Easiest way to solve this system of equations

Question

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!!

Answer 1

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 $$\frac{1}{x} + \frac{1}{y} = \frac{1}{a} + \frac{1}{b}$$

Answer 2

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 $\,k\,$.

Answer 3

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.