Inequality Calculator
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Solving Linear Inequalities: Step-by-Step Method
A linear inequality resembles a linear equation but uses inequality signs (>, <, ≥, ≤) instead of equals. The solution is not a single value but a range (interval) of values. Solving linear inequalities follows the same algebraic rules as equations, with one critical exception.
The sign-flip rule: When you multiply or divide both sides of an inequality by a negative number, the inequality direction reverses. This is the single most important rule — and the most common source of errors.
Example 1: Solve 2x + 3 ≤ 11.
- Subtract 3 from both sides: 2x ≤ 8
- Divide by 2 (positive, so no flip): x ≤ 4
- Solution: x ≤ 4, written in interval notation as (−∞, 4]
Example 2: Solve −3x + 1 > 7.
- Subtract 1 from both sides: −3x > 6
- Divide by −3 (negative! flip the sign): x < −2
- Solution: x < −2, written as (−∞, −2)
Interval Notation Reference Table
Solutions to inequalities are expressed using interval notation, which uses brackets and parentheses to indicate whether endpoints are included or excluded.
| Inequality | Interval Notation | Number Line | Endpoint Included? |
|---|---|---|---|
| x < 5 | (−∞, 5) | Open circle at 5, arrow left | No (5 excluded) |
| x ≤ 5 | (−∞, 5] | Closed circle at 5, arrow left | Yes (5 included) |
| x > −2 | (−2, +∞) | Open circle at −2, arrow right | No (−2 excluded) |
| x ≥ −2 | [−2, +∞) | Closed circle at −2, arrow right | Yes (−2 included) |
| −3 < x < 7 | (−3, 7) | Open circles, shaded between | Neither endpoint |
| −3 ≤ x ≤ 7 | [−3, 7] | Closed circles, shaded between | Both endpoints |
| x < 0 or x > 4 | (−∞, 0) ∪ (4, +∞) | Two separate rays | Neither 0 nor 4 |
The ∪ symbol means "union" (combining both sets). Square brackets [ ] indicate closed intervals (endpoint included). Parentheses ( ) indicate open intervals (endpoint excluded). Infinity always uses parentheses because infinity is not an actual reachable value.
Compound Inequalities: AND and OR
Compound inequalities combine two separate inequalities with "and" or "or," creating solutions that are intersections or unions of two intervals.
"And" compound inequalities (conjunction) require both conditions to be satisfied simultaneously. The solution is the intersection of both solution sets. Example: −2 < x + 1 ≤ 5. Subtract 1 from all parts: −3 < x ≤ 4. Solution: (−3, 4].
"Or" compound inequalities (disjunction) require at least one condition to be satisfied. The solution is the union. Example: 2x − 1 < 3 or 3x + 1 > 10. Solve each: x < 2 or x > 3. Solution: (−∞, 2) ∪ (3, +∞).
| Compound Type | Example | Solution | Graph Shape |
|---|---|---|---|
| AND (both conditions) | x > −1 AND x < 4 | (−1, 4) | Bounded segment |
| OR (either condition) | x < −2 OR x > 3 | (−∞,−2) ∪ (3,+∞) | Two rays outward |
| AND (no overlap) | x > 5 AND x < 2 | ∅ (empty set) | No solution |
| OR (complete overlap) | x > 1 OR x < 8 | (−∞, +∞) | All real numbers |
Absolute Value Inequalities
Absolute value inequalities convert to compound inequalities using the fundamental rules:
- |A| < b (b > 0) → −b < A < b (AND type → bounded interval)
- |A| > b (b > 0) → A < −b OR A > b (OR type → two rays)
- |A| ≤ b → −b ≤ A ≤ b
- |A| ≥ b → A ≤ −b OR A ≥ b
Example 1: |x − 3| < 5. Apply rule: −5 < x − 3 < 5. Add 3: −2 < x < 8. Solution: (−2, 8).
