Find z

$\dfrac{3z+9}{6}+\dfrac{8z-2}{4}=\dfrac{5+4z}{3}$

Multiply both sides by $12$ .

$12\left(\dfrac{3z+9}{6}+\dfrac{8z-2}{4}=\dfrac{5+4z}{3}\right)$

$2(3z+9)+3(8z-2)=4(5+4z)$

Simplify by applying the distributive property and combine like terms.

$6z+18+24z-6=20+16z$

$14z=8$

Divide both sides by $14$ to get

$z=\dfrac{8}{14}=\boxed{\dfrac{4}{7}}$

We can solve this by multiplying both sides by 12 to eliminate all of the fractions. $\begin{aligned} \frac{3z+9}{6} + \frac{8z-2}{4} &= \frac{5+4z}{3} \\ 12\left(\frac{3z+9}{6} + \frac{8z-2}{4}\right) &= 12\left(\frac{5+4z}{3}\right) \\ 6z+18+24z-6 &= 20+16z \\ 14z &= 8 \\ z &= \boxed{\frac{4}{7}} \\ \end{aligned}$ Therefore, the answer is $\displaystyle \frac{4}{7}$ .