Compound inequalities — AND, OR: Visual walkthrough

We will only assume you know what a number line is (a straight line where every point is a number, small numbers on the left, big numbers on the right). If you can point to "3 is to the right of 1", you are ready.

Step 1 — One condition is a shaded ray

WHAT. Take the single inequality $x > 1$ . The symbol $x$ means "some unknown number". The symbol $>$ means "strictly greater than". So $x > 1$ is a question about a number: is it bigger than 1?

WHY. Before we can glue two conditions, we must know what one condition looks like as a shape. A condition is not a single point — it is every number that passes the test.

PICTURE. Every number to the right of $1$ passes. We draw an open circle at $1$ (open = " $1$ itself is NOT included", because $1 > 1$ is false) and shade everything to its right. That shaded ray is the set of solutions.

$x$ ::: the unknown number we are testing
$>$ ::: strictly greater than — the boundary is excluded (open circle)
$\geq$ ::: greater than or equal to — the boundary is included (filled circle)

Step 2 — A second condition is a second ray

WHAT. Now bring in a second condition, $x \leq 5$ . The symbol $\leq$ means "less than or equal to". This is $x \leq 5$ : is the number at most 5?

WHY. A compound inequality is two conditions. We must draw the second one on the same line as the first so we can compare where they agree.

PICTURE. Everything to the left of $5$ passes, and $5$ itself passes too (because $5 \leq 5$ is true). So we draw a filled circle at $5$ (filled = included) and shade to its left. Now we have two rays on one line, drawn on two levels so we can see them separately.

Step 3 — AND: keep only where BOTH rays are lit

WHAT. Glue the two with AND: $x > 1$ and $x \leq 5$ . Call the first set $A = (1, \infty)$ and the second $B = (- \infty, 5]$ . The phrase " $A = (1, \infty)$ " just names the ray from Step 1 (everything past 1); " $B = (- \infty, 5]$ " names the ray from Step 2.

WHY. The word AND means both tests must pass for the same number. A number that only passes one test fails the compound statement. So we ask: where do the two shadings sit on top of each other?

PICTURE. Slide the two rays onto one level. Where they overlap — where the line is shaded by BOTH — is the answer. That overlap runs from just after $1$ up to and including $5$ : the interval $(1, 5]$ . This overlap is exactly what intersection $\cap$ means.

Compact form: $1 < x \leq 5$ . This one-line form is allowed only for AND, and only because the overlap is a single connected block.

Step 4 — OR: keep everywhere EITHER ray is lit

WHAT. Now glue with OR instead: $x < - 2$ or $x \geq 3$ . Let $A = (- \infty, - 2)$ and $B = [3, \infty)$ .

WHY. The word OR (the "inclusive or") means at least one test passes. A number is in the answer if it is in $A$ , or in $B$ , or in both. So instead of hunting for overlap, we keep every piece that is lit by anything.

PICTURE. Both rays stay. Nothing is thrown away. The answer is the two shaded pieces together: everything left of $- 2$ , PLUS everything from $3$ rightward. This "keep everything lit" is exactly union $\cup$.

Step 5 — Edge case: AND with rays pointing apart → empty

WHAT. Solve $x > 4$ and $x < 1$ . Set $A = (4, \infty)$ , $B = (- \infty, 1)$ .

WHY. We test the AND machine on a case where the two rays point away from each other. For AND we still hunt for overlap — but is there any?

PICTURE. The right-pointing ray starts at $4$ ; the left-pointing ray ends at $1$ . There is a whole empty stretch between $1$ and $4$ that neither covers, and the two shadings never sit on top of each other. No overlap = no solution. We write the empty set $\emptyset$ .

Step 6 — Edge case: OR with overlapping rays → the whole line

WHAT. Solve $x < 5$ or $x > 2$ . Set $A = (- \infty, 5)$ , $B = (2, \infty)$ .

WHY. We test the OR machine when the rays overlap and reach past each other. Does keeping "everything lit" ever cover the entire line?

PICTURE. $A$ covers everything below $5$ . $B$ covers everything above $2$ . Pick any number at all: if it is $\leq 2$ it is caught by $A$ ; if it is $\geq 5$ it is caught by $B$ ; anything between $2$ and $5$ is caught by both. Nowhere is dark. The union fills the whole line: $R$ .

Step 7 — Solving a three-part AND, seen as sliding walls

WHAT. Solve $- 1 \leq \frac{3 - x}{2} < 4$ . This is secretly $(- 1 \leq \frac{3 - x}{2})$ AND $(\frac{3 - x}{2} < 4)$ — an AND, so it is one connected band with two walls.

WHY. We want to see why "do the same operation to all three parts" keeps the band correct. Every operation moves both walls so the trapped region stays trapped.

PICTURE. Start with the band $- 1 \leq (stuff) < 4$ .

Multiply all parts by $2$ : $- 2 \leq 3 - x < 8$ — both walls scale.
Subtract $3$ from all parts: $- 5 \leq - x < 5$ — both walls shift left.
Multiply all parts by $- 1$ : this reflects the picture, so the walls swap sides and both signs flip: $5 \geq x > - 5$ . Reading it small-to-large: $- 5 < x \leq 5$ , the interval $(- 5, 5]$ .

The one-picture summary

Here is the entire walkthrough in a single frame: two rays $A$ and $B$ , and the four outcomes stacked so you can compare AND (overlap) against OR (whole coverage) at a glance.

Recall Feynman retelling — say it to a friend

Picture two flashlights aimed at a long dark ruler. Each flashlight lights up one stretch of the ruler — that stretch is one inequality's solution set. AND means "stand where BOTH beams hit you", so you only keep the little patch where the two lights cross; if the beams miss each other entirely, there's nowhere to stand — that's the empty set. OR means "stand where AT LEAST ONE beam hits you", so you keep both patches; if the beams happen to sweep past each other, together they light the whole ruler. The word — AND or OR — is the only thing that decides whether you keep the narrow crossing or the wide coverage. And for a three-part AND, every fair move you make (add, multiply, reflect) slides both walls of the lit band at once, which is why you must always touch all three parts — and why a negative multiplier reflects the ruler and flips both signs together.

Connections

Compound inequalities — AND, OR — the parent note this walkthrough visualises.
Number line representation — every step lives on this line.
Set operations — union and intersection — overlap $= \cap$ , coverage $= \cup$ .
Interval notation — how the shaded regions become $(1, 5]$ , $(- \infty, - 2) \cup [3, \infty)$ , etc.
Linear inequalities in one variable — each single ray comes from one of these.
Absolute value inequalities — where $∣ x ∣ < a$ turns into an AND and $∣ x ∣ > a$ into an OR.