Inequalities containing the absolute value |x| become easier to understand when considered as distances from the origin.
Basic Patterns
When a > 0
・|x| < a ⇔ −a < x < a
・|x| > a ⇔ x < −a or x > a
・|x| ≦ a ⇔ −a ≦ x ≦ a
・|x| ≧ a ⇔ x ≦ −a or x ≧ a
Example ①
Solve |x − 2| < 3.
Solution:−3 < x − 2 < 3. Adding 2 to each part gives −1 < x < 5
Solve |x − 2| < 3.
Solution:−3 < x − 2 < 3. Adding 2 to each part gives −1 < x < 5
Example ②
Solve |2x + 1| ≧ 5.
Solution:2x + 1 ≦ −5 or 2x + 1 ≧ 5。
Former: 2x ≦ −6 → x ≦ −3
Latter: 2x ≧ 4 → x ≧ 2
Therefore, x ≦ −3 or x ≧ 2
Solve |2x + 1| ≧ 5.
Solution:2x + 1 ≦ −5 or 2x + 1 ≧ 5。
Former: 2x ≦ −6 → x ≦ −3
Latter: 2x ≧ 4 → x ≧ 2
Therefore, x ≦ −3 or x ≧ 2
Key Points
- |x| < a → −a < x < a (between)
- |x| > a → x < −a or x > a (outside)
- Drawing the solution as a distance from the origin makes it easier to visualize
Practice Problems
- Solve |x| ≦ 4.
- Solve |3x − 2| < 7.
- Solve |x + 5| > 3.
Solutions and Explanations
- Solution:−4 ≦ x ≦ 4
Explanation: This is in the form |x| ≦ a. - Solution:−5/3 < x < 3
Explanation: −7 < 3x − 2 < 7 → −5 < 3x < 9 → −5/3 < x < 3. - Solution:x < −8 or x > −2
Explanation: x + 5 < −3 or x + 5 > 3.