Lesson 6: Compound Inequalities

Property

A compound inequality is made up of two inequalities connected by the word “and” or the word “or.”
To solve a compound inequality means to find all values of the variable that make the compound inequality a true statement.
We solve each inequality separately and then consider the two solutions.

Examples

• The statement $x > 2$ and $x < 7$ is a compound inequality. A value like $x=5$ is a solution because it satisfies both conditions.

• The statement $y \leq 0$ or $y \geq 4$ is a compound inequality. A value like $y=6$ is a solution because it satisfies the second condition, even though it fails the first.

Property

To solve an "and" compound inequality, solve each inequality separately and find the intersection (overlapping region) of their solution sets.

A chain inequality (e.g., $a < x < b$) is a compact "and" statement. To solve a chain inequality, perform the same inverse operations to all three parts (left, middle, right) simultaneously to isolate the variable in the center.

Examples

• Solving Separate "And" Statements: Solve $4x - 1 < 7$ and $x + 5 \geq 2$.

First, solve each part: $4x < 8 \rightarrow x < 2$, and $x \geq -3$.
The solution is the overlapping region where numbers are both greater than or equal to -3 and less than 2: $-3 \leq x < 2$.

• Solving a Chain Inequality: Solve $-1 < 2x + 3 < 9$.

Subtract 3 from all three parts: $-4 < 2x < 6$.
Divide all three parts by 2: $-2 < x < 3$.

• Chain Inequality with a Negative: Solve $-6 < -2x + 4 < 2$.

Subtract 4 from all parts: $-10 < -2x < -2$.
Divide by -2 and flip all inequality signs: $5 > x > 1$, which is logically rewritten from least to greatest as $1 < x < 5$.

Property

To solve an "or" compound inequality, you must solve the two inequalities completely separately. The final solution is the union of the two solution sets, meaning any number that makes either the first inequality true, the second true, or both true is included in the final answer.

Examples

• Standard "Or" Solution: Solve $3x - 2 > 10$ or $x + 1 < 0$.

Solve each part independently: $3x > 12 \rightarrow x > 4$, and $x < -1$.
The solution is $x < -1$ or $x > 4$.

• Overlapping "Or" Solution: Solve $x - 5 > 2$ or $x - 5 > 0$.

This simplifies to $x > 7$ or $x > 5$. Since any number greater than 7 is already greater than 5, the two rules merge, and the final combined solution is simply $x > 5$.

• All Real Numbers: Solve $x + 3 \leq 5$ or $x - 4 \geq -2$.

This simplifies to $x \leq 2$ or $x \geq 2$. Because this covers all numbers less than 2, equal to 2, and greater than 2, the solution is All Real Numbers.

Explanation

Solving an "or" inequality is about gathering all possible solutions into one big group. You simply solve the two inequalities completely independently of one another. When graphing them on a number line, you will usually draw two separate arrows pointing in opposite directions. As long as a number falls under at least one of those shaded arrows, it is a valid solution.