We are given the equation:

\[
5 \cdot |6 + 3x| - 8 = 52
\]

### Step 1: Isolate the absolute value term
First, add 8 to both sides to eliminate the \(-8\):

\[
5 \cdot |6 + 3x| = 52 + 8
\]
\[
5 \cdot |6 + 3x| = 60
\]

### Step 2: Divide by 5
Now, divide both sides by 5 to simplify:

\[
|6 + 3x| = \frac{60}{5}
\]
\[
|6 + 3x| = 12
\]

### Step 3: Set up two cases for the absolute value
Since \(|6 + 3x| = 12\), this means:
1. \(6 + 3x = 12\)
2. \(6 + 3x = -12\)

#### Case 1: \(6 + 3x = 12\)
\[
3x = 12 - 6
\]
\[
3x = 6
\]
\[
x = \frac{6}{3} = 2
\]

#### Case 2: \(6 + 3x = -12\)
\[
3x = -12 - 6
\]
\[
3x = -18
\]
\[
x = \frac{-18}{3} = -6
\]

### Step 4: Final answer
The solutions are:
\[
x = 2 \quad \text{or} \quad x = -6
\]

Would you like more details on any of these steps?

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Here are 5 related questions for further practice:
1. Solve \(3|2x + 1| - 5 = 13\).
2. Solve \(7|x - 4| + 9 = 30\).
3. Solve \(|4x + 7| + 6 = 15\).
4. Solve \(2|x + 5| - 3 = 7\).
5. Solve \(|x - 3| = 9\).

**Tip:** When solving absolute value equations, always remember to set up two cases: one for the positive and one for the negative value.

Learn how to solve the absolute value equation 5|6 + 3x| − 8 = 52 with a detailed, step-by-step explanation. This algebra problem covers key concepts such as isolating the absolute value, setting up two cases, and solving linear equations. Suitable for students in Grades 8-10.

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