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Compound Inequality

Compound Inequality – definition and example, Algebra 1 math glossary – Mathovore

Key takeaway: A compound inequality joins two inequalities with the word and or the word or, and its solution is every number that makes the combined statement true.

Grade 9 – Algebra 1A-REI.B.3Expressions and Equations

Definition of Compound Inequality

Definition

A compound inequality puts two inequality statements into one. With “and”, a number must satisfy both parts, so the solution is the overlap and looks like a segment between two endpoints. With “or”, a number needs to satisfy at least one part, so the solution is the union and often shows two rays pointing away from each other.

An “and” inequality is frequently written as a chain, such as \(-4 < x \le 6\), meaning \(x\) lies between \(-4\) and \(6\). Solve it by applying each operation to all three parts at once.

A compound inequality appears in everyday limits: a roller coaster may require riders to be at least 48 inches tall and at most 78 inches tall, which is one “and” statement.

Formula and notation

Chain form for “and”, split form for “or”:

\[a < x \le b \qquad\text{and}\qquad x < c \ \text{ or } \ x > d\]

On a number line, an open circle marks an excluded endpoint (\(<\) or \(>\)) and a closed circle marks an included one (\(\le\) or \(\ge\)).

A ticket rule such as “under 5 or over 64 rides free” is an “or” statement, because a person only has to fit one condition. Read the connecting word first.

-6-5-4-10367-4 < x ≤ 6
The open circle at -4 excludes it, the closed circle at 6 includes it, and the shaded segment holds the solutions.

Key facts

  • “And” means the overlap of two solution sets; “or” means everything in either set.
  • In a compound inequality, multiplying or dividing every part by a negative number reverses each symbol.
  • A chain such as 2 < x < 9 is shorthand for x > 2 and x < 9.
  • Some compound inequalities have no solution, for example x < 1 and x > 4.

Compound Inequality example

Problem

Solve \(-3 < 2x + 5 \le 17\) and describe the solution on a number line.

  1. Subtract 5 from all three parts: \(-8 < 2x \le 12\).

  2. Divide all three parts by 2: \(-4 < x \le 6\).

  3. Test \(x = 0\): \(2(0) + 5 = 5\), and \(-3 < 5 \le 17\) is true.

The solution is \(-4 < x \le 6\): an open circle at \(-4\), a closed circle at \(6\), and the segment between them shaded.

Common mistake

Watch out

Pia subtracts 5 only from the middle: \(-3 < 2x \le 17\), then divides to get \(-1.5 < x \le 8.5\).

Why it is wrong

An operation must be applied to all three parts, or the statement changes. Subtract 5 from \(-3\) and from \(17\) too, which gives \(-8\) and \(12\). Then \(x = -4\) and \(x = 6\) are the real endpoints. A quick test of a number outside the answer, such as 8, confirms that it fails the original statement.

Related terms and lessons

Browse the glossary: A–Z list · Algebra 1 vocabulary · Expressions and Equations

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