https://file.aiquickdraw.com/mathbot/file/1726008481786pk0wlv5c.jpg

The inequality provided in the image is:

\[
|5 - 3x| + 2 < 6
\]

Let's solve this step by step.

### Step 1: Isolate the absolute value
First, subtract 2 from both sides:

\[
|5 - 3x| < 6 - 2
\]

\[
|5 - 3x| < 4
\]

### Step 2: Set up two inequalities from the absolute value
The inequality \( |5 - 3x| < 4 \) means that:

\[
-4 < 5 - 3x < 4
\]

### Step 3: Solve the compound inequality
Now, solve each part:

1. For \( -4 < 5 - 3x \):
   - Subtract 5 from both sides:
     \[
     -4 - 5 < -3x
     \]
     \[
     -9 < -3x
     \]
   - Divide by -3 (remember to reverse the inequality):
     \[
     \frac{-9}{-3} > x
     \]
     \[
     3 > x \quad \text{or} \quad x < 3
     \]

2. For \( 5 - 3x < 4 \):
   - Subtract 5 from both sides:
     \[
     -3x < 4 - 5
     \]
     \[
     -3x < -1
     \]
   - Divide by -3 (again, reverse the inequality):
     \[
     x > \frac{-1}{-3}
     \]
     \[
     x > \frac{1}{3}
     \]

### Step 4: Combine the two results
The solution to the inequality is:

\[
\frac{1}{3} < x < 3
\]

Thus, \( x \) lies between \( \frac{1}{3} \) and \( 3 \).

---

Would you like further details or have any questions?

Here are 5 related questions to explore further:
1. What happens if the inequality was \( |5 - 3x| + 2 \leq 6 \)?
2. How do you solve absolute value inequalities with a greater than sign, like \( |5 - 3x| > 4 \)?
3. What are some practical applications of solving absolute value inequalities?
4. How would the solution change if the absolute value expression involved more complex terms?
5. Can absolute value inequalities have no solution? If so, under what conditions?

**Tip:** When solving absolute value inequalities, always remember to split the inequality into two parts—one positive and one negative—then solve each part separately.

Step-by-step guide to solving the absolute value inequality |5 - 3x| + 2 < 6. This problem explains how to isolate the absolute value expression and split it into two linear inequalities. Suitable for high school algebra students.

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