Example 2: |2x + 1| ≥ 7. Apply rule: 2x + 1 ≤ −7 OR 2x + 1 ≥ 7. Case 1: 2x ≤ −8 → x ≤ −4. Case 2: 2x ≥ 6 → x ≥ 3. Solution: (−∞, −4] ∪ [3, +∞).
Absolute value inequalities appear in error analysis (|measured − true| ≤ tolerance), distance problems (|x − center| < radius), and control systems (|error signal| < threshold). Understanding them is essential for applied mathematics and engineering.
Quadratic and Polynomial Inequalities
For inequalities involving x² and higher powers, the approach differs. A quadratic inequality like ax² + bx + c > 0 cannot be solved by simple algebraic manipulation — it requires finding the roots and testing intervals.
Method for quadratic inequalities:
- Move everything to one side: Get the form ax² + bx + c > 0 (or <, ≥, ≤).
- Find the roots by solving ax² + bx + c = 0 using factoring, the quadratic formula, or completing the square.
- Create a sign chart: The roots divide the number line into intervals. Test one point in each interval.
- Identify which intervals satisfy the inequality.
Example: x² − x − 6 > 0. Factor: (x − 3)(x + 2) > 0. Roots: x = 3 and x = −2. Three intervals: x < −2, −2 < x < 3, x > 3. Test x = −3: (−6)(−1) = 6 > 0 ✓. Test x = 0: (−3)(2) = −6 < 0 ✗. Test x = 4: (1)(6) = 6 > 0 ✓. Solution: (−∞, −2) ∪ (3, +∞).
| Inequality Type | Method | Example | Solution |
|---|---|---|---|
| Linear: ax + b > c | Direct solving (mind sign flip) | 2x − 4 > 6 | x > 5 → (5, +∞) |
| Quadratic: ax² + bx + c > 0 | Roots + sign chart | x² − 4 > 0 | x < −2 or x > 2 |
| Rational: p(x)/q(x) > 0 | Critical points + sign chart | (x+1)/(x−2) > 0 | x < −1 or x > 2 |
Inequalities in Real Life: Applications and Modeling
Inequalities model constraints in virtually every quantitative field. Unlike equations that describe exact conditions, inequalities describe feasible regions — ranges of acceptable values.
Personal finance: "I can afford a monthly car payment if my total debt payments stay under 36% of gross income." If gross income = $5,000/month and other debts = $800/month: car payment + 800 ≤ 0.36 × 5000 = 1800. Car payment ≤ $1,000.
Engineering design: A bridge beam must handle load L without failure. Safety factor requires stress σ ≤ σ_yield/1.5. This inequality determines the minimum required beam cross-section.
Medicine and dosing: A drug is safe when blood concentration is between 10 and 20 mg/L: 10 ≤ C(t) ≤ 20. The dosing schedule must maintain concentration in this therapeutic window — too low is ineffective, too high is toxic.
Quality control: A manufacturing process is acceptable when measurements fall within ±2σ of the target: |x − μ| ≤ 2σ. Parts outside this range are rejected. Statistical process control uses inequality monitoring constantly.
Linear programming: Businesses maximize profit P = 3x + 5y subject to constraints: x ≥ 0, y ≥ 0, 2x + y ≤ 100, x + 3y ≤ 90. The optimal solution always occurs at a vertex of the feasible region (the area satisfying all constraints). This is the foundation of operations research and logistics optimization.
Graphing Inequalities on the Number Line and Coordinate Plane
Visualizing inequalities helps build intuition for their solutions. On a number line, the solution to a one-variable inequality is represented by:
- Open circle at the endpoint for strict inequalities (< or >) — the endpoint is not included
- Closed circle (filled dot) for non-strict inequalities (≤ or ≥) — the endpoint is included
- Shaded region (arrow or line) indicating all solution values
For two-variable linear inequalities (2x + 3y ≤ 12), the solution is a half-plane on the coordinate plane. Graphing method: (1) Draw the boundary line 2x + 3y = 12 as a dashed line (strict inequality) or solid line (non-strict). (2) Test a point not on the line (typically the origin): 2(0) + 3(0) = 0 ≤ 12 ✓. Shade the side containing the test point. The shaded region represents all (x, y) pairs satisfying the inequality.
