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.

Learn how to solve the absolute value inequality |2x−10|≥16 with a step-by-step solution. This problem explores key algebraic concepts such as absolute value inequalities, including the breakdown into two linear inequalities and the use of basic algebraic manipulation. Suitable for students in Grades 8-10.

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