Inequality Solver
Type a linear inequality, a double inequality such as −3 < 2x + 1 ≤ 7, or two joined by “and” or “or”. You get the solution as an inequality, in interval notation and in set-builder notation, a shaded number line, and every algebra step — with the moment the sign reverses highlighted. Fractions stay exact. Free, no sign-up.
Use <, >, ≤ or ≥ (or <= and >=). Chain two, as in −3 < 2x + 1 ≤ 7, or join two with “and” / “or”.
Solution
x ≥ −4
Inequality form
[−4, ∞)
Interval notation
{x | x ≥ −4}
Set-builder notation
Number line
Filled circle: the end value is included. Open circle: it is excluded. An arrow means the set continues without end.
Step by step
Start with the inequality.
−2x + 3 ≤ 11
Subtract 3 from both sides.
−2x ≤ 8
Divide both sides by −2. The number is negative, so every inequality sign reverses.
x ≥ −4
Sign reversed: ≤ becomes ≥.
Check with a test value
Substitute x = −3, a number in the solution set:
−2 × (−3) + 3 ≤ 11
9 ≤ 11 — true
How to Use the Inequality Solver
- 1. Type the inequality. Use <, >, ≤ and ≥ (or <= and >=). One letter is the unknown, and brackets, fractions such as x/3 and decimals are fine.
- 2. Combine if you need to. Chain two relations for a double inequality (−3 < 2x + 1 ≤ 7), or join two inequalities with the word “and” or “or”.
- 3. Read the answer three ways. The solution appears as an inequality, in interval notation and in set-builder notation, with the same set shaded on a number line.
- 4. Follow the steps. Each operation is written out, and any step that reverses a sign is highlighted. A test value is substituted at the end as a check.
The One Rule That Differs From Equations
Solving an inequality uses the same steps as solving an equation — expand, collect like terms, move terms across, divide — with one exception. Multiplying or dividing both sides by a negative number reverses the inequality sign. Adding or subtracting any number never does, and neither does multiplying or dividing by a positive one.
The reason is on the number line: 2 < 5, but multiplying by −1 reflects the numbers to −2 and −5, and now −2 > −5. The order has reversed, so the sign must too. Forgetting this is the most common mistake in the topic.
Worked Example: −2x + 3 ≤ 11
- Subtract 3 from both sides. −2x ≤ 8. The sign stays.
- Divide both sides by −2. The divisor is negative, so ≤ becomes ≥: x ≥ −4.
- Write the answer. In interval notation the solution is [−4, ∞). The square bracket includes −4, and ∞ always takes a round bracket.
- Check. Try x = 0, which is in the set: −2 × 0 + 3 = 3, and 3 ≤ 11 is true.
Interval Notation and the Number Line
| Inequality | Interval notation | Number line |
|---|---|---|
| x < 3 | (−∞, 3) | Open circle at 3, shaded to the left |
| x ≥ −4 | [−4, ∞) | Filled circle at −4, shaded to the right |
| −2 < x ≤ 3 | (−2, 3] | Open circle at −2, filled circle at 3, shaded between |
| x < −2 or x > 2 | (−∞, −2) ∪ (2, ∞) | Two shaded rays; the middle is empty |
A round bracket or open circle means the end value is not a solution; a square bracket or filled circle means it is. Infinity is not a number you can reach, so it always takes a round bracket. The symbol ∪ means union — everything in either piece.
Double and Compound Inequalities
A double inequality such as −3 < 2x + 1 ≤ 7 is solved by doing the same thing to all three parts. Subtracting 1 gives −4 < 2x ≤ 6, and dividing by 2 gives −2 < x ≤ 3. If you divide by a negative number, every sign reverses and the order of the parts is reversed too, so −3 < −2x ≤ 7 becomes 3/2 > x ≥ −7/2, which you then rewrite in increasing order as −7/2 ≤ x < 3/2.
| Joined by | Means | Example | Solution |
|---|---|---|---|
| and | Both must be true: the overlap | 2x − 1 ≥ 3 and x + 4 < 10 | [2, 6) |
| or | Either may be true: the union | x + 3 > 5 or 2x < −4 | (−∞, −2) ∪ (2, ∞) |
When “and” leaves nothing in common (x > 5 and x < 2) the answer is the empty set ∅. When the variable cancels, the inequality is either always true — 2x + 1 < 2x + 5 holds for every real number — or never true: x + 1 > x + 3 has no solution.
Common Mistakes
- Not reversing the sign after dividing or multiplying by a negative number.
- Reversing it when it is not needed — for example when subtracting a negative term or dividing by a positive.
- A square bracket at infinity. Write (−4, ∞), never (−4, ∞].
- Treating “or” like “and”. Two rays that point away from each other have no overlap, so “and” gives ∅ but “or” gives both rays.
- Reading the circle backwards. Filled means included (≤, ≥); open means excluded (<, >).
Frequently Asked Questions
- When do you flip the inequality sign?
- Only when you multiply or divide both sides by a negative number. Adding, subtracting, and multiplying or dividing by a positive number leave the sign alone. The solver highlights each step where the sign reverses.
- What is the difference between ( ) and [ ] in interval notation?
- A round bracket excludes its end value (used with < and >), and a square bracket includes it (used with ≤ and ≥). Infinity and negative infinity always take round brackets because they are not numbers that can be reached.
- How do you solve a compound inequality?
- Solve each part separately, then combine. With “and”, keep only the values that satisfy both, which is the intersection. With “or”, keep every value that satisfies at least one, which is the union. A double inequality like −3 < 2x + 1 ≤ 7 is solved by acting on all three parts at once.
- What does an open circle mean on a number line?
- The end value is not included in the solution, so the inequality is strict (< or >). A filled circle means the end value is included (≤ or ≥).
- Can an inequality have no solution, or every number as a solution?
- Yes. If the variable cancels and the leftover statement is false, such as 1 > 3, there is no solution. If the leftover statement is true, such as 1 < 5, every real number is a solution, written (−∞, ∞).
- How do I solve an inequality with fractions?
- Multiply every part by the least common multiple of the denominators. That number is positive, so the sign does not change. For x/3 + 1/2 ≥ 5, multiplying by 6 gives 2x + 3 ≥ 30, so x ≥ 27/2.
- Does this solve quadratic or absolute value inequalities?
- No. It handles linear expressions only and says so when a power above 1 appears. A quadratic inequality is solved from the roots of its equation, which the Quadratic Equation Solver finds.