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Inequalities Explained: Linear, Compound, Quadratic

Master inequalities — linear, compound, and quadratic — with the one rule everyone forgets. Worked examples and how to graph solutions on a number line.
AI-Math Editorial Team

By AI-Math Editorial Team

Published 2026-05-02

Inequalities look identical to equations until you reach the rule that flips you up at midnight: when you multiply or divide by a negative number, the inequality direction flips. This guide walks through linear, compound, and quadratic inequalities with the patterns that solve 95% of homework.

The one rule everyone forgets

For equations: every operation preserves the equality. 5=55 = 5 implies 5⋅(−1)=5⋅(−1)5 \cdot (-1) = 5 \cdot (-1) — both sides equally negated, equality holds.

For inequalities: multiplying or dividing both sides by a negative number flips the direction. 5>35 > 3 is true, but multiply both by −1-1 and we get −5>−3-5 > -3, which is false. The corrected statement is −5<−3-5 < -3.

This single rule is the source of most inequality mistakes. Burn it into your reflexes:

  • Add/subtract anything → no flip.
  • Multiply/divide by positive → no flip.
  • Multiply/divide by negative → flip the inequality.

Linear inequalities

Solve like you solve linear equations, watching out for sign flips.

Example 1: 3x+5>143x + 5 > 14.

  • Subtract 5: 3x>93x > 9.
  • Divide by 33 (positive, no flip): x>3x > 3.
  • Solution set: (3,∞)(3, \infty) — open parenthesis means x=3x = 3 is not included.

Example 2 (with the flip): −2x+7≤1-2x + 7 \leq 1.

  • Subtract 7: −2x≤−6-2x \leq -6.
  • Divide by −2-2 (negative — FLIP): x≥3x \geq 3.
  • Solution set: [3,∞)[3, \infty) — square bracket because of ≤\leq, including 33.

Compound inequalities

A "compound" inequality joins two simple inequalities with AND or OR.

AND is often written as a single chain: −1<2x+3≤7-1 < 2x + 3 \leq 7. Operate on all three parts simultaneously.

  • Subtract 3 everywhere: −4<2x≤4-4 < 2x \leq 4.
  • Divide by 2 everywhere: −2<x≤2-2 < x \leq 2.
  • Solution: (−2,2](-2, 2].

OR stays as two separate inequalities. The solution is the union of both individual solution sets:

x<−3x < -3 or x>5x > 5 → solution (−∞,−3)∪(5,∞)(-\infty, -3) \cup (5, \infty).

Quadratic inequalities

For x2+bx+c>0x^2 + bx + c > 0 (or any inequality ≠0\neq 0):

  1. Find roots of x2+bx+c=0x^2 + bx + c = 0.
  2. Plot roots on the number line — they divide it into intervals.
  3. Test a point in each interval to see whether the quadratic is positive or negative there.
  4. Pick the intervals matching the inequality direction.

Example: x2−5x+6>0x^2 - 5x + 6 > 0.

  • Factor: (x−2)(x−3)>0(x - 2)(x - 3) > 0. Roots at x=2x = 2 and x=3x = 3.
  • Test intervals:
    • x=0x = 0: (0−2)(0−3)=6>0(0-2)(0-3) = 6 > 0 ✓
    • x=2.5x = 2.5: (0.5)(−0.5)=−0.25<0(0.5)(-0.5) = -0.25 < 0 ✗
    • x=4x = 4: (2)(1)=2>0(2)(1) = 2 > 0 ✓
  • Solution: (−∞,2)∪(3,∞)(-\infty, 2) \cup (3, \infty).

For ≤\leq or ≥\geq inequalities, include the roots (closed intervals): (−∞,2]∪[3,∞)(-\infty, 2] \cup [3, \infty).

Graphing solutions on a number line

  • Open circle (○) at a value not included (<< or >>).
  • Closed circle (●) at a value included (≤\leq or ≥\geq).
  • Arrow extending to infinity in the direction of the solution.

Compound AND → bracket between two circles. Compound OR → two separate rays going outward.

Inequalities with absolute value

∣x−a∣<b|x - a| < b unpacks to −b<x−a<b-b < x - a < b, i.e. a−b<x<a+ba - b < x < a + b — a bounded interval.

∣x−a∣>b|x - a| > b unpacks to x−a<−bx - a < -b OR x−a>bx - a > b, i.e. x<a−bx < a - b OR x>a+bx > a + b — two rays going outward.

Common mistakes

  1. Forgetting to flip when dividing by a negative. The single biggest source of wrong inequality answers.
  2. Including endpoints incorrectly. << vs ≤\leq matters — your bracket type depends on it.
  3. Treating compound AND like equals. −2<x<5-2 < x < 5 is a single statement; you can't break it into "x=−2x = -2 or x=5x = 5."
  4. Solving quadratic inequalities like equations. Setting x2−4>0x^2 - 4 > 0 "equal to zero" gives roots ±2\pm 2; the inequality solution isn't {−2,2}\{-2, 2\} but the intervals between/around them.

Try it yourself

Drop any inequality (linear, compound, quadratic, with absolute value) into our free Inequality Solver — the AI flips signs correctly and shows every step, plus a number-line solution graph.

Related material:

Frequently Asked Questions

When you multiply or divide both sides of an inequality by a negative number, you must reverse the inequality sign. All other operations — adding and subtracting — work the same as with equations.

Solve each part separately, then find the intersection (for "and") or union (for "or"). On a number line, closed circles represent ≤ or ≥, and open circles represent < or >.

Factor the quadratic to find the roots, then test a value in each interval between the roots to determine where the inequality holds. Sketching the parabola confirms which intervals give positive or negative values.

AI-Math Editorial Team

By AI-Math Editorial Team

Published 2026-05-02

A small team of engineers, mathematicians, and educators behind AI-Math, focused on making step-by-step math help accessible to every student.