Solving Linear Inequalities

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Skill: linear-inequalities

Solving Linear Inequalities

A linear inequality is like a linear equation, but with <, >, ≤, or ≥ instead of =. Instead of one answer you get a solution set — every number that makes the statement true. You solve them exactly like equations, with one famous exception.

1 Understand: Equations With a Range of Answers

An equation like x + 7 = 12 has one answer (x = 5). An inequality like x + 7 > 12 has infinitely many answers — every number bigger than 5. You graph that whole set as a shaded ray on the number line.

The one famous exception

Solve inequalities exactly the way you solve equations — add, subtract, multiply, divide both sides — with one rule:

The flip rule: whenever you multiply or divide by a negative number, the inequality sign reverses: < becomes >, ≤ becomes ≥, and vice versa.

Why? Because multiplying by −1 mirrors the number line: the order of every pair of numbers reverses. 2 < 5, but −2 > −5. The algebra just reports that mirror.

Where it is used

Budgets (“spend at most $50”) · speed limits · grade cutoffs (“at least 90 for an A”) · manufacturing constraints. Real life is full of “at least,” “at most,” and “under budget” — all inequalities.

−2−10 123 4 x > 3 open circle: 3 is NOT included shade right: every number bigger than 3
x > 3 means every number right of 3 — the open circle says 3 itself is not included. A closed dot would mean ≤ or ≥.

2 See It: The Sign Flip

Watch the one step that makes inequalities different from equations.

−3x ≤ 18 ÷ (−3) FLIP! ≤ becomes ≥ x ≥ −6 Dividing by −3 turns ≤ into ≥. The sign change is the whole trick of this skill. Check: x = −5 gives −3(−5) = 15 ≤ 18 ✓; x = −7 gives 21 ≤ 18, correctly excluded.
Watch the flip: dividing by −3 turns ≤ into ≥. The sign change is the whole trick of this skill.

Open vs. closed dots

  • > and < (strict): open circle — the endpoint is not included.
  • ≥ and ≤ (“or equal to”): closed dot — the endpoint is included.

Memory hook: “or equal to” gets the filled-in dot — the endpoint is invited to the solution set.

Compound inequalities

A compound like −4 < 2x ≤ 10 traps x between two bounds. Solve it by doing the same operation to all three parts at once:

−4 < 2x ≤ 10  →  −2 < x ≤ 5

Divide everything by 2 (positive — no flip). Whatever you do to the middle, do to both ends in the same step.

3 Worked Examples

Follow each step. Ask at every multiplication or division: is it negative? Then flip.

Example 1 Basic: x + 7 > 12
  1. Thinking: same as an equation — subtract 7 from both sides. No multiplication or division, so no flip.
  2. x > 12 − 7, so x > 5.
  3. Check: try 6: 6 + 7 = 13 > 12 ✓. Try 5: 12 is not > 12 ✓ (the boundary never counts for strict >).
  4. Graph: open circle at 5, shade right.
x > 5
Example 2 The flip: −3x ≤ 18
  1. Thinking: divide by −3 — and flip the sign because the divisor is negative.
  2. x ≥ 18 ÷ (−3) = −6.
  3. Check: try −5: −3(−5) = 15 ≤ 18 ✓. Try −7: 21 ≤ 18? No — correctly excluded ✓.
  4. Graph: closed dot at −6, shade right.
x ≥ −6
Example 3 Multi-step: 2x − 5 < 11
  1. Thinking: add 5, then divide by 2 (positive — no flip).
  2. Add 5: 2x < 16. Divide by 2: x < 8.
  3. Check: x = 7: 14 − 5 = 9 < 11 ✓.
  4. Graph: open circle at 8, shade left.
x < 8
Example 4 Compound: −4 < 2x ≤ 10
  1. Thinking: do the same operation to all three parts: divide everything by 2.
  2. −2 < x ≤ 5.
  3. Check: x = 0 works (−4 < 0 ≤ 10 ✓); x = −2 fails the left side ✓.
  4. Graph: open circle at −2, closed dot at 5, shade between.
−2 < x ≤ 5

4 Common Mistakes

These three errors show up on almost every inequalities quiz.

Mistake 1: Forgetting to flip when dividing by a negative
Wrong
−2x > 14 → “x > −7”
The sign was never flipped — but −7 does NOT satisfy −2x > 14 (it gives 14 > 14, false).
Right
−2x > 14 → x < −7
Check: x = −8 gives 16 > 14 ✓.
Rule: every time you multiply or divide by a negative, flip the sign — make it a reflex, not a decision.
Mistake 2: Open vs. closed circle confusion
Wrong
Graphing x ≥ 7 with an open circle.
An open circle excludes 7 — but 7 satisfies x ≥ 7.
Right
Closed dot at 7, shade right.
≥ and ≤ include the endpoint; > and < do not.
Rule: “or equal to” gets the filled-in dot — the endpoint is invited.
Mistake 3: Splitting a compound inequality into unrelated pieces
Wrong
Solving −1 < x + 3 and x + 3 ≤ 8 separately and losing track of which bound goes where.
Right
Operate on all three parts at once: −4 < x ≤ 5.
One step, both bounds, nothing lost.
Rule: whatever you do to the middle, do to both ends in the same step.

5 Quick Check

Try each one on paper first, then reveal the answer.

1. Solve 5x ≥ 35 and describe the graph.
Answer
5 is positive, so no flip: x ≥ 7.
Graph: closed dot at 7 (≥ includes the endpoint), shade right.
2. Solve −2x > 14.
Answer
Divide by −2 and flip: x < −7.
Check: x = −8 gives 16 > 14 ✓. Graph: open circle at −7, shade left.
3. Solve −1 < x + 3 ≤ 8.
Answer
Subtract 3 from all three parts: −4 < x ≤ 5.
Graph: open circle at −4, closed dot at 5, shade between.

Key Points to Remember

  • Solve like an equation — except: multiplying or dividing by a negative flips the sign.
  • The solution is a set: graph it as a shaded ray on the number line.
  • >, <: open circle (endpoint excluded). ≥, ≤: closed dot (endpoint included).
  • Compound inequalities: operate on all three parts at once.
  • Always check with a test value — one inside the shading, one outside.
Practice inequalities Try the inequality tool Explore the number-line lab
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