Compound Inequalities

Compound inequalities join two conditions.

AND ("between"): −3 < 2x + 1 < 7. Do the same operation to all three parts at once. Subtract 1: −4 < 2x < 6. Divide by 2: −2 < x < 3. The answer is the overlap, a single segment.

OR: x < −2 or x > 5. Solve each separately. The answer is both pieces, going outward in two directions.

Quick way to tell which you have: "between" and "and" produce one connected chunk; "or" usually produces two pieces heading opposite ways.

In interval notation, AND gives one interval like (−2, 3), and OR gives a union like (−∞, −2) ∪ (5, ∞).

A conjunction is an intersection of solution sets, a disjunction is a union. Three-part notation a < f(x) < b is legitimate shorthand for a conjunction only; there is no corresponding shorthand for a disjunction, which is why x < −2 or x > 5 must never be compressed into 5 < x < −2.

Operating on all three parts simultaneously is valid because each operation is applied to the whole conjunction. As always, multiplying or dividing by a negative reverses *both* inequality signs: −4 < −2x < 6 becomes 2 > x > −3, conventionally rewritten in increasing order as −3 < x < 2.

Degenerate cases are worth recognising. A conjunction with no overlap (x > 5 and x < 1) is empty. A disjunction covering everything (x < 5 or x > 1) is all of ℝ. Both occur when a problem is set up to test whether you are tracking the logic rather than pattern-matching the notation.