Solve each inequality. Then graph the solution set.

$\left|\frac{4b-2}{3}\right|\mathbf{<}\mathbf{12}$

The solution of the given inequality$\left|\frac{4b-2}{3}\right|<12$ is:

Case 1:$\frac{4b-2}{3}$ is non-negative.

$\begin{array}{c}\frac{4b-2}{3}<12\\ \frac{4b-2}{3}\times 3<12\times 3\\ 4b-2<36\\ 4b-2+2<36+2\\ 4b<38\\ \frac{4b}{4}<\frac{38}{4}\\ b<9.5\\ b\in \left(-\infty ,9.5\right)\end{array}$

Case 2:$\frac{4b-2}{3}$ is negative.

$\begin{array}{c}-\left(\frac{4b-2}{3}\right)<12\\ \left(-1\right)\left(-\left(\frac{4b-2}{3}\right)\right)>\left(-1\right)\left(12\right)\\ \frac{4b-2}{3}>-12\\ \frac{4b-2}{3}\times 3>-12\times 3\\ 4b-2>-36\\ 4b-2+2>-36+2\\ 4b>-34\\ \frac{4b}{4}>\frac{-34}{4}\\ b>-8.5\\ b\in \left(-8.5,\infty \right)\end{array}$

The solution of the inequality$\left|x\right|\le a$ is$x\ge -a$ and $x\le a$.

That implies the solution of the inequality is the intersection of the solutions of the inequalities$x\ge -a$ and $x\le a$.

Find the intersection of the solutions of the inequalities$\frac{4b-2}{3}<12$ and$\frac{4b-2}{3}>-12$ to find the solution of the inequality $\left|\frac{4b-2}{3}\right|<12$.

The intersection of the solutions of the inequalities$\frac{4b-2}{3}<12$ and$\frac{4b-2}{3}>-12$ is:

$b\in \left(-8.5,9.5\right)$

Therefore, the solution of the inequality$\left|\frac{4b-2}{3}\right|<12$ is $b\in \left(-8.5,9.5\right)$.

The graph of the solution set which is $b\in \left(-8.5,9.5\right)$is: