Compound inequalities — AND, OR: Worked examples

Intuition What this page is

The parent note taught you the idea: AND takes the overlap (intersection), OR takes everything covered (union). This page is a shooting range. We list every kind of shot the topic can fire at you — every sign, every empty case, single unbounded rays, "covers-the-whole-line" traps, a real word problem, and an exam twist — then we work one example for each so no scenario can surprise you.

Before we start, four plain-word reminders (so every symbol below is already earned):

A number line is a ruler stretching left (negative) to right (positive), with $0$ in the middle. See Number line representation.
The symbol $\infty$ (infinity) is not a number — it is shorthand for "keeps going forever with no right end". So $(3, \infty)$ means "all numbers past $3$ , never stopping". The round bracket next to $\infty$ is always round because you can never reach forever.
$(a, b)$ means "all numbers strictly between $a$ and $b$ " — round bracket = excluded endpoint. $[a, b]$ with square brackets = endpoint included. See Interval notation.
$\emptyset$ is the empty set — a box with nothing in it, meaning "no number works".
$R$ (the real numbers) is the name for every number on the whole number line at once — positives, negatives, zero, fractions, decimals, all of it. When an answer is $R$ it means "every number works, no exceptions".

Two more symbols we will earn as we go (see Set operations — union and intersection):

$\cap$ (intersection, "cap") is the picture of two shaded regions overlapping — it keeps only the patch shaded by both. Think of the pointed cap ∩ as a narrow tent covering just the shared middle. This is what AND produces.
$\cup$ (union, "cup") is the picture of pouring both regions into one cup — it keeps everything shaded by at least one. This is what OR produces.

The scenario matrix

Every problem this topic throws lives in exactly one of these cells. We will hit all of them.

Cell	Connective	Situation	What the answer looks like
C1	AND	two rays that do overlap (point toward each other)	a single bounded interval $[a, b)$ etc.
C2	AND	two rays pointing apart	$\emptyset$ (no solution)
C3	AND	boundaries touch at one point (degenerate)	a single point ${c}$
C4	OR	two disjoint rays	two pieces $(- \infty, a) \cup (b, \infty)$
C5	OR	rays point toward each other and overlap → flood	all reals $R$
C5b	OR / AND	same-direction (nested) rays → one unbounded ray	$(a, \infty)$ or $(- \infty, a)$
C6	AND	three-part $a < f (x) < b$ , positive coefficient	single interval
C7	AND	three-part, divide by a negative (flip both)	single interval, order reversed
C8	AND	real-world word problem	interval + units
C9	OR	exam twist — hidden absolute value $	x
C10	AND	exam twist — hidden absolute value $	x

The examples

Worked example Example 1 — Cell C1 (AND, rays overlap)

Solve $x \geq - 2$ and $x < 4$ .

Forecast: guess now — will this be one interval, empty, or two pieces?

Write each condition as its own ray: $A = [- 2, \infty)$ , $B = (- \infty, 4)$ . Why this step? Each single inequality is a half-line on the number line; naming them lets us hunt the overlap.
AND = intersection $\cap$ = where both shading colours land on the same spot. That's numbers $\geq - 2$ and $< 4$ . Why this step? Intersection is defined as "in both", which is exactly what AND means (see Set operations — union and intersection).
Read the overlap: from $- 2$ (included, since $\geq$ ) up to $4$ (excluded, since $<$ ) → $[- 2, 4)$ . Why this step? The overlap starts at the larger left-boundary and ends at the smaller right-boundary — that shared middle strip is $A \cap B$ .

Figure below (s01): a chalkboard number line from $- 5$ to $6$ . The blue ray is $A = [- 2, \infty)$ (filled dot at $- 2$ , running right), the pink ray is $B = (- \infty, 4)$ (hollow dot at $4$ , running left). The thick yellow strip drawn on the axis itself is exactly the overlap $[- 2, 4)$ — filled dot at $- 2$ , hollow dot at $4$ — with an arrow labelling it "overlap = [-2, 4)".

Verify: test $x = 0$ → $0 \geq - 2$ ✓ and $0 < 4$ ✓, and $0 \in [- 2, 4)$ ✓. Test $x = 4$ → $4 < 4$ ✗, correctly excluded. Answer $[- 2, 4)$ .

Figure — Compound inequalities — AND, OR

Worked example Example 4 — Cell C4 (OR, disjoint → two pieces)

Solve $x \leq - 1$ or $x > 2$ .

Forecast: one blob, two blobs, or the whole line?

$A = (- \infty, - 1]$ , $B = (2, \infty)$ . Between $- 1$ and $2$ there is a gap neither ray covers. Why this step? OR keeps everything lit by at least one ray; the gap in the middle stays dark.
Union $\cup$ = both pieces kept, nothing merged. Why this step? Union is defined as "in at least one set" — no need for overlap.
Answer: $A \cup B = (- \infty, - 1] \cup (2, \infty)$ . No compact $a < x < b$ form exists (that form is AND-only). Why this step? Because the two rays are separated by a gap, no single interval can name them; we must join two pieces with $\cup$ .

