Solving Inequalities

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Quick Reference

Textbook: OpenStax College Algebra 2e, Chapter 2 (Equations and Inequalities) Section: 2.7, Linear Inequalities and Absolute Value Inequalities Subsection: 2.7 Linear Inequalities and Absolute Value Inequalities Pages: Page numbers pending faculty verification.

Try This First

A number line is drawn from $-5$ to $5$. Shade every point whose distance from $0$ is less than $3$.

A Set of Answers, Not a Single One

An equation such as $2x + 1 = 7$ has one answer: $x = 3$. An inequality such as $2x + 1 < 7$ has a whole range of answers. Any $x$ smaller than $3$ works, because making $x$ smaller makes the left side smaller, which keeps it under $7$.

Think of an inequality as a question about a process: feed an input $x$ through the rule $2x + 1$, and ask which inputs send the output below the target. The solution is every input that passes the test. That set is what the work produces, and a picture of it on a number line is often the clearest form of the answer.

Solving an inequality reuses every step from solving the matching equation. You add, subtract, multiply, and divide both sides to peel the variable bare. One single rule is new, and that one rule is where almost every mistake hides.

Prerequisite Hub

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        A["Linear<br/>Equations"]
        B["Quadratic<br/>Equations"]
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    subgraph ThisSkill["This Skill"]
        C["Solving<br/>Inequalities"]
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    subgraph Unlocks["Unlocks"]
        D["Absolute<br/>Value"]
        E["Graphing<br/>Functions (Basic)"]
        F["Domain<br/>and Range"]
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Builds on (master these first):

Unlocks (this skill is a gateway to):

No cross-course prerequisites are required for this node.

Official Definition

Interval notation. A method of writing the set of solutions to an inequality, using brackets to indicate that an endpoint is included and parentheses to indicate that an endpoint is excluded.

Source: OpenStax College Algebra 2e, Section 2.7 Linear Inequalities and Absolute Value Inequalities.

Three ways to record the same solution set sit side by side. Translating between them is part of the skill.

A square bracket means the endpoint is part of the set. A parenthesis means it is left out. Infinity is never reached, so it always gets a parenthesis.

The One Rule That Is New

Adding or subtracting the same amount on both sides never disturbs an inequality. Multiplying or dividing both sides by a positive number never disturbs it either. The single move that changes things is multiplying or dividing both sides by a negative number.

The reversal rule. When both sides of an inequality are multiplied or divided by a negative number, the direction of the inequality sign reverses.

Why is this forced rather than optional? Start with a true statement and watch what a negative multiplier does.

$$ 2 < 5 \quad\text{(true)} $$

Multiply both sides by $-1$. The numbers become $-2$ and $-5$. On the number line, $-2$ sits to the right of $-5$, so $-2$ is the larger one.

$$ -2 > -5 \quad\text{(true, with the sign flipped)} $$

If the sign had been left as $<$, the result $-2 < -5$ would be false. Reversing the sign is the only way to keep a true statement true. The rule is not a convention to memorize blindly; it is what the number line demands.

Worked Examples

Example 1: A one step linear inequality

Solve $-4x \ge 12$ and write the answer in interval notation.

Predict first. The variable is multiplied by a negative number, so isolating $x$ will require dividing by $-4$, and that triggers the reversal rule. Predict that the final sign points the opposite way from the one given.

Solve. Divide both sides by $-4$. Because $-4$ is negative, reverse the sign.

$$ -4x \ge 12 $$ $$ \frac{-4x}{-4} \le \frac{12}{-4} $$ $$ x \le -3 $$

Check the prediction. The original sign was $\ge$ and the final sign is $\le$. The direction reversed, exactly as predicted, because of the negative divisor.

Check a value. Pick $x = -5$, which is inside the claimed set $x \le -3$. Then $-4(-5) = 20$, and $20 \ge 12$ holds. Pick $x = 0$, which is outside the set. Then $-4(0) = 0$, and $0 \ge 12$ is false. The boundary behaves correctly.

Answer. $x \le -3$, or in interval notation $(-\infty,\, -3]$. The bracket on $-3$ records that $-3$ itself satisfies $\ge$.

Example 2: A multi step linear inequality

Solve $5 - 3x < 2x + 20$.

Predict first. Gathering the $x$ terms can be done in a way that keeps the coefficient positive, which avoids the reversal rule. Predict that the final sign will match whatever sign survives once $x$ stands alone.

