How do I reach this form I just want to check with you guys how to do this correctly:
I have to reduce:
$$ \frac{1}{N}+3=\frac{2R+2}{5R+2}$$
...to the following form:
$$ N=\frac{aR+b}{cR+d}$$
Any leads on how I reach this form? Should I flip all terms, to get N=, then bring 1/3 to the other side and subtract it there?
Thank you in advance.

First, subtract $3$ from each side of the equation. Then find a common denominator: $$\begin{align} \frac{1}{N}+3=\frac{2R+2}{5R+2} &\iff  \frac{1}{N} =\frac{2R+2}{5R+2} -3 \\ \\ & \iff  \dfrac 1N = \dfrac{(2R+2) - 3(5R+2)}{5R + 2} \\ \\ & \iff \frac 1N = \frac{-13 R- 4}{5R + 2}\end{align}$$
Simplify, and "flip"/invert":
$$\iff N =\frac {5R + 2}{-13 R- 4} = \frac {-5R - 2}{13 R + 4} =-\frac {5R + 2}{13 R+  4}$$