Figure below (s02): the same chalkboard number line. The blue ray $A = (- \infty, - 1]$ runs left with a filled dot at $- 1$ ; the pink ray $B = (2, \infty)$ runs right with a hollow dot at $2$ . On the axis, two separate yellow pieces sit with a dark gap between $- 1$ and $2$ — an arrow points at that gap labelled "dark gap: neither ray", i.e. exactly the numbers no ray lit.

Verify: $x = 0$ → $0 \leq - 1$ ✗ and $0 > 2$ ✗, so $0$ is excluded, matching the gap. $x = - 1$ → $- 1 \leq - 1$ ✓ included. Answer $(- \infty, - 1] \cup (2, \infty)$ .

Worked example Example 5 — Cell C5 (OR, rays point toward each other → everything)

Solve $x > - 3$ or $x < 7$ .

Forecast: careful — this OR hides a trap. What's the answer?

$A = (- 3, \infty)$ points right, $B = (- \infty, 7)$ points left — they aim toward each other and overlap on $(- 3, 7)$ . Why this step? Direction is everything: when rays face each other and overlap, their union has no gaps.
Pick any real number and ask "is it $> - 3$ OR $< 7$ ?". A number like $100$ is $> - 3$ ✓. A number like $- 100$ is $< 7$ ✓. Everything passes at least one test. Why this step? OR only needs one condition true, and no number can fail both.
Union $A \cup B = R$ (all real numbers — every point on the line). Why this step? Since the two rays cover every point between and beyond them, the cup $\cup$ overflows to fill the entire line.

Verify: try to find a number failing both: it would need $x \leq - 3$ and $x \geq 7$ — impossible. Answer $R$ . (Contrast Example 2: same-looking setup, opposite connective, opposite answer.)

Worked example Example 5b — Cell C5b (same-direction rays → one unbounded ray)

Solve (a) $x > 1$ or $x > 3$ , then (b) $x > 1$ and $x > 3$ .

Forecast: neither answer is $R$ and neither is empty — guess the two rays first.

Both rays point right: $A = (1, \infty)$ , $B = (3, \infty)$ , and $B$ sits entirely inside $A$ . Why this step? When one ray is nested in the other, "overlap" and "at least one" become trivial to read off.
(a) OR: "in at least one" — every number of $B$ is already in $A$ , so the union is just the bigger ray. $A \cup B = (1, \infty)$ . Why this step? Union can never be smaller than its larger piece, and here the larger piece already swallows the smaller.
(b) AND: "in both" — a number must clear the stricter cut $> 3$ to be in both. $A \cap B = (3, \infty)$ . Why this step? Intersection keeps only what survives both filters, so the tighter bound wins.

Verify: $x = 2$ → OR: $2 > 1$ ✓ so included in $(1, \infty)$ ✓; AND: $2 > 3$ ✗ so excluded from $(3, \infty)$ ✓. $x = 4$ → in both answers ✓. Answers: (a) $(1, \infty)$ , (b) $(3, \infty)$ — unbounded rays, not $R$ and not $\emptyset$ .

Worked example Example 7 — Cell C7 (three-part AND, divide by negative → flip both)

Solve $1 < 4 - 3 x \leq 10$ .

Forecast: which way will the signs end up pointing?

Subtract $4$ from all three parts: $- 3 < - 3 x \leq 6$ . Why this step? Isolate the $x$ -term; same operation on every part keeps the chain honest.
Divide all three by $- 3$ . Dividing by a negative flips BOTH relation signs at once: $1 > x \geq - 2$ . Why this step? This is the classic trap from the parent note — flipping only one arrow gives nonsense.
Rewrite in increasing order: $- 2 \leq x < 1$ → $[- 2, 1)$ . Why this step? Left-to-right on a number line should read smallest-to-largest, and re-ordering keeps each endpoint's open/closed status attached to its own number.

Verify: $x = - 2$ → $4 - 3 (- 2) = 10$ , and $1 < 10 \leq 10$ ✓ (endpoint included). $x = 1$ → $4 - 3 (1) = 1$ , and $1 < 1$ ✗ (correctly excluded). Answer $[- 2, 1)$ .

Worked example Example 8 — Cell C8 (real-world AND with units)

A vaccine must be stored at or above $2^{\circ} C$ and strictly below $8^{\circ} C$ . A fridge's temperature is modelled as $T = 5 + 2 h$ where $h$ is hours since power-on ( $h \geq 0$ ). For which $h$ is storage safe?

Forecast: guess the time-window before computing.

Safe means $2 \leq T < 8$ , i.e. $2 \leq 5 + 2 h < 8$ — an AND (both bounds required). Why this step? "at or above and below" is a conjunction → intersection → three-part form.
Subtract $5$ from all parts: $- 3 \leq 2 h < 3$ . Why this step? Same operation on every part; peels the constant off the middle.
Divide all parts by $2$ (positive): $- 1.5 \leq h < 1.5$ . Why this step? Same positive division everywhere; signs unchanged.
Intersect with the physical constraint $h \geq 0$ : $0 \leq h < 1.5$ . Why this step? Physical reality ("hours can't be negative") is itself an extra AND-condition, so we take one more intersection $\cap$ to discard impossible times.

