How do I know whether to write my answer as a three-part inequality or with the word "or"?

Look at the graph. If the solution is one connected piece of the number line with two endpoints, it is an AND statement and can be written in three-part form such as -2 ≤ x 6. Never write an OR answer in three-part form; the three-part form always means AND, so 6 < x < -2 would claim a number is both greater than 6 and less than -2, which nothing satisfies.

When I divide a three-part inequality by a negative number, do I flip both symbols?

Yes. Multiplying or dividing by a negative reverses every inequality relationship in the statement, so both symbols flip at the same time. For instance, dividing -8 ≤ -2x -3. It is then good practice to rewrite it with the smaller number first as -3 < x ≤ 4 so the order matches the number line.

Can an AND compound inequality ever have all real numbers as its solution?

No. AND keeps only the overlap of the two solution sets, and the overlap of two half-lines never covers the entire number line. When both parts point the same direction, the AND simply collapses to the stricter one: x > 3 AND x > 8 is just x > 8, and x > -100 AND x > -5 is just x > -5. True "all real numbers" answers come from OR statements whose two rays overlap or leave no gap, such as x ≤ 5 or x > 1.

How can I check a compound-inequality answer quickly?

Test three numbers: one inside your shaded region, one outside it, and one right at a boundary. Substitute each into the original inequality. The inside value should make the statement true, the outside value should make it false, and the boundary value tells you whether the circle should be open or closed. This catches both arithmetic slips and forgotten symbol flips.

ALG1-3.2

Compound Inequalities

Q: When I divide a three-part inequality by a negative number, do I flip both symbols?

Yes. Multiplying or dividing by a negative reverses every inequality relationship in the statement, so both symbols flip at the same time. For instance, dividing -8 ≤ -2x -3. It is then good practice to rewrite it with the smaller number first as -3 < x ≤ 4 so the order matches the number line.

Q: Can an AND compound inequality ever have all real numbers as its solution?

No. AND keeps only the overlap of the two solution sets, and the overlap of two half-lines never covers the entire number line. When both parts point the same direction, the AND simply collapses to the stricter one: x > 3 AND x > 8 is just x > 8, and x > -100 AND x > -5 is just x > -5. True "all real numbers" answers come from OR statements whose two rays overlap or leave no gap, such as x ≤ 5 or x > 1.

Learn to solve AND (intersection) and OR (union) compound inequalities, work three-part inequalities on all three parts, and graph solution sets on a number line.

What you'll do in this lesson

A voice-first session with the Crimsora tutor on Compound Inequalities, then targeted practice and FRQs — with the tutor adapting to where you get stuck.

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What this lesson covers

A single inequality like

2x + 1 > 9

describes one condition. But real situations usually stack conditions: a thermostat must keep a room warmer than 68 degrees and cooler than 74 degrees; a roller coaster rejects riders shorter than 48 inches or taller than 78 inches. Those pairs of conditions are called compound inequalities, and the little words AND and OR completely change what the answer looks like.

In this lesson you will solve both types, learn why AND problems can be compressed into a three-part form like

-3 \le 2x + 1 < 7

, and learn how to read a number-line graph so you can check your own work. You will also meet the two surprising cases: compound inequalities with no solution at all, and ones that every real number satisfies. By the end you should be able to move fluently between the symbolic solution and the picture.

AND Means Intersection, OR Means Union

A compound inequality is two inequalities joined by AND or OR. The joining word tells you how to combine the two solution sets.

AND (intersection) keeps only the numbers that satisfy both parts. If

x > 2

AND

x \le 6

, a number must clear both hurdles, so the solution is the overlap:

2 < x \le 6

. On a number line this is a single segment, shaded between the two endpoints.

OR (union) keeps every number that satisfies at least one part. If

x < -1

x \ge 4

, a number only needs to pass one test. The graph is two separate rays shooting outward in opposite directions, with a gap in the middle.

Feature	AND (intersection)	OR (union)
Requirement	both parts true	at least one part true
Typical graph	one shaded segment between endpoints	two rays pointing away from each other
Typical solution	$-2 \le x < 5$	$x < -2$ or $x \ge 5$
Can be written in three parts	yes	no

The single most common error in this lesson is treating OR like AND. Students see

x < 6

x > 1

and shade only the middle. But every real number is either less than 6 or greater than 1 (most numbers are both), so that solution set is all real numbers. Test a value before you trust a graph: try

x = 100

. It fails

x < 6

but passes

x > 1

, and for OR that is enough, so 100 belongs in the solution.

Use open circles for

<

and

>

, closed circles for

\le

and

\ge

, exactly as with simple inequalities.

Three-Part Inequalities: Operate on All Three Parts

When an AND statement has the same expression in both halves, you can compress it. Instead of writing

2x + 1 \ge -3

AND

2x + 1 < 7

, write

-3 \le 2x + 1 < 7.

