah. That's easy. I think you should make that an answer. Jul 16 '18 at 3:40
• 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 $y^{-1}$. Jul 16 '18 at 3:42
• @iamwhoiam This was what I was looking for!! Thanks a lot! Jul 16 '18 at 3:43
• @iamwhoiam I think you should post it as an answer! 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 $$\frac{1}{x} + \frac{1}{y} = \frac{1}{a} + \frac{1}{b}$$

• Nicely done (+1).
– dxiv
Jul 16 '18 at 4:00
• What step am I missing to get to $\frac{1}{a} + \frac{1}{b}$?$\frac{1}{x} + \frac{1}{y} = \frac{a + ak + b + kb}{ab(1 + k)} \iff \frac{a (1 + k) + b (1 + k)}{ab(1 + k)} \iff \frac{(a + b) (1 + k)}{ab(1 + k)} \iff \frac{(a + b)}{ab} \iff ???$ Jul 16 '18 at 20:43
• @PhilPatterson $\frac{(a+b)}{ab}$ <-> $\frac{a}{ab} + \frac{b}{ab}$ <-> $\frac{1}{b} + \frac{1}{a}$ <-> $\frac1a+\frac1b$ Jul 16 '18 at 21:18
• @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! Jul 17 '18 at 4:28
• 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 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 $\,k\,$.

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