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To solve the inequality \( |3x - 2| \leq 4 \), we need to remove the absolute value by breaking it down into two separate inequalities.

The inequality \( |3x - 2| \leq 4 \) means:
\[
-4 \leq 3x - 2 \leq 4
\]

Now, we will solve for \( x \) by isolating it in both inequalities.

### Step 1: Solve the Left Inequality
\[
-4 \leq 3x - 2
\]
Add 2 to both sides:
\[
-2 \leq 3x
\]
Now, divide by 3:
\[
-\frac{2}{3} \leq x
\]

### Step 2: Solve the Right Inequality
\[
3x - 2 \leq 4
\]
Add 2 to both sides:
\[
3x \leq 6
\]
Now, divide by 3:
\[
x \leq 2
\]

### Step 3: Combine the Results
The solution to the inequality is:
\[
-\frac{2}{3} \leq x \leq 2
\]

### Final Answer in Interval Notation
In interval notation, the solution is:
\[
\left[ -\frac{2}{3}, 2 \right]
\]

So, the correct choice is **A**: The solution set is \( \left[ -\frac{2}{3}, 2 \right] \).

---

Would you like further details or have any questions?

Here are some related questions for further practice:
1. How would you solve \( |4x + 5| \leq 7 \) in interval notation?
2. What are the steps for solving \( |2x - 3| > 5 \)?
3. How do you interpret absolute value inequalities geometrically?
4. What are the differences between strict inequalities \( <, > \) and inclusive inequalities \( \leq, \geq \) in absolute value equations?
5. How would you convert the interval notation \( (-\infty, -1] \cup [3, \infty) \) into inequality form?

**Tip:** Always remember that absolute value inequalities \( |ax + b| \leq c \) are solved by creating two separate inequalities: \( -c \leq ax + b \leq c \).

Solve the inequality. Write the solution in interval notation: |3x - 2| ≤ 4.

Learn how to solve the absolute value inequality |3x - 2| ≤ 4 and express the solution in interval notation. This problem covers important concepts in algebra, including inequalities, absolute values, and interval notation, suitable for Grades 8-10.

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