Solve. Collect the variable terms on the side that keeps the coefficient positive. Add $3x$ to both sides.

$$ 5 - 3x < 2x + 20 $$ $$ 5 < 5x + 20 $$

Subtract $20$ from both sides.

$$ -15 < 5x $$

Divide both sides by $5$. Because $5$ is positive, the sign stays the same.

$$ -3 < x $$

Check a value. Pick $x = 0$, inside the set $x > -3$. The original reads $5 - 0 < 0 + 20$, that is $5 < 20$, true. Pick $x = -4$, outside the set. The original reads $5 + 12 < -8 + 20$, that is $17 < 12$, false. The boundary behaves correctly.

Answer. $x > -3$, or in interval notation $(-3,\, \infty)$. A parenthesis on $-3$ records that $-3$ gives equality, not strict inequality, so it is excluded.

Another way to see this. The same problem can be solved by moving the variable terms to the left instead. Subtract $2x$ from both sides at the start to get $5 - 5x < 20$, then subtract $5$ to get $-5x < 15$, then divide by $-5$ and reverse the sign to get $x > -3$. Both paths reach the identical set. The second path forces a reversal that the first path avoided. Which path do you prefer, and why? Choosing the side that keeps the coefficient positive is one habit that sidesteps the most common error on this page.

Example 3: A compound inequality

Solve $-1 \le 2x + 3 < 9$ and write the answer in interval notation.

Predict first. Two inequalities are stitched together. Whatever is done to the middle must be done to all three parts at once. No negative multiplier appears, so predict no sign reversal.

Solve. Subtract $3$ from all three parts.

$$ -1 - 3 \le 2x + 3 - 3 < 9 - 3 $$ $$ -4 \le 2x < 6 $$

Divide all three parts by $2$. Because $2$ is positive, both signs stay.

$$ -2 \le x < 3 $$

Check a value. Pick $x = 0$, inside the set. The original reads $-1 \le 3 < 9$, true. Pick $x = 3$, the excluded right endpoint. The original reads $-1 \le 9 < 9$, and $9 < 9$ is false, so $3$ is correctly left out.

Answer. $-2 \le x < 3$, or in interval notation $[-2,\, 3)$. The bracket includes $-2$; the parenthesis excludes $3$.

Example 4: An absolute value inequality

Solve $|x - 4| < 5$ and write the answer in interval notation.

Predict first. Absolute value measures distance from a point. The inequality asks for every $x$ whose distance from $4$ is less than $5$. Predict a single bounded stretch centered on $4$.

Solve. A statement $|A| < b$ with $b > 0$ means $A$ sits between $-b$ and $b$. Rewrite as a compound inequality.

$$ -5 < x - 4 < 5 $$

Add $4$ to all three parts.

$$ -1 < x < 9 $$

Check the prediction. The set is a single stretch centered on $4$, reaching from $-1$ to $9$, which is exactly five units on each side of $4$. The prediction holds.

Answer. $-1 < x < 9$, or in interval notation $(-1,\, 9)$. Both endpoints are excluded because the inequality is strict.

Common Misconceptions

Practice Problems

Extension Prompt

A “less than” absolute value such as $|x| < b$ gives a single bounded stretch, while a “greater than” absolute value such as $|x| \ge b$ gives two stretches reaching outward. Sketch both on a number line and explain, in terms of distance from zero, why one produces a single piece and the other produces two. How would you convince a classmate that $|x| \ge b$ can never be a single bounded interval when $b$ is positive?

Mastery Checklist

Novice (Level 1-2):

• I can read and write a solution set in interval notation, choosing brackets and parentheses correctly.
• I can state the reversal rule and explain why a negative multiplier forces it.
• I can solve a one step linear inequality and check the result against a test value.

Competent (Level 3-4):

• I can solve a multi step linear inequality and choose a path that keeps the coefficient positive.
• I can solve a compound inequality by operating on all three parts at once.
• I can translate a solution among inequality form, number line, and interval notation.

Proficient (Level 5):

• I can solve an absolute value inequality of both “less than” and “greater than” type.
• I can explain why a “greater than” absolute value produces a union of two intervals.
• I can justify each step by tying it back to distance on the number line.

Connections

Looking back:

• linear-equations supplies every isolation step reused here, minus the reversal rule.
• quadratic-equations builds the algebraic fluency that compound and absolute value work assumes.

Looking ahead:

• absolute-value formalizes the distance idea that drives Example 4 and the Level 5 practice.
• graphing-functions-basic reads inequalities visually, as the region where one graph sits above or below another.
• m141-domain-and-range records domains as solution sets of inequalities, for example where an expression under a square root stays at or above zero.

Real world connections:

• A budget cap is an inequality: total spending must stay at or below a fixed amount.
• A tolerance in manufacturing is an absolute value inequality: a part is acceptable when its measured size differs from the target by less than a set amount.
• A speed limit and a minimum height for a ride are everyday inequalities with a single boundary.

Resources

Last updated: 2026-06-16