To solve the inequality $|5x - 8| > 2x + 1$, we need to break it into two cases since it involves an absolute value. The definition of absolute value leads to two scenarios: one where the expression inside the absolute value is positive and one where it is negative.

Case 1: $5x - 8 > 0$

Here, $|5x - 8| = 5x - 8$, so the inequality becomes:

$5x - 8 > 2x + 1$

Now, solve for $x$:

$5x - 2x > 1 + 8$

$3x > 9$

$x > 3$

Case 2: $5x - 8 < 0$

In this case, $|5x - 8| = -(5x - 8) = -5x + 8$, so the inequality becomes:

$-5x + 8 > 2x + 1$

Now, solve for $x$:

$-5x - 2x > 1 - 8$

$-7x > -7$

$x < 1 \quad \text{(Remember to flip the inequality when dividing by a negative number)}$

Combine the results:

• From Case 1, we found $x > 3$.
• From Case 2, we found $x < 1$.

Thus, the solution to the inequality $|5x - 8| > 2x + 1$ is:

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

This can be written in interval notation as:

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

Would you like further details or clarifications on any of the steps?

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Here are some related questions you might find interesting:

1. How do you solve absolute value equations like $|x - 5| = 3$?
2. What happens if the inequality is $|5x - 8| \leq 2x + 1$?
3. Can this method be applied to inequalities with more complex expressions inside the absolute value, like $|2x + 3| > 4x - 1$?
4. How do you graph absolute value inequalities on a number line?
5. What is the importance of considering both positive and negative cases when solving absolute value inequalities?

Tip: Always remember to flip the inequality when dividing or multiplying by a negative number.