Linear, compound, polynomial, rational, and absolute value inequalities — with sign charts, interval notation, and step-by-step methods for every type.
Every inequality can be written three ways: as an inequality, in interval notation, or on a number line. Mastering all three is essential for precalculus and calculus.
| Inequality | Meaning | Interval Notation | Number Line |
|---|---|---|---|
| x > 3 | x is greater than 3 | (3, ∞) | Open circle at 3, shade right |
| x ≥ 3 | x is at least 3 | [3, ∞) | Closed circle at 3, shade right |
| x < −2 | x is less than −2 | (−∞, −2) | Open circle at −2, shade left |
| x ≤ −2 | x is at most −2 | (−∞, −2] | Closed circle at −2, shade left |
| −1 < x ≤ 5 | x between −1 and 5 | (−1, 5] | Open at −1, closed at 5 |
| x < 0 or x ≥ 4 | x outside [0, 4) | (−∞, 0) ∪ [4, ∞) | Two rays, union |
Bracket Rules
Square bracket [ ] means the endpoint is included (≤ or ≥). Parenthesis ( ) means excluded (< or >). Infinity always uses parentheses.
Union vs. Intersection
∪ (union) combines two sets — use for OR inequalities. ∩ (intersection) takes the overlap — use for AND inequalities. Most compound inequalities produce one of these.
A linear inequality has degree 1. Solve it exactly like a linear equation with one critical exception: flip the inequality sign when multiplying or dividing by a negative.
Simplify both sides
Distribute, combine like terms, and clear fractions by multiplying by the LCD. Do this on both sides before moving variables.
Isolate the variable term
Use addition and subtraction to get all variable terms on one side. These operations never flip the inequality sign.
Divide to solve — and watch the sign
Divide both sides by the coefficient of the variable. If that coefficient is negative, FLIP the inequality symbol. This is the #1 source of errors.
Write in interval notation
Convert the answer to interval notation. Use a bracket at any endpoint that is included and a parenthesis at any endpoint that is excluded or at infinity.
The Golden Rule
−4x ≥ 20 → divide by −4 and flip: x ≤ −5
Dividing by −4 reverses the inequality. Solution: (−∞, −5].
Written as a three-part inequality: a < expression < b. Solve by operating on all three parts simultaneously. The solution is an intersection — values satisfying both conditions at once.
−5 < 3x + 1 ≤ 10
−6 < 3x ≤ 9 (subtract 1)
−2 < x ≤ 3 (divide by 3)
Answer: (−2, 3]
Two separate inequalities joined by “or.” Solve each independently, then take the union (∪) of both solution sets. The solution includes values satisfying either condition.
2x − 3 < −7 or 2x − 3 > 5
2x < −4 or 2x > 8
x < −2 or x > 4
Answer: (−∞, −2) ∪ (4, ∞)
No Solution
An AND inequality like x > 5 AND x < 2 has no solution — no number is simultaneously greater than 5 and less than 2.
Answer: ∅ (empty set)
All Real Numbers
An OR inequality like x > 5 OR x < 7 is satisfied by every real number — every x is either greater than 5 or less than 7 (most are both).
Answer: (−∞, ∞)
A polynomial inequality has degree ≥ 2. You cannot simply divide out the variable because it might be zero or negative. The sign chart method systematically tracks the sign of the polynomial across every interval.
Move everything to one side
Rewrite so one side is zero: p(x) > 0, p(x) < 0, p(x) ≥ 0, or p(x) ≤ 0. Never leave terms on both sides.
Factor completely
Factor out any GCF, then factor the remaining polynomial. For quadratics, use factoring, completing the square, or the quadratic formula to find real zeros.
Find all critical points
Set each factor equal to zero and solve. These are the x-values where the polynomial can change sign. Mark them on a number line in order.
Test each interval
Pick one test value inside each interval and evaluate the sign of each factor. Record + or − for each factor in each interval.