Verify: $h = 0$ → $T = 5$ , and $2 \leq 5 < 8$ ✓. $h = 1.5$ → $T = 8$ , fails $T < 8$ ✗ (correctly the cut-off). Answer $0 \leq h < 1.5$ hours, i.e. $[0, 1.5)$ hours.

Worked example Example 9 — Cell C9 (exam twist, hidden OR:

$∣ x ∣ > k$ ) Solve $∣ x - 1∣ > 3$ .

Forecast: will this be an AND (interval) or an OR (two pieces)?

Read $∣ x - 1∣$ as distance of $x$ from $1$ on the number line. The condition says "distance from $1$ is more than $3$ ". Why this step? Absolute value = distance; that picture tells us which connective appears (see Absolute value inequalities).
Being more than $3$ away means you're far to the left OR far to the right — an OR, hence a union $\cup$ : $x - 1 < - 3$ or $x - 1 > 3$ . Why this step? $∣ \cdot ∣ > k$ always splits into two outward rays.
Solve each and union: $x < - 2$ or $x > 4$ → $(- \infty, - 2) \cup (4, \infty)$ . Why this step? The two outward rays never meet (the region near $1$ is too close), so the answer stays as two $\cup$ -joined pieces.

Figure below (s03): a chalkboard number line from $- 4$ to $6$ . A yellow square marks the centre $1$ ; two dashed chalk lines mark the distance- $3$ boundaries at $- 2$ and $4$ . A blue double-arrow between $1$ and $- 2$ is labelled "distance 3". The two pink rays — one running left from $- 2$ (hollow dot), one running right from $4$ (hollow dot) — are the "too far" region, labelled "x < -2" and "x > 4". The middle strip near $1$ stays dark because those points are too close.

Verify: midpoint $x = 1$ → $∣1 - 1∣ = 0 > 3$ ? No, correctly excluded (it sits in the gap). $x = 5$ → $∣5 - 1∣ = 4 > 3$ ✓. Answer $(- \infty, - 2) \cup (4, \infty)$ .

Worked example Example 10 — Cell C10 (exam twist, hidden AND:

$∣ x ∣ < k$ ) Solve $∣2 x + 4∣ \leq 6$ .

Forecast: one interval or two pieces this time?

$∣2 x + 4∣$ is the distance of $2 x + 4$ from $0$ ; " $\leq 6$ " means "within $6$ of $0$ ". Why this step? Being within a distance is a squeeze from both sides → an AND (see Absolute value inequalities).
Within-distance unpacks to a three-part AND: $- 6 \leq 2 x + 4 \leq 6$ . Why this step? $∣ \cdot ∣ \leq k$ always becomes $- k \leq (\cdot) \leq k$ — the two ends of the squeeze.
Subtract $4$ from all parts: $- 10 \leq 2 x \leq 2$ . Why this step? Same operation to all three parts keeps the AND-chain valid.
Divide all parts by $2$ (positive): $- 5 \leq x \leq 1$ . Why this step? Positive division leaves the $\leq$ signs untouched, and the closed bounds give the single interval $[- 5, 1]$ .

Verify: $x = - 5$ → $∣2 (- 5) + 4∣ = ∣ - 6∣ = 6 \leq 6$ ✓. $x = 1$ → $∣2 (1) + 4∣ = 6 \leq 6$ ✓. $x = 0$ → $∣4∣ = 4 \leq 6$ ✓ (inside). Answer $[- 5, 1]$ .

Active recall

Which set operation gives the narrow overlap?

Intersection

\cap

(the AND case).

What does

\infty

mean and why does it always take a round bracket?

It means "goes on forever, no end"; it's not a reachable number, so the endpoint can never be included.

What does

R

stand for?

The real numbers — every number on the whole line at once.

∣ x ∣ < k

becomes which kind of compound inequality?

An AND — the single interval

- k < x < k

∣ x ∣ > k

becomes which kind of compound inequality?

An OR — two rays

x < - k

x > k

Solve

x \geq 3

AND

x \leq 3

The single point

{3}

Solve

x \leq - 1

x > 2

(- \infty, - 1] \cup (2, \infty)

Why does

x > - 3

x < 7

equal all reals?

The rays point toward each other and overlap, so no number can fail both — the union floods

R

Solve

x > 1

x > 3

(1, \infty)

— the bigger ray swallows the smaller.

Solve

x > 1

AND

x > 3

(3, \infty)

— the stricter bound wins.

Dividing a three-part inequality by

- 3

does what?

Flips BOTH relation signs at once.

Connections

Parent: Compound inequalities — AND, OR — the core idea this page drills.
Linear inequalities in one variable — each ray here is one of these.
Set operations — union and intersection — the meaning of $\cap$ and $\cup$ .
Interval notation — how we wrote every answer.
Absolute value inequalities — the source of Examples 9 and 10.
Number line representation — every figure lives here.