What method should I use to solve rational equations like this for a different quantity? With electronics, various characteristics of a device can often be described by solving one equation for different quantities. The problem that I run into a lot with my textbooks is that I can't figure out how they manipulate (most often) rational equations to do so.
For example, this equation describes the output sensitivity of a Wheatstone Bridge with variable resistances as a function of $\epsilon$:
$$\frac{V_o}{V_i}=\frac{\epsilon G(\nu+1)}{2-\epsilon G(\nu-1)}$$
If $G=2,\,\nu=0.3$, and $\epsilon=10^{-3}$, then the sensitivity is $1.3\;mV/V$.
The text then goes on to say that the equation can be used to determine the output per unit $\,\epsilon\,$ i.e., $\,V_o/\epsilon\,$, as a function of $\,V_i\,$ and that the result, given $G=2,\,\nu=0.3$, and $V_i=10$, is $13\;{\mu}V/\mu\epsilon$.
I tried whatever I could think of to arrive at an equation for $\,V_o/\epsilon\,$ in terms of $\,V_i\,$ but now I've pretty much resigned myself to computing it with a script.
Are there are any methods I can use to solve this equation (and equations like it) for $\,V_o/\epsilon\,$ or a different quantity by hand?
 A: You have to isolate $\epsilon$ on one side.
The trick is factoring $\epsilon$ out. It looks like this:
$$\frac{V_o}{V_i}=\frac{\epsilon G(\nu+1)}{2-\epsilon G(\nu-1)}$$
Multiply with the denominators, we get
$$V_o(2-\epsilon G(\nu-1)=V_i\epsilon G(\nu+1)$$
Now factor out the left-hand-side.
$$2V_o-V_o\epsilon (G\nu-1)=V_i\epsilon G(\nu+1)$$
Separate the $\epsilon$-terms from everything else:
$$2V_o=V_o\epsilon (G\nu-1)+V_i\epsilon G(\nu+1)$$
As I said now the "trick" is factoring $\epsilon$ out.
Then
$$2V_o=\epsilon(V_o G(\nu-1)+V_i G(\nu+1))$$
Now divide by $\epsilon$.
You get $$\frac{2V_o}{\epsilon}= V_o G(\nu-1)+V_i G(\nu+1)$$.
So this would be the standard method on solving equations like this for $\epsilon$, as you could isolate $\epsilon$ completly.
However I am unsure if I interpreted your question correctly, and what your textbook says.
It should not be possible, to write this equation into the form
$V_o/\epsilon=\dotso$ without having $V_o$ appearing on the right-hand-side of the equation.
