91Ó°ÊÓ

The solution of the given inequality$\left|m+19\right|≤1$ is:

Case 1:$m+19$ is non-negative.

$\begin{array}{c}m+19≤1\\ m+19−19≤1−19\\ m≤−18\\ m∈\left(−∞,−18\right]\end{array}$

Case 2:$m+19$ is negative.

$\begin{array}{c}−\left(m+19\right)≤1\\ \left(−1\right)\left(−\left(m+19\right)\right)â‰\yen \left(−1\right)\left(1\right)\\ m+19â‰\yen −1\\ m+19−19â‰\yen −1−19\\ mâ‰\yen −20\\ m∈\left[−20,∞\right)\end{array}$

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$m+19≤1$ and$m+19â‰\yen −1$ to find the solution of the inequality $\left|m+19\right|≤1$.

The intersection of the solutions of the inequalities$m+19≤1$ and$m+19â‰\yen −1$ is:

$m∈\left[−20,−18\right]$

Therefore, the solution of the inequality$\left|m+19\right|≤1$ is $m∈\left[−20,−18\right]$.

The graph of the solution set which is$m∈\left[−20,−18\right]$ is: