Algebra 1 · Algebra 2 · Grades 9, 10

How to Solve Compound Inequalities (AND and OR)

Quick answer

A compound inequality joins two inequalities with AND or OR. AND means both must be true, so the solution is the overlap. OR means either can be true, so the solution is everything both parts cover. Solving 3 < x + 1 < 7 gives 2 < x < 6, the numbers between 2 and 6.

What you'll learn

  • Solve compound inequalities joined by AND or by OR
  • Graph the solution set of a compound inequality
  • Identify when a compound inequality has no solution or all real numbers

What a compound inequality is

A compound inequality is two inequalities joined by the word and or the word or:

x>2   and   x<6x<−1   or   x>4x > 2 \;\text{ and }\; x < 6 \qquad\qquad x < -1 \;\text{ or }\; x > 4

The joining word is not decoration. It decides the answer completely, and the two words pull in opposite directions.

Why AND narrows and OR widens

Think about what each word demands of a number applying to join the solution set.

AND demands that the number pass both tests. Every extra condition can only rule numbers out, never let new ones in, so an AND is always at most as large as either piece on its own. On a number line the answer is the overlap.

OR demands only that the number pass one test. A number that fails the first still gets in on the second, so an OR is always at least as large as either piece. On a number line the answer is both pieces together.

WordA number qualifies ifNumber lineEffect on the answer
andboth parts are trueoverlap of the twonarrows
orat least one part is trueboth parts combinedwidens

That single distinction is the whole lesson. Everything below is applying it.

The three-part shorthand

An AND where xx is trapped between two numbers has a compact form:

2<x<6meansx>2   and   x<62 < x < 6 \qquad\text{means}\qquad x > 2 \;\text{ and }\; x < 6

Read it from the middle outwards: ”xx is greater than 22 and less than 66.”

When you solve a three-part inequality, whatever you do must be done to all three parts, not only the two you happen to be looking at.

Worked examples

Common mistakes

Practice problems

  1. Solve 4<x+2<94 < x + 2 < 9.

    Hint

    Subtract from all three parts at once.

    Answer

    2<x<72 < x < 7

    Full solution

    Subtract 22 from every part: 2<x<72 < x < 7.

    Check: x=5x = 5 gives 4<7<94 < 7 < 9 ✓

  2. Solve −3≤2x−1≤7-3 \le 2x - 1 \le 7.

    Answer

    −1≤x≤4-1 \le x \le 4

    Full solution

    Add 11 to all three parts: −2≤2x≤8-2 \le 2x \le 8. Divide all three by 22: −1≤x≤4-1 \le x \le 4.

    Check: x=0x = 0 gives −3≤−1≤7-3 \le -1 \le 7 ✓

  3. Solve x+5>8x + 5 > 8 and x−2<4x - 2 < 4.

    Answer

    3<x<63 < x < 6

    Full solution

    First part: x>3x > 3. Second part: x<6x < 6. The overlap is 3<x<63 < x < 6.

  4. Solve x<−3x < -3 or x≥2x \ge 2.

    Hint

    This is already solved. Describe the graph.

    Answer

    Two pieces: everything left of −3-3, and everything from 22 rightwards.

    Full solution

    Open circle at −3-3 with an arrow left; closed circle at 22 with an arrow right. Nothing between −3-3 and 22 is shaded, because a number there satisfies neither part.

  5. Solve 2x+1<−52x + 1 < -5 or 3x≥123x \ge 12.

    Answer

    x<−3x < -3 or x≥4x \ge 4

    Full solution

    First: 2x<−62x < -6, so x<−3x < -3. Second: x≥4x \ge 4.

    Joined with OR, the solution is both stretches: x<−3x < -3 or x≥4x \ge 4.

  6. Solve x>7x > 7 and x<3x < 3.

    Answer

    No solution.

    Full solution

    The two stretches never overlap, so no number satisfies both conditions at once.

  7. Solve x≥0x \ge 0 or x≤10x \le 10.

    Hint

    Test a number from well outside the range 00 to 1010.

    Answer

    All real numbers.

    Full solution

    Test x=−50x = -50: it fails the first part but satisfies x≤10x \le 10, so it qualifies through OR. Test x=500x = 500: it satisfies x≥0x \ge 0.

    Every number satisfies at least one part, so the solution is all real numbers.

  8. A ride requires riders to be at least 42 inches tall and no more than 76 inches tall. Write a compound inequality for the allowed heights hh, and say whether someone 76 inches tall can ride.

    Hint

    “At least” and “no more than” both include their boundary value.

    Answer

    42≤h≤7642 \le h \le 76, and yes.

    Full solution

    Both conditions must hold at once, so this is an AND, written 42≤h≤7642 \le h \le 76.

    Both symbols include equality, so a rider of exactly 7676 inches satisfies 76≤7676 \le 76 and can ride.

Frequently asked questions

What is the difference between AND and OR?

AND means a number has to satisfy both parts at once, so the solution is the overlap of the two pieces. OR means a number only has to satisfy one part, so the solution is both pieces put together. AND narrows the answer; OR widens it.

Can I write an OR inequality in the three-part form?

No. The form a < x < b is shorthand for two conditions that are both true, so it only ever means AND. Writing x < 2 or x > 7 as 7 < x < 2 says x is bigger than 7 and smaller than 2, which nothing satisfies.

When does a compound inequality have no solution?

An AND has no solution when the two pieces never overlap, like x > 5 and x < 1. An OR has no solution only if neither piece has one, which is rare, since OR normally makes the answer larger.

What to learn next

Standards alignment

This lesson covers the following Common Core State Standards for Mathematics.

  • CCSS.MATH.CONTENT.HSA.REI.B.3Reasoning with Equations and InequalitiesSolve linear equations and inequalities in one variable, including equations with coefficients represented by letters.
For teachersTick to print a blank worksheet; leave it clear to print the lesson with the full answer key.