Inequalities • Topic 3 of 4

Compound Inequalities

A compound inequality traps a variable between two bounds, like −3 < 2x + 1 ≤ 7. Solve it by doing the same operation to all three parts at once: subtract 1 everywhere to get −4 < 2x ≤ 6, then divide everything by 2 to get −2 < x ≤ 3. The variable now lives in an interval. If a step involves multiplying or dividing all parts by a negative, both inequality signs flip and the bounds swap ends. Distinguish "AND" from "OR": a < x < b is an AND — x must satisfy both bounds simultaneously — while an OR condition (x < a or x > b) describes two separate regions. On the GRE, compound inequalities often set the range for a Quantitative Comparison; knowing the full interval tells you whether a quantity is pinned down or genuinely varies.

A number line showing the bounded interval from -1 up to 4, with a closed endpoint at -1 and an open endpoint at 4-1 ≤ x < 4-5-4-3-2-1012345Closed at -1 (included), open at 4 (excluded).

✅ Solved examples

1. Solve −3 < 2x + 1 ≤ 7.
Subtract 1 throughout: −4 < 2x ≤ 6. Divide by 2: −2 < x ≤ 3.
2. Solve 1 ≤ 3x − 2 < 10.
Add 2 throughout: 3 ≤ 3x < 12. Divide by 3: 1 ≤ x < 4.
3. Solve −6 ≤ −2x < 4.
Divide all parts by −2 and FLIP both signs (and swap ends): 3 ≥ x > −2, i.e. −2 < x ≤ 3.
4. If −1 < x < 4, what is the range of 2x + 3?
Multiply the interval by 2: −2 < 2x < 8. Add 3: 1 < 2x + 3 < 11.

✏️ Practice — try these, take hints as needed

1. Solve −5 < x + 2 < 5.
Subtract 2 from all parts.
−7 < x < 3.
That is the interval.
−7 < x < 3
2. Solve 4 ≤ 2x ≤ 10.
Divide all parts by 2.
Keep signs (positive divisor).
2 ≤ x ≤ 5.
2 ≤ x ≤ 5
3. Solve 0 < 3x − 6 ≤ 9.
Add 6 to all parts.
6 < 3x ≤ 15.
Divide by 3.
2 < x ≤ 5
4. Solve −4 ≤ −x < 2.
Multiply all parts by −1.
Flip both signs and swap ends.
4 ≥ x > −2.
−2 < x ≤ 4
5. If 2 < x < 6, find the range of 3x − 1.
Multiply interval by 3.
6 < 3x < 18.
Subtract 1.
5 < 3x − 1 < 17

📝 Topic test — 8 questions

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