Let's solve the inequality:
$4 \cdot |x + 4| + 6 > 18$

Step 1: Isolate the absolute value expression

First, subtract 6 from both sides: $4 \cdot |x + 4| > 12$

Step 2: Divide both sides by 4

$|x + 4| > 3$

Step 3: Split into two cases (definition of absolute value)

The expression $|x + 4| > 3$ translates into two inequalities:

1. $x + 4 > 3$
2. $x + 4 < -3$

Step 4: Solve both inequalities

Case 1: $x + 4 > 3$

$x > 3 - 4$ $x > -1$

Case 2: $x + 4 < -3$

$x < -3 - 4$ $x < -7$

Step 5: Combine the solutions

The solution to the inequality $|x + 4| > 3$ is: $x > -1 \quad \text{or} \quad x < -7$

Step 6: Write the final solution in interval notation

$x \in (-\infty, -7) \cup (-1, \infty)$

This is the solution to the original inequality.

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Do you want any clarifications or further details?
Here are five related questions:

1. What if the inequality was $|x + 4| < 3$?
2. How does the solution change if we replace $>$ with $\geq$?
3. Can you solve a similar problem with two absolute values, like $|x - 2| + |x + 3| > 7$?
4. What happens if the absolute value is multiplied by a negative number?
5. How do we handle absolute value inequalities when dealing with quadratic terms?

Tip: Always be careful with the direction of inequalities when working with absolute values—splitting into two cases is essential!