Inequalities look a lot like equations, but instead of asking “what value makes both sides equal?” they ask “what values make one side bigger or smaller than the other?” That tiny shift in the question changes the answer from a single number to a whole range of numbers. Once you understand that idea, solving inequalities becomes almost as routine as solving equations, with one famous twist you will learn to watch for.
In this guide you will learn exactly how to solve inequalities from the ground up: what the symbols mean, the one golden rule that trips up most students, how to handle two-step and multi-step problems, how to work with compound inequalities, and how to graph your answers on a number line and write them in interval notation. We will work through several fully worked examples, checking every single step so you can see precisely where each number comes from.
If you ever want to verify your work or see a clean, structured solution while you study, our free algebra solver walks through inequalities one move at a time. Use it to confirm your answers as you practice the examples below.
What Is an Inequality?
An inequality is a mathematical statement that compares two expressions using a relationship other than “equals.” Instead of an equals sign, it uses one of four comparison symbols. For example, \( x > 3 \) is read “x is greater than 3,” and it is true for infinitely many numbers: \( 3.1 \), \( 4 \), \( 100 \), and so on.
Because the solution is usually a set of numbers rather than a single value, we describe the answer in one of three ways:
- Inequality notation: \( x > 3 \)
- A number-line graph: an arrow shaded to the right of 3
- Interval notation: \( (3, \infty) \)
The four inequality symbols
Knowing each symbol cold is the first step. Notice how the “or equal to” versions decide whether the endpoint is included.
| Symbol | Meaning | Number-line dot | Interval bracket |
|---|---|---|---|
| \( < \) | less than | open circle (not included) | parenthesis ( ) |
| \( > \) | greater than | open circle (not included) | parenthesis ( ) |
| \( \le \) | less than or equal to | closed (filled) circle | square bracket [ ] |
| \( \ge \) | greater than or equal to | closed (filled) circle | square bracket [ ] |
Quick memory trick
The inequality symbol always “opens” toward the larger quantity, like a hungry mouth that wants the bigger number. In \( 5 > 2 \), the wide opening faces the 5.
How to Solve Inequalities Step by Step
Here is the most reassuring fact in this whole topic: you solve an inequality almost exactly the way you solve an equation. Your goal is the same: to isolate the variable on one side by undoing operations with their inverses. If you are comfortable with our guide on How to Solve Linear Equations Step by Step, you already know most of the moves.
The reliable, repeatable process looks like this:
- Simplify each side. Distribute through parentheses and combine like terms.
- Collect variable terms on one side and constant terms on the other, using addition or subtraction.
- Isolate the variable by multiplying or dividing.
- Apply the flip rule (explained next) whenever you multiply or divide by a negative number.
- Graph or write the solution in interval notation.
- Check with a test value to confirm the direction is right.
The one golden rule: flip the sign for negatives
There is exactly one rule that makes inequalities different from equations, and it is the source of nearly every mistake:
The flip rule
Whenever you multiply or divide both sides by a negative number, you must reverse the direction of the inequality symbol. So \( < \) becomes \( > \), and \( \ge \) becomes \( \le \).
Why does this happen? Think about a true statement like \( 2 < 6 \). Multiply both sides by \( -1 \) and you get \( -2 \) and \( -6 \). On the number line, \( -2 \) sits to the right of \( -6 \), so \( -2 > -6 \). The relationship literally turned around, which is why the symbol must turn around too. Adding and subtracting never trigger this; only multiplying or dividing by a negative does.
Worked Example 1: A One-Step Inequality
Let’s start simple to build confidence.
Solve: \( x + 7 < 12 \)
Step 1 — Undo the addition. The variable has 7 added to it, so subtract 7 from both sides:
$$ x + 7 - 7 < 12 - 7 $$Step 2 — Simplify.
$$ x < 5 $$Because we only subtracted, no flip was needed. The solution is every number less than 5.
Step 3 — Check. Pick a test value that should work, such as \( x = 0 \): \( 0 + 7 = 7 \), and \( 7 < 12 \) is true. Now test a value that should fail, like \( x = 10 \): \( 10 + 7 = 17 \), and \( 17 < 12 \) is false. Perfect, our boundary is correct.
On a number line, draw an open circle at 5 (because 5 itself is not included) and shade everything to the left.
Worked Example 2: A Two-Step Inequality With a Flip
This example shows the flip rule in action, so read it carefully.
