The inequality given is:

$|3x + 2| + 1 \leq 9$

To solve for $x$, we can isolate the absolute value term and then split it into two cases.

Step 1: Isolate the Absolute Value

Subtract 1 from both sides:

$|3x + 2| \leq 8$

Step 2: Remove the Absolute Value by Creating Two Cases

For an absolute value inequality $|A| \leq B$, we have:

$-B \leq A \leq B$

So, we set up two inequalities:

1. $3x + 2 \leq 8$
2. $3x + 2 \geq -8$

Step 3: Solve Each Inequality

Inequality 1: $3x + 2 \leq 8$

Subtract 2 from both sides:

$3x \leq 6$

Then divide by 3:

$x \leq 2$

Inequality 2: $3x + 2 \geq -8$

Subtract 2 from both sides:

$3x \geq -10$

Then divide by 3:

$x \geq -\frac{10}{3}$

Step 4: Combine the Results

The solution to the inequality is:

$-\frac{10}{3} \leq x \leq 2$

Final Answer

$x \in \left[ -\frac{10}{3}, 2 \right]$

Would you like more details on any step, or do you have any questions?

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Here are some related questions for further practice:

1. How would you solve an inequality with a different absolute value term, such as $|2x - 5| + 3 \leq 7$?
2. What is the process for solving an inequality involving two absolute values, like $|x - 1| + |x + 3| \leq 10$?
3. How do we solve a strict inequality like $|4x - 1| < 5$?
4. What if the inequality was $|3x + 2| - 1 \geq 9$? How would the solution change?
5. Can you explain why we split the absolute value inequality into two cases?

Tip: Remember that for absolute value inequalities of the form $|A| \leq B$, you can split it into $-B \leq A \leq B$ as long as $B \geq 0$.