To solve the inequality $|2x - 10| \geq 16$, we need to consider the definition of absolute value. Recall that for any expression $|A| \geq B$, where $B$ is a positive number, the solution breaks into two cases:

1. $A \geq B$
2. $A \leq -B$

In this case, $A = 2x - 10$ and $B = 16$. Thus, we have two inequalities:

1. $2x - 10 \geq 16$
2. $2x - 10 \leq -16$

Solving the first inequality:

$2x - 10 \geq 16$

Add 10 to both sides:

$2x \geq 26$

Now, divide by 2:

$x \geq 13$

Solving the second inequality:

$2x - 10 \leq -16$

Add 10 to both sides:

$2x \leq -6$

Now, divide by 2:

$x \leq -3$

Final Solution:

The solution to the inequality is:

$x \geq 13 \quad \text{or} \quad x \leq -3$

In interval notation, the solution is:

$(-\infty, -3] \cup [13, \infty)$

Would you like further details or have any questions?

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Here are 5 related questions:

1. What is the difference between solving an absolute value inequality and a regular inequality?
2. How would the solution change if the inequality were $|2x - 10| \leq 16$?
3. What happens if the absolute value inequality involves a quadratic expression inside the absolute value?
4. Can you graph the solution set of this inequality on a number line?
5. How do you solve compound inequalities that involve absolute values?

Tip: Always remember to break absolute value inequalities into two cases when solving.