Answer
$\dfrac{3(3x - 4)}{(x + 2)(x - 2)}, \space x \ne -2, 2$
Work Step by Step
Factor all expressions in the original exercise:
$\dfrac{3x}{(x + 2)(x - 2)} + \dfrac{6}{x + 2}$
The least common denominator, or LCD, is $(x + 2)(x - 2)$.
Convert each rational expression to an equivalent one by multiplying its numerator with whatever factor is missing between its denominator and the LCD:
$\dfrac{3x}{(x + 2)(x - 2)} + \dfrac{6(x - 2)}{(x + 2)(x - 2)}$
Multiply to simplify:
$\dfrac{3x}{(x + 2)(x - 2)} + \dfrac{6x - 12}{(x + 2)(x - 2)}$
Add:
$\dfrac{3x + (6x - 12)}{(x + 2)(x - 2)}$
Simplify:
$\dfrac{9x - 12}{(x + 2)(x - 2)}$
Factor the numerator:
$\dfrac{3(3x - 4)}{(x + 2)(x - 2)}$
Restrictions on $x$ occur when the value of $x$ makes the denominator equal $0$, which means that the rational expression becomes undefined.
Set the factors in the denominators equal to $0$ to find restrictions:
First factor:
$\begin{align*}
x + 2 &= 0\\
x&=-2
\end{align*}$
Second factor:
$\begin{align*}
x - 2 &= 0\\
x&=2
\end{align*}$
Restriction: $x \ne -2, 2$