Determine the overall sign
Multiply the signs across each interval. (+)(+) = +; (−)(−) = +; (+)(−) = −. The product gives the sign of p(x) in that interval.
Write the solution
Select the intervals where p(x) has the correct sign. Include endpoints (use [ ]) if the inequality is ≤ or ≥; exclude them (use ( )) if it is strict.
x² − x − 6 = (x − 3)(x + 2) > 0
Critical points: x = 3 and x = −2
Intervals: (−∞, −2), (−2, 3), (3, ∞)
| Interval | Test x | (x − 3) | (x + 2) | Product Sign |
|---|---|---|---|---|
| (−∞, −2) | x = −3 | − | − | + ✓ |
| (−2, 3) | x = 0 | − | + | − ✗ |
| (3, ∞) | x = 4 | + | + | + ✓ |
A rational inequality contains a variable in the denominator. The key principle: never cross-multiply — the denominator's sign is unknown. Always bring everything to one side and analyze numerator and denominator separately.
Get zero on one side
Move all terms to one side to get a single rational expression compared to zero. Combine fractions over a common denominator if needed.
Find critical points from numerator AND denominator
Set the numerator = 0 to find where the expression equals zero. Set the denominator = 0 to find where the expression is undefined. Both types are critical points.
Mark critical points and test intervals
Plot all critical points on a number line. Choose a test value in each interval. Evaluate the sign of the numerator and denominator separately, then divide signs.
Exclude denominator zeros from the solution
Values that make the denominator zero are never in the solution, even with ≤ or ≥. Always use a parenthesis at denominator zeros.
Numerator zero: x − 1 = 0 → x = 1
Denominator zero: x + 4 = 0 → x = −4 (excluded)
Intervals: (−∞, −4), (−4, 1), (1, ∞)
| Interval | Numerator | Denominator | Fraction Sign |
|---|---|---|---|
| (−∞, −4) | − | − | + ✓ |
| (−4, 1) | − | + | − ✗ |
| (1, ∞) | + | + | + ✓ |
x = 1: numerator = 0, denominator ≠ 0 → expression = 0, which satisfies ≥ 0 ✓ include
x = −4: denominator = 0 → undefined, never include ✗
Absolute value measures distance from zero. |expression| = k means the expression is k units from zero, so the expression is either k or −k. Inequalities extend this to “within k units” or “farther than k units.”
The expression must be within k units of zero. Rewrite as a three-part AND inequality.
|u| < k ⟺ −k < u < k
Same rule for ≤: use −k ≤ u ≤ k
|2x − 3| < 7
−7 < 2x − 3 < 7
−4 < 2x < 10
−2 < x < 5
Answer: (−2, 5)
The expression must be more than k units from zero. Rewrite as two separate OR inequalities.
|u| > k ⟺ u < −k or u > k
Same rule for ≥: use u ≤ −k or u ≥ k
|3x + 1| ≥ 8
3x + 1 ≤ −8 or 3x + 1 ≥ 8
3x ≤ −9 or 3x ≥ 7
x ≤ −3 or x ≥ 7/3
Answer: (−∞, −3] ∪ [7/3, ∞)
|u| < 0
No Solution
Absolute value is never negative, so |u| can never be less than 0.
|u| ≥ 0
All Real Numbers
Absolute value is always ≥ 0, so every value of x satisfies this.
|u| > 0
x ≠ (zero of u)
True everywhere except where u = 0. Example: |x − 2| > 0 is all x ≠ 2.
