Can this be done, and if so, how:
Given four constants $(K_i)$ and variables $x$ and $y$, is there a way for me to find the values of $x$ and $y$ when $K_{1-4}$ are known for these two simultaneous equations:
$$y = \frac{K_1}{\sqrt{\frac{K_2}{x^2} + 1} - 1}$$
and
$$y = \frac{K_3}{\sqrt{\frac{K_4}{x^2} + 1} - 1}$$
As you can see, they are really the same equation, just with different constants $K$.
$$y = \frac{K_1}{\sqrt{\frac{K_2}{x^2} + 1} - 1}= \frac{K_3}{\sqrt{\frac{K_4}{x^2} + 1} - 1}$$ Let $\frac{1}{x^2}=X$ $$\frac{K_1}{\sqrt{K_2 X + 1} - 1}= \frac{K_3}{\sqrt{K_4 X + 1} - 1}$$ $$K_1 (\sqrt{K_4 X + 1} - 1)=K_3 (\sqrt{K_4 X + 1} - 1)$$ $$\sqrt{K_4 X + 1}=\frac{K_3}{K_1} (\sqrt{K_4 X + 1} - 1)+1$$ $$K_4 X + 1=\left(\frac{K_3}{K_1} (\sqrt{K_4 X + 1} - 1)+1 \right) ^2$$ Expand and isolate $\sqrt{K_4 X + 1}$ $$\sqrt{K_4 X + 1}=\frac{(K_2K_3^2-K_4K_1^2)X+2K_3(K_3-K_1)}{2K_3(K_3-K_1)}$$ $$K_4 X + 1=\left(\frac{(K_2K_3^2-K_4K_1^2)X+2K_3(K_3-K_1)}{2K_3(K_3-K_1)}\right)^2$$ Expand and solve the quadratic equation obtained : $$X=4K_1K_3(K_3-K_1)\frac{K_1K_4-K_2K_3}{K_1^2K_4-K_3^2K_2}$$ $$x=2\space\sqrt{\frac{K_1^2K_4-K_3^2K_2}{K_1K_3(K_3-K_1)(K_1K_4-K_2K_3)}}$$ Then bing $X$ back into $y = \frac{K_1}{\sqrt{K_2X + 1} - 1}$ and find $y$ $$y=\frac{K_1^2K_4-K_3^2K_2}{2K_2K_3(K_3-K_1)}$$