How do you solve −3x + 6 > 15?
Subtract 6 from both sides, then divide by −3 and flip the sign. The answer is x < −3, which is (−∞, −3) in interval notation.
A quick check: pick a number that should work, like x = −4. Then −3(−4) + 6 = 18, and 18 > 15 is true. Pick one that shouldn't, like x = 0: 6 > 15 is false, as expected.
When do you flip the inequality sign?
Flip it only when you multiply or divide both sides by a negative number. Adding or subtracting, even a negative, never flips it.
The reason shows up with plain numbers. 2 < 5 is true. Multiply both sides by −1 and you get −2 and −5, and −2 is the larger one, so the sign has to turn around: −2 > −5.
| Move | Flip the sign? |
|---|---|
| Add or subtract any number | No |
| Multiply or divide by a positive | No |
| Multiply or divide by a negative | Yes |
| Swap the two sides | Yes (5 > x becomes x < 5) |
You can also dodge the flip by collecting x on the side where its coefficient is positive. For −x + 4 ≥ 2x − 5, add x to both sides to get 4 ≥ 3x − 5, which leads to x ≤ 3 without dividing by a negative.
When x is on both sides and its coefficient ends up positive, there's nothing to flip. For 2x − 1 > x + 4, subtract x to get x − 1 > 4, then add 1 for x > 5.
How do you write the answer in interval notation?
Interval notation lists the start and end of the solution set. A parenthesis means the endpoint is left out (for < and >) and a square bracket means it's included (for ≤ and ≥). Infinity always gets a parenthesis.
| Inequality | Interval notation |
|---|---|
| x < −3 | (−∞, −3) |
| x ≥ −4 | [−4, ∞) |
| −2 < x < 3 | (−2, 3) |
| x ≤ 2 or x ≥ 3 | (−∞, 2] ∪ [3, ∞) |
On a number line, a parenthesis matches an open circle and a bracket matches a filled-in dot.
How does the compound inequality calculator work?
A compound inequality like −3 < 2x + 1 < 7 has three parts, and you do each move to all three at once. The goal is to get x alone in the middle.
In interval notation that's (−2, 3). Mixed signs work the same way: −4 ≤ 3x + 2 < 11 gives −2 ≤ x < 3, or [−2, 3). If you divide all three parts by a negative, both signs flip.
How do you solve a quadratic inequality?
Get zero on one side, find where the quadratic equals zero, then test a number in each interval between those points. This is called a sign chart, and it's exactly what the quadratic inequality calculator shows.
For x² − 5x + 6 ≥ 0, the boundary equation x² − 5x + 6 = 0 has roots 2 and 3. Those split the number line into three intervals.
| Interval | Test value | x² − 5x + 6 | Keep it? |
|---|---|---|---|
| (−∞, 2) | x = 1 | 2 (positive) | Yes |
| (2, 3) | x = 5/2 | −1/4 (negative) | No |
| (3, ∞) | x = 4 | 2 (positive) | Yes |
Because the sign is ≥, the roots themselves count. The answer is x ≤ 2 or x ≥ 3, or (−∞, 2] ∪ [3, ∞). With a strict sign like x² − 9 < 0 the roots are left out, giving −3 < x < 3.
What are the common mistakes with inequalities?
Forgetting to flip after dividing by a negative is the classic one. From −3x > 9, writing x > −3 gives the exact opposite of the right set.
Flipping when you shouldn't comes a close second. Subtracting 6, or dividing by a positive 3, leaves the sign alone even if the numbers are negative.
With quadratics, students often solve x² − 9 < 0 as x < 3 by taking a square root. That misses the left side of the interval. The sign chart avoids this.
How do you show an inequality on a number line?
Put a circle at the boundary number, then shade the side that works. Use an open circle for < or > and a filled dot for ≤ or ≥.
With x < −3, draw an open circle at −3 and shade everything to the left. For −2 < x < 3, draw open circles at −2 and 3 and shade between them. For x ≤ 2 or x ≥ 3, shade two separate rays pointing outward, with filled dots at 2 and 3.
Reading the sign backwards is another trap. 5 > x and x < 5 are the same set, so if the variable is on the right, read the sign from the variable's side or rewrite it first.
What is the difference between "and" and "or" inequalities?
An "and" inequality keeps only the numbers that satisfy both conditions, so the answer is one connected piece. An "or" inequality keeps numbers that satisfy either condition, so the answer is often two separate pieces.
The compound inequality −2 < x < 3 is an "and" statement: x > −2 and x < 3 at the same time. The quadratic answer x ≤ 2 or x ≥ 3 is an "or" statement, and no single number can be in both parts. In interval notation, the ∪ symbol joins the pieces of an "or" answer.