Lesson · Linear Equations and Inequalities · Inequalities

Compound Inequalities

Two inequalities joined by a word. And keeps the numbers satisfying both, which is the intersection of the two solution sets; or keeps the numbers satisfying at least one, which is the union. Everything else is “Linear Inequalities” done twice.

Open the interactive lesson Practice

A and B both true intersection ∩ A or B at least one union ∪ a < x < b means a < x AND x < b never an or x < 4 or x > 1 is every real number

Two connectors, two set operations

Core ideas

What this lesson establishes

The solving is “Linear Inequalities”. The lesson is what the connecting word does to two answers once you have them.

And is intersection

x ≥ 2 and x < 4 is true at exactly the numbers lying in both solution sets. Solve the two statements separately, then keep the overlap. That overlap can be an interval, one of the two sets entire, a single number, or nothing at all.

Or is inclusive

A or B is true when A holds, when B holds, and when both hold. Mathematical or is never exclusive, and that is why x < 4 or x > 1 is every real number rather than the numbers outside the overlap.

The three-part form always means and

3 < 2x + 1 ≤ 9 is a compact way of writing two statements that hold together. There is no corresponding compact form for or, and inventing one produces a statement describing nothing at all.

The material

Both, or at least one

Solving two inequalities at once, combining the two answers, and the shapes the combination can take.

Definition · Compound inequality

A compound inequality is two inequalities joined by and or by or. Its solution set is the intersection of the two solution sets when the connector is and, and their union when the connector is or.

∩ denotes intersection, the numbers lying in both sets. ∪ denotes union, the numbers lying in at least one.

The method is the definition. Solve each inequality on its own by “Linear Inequalities”, draw both solution sets against the same number line, and read off the overlap or the combination. Almost every error in this lesson is an error made before the drawing or instead of it.

The three-part form

When both statements trap the variable between two numbers they are usually written as one. 3 < 2x + 1 ≤ 9 means 3 < 2x + 1 and 2x + 1 ≤ 9. It is solved by operating on all three parts at once.

3 < 2x + 1 ≤ 9 subtract 1 2 < 2x ≤ 8 divide by 2 1 < x ≤ 4 check x = 2 2(2) + 1 = 5 3 < 5 ≤ 9 ✓ x = 1 2(1) + 1 = 3 3 < 3 false ✓ x = 4 2(4) + 1 = 9 9 ≤ 9 ✓

Every operation applies to all three parts, because each of the two statements the compact form stands for needs it. Operating on two parts and forgetting the third breaks the equivalence, and there is nothing on the page to warn you.

The compact form is usable only when the connector is and and the variable sits alone in the middle. x < −2 or x ≥ 3 cannot be compressed: writing 3 ≤ x < −2 asks for a number that is at once at least 3 and below −2, and there is none. The compact form for that solution set is (−∞, −2) ∪ [3, ∞), and “Interval and Set-Builder Notation” is about it.

The four shapes an intersection can take

An interval. x ≥ 2 and x < 4 gives the numbers from 2 up to but not including 4.
One of the two, entire. x > 1 and x > 5 gives x > 5. The stronger condition already implies the weaker one, so the weaker one adds nothing.
A single number. x ≥ 3 and x ≤ 3 is true only at x = 3.
Nothing. x > 5 and x < 2 has no solutions. The empty set is a legitimate answer and is written ∅.

The three shapes a union can take

Two separated pieces. x < −2 or x ≥ 3. Most absolute-value inequalities land here, which is what “Absolute Value Inequalities” is about.
One piece. x < 4 or x < 1 is x < 4, because everything in the second set is already in the first.
Every real number. x < 4 or x > 1. Take any number: if it is below 4 the first statement holds, and if it is not below 4 then it is at least 4, so it is above 1 and the second holds. Nothing is left out.

Example · Two statements, both connectors

2x + 1 ≥ 5 solves to x ≥ 2, and x − 4 < 0 solves to x < 4. Joined by and the answer is 2 ≤ x < 4. Joined by or the answer is every real number, because a number below 2 satisfies the second and a number at or above 2 satisfies the first. Same two statements; the connector alone decides.

