Question

Find the solution to −3|x+1|≥−12. Provide your answer using interval notation.(1 point)%0D%0AThe solution to −3|x+1|≥−12%0D%0A−%0D%0A3%0D%0A|%0D%0A𝑥%0D%0A+%0D%0A1%0D%0A|%0D%0A≥%0D%0A−%0D%0A12%0D%0A is [−9,9

Answer 1

To solve the inequality \(-3|x+1| \geq -12\), we can start by dividing both sides by \(-3\). Remember that when we divide an inequality by a negative number, we must reverse the inequality sign:

\[
|x + 1| \leq 4
\]

Next, we interpret the absolute value inequality \( |x + 1| \leq 4 \). This means that the expression inside the absolute value lies between \(-4\) and \(4\):

\[
-4 \leq x + 1 \leq 4
\]

We can solve this compound inequality by separating it into two inequalities:

1. From \( -4 \leq x + 1 \):
\[
-4 - 1 \leq x \implies -5 \leq x
\]

2. From \( x + 1 \leq 4 \):
\[
x \leq 4 - 1 \implies x \leq 3
\]

Combining the results, we get:

\[
-5 \leq x \leq 3
\]

In interval notation, the solution is:

\[
[-5, 3]
\]