Answer
$\dfrac{3}{x-3}, \space x\ne-3$
Work Step by Step
Note that:
$\dfrac{3x+9}{x^2-9}=\dfrac{3x+3(3)}{x^2-3^2}$
Factor out $3$ in the numerator and use special formula $a^2-b^2=(a+b)(a-b)$ to factor the denominator.
$=\dfrac{3(x+3)}{(x+3)(x-3)}$
Cancel common factor $x+3$:.
$=\dfrac{3}{x-3}, \space x\ne -3$
Hence, the lowest term is $\dfrac{3}{x-3}, x\ne -3$.
