How to add/subtract complex rational expressions? I'm studying for my Precalculus final and have noticed I still don't fully grasp performing basic operations on complex rational expressions, or finding if any values must be restricted from the domain. For example:
$$\frac{3}{5x+2}+\frac{5x}{25x^2-4}.$$
I know if this was a division problem, I would first factor out any exponential values of x and then multiply the first expression by the reciprocal of the second complex rational expression and then simplify.
Is there a similar process for addition/subtraction?

Find the (least) common denominator for $$\frac{3}{5x+2}+\frac{5x}{25x^2-4}.$$ Note that $$25x^2 - 4 = (5x-2)(5x+2),$$ and so, $$d = 25x^2 - 4 = (5x-2)(5x+2), \quad d\neq \pm \frac 25$$ is the common denominator, because each of $5x+2$ and $(5x+2)(5x-2)$ divide $d$.
Write each fraction using the common denominator (let's just call it $d$ for now). Once you've done that, find the sum:
$$\text{E.g.,}\;\;\frac ad + \frac cd = \frac {a+c}{d}$$
That gives us $$\begin{align} \frac{3}{5x+2}+\frac{5x}{25x^2-4} & =\frac{3(5x-2)}{(5x - 2)(5x+2)}+\frac{5x}{(5x-2)(5x+2)}\\ \\& = \dfrac{3(5x-2) + 5x}{(5x-2)(5x+2)}\end{align}$$
Now simplify.
Added note: Don't forget that we need to keep in mind that the sum is not defined when $(5x+2)(5x-2) = 0$, i.e., it is not defined when $x = \pm \frac 25$. But that should not be surprising, neither of the original summands is defined at $x = -\frac 25$, and additionally, the second summand is not defined at $x = \frac 25$. So it makes sense that the function, as a sum, is not defined at $\pm \frac{2}{5}$.
