Solve each inequality. Then graph the solution set.

$\left|-4y-3\right|\mathbf{<}\mathbf{13}$

The solution of the given inequality$\left|-4y-3\right|<13$ is:

Case 1:$-4y-3$ is non-negative.

$\begin{array}{c}-4y-3<13\\ -4y-3+3<13+3\\ -4y<16\\ \frac{-4y}{-4}>\frac{16}{-4}\\ y>-4\\ y\in \left(-4,\infty \right)\end{array}$

Case 2:$-4y-3$ is negative.

$\begin{array}{c}-\left(-4y-3\right)<13\\ \left(-1\right)\left(-\left(-4y-3\right)\right)>\left(-1\right)\left(13\right)\\ -4y-3>-13\\ -4y-3+3>-13+3\\ -4y>-10\\ \frac{-4y}{-4}<\frac{-10}{-4}\\ y<\frac{5}{2}\\ y<2.5\\ y\in \left(-\infty ,2.5\right)\end{array}$

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$\left|x\right|\le a$ 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$-4y-3<13$ and$-4y-3>-13$ to find the solution of the inequality $\left|-4y-3\right|<13$.

The intersection of the solutions of the inequalities$-4y-3<13$ and$-4y-3>-13$ is:

$y\in \left(-4,2.5\right)$

Therefore, the solution of the inequality$\left|-4y-3\right|<13$ is $y\in \left(-4,2.5\right)$.

The graph of the solution set which is $y\in \left(-4,2.5\right)$is: