To solve the inequality $|3x - 6| > 3$, we need to break it into two cases based on the definition of absolute values.

Case 1: $3x - 6 > 3$

Solve for $x$:

$3x - 6 > 3$

Add 6 to both sides:

$3x > 9$

Now, divide both sides by 3:

$x > 3$

Case 2: $3x - 6 < -3$

Solve for $x$:

$3x - 6 < -3$

Add 6 to both sides:

$3x < 3$

Now, divide both sides by 3:

$x < 1$

Final solution:

The solution to $|3x - 6| > 3$ is:

$x < 1 \quad \text{or} \quad x > 3$

In interval notation, the solution is:

$(-\infty, 1) \cup (3, \infty)$

Would you like more details or have any questions about this?

Here are 5 related questions:

1. What is the solution to $|2x + 5| > 7$?
2. How would the solution change if the inequality was $|3x - 6| \geq 3$?
3. What is the general approach for solving absolute value inequalities?
4. How do you solve $|4x - 1| \leq 9$?
5. How would you graph the solution to $|3x - 6| > 3$?

Tip: When solving absolute value inequalities, remember that they split into two separate cases—one positive and one negative!