Systems of linear inequalities create feasible regions that are intersections of multiple half-planes. These convex polygonal regions are the foundation of linear programming — the optimal value of any linear objective function over a feasible region always occurs at one of the vertices (corner points).
Frequently Asked Questions
What happens when you multiply both sides of an inequality by a negative number?
The inequality direction reverses. If a > b and c < 0, then ac < bc. Example: 3 > 1; multiply by −2: −6 < −2 ✓. This is the most important rule in inequality algebra. Forgetting to flip the sign is the most common error. When dividing by a negative (e.g., to isolate x with a negative coefficient), always flip the inequality.
What is interval notation?
Interval notation describes the solution set of an inequality using brackets and parentheses. Parentheses ( ) indicate an open boundary (endpoint excluded); brackets [ ] indicate a closed boundary (endpoint included). Infinity always uses parentheses. Examples: x > 3 → (3, +∞); x ≤ 7 → (−∞, 7]; 2 ≤ x < 9 → [2, 9).
Can a linear inequality have no solution?
Yes. If the coefficient of x is 0 and the resulting statement is false, there is no solution. Example: 0·x + 5 < 3 simplifies to 5 < 3, which is always false — no solution (empty set). Conversely, if the simplified statement is always true (5 > 3), all real numbers are the solution.
How is solving an inequality different from solving an equation?
The process is nearly identical, except: (1) the solution is an interval (or union of intervals) rather than specific values; (2) multiplying/dividing by a negative number flips the inequality sign. An equation ax + b = c has at most one solution (for a≠0); an inequality ax + b < c has infinitely many solutions forming an interval.
What does "strict" vs. "non-strict" inequality mean?
Strict inequalities (<, >) exclude the boundary value; the endpoint is not part of the solution. Non-strict inequalities (≤, ≥) include the boundary value. On a number line, strict → open circle (hollow dot); non-strict → closed circle (filled dot). In interval notation, strict → parenthesis; non-strict → bracket.
How do you solve an absolute value inequality?
|A| < b → −b < A < b (bounded interval). |A| > b → A < −b OR A > b (two rays). Always check that b > 0 first: if b ≤ 0, |A| < b has no solution (absolute values are non-negative); |A| > b (with b < 0) has all real numbers as solution.
What is the solution set of x² < 4?
x² < 4 means |x| < 2, so −2 < x < 2. Solution: (−2, 2). Verify: at x = 1.5, 1.5² = 2.25 < 4 ✓. At x = 2, 4 < 4 is false ✗ (strict inequality, endpoint excluded). At x = 3, 9 < 4 is false ✗.
How do you graph a system of inequalities?
Graph each inequality separately, shading the feasible half-plane for each. The solution to the system is the region shaded by ALL inequalities simultaneously (intersection). For a system of 3 or more inequalities, the feasible region may be a polygon with vertices at the intersections of boundary lines. These vertices are critical for linear programming optimization.
What is a rational inequality and how do I solve it?
A rational inequality has the form p(x)/q(x) > 0 (or <, ≥, ≤). Critical points are where p(x) = 0 (numerator zero) or q(x) = 0 (denominator zero — excluded from domain). These points divide the number line into intervals. Test each interval: a rational expression has constant sign within each interval. Collect intervals where the expression satisfies the inequality. Note: denominator zeros are never included, even with ≥ or ≤.
Can inequalities have no solution or infinitely many solutions?
Yes to both. A linear inequality typically has infinitely many solutions (an interval). Special cases: (1) No solution: when the inequality simplifies to a false statement like 3 < 1. This happens with contradictory compound AND inequalities (x > 5 AND x < 2 → empty set). (2) All real numbers: when it simplifies to an always-true statement like 1 < 3. OR inequalities can cover all reals: x > 1 OR x < 2 → all reals, since every real number satisfies at least one condition.