Given the inequality::
[tex]3|x+1|<6[/tex]Let's solve the inequality for x and graph.
To solve for x, first divide both sides of the inequality by 3:
[tex]\begin{gathered} \frac{3|x+1|}{3}<\frac{6}{3} \\ \\ |x+1|<2 \end{gathered}[/tex]Since the left side is an absolute value, we have two possible solutions:
[tex]\begin{gathered} x+1<2 \\ \\ AND \\ -(x+1)<2 \end{gathered}[/tex]Let's solve each inequality for x:
[tex]\begin{gathered} x+1<2 \\ \text{ Subtract 1 from both sides:} \\ x+1-1<2-1 \\ x<1 \end{gathered}[/tex]For the second inequality:
[tex]\begin{gathered} -(x+1)<2 \\ -x-1<2 \\ \text{ Add 1 to both sides:} \\ -x-1+1<2+1 \\ -x<3 \\ Divide\text{ both sides by -1:} \\ \frac{-x}{-1}<\frac{3}{-1} \\ \\ x>-3 \end{gathered}[/tex]Hence, we have the solutions:
x < 1 and x > -3
Therefore, the solution is:
-3 < x < 1
The graph of the inequality is shown below:
ANSWER:
-3 < x < 1
