CALTRX

Inequality Solver

Find the solution set for linear inequalities. See the complete solution with step-by-step working and formula explanations.

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Tips & Notes

  • ✓CRITICAL: flip the inequality sign when multiplying or dividing by a NEGATIVE number.
  • ✓Interval notation: () for strict (<,>), [] for inclusive (≤,≥). Infinity always uses ().
  • ✓Check solution: substitute a value from inside your interval — must satisfy the original inequality.
  • ✓|x−a|<b means x is within b units of a: solution is (a−b, a+b).
  • ✓Graph on number line: open circle for strict, filled for inclusive. Shade the solution region.

Common Mistakes

  • ✗Forgetting to flip the inequality sign when multiplying or dividing both sides by a negative number. −2x > 6 → x < −3 (sign flips). Not flipping gives x > −3, which is the opposite solution.
  • ✗Using the wrong bracket type in interval notation. Strict inequalities (< and >) use parentheses (). Non-strict (≤ and ≥) use square brackets []. Writing [3, ∞) for x > 3 incorrectly includes 3 in the solution.
  • ✗Adding or subtracting incorrectly when isolating the variable. In 3x − 5 > 7, adding 5 to both sides gives 3x > 12, then x > 4. A common error is adding 5 only to one side.
  • ✗Failing to check the solution by substituting a test value. Always pick a number inside the solution interval and verify it satisfies the original inequality — this catches sign-flip errors immediately.
  • ✗Treating a compound AND inequality as an OR. −2 < x AND x < 5 means both must hold simultaneously: x is between −2 and 5. Writing x < −2 OR x > 5 is the complement — the opposite of the correct solution.

Inequality Solver Overview

An inequality is a mathematical statement comparing two expressions with one of four relations: less than (<), greater than (>), less than or equal to (≤), or greater than or equal to (≥). Unlike equations which have isolated solutions, inequalities typically have infinite solution sets — entire intervals of numbers that satisfy the condition. Expressing these solutions correctly using interval notation or graphs is as important as solving the inequality itself.

Linear inequalities — solved like equations with one crucial exception: multiplying or dividing both sides by a negative number reverses the inequality direction.

EX: 2x + 3 < 11 → subtract 3: 2x < 8 → divide by 2 (positive, no flip): x < 4 → solution: (−∞, 4)
EX: −3x ≥ 9 → divide by −3 (negative → FLIP ≥ to ≤): x ≤ −3 → solution: (−∞, −3]
Why the sign flips: multiplying by −1 reverses number line order. Since 3 > 1, multiplying both by −2 gives −6 < −2 — the relationship reverses. Forgetting this flip is the most common error in inequality solving. Interval notation: - Parenthesis () = strict inequality, endpoint excluded: (2, 7) means 2 < x < 7 - Bracket [] = inclusive inequality, endpoint included: [2, 7] means 2 ≤ x ≤ 7 - Infinity always uses parenthesis: [3, ∞) means x ≥ 3. ∞ is not a number — it cannot be included. Compound inequalities:
EX (AND): −1 ≤ 2x+1 < 5 → subtract 1: −2 ≤ 2x < 4 → divide by 2: −1 ≤ x < 2 → solution: [−1, 2)
EX (OR): x < −3 or x > 5 → solution: (−∞, −3) ∪ (5, ∞) — two disconnected intervals
Absolute value inequalities — the key patterns: - |x| < a → −a < x < a (one connected interval — within a of zero) - |x| > a → x < −a OR x > a (two separate intervals — farther than a from zero)
EX: |2x−1| ≤ 5 → −5 ≤ 2x−1 ≤ 5 → −4 ≤ 2x ≤ 6 → −2 ≤ x ≤ 3 → solution: [−2, 3]
Inequalities describe regions rather than points — the solution is an interval or union of intervals, not a single value. The most important rule is the sign flip: whenever you multiply or divide both sides by a negative number, the direction of the inequality reverses. This is the step most often missed in manual solving, and it is why checking a value from the solution region against the original inequality is always worth doing. Absolute value inequalities split into two cases depending on direction: |f(x)| < k becomes −k < f(x) < k (a bounded interval), while |f(x)| > k becomes f(x) < −k or f(x) > k (an unbounded union). Compound inequalities combine two conditions: AND requires both to hold simultaneously, producing the intersection; OR requires at least one, producing the union. Sketching a number line before writing the final solution prevents notation errors in the final interval expression.

Frequently Asked Questions

Isolate the variable using the same rules as equations — with one critical exception: multiplying or dividing both sides by a negative number reverses the inequality direction. Example: −2x > 6. Divide by −2 and flip: x < −3. Standard steps: simplify each side, move variable terms to one side, move constants to the other, divide by the coefficient (flipping if negative). Always verify by substituting a test value from the solution into the original inequality.

Multiplying or dividing by a negative reverses the inequality because it reflects the number line. If 3 > 1, then −3 < −1 (negating flips the order). Example: −x > 5. Multiply both sides by −1: x < −5 (sign flips). The algebraic reason: inequality a > b means a − b > 0. Multiply by −1: (−a) − (−b) = b − a > 0 means −a < −b. This reversal applies to division by negatives as well — the direction change is not optional.

Write the solution set using interval notation. Strict inequalities (< and >) use parentheses — the endpoint is excluded. Non-strict (≤ and ≥) use square brackets — the endpoint is included. Examples: x > 3 → (3, ∞). x ≤ 7 → (−∞, 7]. −2 ≤ x < 5 → [−2, 5). Infinity always uses parentheses because infinity is not a real number and cannot be included. The union symbol ∪ combines disconnected intervals: x < −2 or x > 3 → (−∞,−2) ∪ (3,∞).

Compound inequalities combine two conditions. AND (intersection): both must hold. −3 < 2x + 1 ≤ 7 → solve each part: −4 < 2x and 2x ≤ 6 → −2 < x ≤ 3. Solution: (−2, 3]. OR (union): either condition holds. x < −1 or x > 4 → solution: (−∞,−1) ∪ (4,∞). AND gives a connected interval; OR gives two disconnected regions. Always verify endpoint behavior — check whether boundaries are included or excluded.

Absolute value inequalities split into two cases. |x − a| < b means −b < x−a < b → a−b < x < a+b (connected interval). |x − a| > b means x−a < −b or x−a > b → x < a−b or x > a+b (two disconnected intervals). Example: |2x−1| ≤ 5 → −5 ≤ 2x−1 ≤ 5 → −4 ≤ 2x ≤ 6 → −2 ≤ x ≤ 3 → [−2, 3]. Rule: less-than absolute value → AND → interval. Greater-than absolute value → OR → two rays.

Quadratic inequalities: find roots first, then test sign in each interval. x² − 5x + 6 > 0. Factor: (x−2)(x−3) > 0. Roots at x=2 and x=3 divide the number line into three regions: (−∞,2), (2,3), (3,∞). Test x=0: (−2)(−3)=6>0 ✓. Test x=2.5: (0.5)(−0.5)=−0.25<0 ✗. Test x=4: (2)(1)=2>0 ✓. Solution: (−∞,2) ∪ (3,∞). The product is positive when both factors share the same sign.