Read it as "

2x+1

is at least

-3

and less than 7." To solve, isolate

x

in the middle by doing the same operation to all three parts.

Subtract 1 everywhere:

-4 \le 2x < 6

. Divide by 2 everywhere:

-2 \le x < 3

. The solution is every number from

-2

(included) up to 3 (excluded), graphed as a closed circle at

-2

, an open circle at 3, and shading between.

Two rules keep three-part work honest. First, whatever you do, do it to the left part, the middle, and the right part — forgetting one end is the number-one source of wrong answers here. Second, if you multiply or divide all three parts by a negative number, both inequality symbols reverse. Solving

-6 < -3x \le 9

means dividing by

-3

to get

2 > x \ge -3

, which is more readable rewritten with the smaller number on the left as

-3 \le x < 2

.

A three-part inequality describes a band between two bounds only when the two inequality symbols point the same direction. Something like

5 < x > 9

is still an AND — it says

x > 5

and

x > 9

— but the mixed symbols disguise the fact that it collapses to the single condition

x > 9

, with no upper bound at all. Write it as

x > 9

rather than leaving it in a form that looks like a bounded interval and is not one.

No Solution, All Reals, and Reading the Graph Back

Not every compound inequality has a normal-looking answer, and recognizing the two extreme cases saves a lot of confusion.

An AND statement has no solution when the two conditions never overlap. Consider

x > 7

AND

x < 2

. No number is simultaneously bigger than 7 and smaller than 2, so the solution set is empty; the number line stays blank. Written in three-part form this would be the impossible

7 < x < 2

.

An OR statement covers all real numbers when the two rays overlap or at least meet with no gap. Consider

x \ge 0

x < 5

. Every number falls into at least one of those categories, so the entire line is shaded.

The reverse cases are milder but worth noting: an AND whose parts point the same way collapses to the stricter one (

x > 3

AND

x > 8

is just

x > 8

), and an OR whose parts point the same way collapses to the looser one (

x > 3

x > 8

is just

x > 3

).

When you read a graph back into symbols, ask two questions. Is the shading one connected piece or two? One piece means AND, and you may write it in three-part form. Two pieces means OR. Then check each circle: hollow means the endpoint is excluded (

<

>

), filled means included (

\le

\ge

). A graph shaded left from a filled dot at

-4

and right from a hollow dot at 1 translates to

x \le -4

x > 1

Translating Word Problems into Compound Inequalities

Most compound inequalities you meet outside of a homework set start as sentences, so the translation step matters.

Phrases that signal AND include "between," "from ... to ...," "at least ... but no more than ...," and "within." A lifeguard chair rated for a load between 150 and 400 pounds inclusive becomes

150 \le w \le 400

. Note that plain "between" usually means strict inequalities and "inclusive" or "at least/at most" means the endpoints count.

Phrases that signal OR include "outside of," "either ... or ...," "less than ... or more than ...," and safety-style rules that reject extremes. A machine part is defective if its length is under 9.8 cm or over 10.2 cm:

L < 9.8

L > 10.2

.

Here is a full translation. A summer camp accepts campers who are at least 7 years old and younger than 13. Let

a

be the age:

7 \le a < 13

. If a sibling discount applies to campers younger than 7 or 13 and older, that is the complement:

a < 7

a \ge 13

. Notice how the AND set and the OR set fit together like puzzle pieces covering the whole line — that pairing shows up again when you study absolute value.

One caution about context: after solving, check whether the variable can realistically take every value in your interval. Ages, numbers of tickets, and counts of people are often restricted to whole numbers, so a solution of

7 \le a < 13

may describe only the integers 7 through 12 in practice. Say so when the situation calls for it.

Key terms

Compound inequality.: Two inequalities joined by the word AND or the word OR, describing a combined condition on one variable.
Intersection (AND).: The set of values that satisfy both inequalities at the same time; graphed as the overlap of the two individual solution sets.
Union (OR).: The set of values that satisfy at least one of the two inequalities; graphed as everything shaded by either solution set.
Three-part inequality.: A compact form of an AND statement, such as $-1 < 3x + 2 \le 8$ , in which the variable expression sits between two bounds and every operation is applied to all three parts.
Open circle.: A hollow endpoint on a number line showing the boundary value is not included, used for $<$ and $>$ .
Closed circle.: A filled endpoint on a number line showing the boundary value is included, used for $\le$ and $\ge$ .
Empty solution set.: The result when an AND statement asks for values in two regions that never overlap, so no number works and nothing is shaded.
Boundary value.: A number where the inequality changes from true to false; it becomes an endpoint of the graph.

Worked example

Solve and graph:

-5 \le 3 - 2x < 9

. Then state whether

x = -3

is a solution.

This is a three-part (AND) inequality, so isolate

x

in the middle while doing the same thing to all three parts.

Step 1: Subtract 3 from every part.

-5 - 3 \le 3 - 2x - 3 < 9 - 3

, which simplifies to

-8 \le -2x < 6

.