Interactive

And, or, and the line

Two statements at once: overlap or gather

the first statement

the second statement

the two of them together

Two statements, three number lines

Solve both statements on paper and predict the combined shape before choosing and or or. The lab draws each set and the combination, using inequalities and words only; interval and set-builder notation wait for “Interval and Set-Builder Notation”.

Statement

First: left side

First: relation

First: right side

Joined by

Second: left side

Second: relation

Second: right side

first—

second—

combined—

and keeps the numbers in both sets, or keeps the numbers in either. Two of the presets show what that does when the sets miss each other entirely.

Method

Solving a compound inequality

The two halves are solved independently. The connector is applied once, at the end, to the two answers.

Write down which two statements you have

A three-part statement a < f(x) < b is two statements. You may keep it compact and operate on all three parts at once, but you must know which two statements those parts stand for.

Solve each one on its own, by “Linear Inequalities”

Each half gets its own reversal check. A negative coefficient in one half says nothing whatever about the other, and the two halves may well need different treatment.

Draw both solution sets against one number line

Two marked regions above the same axis. Filled endpoints for ≤ and ≥, hollow for < and >. This is the step that turns the combination into something you read off instead of something you reason about.

Take the overlap for and, everything covered for or, then check the endpoints

For an and, an endpoint belongs only if the statement that produced it includes it. For an or, test it against both statements, because the other one may cover it even when its own does not. Then test a number from each region against the original pair.

And, or, and what is left

Score 0 / 0

Choose the answer you can justify.

The same pair, joined two ways

2x + 1 ≥ 5 and x − 4 < 0, then the identical two statements joined by or.

first 2x + 1 ≥ 5 second x - 4 < 0 2x ≥ 4 x < 4 x ≥ 2 sets A = { x : x ≥ 2 } B = { x : x < 4 } A and B the numbers in both 2 ≤ x < 4 A or B the numbers in at least one every real number check the and answer x = 2 2(2) + 1 = 5 ≥ 5 true 2 - 4 = -2 < 0 true in x = 4 2(4) + 1 = 9 ≥ 5 true 4 - 4 = 0 < 0 FALSE out x = 1 2(1) + 1 = 3 ≥ 5 FALSE 1 - 4 = -3 < 0 true out check the or answer x = 0 first FALSE second true in x = 9 first true second FALSE in

In the and answer the two endpoints behave differently, and each is settled by the statement it came from. 2 came from x ≥ 2, which includes it, so 2 is in. 4 came from x < 4, which excludes it, so 4 is out. In the or answer 4 is back in, because 4 ≥ 2 makes the first statement true there: an endpoint is settled by the statement that produced it only when it is an endpoint of the combined set.

The or answer is worth dwelling on. Every real number is either less than 2 or at least 2. If it is less than 2 it is certainly less than 4, so the second statement holds; if it is at least 2, the first holds. Nothing is left out, and the union is the whole line even though neither statement is true everywhere on its own.

For a faded intersection, use x + 2 ≥ 7 and 3x < 18. The first statement has been solved for you as x ≥ 5; solve the second, draw both against one line, combine them, and test both resulting endpoints before using the lab.

Common mistakes

Writing an or as a three-part statementx < −2 or x ≥ 3 is not 3 ≤ x < −2. The three-part form always means and, so that line asks for numbers at once at least 3 and below −2. A solution set with two infinite pieces has been written down as the empty set.

Operating on two of the three partsIn 3 < 2x + 1 ≤ 9, subtracting 1 gives 2 < 2x ≤ 8 — every part changes. The compact form stands for two statements, and both of them need the subtraction if the result is to be equivalent.

Reading or as exclusiveIn x < 4 or x > 1 the numbers between 1 and 4 satisfy both statements, and mathematical or keeps them. The solution set is every real number, not the numbers outside the overlap and not the overlap itself.

Check your understanding

✓

Finish when you can predict the shape of the answer before you combine.

Solve 3x − 2 < 7 and 5 − x ≤ 3, draw both sets and report their overlap with correct endpoint inclusion. Then solve 2x + 1 ≤ −3 or x − 4 > 0 and report both rays. Test every finite boundary against the original pair.

Note: The connector decides everything here, and “Absolute Value Equations” and “Absolute Value Inequalities” supply it automatically: |X| < k produces an and and |X| > k produces an or. That is the reason absolute value is treated after this lesson rather than before it.