Inequalities • Topic 4 of 4

Absolute-Value Inequalities

Absolute value measures distance from zero, so an absolute-value inequality is really a statement about distance — and it splits into cases. |x| < k (with k > 0) means x is within k of zero, which is the compound band −k < x < k. |x| > k means x is more than k from zero, giving two separate regions: x < −k OR x > k. Shift the centre for expressions: |x − 3| < 5 means x is within 5 of 3, i.e. −2 < x < 8. The two directions behave oppositely — "less than" collapses to a single interval (AND), while "greater than" opens up two rays (OR). One boundary condition worth remembering: if k is negative, |x| < k has no solution (a distance can't be negative) and |x| > k is true for all x. The GRE tests exactly these edge cases.

A number line showing the absolute value inequality x is at most 3 as the closed interval from -3 to 3|x| ≤ 3-5-4-3-2-1012345Same as -3 ≤ x ≤ 3 (both ends included).

✅ Solved examples

1. Solve |x| < 6.
Distance from 0 under 6: −6 < x < 6.
2. Solve |x − 3| ≤ 5.
Within 5 of 3: 3 − 5 ≤ x ≤ 3 + 5, so −2 ≤ x ≤ 8.
3. Solve |2x + 1| > 7.
Split: 2x + 1 > 7 OR 2x + 1 < −7. First gives x > 3; second gives 2x < −8 ⇒ x < −4. So x > 3 or x < −4.
4. How many solutions does |x + 2| < −3 have?
An absolute value is never negative, so it can never be less than −3. No solution.

✏️ Practice — try these, take hints as needed

1. Solve |x| ≤ 4.
Within 4 of zero.
Compound band.
−4 ≤ x ≤ 4.
−4 ≤ x ≤ 4
2. Solve |x − 1| < 3.
Within 3 of 1.
1 − 3 < x < 1 + 3.
Simplify bounds.
−2 < x < 4
3. Solve |x + 4| ≥ 6.
Two regions.
x + 4 ≥ 6 OR x + 4 ≤ −6.
Solve each.
x ≥ 2 or x ≤ −10
4. Solve |3x| < 12.
−12 < 3x < 12.
Divide all parts by 3.
Keep the band.
−4 < x < 4
5. For how many real x is |x − 5| > −1 true?
Absolute value is ≥ 0.
Always greater than −1.
Every real number works.
All real numbers

📝 Topic test — 8 questions

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