3(x − 2) − 4 > 5x + 6
Step 1: Distribute: 3x − 6 − 4 > 5x + 6
Step 2: Combine: 3x − 10 > 5x + 6
Step 3: Subtract 5x: −2x − 10 > 6
Step 4: Add 10: −2x > 16
Step 5: Divide by −2 (FLIP): x < −8
Answer: (−∞, −8)
−1 ≤ (4 − 2x)/3 < 3
Multiply all parts by 3: −3 ≤ 4 − 2x < 9
Subtract 4: −7 ≤ −2x < 5
Divide by −2 (FLIP both symbols): 7/2 ≥ x > −5/2
Rewrite in increasing order: −5/2 < x ≤ 7/2
Answer: (−5/2, 7/2]
x³ − 4x² − 12x ≤ 0
Factor: x(x² − 4x − 12) = x(x − 6)(x + 2) ≤ 0
Critical points: x = 0, x = 6, x = −2
Intervals: (−∞, −2), (−2, 0), (0, 6), (6, ∞)
Test x = −3: (−3)(−9)(−1) = −27 ≤ 0 ✓
Test x = −1: (−1)(−7)(1) = 7 > 0 ✗
Test x = 1: (1)(−5)(3) = −15 ≤ 0 ✓
Test x = 7: (7)(1)(9) = 63 > 0 ✗
Include all endpoints (≤): x = −2, 0, 6 all give 0 ≤ 0 ✓
Answer: (−∞, −2] ∪ [0, 6]
(x² − 4) / (x − 1) < 0
Factor numerator: (x − 2)(x + 2) / (x − 1) < 0
Critical points: x = 2, x = −2 (num zeros); x = 1 (den zero, excluded)
Intervals: (−∞, −2), (−2, 1), (1, 2), (2, ∞)
Test x = −3: (−5)(−1)/(−4) = −5/4 < 0 ✓
Test x = 0: (−2)(2)/(−1) = 4 > 0 ✗
Test x = 1.5: (−0.5)(3.5)/(0.5) = −3.5 < 0 ✓
Test x = 3: (1)(5)/(2) = 2.5 > 0 ✗
Strict inequality → exclude all critical points; x = 1 always excluded
Answer: (−∞, −2) ∪ (1, 2)
|5 − 2x| ≤ 9 → AND case (less than or equal)
Rewrite: −9 ≤ 5 − 2x ≤ 9
Subtract 5: −14 ≤ −2x ≤ 4
Divide by −2 (FLIP): 7 ≥ x ≥ −2
Rewrite: −2 ≤ x ≤ 7
Answer: [−2, 7]
✗ Not flipping the inequality sign
Flip whenever you multiply or divide both sides by a NEGATIVE number. Adding or subtracting never changes the direction.
✗ Cross-multiplying in rational inequalities
Never cross-multiply — you don't know the sign of the denominator. Bring everything to one side and use a sign chart.
✗ Including denominator zeros in the solution
Values that make the denominator zero are UNDEFINED — they can never be in the solution, even with ≤ or ≥.
✗ Wrong case for absolute value
Less than (< or ≤) gives an AND (between) inequality. Greater than (> or ≥) gives an OR (outside) inequality. Don't mix them up.
✗ Using wrong bracket type in interval notation
Square bracket [ ] for included endpoints (≤ or ≥). Parenthesis ( ) for excluded endpoints (< or >) and always for ±∞.
✗ Forgetting to factor before using the sign chart
The sign chart only works on a factored polynomial. Factor completely — including pulling out any GCF — before marking critical points.
Solving a linear inequality is almost identical to solving a linear equation — isolate the variable using the same addition, subtraction, multiplication, and division steps. The critical difference: whenever you multiply or divide both sides by a negative number, you must flip the inequality symbol. For example, −3x > 12 becomes x < −4 after dividing both sides by −3 (symbol flips). Express the answer in interval notation or on a number line. Example: 2x − 5 ≤ 7 → 2x ≤ 12 → x ≤ 6, written as (−∞, 6].