Step 2: Divide every part by

-2

. Because the divisor is negative, both inequality symbols reverse:

\frac{-8}{-2} \ge x > \frac{6}{-2}

, giving

4 \ge x > -3

.

Step 3: Rewrite with the smaller number on the left so it reads naturally:

-3 < x \le 4

.

Step 4: Graph. Put an open circle at

-3

(strict inequality, not included), a closed circle at 4 (included), and shade the segment between them.

Step 5: Check

x = -3

. Substitute into the original:

3 - 2(-3) = 3 + 6 = 9

. The right-hand condition requires the middle to be strictly less than 9, and

9 < 9

is false, so

x = -3

is not a solution. That matches the open circle at

-3

in the graph.

Quick verification with an interior point: try

x = 0

. Then

3 - 2(0) = 3

, and

-5 \le 3 < 9

is true, so the shaded region is in the right place.

Practice questions

Which number line graph matches the solution of

2x + 5 < 3

x - 1 \ge 4

A single shaded segment from $-1$ to 5 with open circles at both ends
Two rays: shading left from an open circle at $-1$ and right from a closed circle at 5
Two rays: shading left from a closed circle at $-1$ and right from an open circle at 5
The entire number line shaded

Answer: Two rays: shading left from an open circle at $-1$ and right from a closed circle at 5

Solve each part separately. From

2x + 5 < 3

, subtract 5 to get

2x < -2

, then divide by 2 to get

x < -1

— strict, so an open circle at

-1

with shading to the left. From

x - 1 \ge 4

, add 1 to get

x \ge 5

— inclusive, so a closed circle at 5 with shading to the right. The joining word is OR, so the solution is the union of the two rays, not the region between them. Choosing the middle segment is the classic mistake of treating OR as AND.

Solve

-4 < \frac{x}{3} + 1 \le 2

and describe the graph in words.

Answer: $-15 < x \le 3$ ; open circle at $-15$ , closed circle at 3, shading between them.

Subtract 1 from all three parts:

-5 < \frac{x}{3} \le 1

. Multiply all three parts by 3; since 3 is positive, the symbols keep their direction:

-15 < x \le 3

. The left symbol is strict, so

-15

gets an open circle and is excluded; the right symbol includes equality, so 3 gets a closed circle. Shade the segment connecting them. Check an interior value such as

x = 0

\frac{0}{3} + 1 = 1

, and

-4 < 1 \le 2

is true.

Explain why

x < 4

AND

x > 9

has no solution, but

x < 4

x > 9

has infinitely many solutions. Include one specific number in your explanation.

Answer: No number is simultaneously below 4 and above 9, so the AND set is empty; but any number below 4 or above 9 satisfies the OR statement, for example $x = 100$ .

For AND, a value must make both parts true at once. Being less than 4 places a number to the left of 4, while being greater than 9 places it to the right of 9. Those two regions never overlap, so the intersection is empty and nothing is shaded. For OR, only one part needs to be true. The number 100 fails

x < 4

but satisfies

x > 9

, so it belongs to the union. So does

x = 0

, which satisfies

x < 4

. The graph is two rays with a gap between 4 and 9.

FAQ

How do I know whether to write my answer as a three-part inequality or with the word "or"?: Look at the graph. If the solution is one connected piece of the number line with two endpoints, it is an AND statement and can be written in three-part form such as $-2 \le x < 6$ . If the solution is two separate pieces with a gap, you must write it as two inequalities joined by "or," such as $x \le -2$ or $x > 6$ . Never write an OR answer in three-part form; the three-part form always means AND, so $6 < x < -2$ would claim a number is both greater than 6 and less than $-2$ , which nothing satisfies.
When I divide a three-part inequality by a negative number, do I flip both symbols?: Yes. Multiplying or dividing by a negative reverses every inequality relationship in the statement, so both symbols flip at the same time. For instance, dividing $-8 \le -2x < 6$ by $-2$ gives $4 \ge x > -3$ . It is then good practice to rewrite it with the smaller number first as $-3 < x \le 4$ so the order matches the number line.
Can an AND compound inequality ever have all real numbers as its solution?: No. AND keeps only the overlap of the two solution sets, and the overlap of two half-lines never covers the entire number line. When both parts point the same direction, the AND simply collapses to the stricter one: $x > 3$ AND $x > 8$ is just $x > 8$ , and $x > -100$ AND $x > -5$ is just $x > -5$ . True "all real numbers" answers come from OR statements whose two rays overlap or leave no gap, such as $x \le 5$ or $x > 1$ .
How can I check a compound-inequality answer quickly?: Test three numbers: one inside your shaded region, one outside it, and one right at a boundary. Substitute each into the original inequality. The inside value should make the statement true, the outside value should make it false, and the boundary value tells you whether the circle should be open or closed. This catches both arithmetic slips and forgotten symbol flips.

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