MISSION · Algebra
Compound Inequalities
Two conditions at once, and whether you need both or either
LEVEL 1 · EXPLORE
Make the idea move
Change the controls and watch what happens. Follow your curiosity and see what you notice.
Loading the activity…
LEVEL 2 · LEARN
Unlock the explanation
Open the chapters in any order. Try the examples as you go.
The idea in one sentence
A compound inequality is two inequalities glued together, and the glue — and or or — decides whether you keep the overlap of the two answers or both of them.
AND keeps the overlap
”\(x > -2\) and \(x < 4\)” asks for numbers that satisfy both at once. Graph each on a number line and keep only the stretch where they agree: everything between \(-2\) and \(4\).
Because this happens so often, it gets its own shorthand:
\[-2 < x < 4\]Read it as “\(x\) is between \(-2\) and \(4\)”. The small number goes on the left and both inequality signs point the same way — writing \(4 < x < -2\) describes nothing, since no number is both.
In interval notation it is \((-2, 4)\), and in set language it is the intersection of the two solution sets.
OR keeps both pieces
”\(x < -2\) or \(x > 4\)” asks for numbers satisfying at least one. The graph is two separate arrows pointing away from each other, with a gap in the middle.
There is no three-part shorthand for this one, and trying to write \(4 < x < -2\) for it is the classic error. Leave it as two statements joined by “or”, or write \((-\infty, -2) \cup (4, \infty)\) — the union.
A quick way to tell them apart on sight: “and” makes a segment, “or” makes everything except a segment.
Solving a three-part one
Whatever you do, do it to all three parts.
\[1 < x + 3 < 7\]Subtract 3 everywhere:
\[-2 < x < 4\]That is the full solution — no need to split it up. The three-part form works whenever the variable appears only in the middle.
With a coefficient, the one extra rule still applies:
\[-6 \le -2x < 8\]Dividing all three parts by \(-2\) flips both signs, and the two ends swap places:
\(3 \ge x > -4\), which is tidier written as \(-4 < x \le 3\).
Flipping only one of the two signs is the most common slip on this topic, and it produces an answer that is quietly impossible.
Absolute value gives you one of each
This is where the two kinds turn up naturally, and the rule follows from reading \(|x|\) as a distance.
\(|x - 3| < 5\) says “\(x\) is within 5 of 3” — a single stretch, so it is an and: \(-5 < x - 3 < 5\), giving \(-2 < x < 8\).
\(|x - 3| > 5\) says “\(x\) is more than 5 away from 3” — you can be far off in either direction, so it is an or: \(x - 3 < -5\) or \(x - 3 > 5\), giving \(x < -2\) or \(x > 8\).
Less-than gives and; greater-than gives or. If you forget, sketch the distance on a number line and read it off.
Two answers worth recognising
Some compound inequalities collapse.
”\(x > 5\) and \(x < 2\)” has no overlap at all, so the solution is the empty set — written \(\varnothing\). Nothing is wrong with your algebra; the answer really is “no solutions”.
”\(x > 5\) or \(x < 2\)“… still leaves a gap, but “\(x < 5\) or \(x > 2\)” covers every real number, because every number satisfies at least one. The answer is all reals.
Getting one of these should prompt a check, not a panic.
Common mix-ups
- Writing an “or” as a three-part inequality. Only “and” compresses.
- Flipping one sign of three when multiplying or dividing by a negative.
- Putting the bigger number on the left in the compressed form.
- Operating on only two of the three parts.
-
**Reading $$ x > 5$$ as an “and”.** Greater-than splits outward.
Check yourself
Solve \(1 < x + 3 < 7\), and say what kind of solution set an “or” inequality has.
Subtracting 3 from all three parts gives \(-2 < x < 4\), so the bounds are \(-2\) and \(4\). An “or” keeps both pieces, so its solution set is the union.
LEVEL 3 · REFLECT
Add this quest to your trail
When you feel ready, mark this idea as explored. You can always come back and try it again.
EXPLORER STRATEGIES