Solve: \( -3x + 4 \ge 19 \)
Step 1 — Subtract 4 from both sides to start isolating the variable term:
$$ -3x + 4 - 4 \ge 19 - 4 $$ $$ -3x \ge 15 $$Step 2 — Divide both sides by \( -3 \). Because we are dividing by a negative number, we must flip \( \ge \) to \( \le \):
$$ \frac{-3x}{-3} \le \frac{15}{-3} $$ $$ x \le -5 $$Step 3 — Check. Test \( x = -6 \) (which should work): \( -3(-6) + 4 = 18 + 4 = 22 \), and \( 22 \ge 19 \) is true. Test the boundary \( x = -5 \): \( -3(-5) + 4 = 15 + 4 = 19 \), and \( 19 \ge 19 \) is true, so the endpoint is included. Finally test \( x = -4 \) (which should fail): \( -3(-4) + 4 = 12 + 4 = 16 \), and \( 16 \ge 19 \) is false. Everything lines up.
Graph it with a closed (filled) circle at \( -5 \) because of the “or equal to,” then shade to the left.
Common mistake
Many students forget to flip and incorrectly write \( x \ge -5 \). If you ever feel unsure, plug a number from your answer back into the original inequality. The test step in Example 2 instantly reveals a missed flip.
Worked Example 3: Variables on Both Sides
When the variable appears on both sides, gather the variable terms together first, exactly as you would with an equation.
Solve: \( 5x - 3 > 2x + 9 \)
Step 1 — Move the variable terms to one side. Subtract \( 2x \) from both sides:
$$ 5x - 3 - 2x > 2x + 9 - 2x $$ $$ 3x - 3 > 9 $$Step 2 — Move the constants to the other side. Add 3 to both sides:
$$ 3x - 3 + 3 > 9 + 3 $$ $$ 3x > 12 $$Step 3 — Divide by 3. Since 3 is positive, the symbol stays the same:
$$ \frac{3x}{3} > \frac{12}{3} $$ $$ x > 4 $$Step 4 — Check. Test \( x = 5 \): left side \( 5(5) - 3 = 22 \), right side \( 2(5) + 9 = 19 \), and \( 22 > 19 \) is true. Test the boundary \( x = 4 \): left side \( 5(4) - 3 = 17 \), right side \( 2(4) + 9 = 17 \), and \( 17 > 17 \) is false, confirming 4 is not included.
Want to check a tricky one fast?
Solve it instantly and see each step laid out with our step-by-step inequality solver, then compare it to your own work to find any slip.
Worked Example 4: A Compound Inequality
A compound inequality traps the variable between two values, like \( -4 \le 2x - 6 < 8 \). The key idea is that whatever you do, you must do it to all three parts at once.
Solve: \( -4 \le 2x - 6 < 8 \)
Step 1 — Add 6 to all three parts to undo the subtraction in the middle:
$$ -4 + 6 \le 2x - 6 + 6 < 8 + 6 $$ $$ 2 \le 2x < 14 $$Step 2 — Divide all three parts by 2. Since 2 is positive, no flips:
$$ \frac{2}{2} \le \frac{2x}{2} < \frac{14}{2} $$ $$ 1 \le x < 7 $$Step 3 — Check a value in the middle. Try \( x = 6 \): \( 2(6) - 6 = 6 \), and \( -4 \le 6 < 8 \) is true. The left endpoint \( x = 1 \) gives \( 2(1) - 6 = -4 \), which satisfies \( -4 \le -4 \), so it is included. The right endpoint \( x = 7 \) gives \( 2(7) - 6 = 8 \), and \( 8 < 8 \) is false, so 7 is excluded.
The graph is a segment from a closed circle at 1 to an open circle at 7, with everything between shaded.
“And” versus “or” compound inequalities
The example above is an “and” compound inequality: a number must satisfy both conditions at the same time, producing a single connected interval. An “or” compound inequality, such as \( x < -2 \) or \( x \ge 5 \), is satisfied if a number meets either condition, so its graph is two separate pieces and its interval notation uses a union symbol: \( (-\infty,\ -2) \cup [5,\ \infty) \).
Worked Example 5: A Real-World Word Problem
Inequalities shine in real situations that involve “at least,” “at most,” “no more than,” or “minimum.” Watch how the words translate into a symbol.
Problem: Maria has scored 88, 92, and 85 on her first three math tests. She wants her average across four tests to be at least 90. What score does she need on the fourth test?