Interval notation is a concise way to write a set of real numbers. Use parentheses ( ) for strict inequalities (the endpoint is NOT included) and square brackets [ ] for non-strict inequalities (the endpoint IS included). The symbol ∞ always uses a parenthesis because infinity is not a real number you can reach. Examples: x > 3 → (3, ∞); x ≤ −1 → (−∞, −1]; −2 < x ≤ 5 → (−2, 5]; x = all reals → (−∞, ∞). For unions of intervals, use the ∪ symbol: x < −1 or x > 4 → (−∞, −1) ∪ (4, ∞).
A compound inequality joins two inequalities with 'and' or 'or'. For an AND compound inequality like −3 < 2x + 1 ≤ 7, work on all three parts simultaneously: subtract 1 from all parts: −4 < 2x ≤ 6, then divide all parts by 2: −2 < x ≤ 3, giving interval notation (−2, 3]. For an OR inequality, solve each part separately and take the union: x < −1 or x > 4 is the set (−∞, −1) ∪ (4, ∞). AND inequalities produce an intersection (overlap); OR inequalities produce a union.
The sign chart method solves polynomial inequalities by analyzing the sign of each factor in each interval between zeros. Step 1: move all terms to one side and factor completely. Step 2: find all real zeros (where each factor equals zero) and mark them on a number line — these are your critical points. Step 3: these points divide the number line into intervals. Step 4: pick a test value in each interval, plug it into each factor, and record the sign (+/−). Step 5: multiply signs to get the overall sign of the polynomial in each interval. Include or exclude endpoints based on whether the inequality is strict (< >) or non-strict (≤ ≥).
To solve a rational inequality like (x + 2)/(x − 3) > 0: Step 1 — move everything to one side so you have a single fraction compared to zero (do NOT cross-multiply since the sign of the denominator is unknown). Step 2 — find zeros of the numerator (where the expression equals zero) and zeros of the denominator (where the expression is undefined). These are the critical points. Step 3 — mark all critical points on a number line. Step 4 — test a value in each interval. Step 5 — identify which intervals make the inequality true. Important: values that make the denominator zero are NEVER included in the solution, even with ≤ or ≥.
There are two cases based on the direction of the inequality. For |expression| < k (less than): this means −k < expression < k — the expression must be between −k and k. Solve the compound AND inequality. Example: |2x − 1| < 5 → −5 < 2x − 1 < 5 → −4 < 2x < 6 → −2 < x < 3, so (−2, 3). For |expression| > k (greater than): this means expression < −k or expression > k — the expression must be outside the range. Example: |3x + 2| ≥ 7 → 3x + 2 ≤ −7 or 3x + 2 ≥ 7 → x ≤ −3 or x ≥ 5/3, so (−∞, −3] ∪ [5/3, ∞).
Cross-multiplication works reliably when you know both sides are positive. In a rational inequality, the denominator might be negative for some values of x. If you multiply both sides by a negative quantity, the inequality symbol must flip — but you don't know ahead of time which values of x make the denominator negative. Cross-multiplying without knowing the sign leads to incorrect answers. The safe approach is to bring everything to one side, get a single fraction ≥ 0 or > 0, then use the sign chart on both the numerator and denominator separately.
On a number line: use an open circle (○) at a value that is NOT included (strict inequality < or >) and a closed circle (●) at a value that IS included (non-strict ≤ or ≥). Shade (draw a ray or segment) over all values in the solution set. For x > 2: open circle at 2, shade to the right. For x ≤ −1: closed circle at −1, shade to the left. For −3 ≤ x < 5: closed circle at −3, open circle at 5, shade between them. The number line picture mirrors the interval notation: (2, ∞) corresponds to open circle at 2 with rightward shading.
An equation has one or finitely many solutions (specific values where both sides are equal). An inequality has infinitely many solutions — an entire interval or union of intervals. When you solve an equation like 2x + 3 = 7, you get one answer: x = 2. When you solve an inequality like 2x + 3 < 7, you get a range: x < 2, meaning every real number less than 2 is a solution. Additional algebraic difference: multiplying or dividing by a negative number flips the inequality symbol but does nothing special to an equation.
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