Answer
The equation has no solutions.
Work Step by Step
STEP 1: Restrictions. There must be no zeros in the denominator, so $x\neq\pm 3$
STEP 2: Fractions. The LCM of all denominators is $x^{2}-9=(x+3)(x-3).$
After multiplying both sides with the LCM,
$x+4(x-3)=3$
STEP 3: Remove all parentheses and simplify.
$x+4x-12=3$
$5x-12=3$
STEP 4: Variables on one side, all other terms on the other...
$5x-12=3\qquad/+12$
$5x=15$
STEP 5: Simplify and solve.
$5x=15\qquad /\div 5$
$x=3$
In step 1, the number $3$ is excluded as a possible solution.
Thus, the equation has no solutions.
