How to solve a linear inequality
A linear inequality compares two expressions in which x appears only to the first power. The calculator treats each side as ax + b and cx + d, then follows these steps:
- Write the left side as ax + b and the right side as cx + d.
- Collect the x terms on one side and the constants on the other. This gives (a − c)x compared with d − b. Adding or subtracting the same amount on both sides never changes the sign.
- Divide both sides by (a − c). If that number is negative, reverse the sign: for example, −2x ≤ 6 becomes x ≥ −3.
- If (a − c) is zero, x has cancelled out. Check whether the remaining statement about constants is true or false.
- Write the answer as an inequality and, if you need it, in interval notation.
Interval notation describes the same solution set. A parenthesis excludes a finite endpoint, a square bracket includes it, and infinity always takes a parenthesis because it is not a number that can be reached. For example, x ≥ −3 becomes [−3, ∞). An empty set, ∅, means no real value works.
Worked examples
| Inequality | Working | Solution | Interval |
|---|---|---|---|
| 3x + 5 < 20 | 3x < 15, then divide by 3 | x < 5 | (−∞, 5) |
| −2x + 3 ≤ 9 | −2x ≤ 6, then divide by −2 and reverse the sign | x ≥ −3 | [−3, ∞) |
| 4x + 12 < 2x + 24 | 2x < 12, then divide by 2 | x < 6 | (−∞, 6) |
| 2x + 3 < 2x + 8 | x cancels, leaving 3 < 8, which is true | All real numbers | (−∞, ∞) |
| 2x + 3 > 2x + 8 | x cancels, leaving 3 > 8, which is false | No solution | ∅ |
Reading the number line
The four inequality signs map to two circle styles, a shading direction and an interval form.
| Inequality | Endpoint circle | Shading | Interval |
|---|---|---|---|
| x < 5 | Open (5 excluded) | To the left | (−∞, 5) |
| x ≤ 5 | Filled (5 included) | To the left | (−∞, 5] |
| x > 5 | Open (5 excluded) | To the right | (5, ∞) |
| x ≥ 5 | Filled (5 included) | To the right | [5, ∞) |
Common mistakes to avoid
- Forgetting to reverse the sign when multiplying or dividing by a negative number.
- Reversing the sign unnecessarily when only adding or subtracting.
- Putting a square bracket on infinity. Write [3, ∞), never [3, ∞].
- Giving the real-number boundary as a count. If x < 5 and x must be a whole number, the largest value is 4, not 5.
- Reading the sign backwards. "x is less than 5" means x < 5, and the symbol's open end faces the larger value.
- Multiplying both sides by an expression that contains x. Its sign is unknown, so this calculator's method does not apply.
Use inequalities in everyday decisions
Stay within a shopping budget
Each item costs 18, delivery adds 7 and your budget is 100 (in any currency). Enter 18x + 7 ≤ 100. The solution is x ≤ 31/6, or about 5.17. Because an item count must be a non-negative whole number, you can buy at most 5 items: the total is 97. Six would cost 115. The calculator gives a real-number range; apply practical restrictions separately.
Compare two rental plans
Plan A costs 12 plus 4 per hour. Plan B costs 24 plus 2 per hour. To find when A is cheaper, solve 4x + 12 < 2x + 24. The result is x < 6. For non-negative rental hours, A is cheaper below six hours; at six hours both plans cost 36. Check billing increments before using a fractional duration.
Reach a production target
You already have 40 units and produce 15 units each hour. To reach at least 100 units, solve 15x + 40 ≥ 100. You need x ≥ 4 hours, assuming a constant production rate and no losses.
Frequently asked questions
Why does dividing by a negative number reverse the sign?
Multiplying or dividing by a negative number flips the order of numbers on the number line. For example, 2 < 3, but after multiplying by −1 you get −2 > −3. Adding or subtracting the same quantity on both sides moves both numbers equally and keeps their order, so it never reverses the sign.
Why is the endpoint included for ≤ and ≥ but not for < and >?
Because the sign decides whether the boundary value itself makes the statement true. In 3x + 5 ≤ 20, the value x = 5 gives exactly 20, so it works and is included. In 3x + 5 < 20, the same value gives 20 < 20, which is false, so it is excluded. A filled circle or square bracket means included; an open circle or parenthesis means excluded.
Why does infinity always get a parenthesis?
A square bracket says the endpoint belongs to the set, and infinity is not a number that any set can contain. It describes a set that continues without end. That is why intervals are written (−∞, 5] or [3, ∞), never with a bracket against the infinity symbol.
What does it mean when x cancels out?
The x terms on both sides are equal, so the answer no longer depends on x. If the remaining statement is true, every real number works. If it is false, no number works. For example, 2x + 3 < 2x + 8 reduces to 3 < 8, which is always true, while 2x + 3 > 2x + 8 reduces to 3 > 8, which is always false.
Why can a correct solution still be unusable in real life?
The calculator solves over all real numbers, but real situations often have extra limits. A count of items must be a non-negative whole number, and a rental may be billed in whole hours. For x < 5 the largest whole number allowed is 4; for x ≤ 5 it is 5. Always apply those limits after solving, and check the boundary carefully.
How can I check my answer without a calculator?
Test three values: one inside the solution set, one outside it and the boundary itself. For 3x + 5 < 20 with solution x < 5, try x = 4 (17 < 20, true), x = 6 (23 < 20, false) and x = 5 (20 < 20, false). If the results match your answer, including at the boundary, the sign and endpoint are probably right.
Why is 3 < x the same as x > 3?
Both say that x is a number larger than 3. Swapping the two sides of an inequality reverses the direction of the sign so the meaning stays the same. This is different from dividing by a negative, which changes the numbers themselves and therefore needs the sign to be reversed to keep the statement true.
Why can't this solver handle quadratic or compound inequalities?
Their solutions work differently. A quadratic such as x² > 4 gives two separate pieces, x < −2 or x > 2, so a single boundary and direction are not enough. A compound inequality combines two conditions that must both hold, or at least one must hold. These need extra steps, such as finding where each part changes sign, which a single-boundary linear method does not cover.