A compound inequality combines two conditions. An "and" inequality like −3 ≤ 2x − 1 < 5 means both hold at once; solve by doing the same operation to all three parts, giving a bounded range such as −1 ≤ x < 3. An "or" inequality is satisfied when either part holds and usually gives two separate rays. For "and" problems keep the variable in the middle and isolate it by adding then dividing across all parts. The SAT uses compound inequalities for "between" situations and for combining constraints in word problems.
✅ Solved examples
1. Solve −3 ≤ 2x − 1 < 5.
Add 1 to all parts: −2 ≤ 2x < 6; divide by 2: −1 ≤ x < 3.
2. Solve 1 < x + 4 < 7.
Subtract 4 from all parts: −3 < x < 3.
3. Solve 4 ≤ 2x ≤ 10.
Divide all parts by 2: 2 ≤ x ≤ 5.
4. Solve 0 < 3x < 9.
Divide all parts by 3: 0 < x < 3.
✏️ Practice — try these, take hints as needed
1. Solve −2 ≤ x + 1 ≤ 4.
Subtract 1 from all parts.
−3 ≤ x ≤ 3.
—
−3 ≤ x ≤ 3.
2. Solve 6 < 2x < 14.
Divide all parts by 2.
3 < x < 7.
—
3 < x < 7.
3. Solve −5 < 2x − 1 < 3.
Add 1 to all parts: −4 < 2x < 4.
Divide by 2.
—
−2 < x < 2.
4. Solve 2 ≤ x − 3 ≤ 6.
Add 3 to all parts.
5 ≤ x ≤ 9.
—
5 ≤ x ≤ 9.
5. Solve 9 < 3x ≤ 18.
Divide all parts by 3.
3 < x ≤ 6.
—
3 < x ≤ 6.
📝 Topic test — 8 questions
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multiplying or dividing by a negative reverses the inequality
Symbols
< ≤ > ≥
Notation
Open interval
(a, b): endpoints excluded (< >)
Closed interval
[a, b]: endpoints included (≤ ≥)
Compound (and)
a < x < b
Number line
open circle = strict, closed = inclusive
Digital SAT reference
Area & Circumference
Circle area
A = πr²
Circle circumference
C = 2πr
Rectangle
A = ℓw
Triangle
A = ½ b h
Volume
Rectangular box
V = ℓwh
Cylinder
V = πr²h
Sphere
V = 4⁄3 πr³
Cone
V = 1⁄3 πr²h
Pyramid
V = 1⁄3 ℓwh
Right triangles
Pythagorean theorem
a² + b² = c²
30°–60°–90°
sides x : x√3 : 2x
45°–45°–90°
sides s : s : s√2
Constants
Degrees in a circle
360°
Radians in a circle
2π
Angles of a triangle
sum = 180°
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