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:


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 $\pm \frac{2}{5}$.

  • $\begingroup$ This is a great answer as it would be common to find the [least] common denominator of the rational expressions, but I prefer my answer because it may save a little work (you don't have to "waste your time" finding the LCD of the denominators). $\endgroup$ – Obinna Nwakwue Mar 18 '16 at 21:38

You can use the algebraic formula for adding fractions: $$\frac ab + \frac cd = \frac {ad + bc}{bd}$$ with: $a = 3$, $b = 5x + 2$, $c = 5x$, and $d = 25x^2 - 4$: $$\frac {3}{5x} + \frac {5x +2}{25x^2 - 4} = \frac {3(25x^2 - 4) + (5x + 2)(5x)}{5x(25x^2 - 4)}$$ When you simplify both sides, you get: $$\frac {(75x^2 - 12) + (25x^2 + 10x)}{125x^3 - 20x}$$ Simplify down further and you get: $$\frac {100x^2 + 10x - 12}{125x^3 - 20x}$$ Simplify down even further and you get: $$\frac {20x - 6}{25x^2 - 10x}$$ You can't simplify further than that, so that is your final answer.


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