Solving Linear Inequalities
Linear inequalities are solved almost exactly like linear equations, with one critical difference: when you multiply or divide both sides by a negative number, you must reverse the inequality sign.
Solve 3x − 7 > 2.
Add 7: 3x > 9.
Divide by 3: x > 3.
The solution is all real numbers greater than 3, written in interval notation as (3, ∞).
(3, ∞)Now consider −2x + 5 ≤ 11.
Subtract 5: −2x ≤ 6.
Divide by −2 and flip the sign: x ≥ −3.
[−3, ∞)Why does the sign flip? Because multiplying by a negative reverses the order of numbers on the number line. If a < b, then −a > −b. This rule is the single most common source of errors in inequality problems.
A useful check: after solving, pick a number from your solution set and substitute it into the original inequality to verify it works. Then pick a number outside the solution set and verify it does not work. Solver AI highlights the sign-flip step clearly so you never miss it.
Compound Inequalities
A compound inequality combines two inequalities using "and" or "or."
"And" compound inequalities
An "and" compound inequality (also called a conjunction) represents the intersection of two solution sets.
−1 < 2x + 3 ≤ 7 means 2x + 3 > −1 AND 2x + 3 ≤ 7. Solve by working on all three parts simultaneously:
Subtract 3 from all parts: −4 < 2x ≤ 4.
Divide by 2: −2 < x ≤ 2.
(−2, 2]"Or" compound inequalities
An "or" compound inequality (disjunction) represents the union of two solution sets.
x − 1 < −3 OR x − 1 > 5. Solve each separately:
x < −2 or x > 6.
(−∞, −2) ∪ (6, ∞)The key difference: "and" gives you the overlap (often a bounded interval), while "or" gives you the combined region (often two separate rays). Graphically, "and" is the intersection of two shaded regions on a number line, and "or" is their union.
When writing answers, use square brackets [ ] to include an endpoint (≤ or ≥) and parentheses ( ) to exclude it (< or >).
Absolute Value Inequalities
Absolute value inequalities come in two flavors, and recognizing which type you have determines the solution method.
| Type | Form | Translates To | Memory Trick |
|---|---|---|---|
| 1 | |expression| < k | Compound "and": −k < expression < k | "less than" means "between" |
| 2 | |expression| > k | Compound "or": expression < −k OR expression > k | "greater than" means "outside" |
Type 1 — Less than: |expression| < k (where k > 0) translates to a compound "and" inequality: −k < expression < k.
|2x − 3| < 5 becomes −5 < 2x − 3 < 5.
Add 3: −2 < 2x < 8.
Divide by 2: −1 < x < 4.
(−1, 4)Type 2 — Greater than: |expression| > k translates to a compound "or" inequality: expression < −k OR expression > k.
|x + 4| ≥ 7 becomes x + 4 ≤ −7 or x + 4 ≥ 7.
Solve: x ≤ −11 or x ≥ 3.
(−∞, −11] ∪ [3, ∞)A helpful memory trick: "less than" means "between" (and), "greater than" means "outside" (or).
Special cases
| Case | Condition | Solution |
|---|---|---|
| 1 | |expression| < 0 | No solution (absolute value is never negative). |
| 2 | |expression| ≥ 0 | True for all real numbers. |
| 3 | |expression| > 0 | True for all values except where the expression equals zero. |
Quadratic Inequalities
A quadratic inequality like x² − 5x + 6 > 0 requires a different approach than linear inequalities.
- Step 1 — Solve the corresponding equation
Solve
x² − 5x + 6 = 0to find the critical values. Factoring:(x − 2)(x − 3) = 0, so x = 2 and x = 3. - Step 2 — Divide the number line
These critical values divide the number line into three intervals:
(−∞, 2),(2, 3), and(3, ∞). - Step 3 — Test a value from each interval
Test a value from each interval in the original inequality. Test x = 0:
0 − 0 + 6 = 6 > 0. True. Test x = 2.5:6.25 − 12.5 + 6 = −0.25 > 0. False. Test x = 4:16 − 20 + 6 = 2 > 0. True. - Step 4 — Assemble the solution
The solution includes intervals where the test is true:
(−∞, 2) ∪ (3, ∞). If the inequality were ≥ instead of >, include the endpoints:(−∞, 2] ∪ [3, ∞).
The sign-chart method works because a polynomial can only change sign at its zeros. Between consecutive zeros, the polynomial is entirely positive or entirely negative.
This technique generalizes to any polynomial inequality.
Rational Inequalities
Rational inequalities involve fractions with the variable in the numerator, denominator, or both.
Never multiply both sides by an expression containing x, because you do not know whether it is positive or negative. Instead, use the sign-chart method.
Example: Solve (x − 1)/(x + 3) ≥ 0.
- Step 1 — Find critical values
Find critical values where the numerator or denominator equals zero: x = 1 (numerator) and x = −3 (denominator).
- Step 2 — Create intervals
Create intervals:
(−∞, −3),(−3, 1),(1, ∞). - Step 3 — Determine the sign in each interval
Determine the sign of the expression in each interval. Test x = −4:
(−5)/(−1) = 5 > 0. Positive. Test x = 0:(−1)/(3) < 0. Negative. Test x = 2:(1)/(5) > 0. Positive. - Step 4 — Assemble the solution
We need ≥ 0, so include positive intervals. x = 1 makes the expression 0 (include it), but x = −3 makes the denominator 0 (exclude it). Solution:
(−∞, −3) ∪ [1, ∞).
The critical point to remember: always exclude values that make any denominator zero, regardless of the inequality symbol.
Solver AI handles rational inequalities with a clear sign chart and flags excluded points.
Graphing Solutions on the Number Line
Visualizing inequality solutions on a number line reinforces understanding and helps you communicate answers clearly.
Graphing common inequality types
| Case | Inequality | How to Graph |
|---|---|---|
| 1 | x > 3 | Place an open circle at 3 and shade to the right. |
| 2 | −2 ≤ x < 4 | Place a closed circle at −2, an open circle at 4, and shade between them. |
| 3 | x ≤ −1 or x > 5 | Shade to the left of −1 (closed) and to the right of 5 (open). |
For quadratic and rational inequalities, the number line naturally shows how critical values divide the line into regions.
Graphing is not just a presentation tool — it is a problem-solving tool. When you sketch the sign chart on a number line, you can visually identify which intervals satisfy the inequality.
On exams, a well-labeled number line graph alongside your interval notation answer demonstrates complete understanding and can earn partial credit even if you make an arithmetic error elsewhere.