Step 1 — Translate the words. “Average of four tests is at least 90” means the average must be \( \ge 90 \). Let \( x \) be the fourth score:
$$ \frac{88 + 92 + 85 + x}{4} \ge 90 $$Step 2 — Add the known scores. Since \( 88 + 92 + 85 = 265 \):
$$ \frac{265 + x}{4} \ge 90 $$Step 3 — Multiply both sides by 4 (a positive number, so no flip):
$$ 265 + x \ge 360 $$Step 4 — Subtract 265 from both sides:
$$ x \ge 95 $$Step 5 — Check. If Maria scores exactly 95, her total is \( 265 + 95 = 360 \), and \( 360 \div 4 = 90 \), which meets the “at least 90” goal exactly.
Graphing Solutions and Interval Notation
Translating between the three formats is a skill teachers test often, so keep this quick reference handy.
| Inequality | Number-line endpoint | Interval notation |
|---|---|---|
| \( x < a \) | open circle at \( a \), shade left | \( (-\infty,\ a) \) |
| \( x \le a \) | closed circle at \( a \), shade left | \( (-\infty,\ a] \) |
| \( x > a \) | open circle at \( a \), shade right | \( (a,\ \infty) \) |
| \( x \ge a \) | closed circle at \( a \), shade right | \( [a,\ \infty) \) |
| \( a \le x < b \) | closed at \( a \), open at \( b \) | \( [a,\ b) \) |
Key idea about infinity
Infinity \( (\infty) \) is a direction, not a number, so it always gets a parenthesis, never a square bracket. You can never “reach and include” infinity.
Special Cases You Might Meet
When the variable cancels out
Sometimes simplifying makes the variable disappear, leaving a statement that is either always true or always false.
- If you end with something true like \( 3 < 7 \), the inequality is true for all real numbers. The solution is \( (-\infty,\ \infty) \).
- If you end with something false like \( 5 < 1 \), there is no solution. We write \( \varnothing \), the empty set.
Inequalities with parentheses
Distribute first, then proceed as usual. For \( 2(x - 3) \le 4x + 2 \), distribute to get \( 2x - 6 \le 4x + 2 \). Subtract \( 4x \): \( -2x - 6 \le 2 \). Add 6: \( -2x \le 8 \). Divide by \( -2 \) and flip: \( x \ge -4 \). A quick check with \( x = 0 \) gives \( 2(-3) = -6 \le 2 \), which is true, confirming the result.
Common Mistakes to Avoid
- Forgetting to flip the symbol when multiplying or dividing by a negative. This is the number one error, so circle every negative coefficient as a reminder.
- Flipping when you shouldn’t. Adding or subtracting a negative number does not flip the symbol. Only multiplying or dividing by a negative does.
- Mixing up open and closed circles. Strict symbols \( (<, >) \) use open circles; “or equal to” symbols \( (\le, \ge) \) use closed circles.
- Dropping infinity’s parenthesis. Writing \( [3, \infty] \) is incorrect; it must be \( [3, \infty) \).
- Only operating on part of a compound inequality. Every move must be applied to all three sections.
Study tip
After solving, always substitute one number from your solution region and one from outside it back into the original inequality. If your inside value works and your outside value fails, your answer (and your flip decision) is almost certainly correct.
How Inequalities Connect to the Rest of Algebra
The mechanics you just practiced carry directly into other topics. The balancing moves are identical to those in How to Solve Linear Equations Step by Step, the only addition being the flip rule. When you reach quadratic inequalities such as \( x^2 - x - 6 > 0 \), you will first factor exactly as shown in How to Solve Quadratic Equations (4 Methods), then test intervals around the roots. And systems of inequalities, where two or more conditions must hold together, build on the same reasoning as How to Solve Systems of Equations (Substitution, Elimination & Graphing). Mastering the basics here pays off across all of them.
Quick Practice
Try these on your own, then verify each with the solver:
- Solve \( x - 5 > 2 \). (Answer: \( x > 7 \))
- Solve \( -2x \le 10 \). (Answer: \( x \ge -5 \) — remember the flip)
- Solve \( 4x + 1 < 2x + 9 \). (Answer: \( x < 4 \))
- Solve \( -1 < 3x + 2 \le 11 \). (Answer: \( -1 < x \le 3 \))
Conclusion
Solving inequalities comes down to a small, dependable routine: simplify, isolate the variable using inverse operations, flip the symbol only when you multiply or divide by a negative, and then graph or write your answer in interval notation. Add a quick substitution check at the end and you will catch nearly any error before it costs you points. With the five worked examples above, you have seen one-step, two-step, both-sides, compound, and real-world problems solved from start to finish.
Ready to put it into practice? Work through a few problems, then confirm every step with our free algebra solver. For more clear, example-driven walkthroughs across algebra and beyond, explore the guides on our math blog. The more you practice, the more these moves become second nature.
