Answer:
Graph D
Step-by-step explanation:
Given inequality:
[tex]3|x+1| < 9[/tex]
Divide both sides by 3:
[tex]\implies |x+1| < 3[/tex]
[tex]\textsf{Apply the absolute rule: \quad If $|u| < a$ when $a > 0$, then $-a < u < a$}[/tex]
[tex]\begin{aligned} \underline{\sf Case\; 1} && \underline{\sf Case\; 2}\\-3& < x+1 & \quad \quad \quad x+1& < 3\\-4& < x & x& < 2\end{aligned}[/tex]
Therefore, the solution set for the given inequality is:
[tex]\{x|-4 < x < 2\}[/tex]
When graphing inequalities on a number line:
- < or > : open circle
- ≤ or ≥ : close circle.
- < or ≤ : shade to the left of the circle.
- > or ≥ : shade to the right of the circle.
Place an open circle at x = -4 and x = 2.
Shade between the two values.
