Answer
$\dfrac{7x - 17}{(x - 3)(x + 3)}$
Restriction: $x \ne -3, 3$
Work Step by Step
Factor all expressions in the original exercise:
$\frac{4}{(x - 3)(x + 3)} + \frac{7}{x + 3}$
Before performing the addition, find the least common denominator (LCD) of the two fractions. The LCD is $(x-3)(x+3)$. Convert the original fractions to equivalent fractions using the LCD:
$=\dfrac{4}{(x - 3)(x + 3)} + \dfrac{7(x - 3)}{(x + 3)(x - 3)}$
Multiply to simplify:
$=\dfrac{4}{(x - 3)(x + 3)} + \dfrac{7x - 21}{(x - 3)(x + 3)}$
Add the fractions:
$=\dfrac{4 + 7x - 21}{(x - 3)(x + 3)}$
Combine like terms:
$=\dfrac{7x - 17}{(x - 3)(x + 3)}$
Restrictions on $x$ occur when the value of $x$ makes the fraction undefined, which means that the denominator becomes $0$.
Set the factors in the denominators equal to $0$ to find restrictions:
First factor of the first denominator:
$x + 3 = 0$
$x = -3$
Second factor of the first denominator:
$x - 3 = 0$
$x = 3$
The second denominator was already covered in the work above.
Thus, the restrictions are: $x \ne -3, 3$
