91Ó°ÊÓ

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

Case 1:$−4y−3$ is non-negative.

$\begin{array}{c}−4y−3<13\\ −4y−3+3<13+3\\ −4y<16\\ \frac{−4y}{−4}>\frac{16}{−4}\\ y>−4\\ y∈\left(−4,∞\right)\end{array}$

Case 2:$−4y−3$ is negative.

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

The solution of the inequality$\left|x\right|≤a$ is$xâ‰\yen −a$ and $x≤a$.

That implies the solution of the inequality$\left|x\right|≤a$ is the intersection of the solutions of the inequalities$xâ‰\yen −a$ and $x≤a$.

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:

$y∈\left(−4,2.5\right)$

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

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