To solve the inequality:

$|x| + 3 > 9$

we'll isolate the absolute value expression and consider both possible cases for $|x|$ being greater than a value.

Step 1: Isolate $|x|$

Subtract 3 from both sides:

$|x| > 6$

Step 2: Break into Two Cases

The inequality $|x| > 6$ means that $x$ is either greater than 6 or less than -6. So we have:

1. $x > 6$
2. $x < -6$

Solution

The solution to the inequality is:

$x > 6 \quad \text{or} \quad x < -6$

In interval notation, this can be written as:

$x \in (-\infty, -6) \cup (6, \infty)$

Would you like further details or have any questions?

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Here are some related questions that expand on this concept:

1. How do we solve inequalities that involve both absolute values and quadratic expressions?
2. What changes if we have $|x| + 3 < 9$ instead?
3. How do we solve compound inequalities involving absolute values, such as $|x - 2| + 3 > 9$?
4. What is the graphical interpretation of absolute value inequalities?
5. How can absolute value inequalities be applied in real-world contexts?

Tip: Remember that absolute values produce two cases in inequalities, corresponding to both positive and negative scenarios.