Answer
$\color{blue}{\dfrac{2}{9}}$
Work Step by Step
Use the rule $\dfrac{a}{b} \div \dfrac{c}{d} = \dfrac{a}{b} \times \dfrac{d}{c}$ to obtain:
$=\dfrac{2k+8}{6} \times \dfrac{2}{3k+12}$
Factor the binomials then cancel common factors to obtain:
$\require{cancel}
=\dfrac{2(k+4)}{3(2)} \times \dfrac{2}{3(k+4)}
\\=\dfrac{\cancel{2}\cancel{(k+4)}}{3\cancel{(2)}} \times \dfrac{2}{3\cancel{(k+4)}}
\\=\dfrac{1}{3} \times \dfrac{2}{3}
\\=\color{blue}{\dfrac{2}{9}}$
