01 · A negative divisor
4 − 5x ≥ 19
Hint
Subtract 4, then divide by −5.
Worked solution
−5x ≥ 15
x ≤ −3
Understand · explore · practise
Linear inequalities
Solve linear inequalities, reverse the sign correctly, and express combined conditions using number lines, intervals and set notation.
Before you startLinear equations and signed numbers
01 / The rules
An inequality describes a set of values.
An equation asks where two expressions are equal. An inequality asks where one is smaller or larger. Its answer is usually an interval or a union of intervals.
You may add or subtract the same quantity on both sides. Multiplying or dividing by a positive number preserves the direction. A negative multiplier or divisor reverses it.
−3 < 2
Multiply by −2: 6 > −4
Negative multiplication reflects positions across zero, reversing their order on a number line. The sign reversal also applies to ≤ and ≥.
Do not divide by zero. Multiplying by zero destroys information. If a variable’s sign is unknown, do not multiply an inequality by it without a case split or another valid method.
Keep the inequality equivalentWorked example
7 − 3x > 16
Subtract 7 on both sides.
−3x > 9
Divide by the negative number −3.
x < −3
The direction reverses.
Check x = −4: 19 > 16 ✓
The boundary x = −3 gives equality and is excluded.
Watch negative multiplication reverse the order
Pause, replay or seek freely. The notes explain the same idea and stay in view.
02 / Solve
Expand, collect and isolate.
Use the same algebraic operations as for a linear equation, while tracking any multiplication or division by a negative number.
Positive denominators can be cleared
(x − 2)/3 + (x + 1)/2 < 4
2(x − 2) + 3(x + 1) < 24
5x − 1 < 24
x < 5
Multiplying by 6 is safe because it is positive. Both fractions and the right side are multiplied.
Higher powers can cancel
x(x − 4) ≤ x² + 6
x² − 4x ≤ x² + 6
−4x ≤ 6
x ≥ −3/2
Classify the inequality after simplifying. The x² terms cancel here.
Keep the outside minus signWorked example
5 − 2(3x − 1) ≥ x − 7
Expand −2 times both terms.
7 − 6x ≥ x − 7
Subtract x and subtract 7.
−7x ≥ −14
Divide by −7 and reverse the direction.
x ≤ 2
At x = 2 the sides are equal, so this endpoint is included.
03 / Notation
The endpoint symbols carry information.
Strict inequalities < and > exclude equality. Inclusive inequalities ≤ and ≥ allow it.
- On a number line, an open circle excludes an endpoint; a filled circle includes it.
- In interval notation, ( or ) excludes the endpoint; [ or ] includes it.
- Infinity is not a real-number endpoint. Always use a parenthesis next to ∞ or −∞.
−2 < x ≤ 5 ↔ (−2, 5]
x ≥ 3 ↔ [3, ∞)
Set-builder notation {x ∈ ℝ : x ≥ 3} means “the real numbers x for which x ≥ 3”. ℝ is the real-number set; ∅ is the empty set.
Unless a question restricts x to integers, include all real values in the intervals, not just the whole numbers.
A chained inequalityWorked example
−1 < 2x + 3 ≤ 9
All three parts of the chain must remain consistent.
−4 < 2x ≤ 6
Subtract 3 from each part.
−2 < x ≤ 3
Divide each part by positive 2.
Interval: (−2, 3]
Left endpoint open; right endpoint filled on a number line.
04 / And / or
Keep the overlap or keep either set.
And means both conditions must hold: take their intersection, written A ∩ B. Or means at least one condition holds: take their union, written A ∪ B. It includes values that satisfy both.
In the diagram, A is −3 < x ≤ 2 and B is 0 ≤ x < 5. The overlap is 0 ≤ x ≤ 2, while their union runs from −3 to 5 with both outer endpoints excluded.
For separated intervals, write “or” or use ∪. Do not turn x < −2 or x > 4 into the impossible chain 4 < x < −2.
The filled endpoint at 2 in A does not create a boundary in the union: B already contains numbers on both sides of 2.
A: −3 < x ≤ 2B: 0 ≤ x < 5
Intersection means both conditions hold: 0 ≤ x ≤ 2. Both endpoints are included.
05 / Nested conditions
Solve each set before combining it.
Brackets in a set expression tell you the order. For A ∩ (B ∪ C), first take the union inside the brackets, then retain only values also in A.
No overlap versus every real number
x ≥ 4 and x < 3 has no solutions: ∅. By contrast, x ≤ 2 or x > −1 covers every real number: ℝ. “Or” does not require the two intervals to be separate.
Check an endpoint against the original conditions if its inclusion is unclear. For an intersection, it must pass every condition; for a union, passing just one is enough.
A ∩ (B ∪ C)Worked example
A: 2x + 3 > −5 ⇒ x > −4
Solve each inequality first.
B: 3x + 2 ≤ −1 ⇒ x ≤ −1
C: 7 − 2x ≤ 3 ⇒ x ≥ 2
The negative divisor in C reverses its sign.
B ∪ C: x ≤ −1 or x ≥ 2
Keep either interval.
A ∩ (B ∪ C):
−4 < x ≤ −1 or x ≥ 2
In interval notation: (−4, −1] ∪ [2, ∞).
06 / Your turn
State which endpoints are included.
Write real solution sets clearly. A number line is a useful check when you combine conditions.
02 · A chain and an integer restriction
Solve −2 ≤ 3x + 4 < 10 over the reals. Then list the integer solutions.
Hint
Subtract 4 from all parts, then divide by 3.
Worked solution
−6 ≤ 3x < 6
−2 ≤ x < 2
The real interval is [−2, 2). Integer solutions: −2, −1, 0, 1.
03 · Cancellation
2x(x + 1) ≤ 2x² + 7
Hint
Expand before deciding the degree of the inequality.
Worked solution
2x² + 2x ≤ 2x² + 7
2x ≤ 7
x ≤ 7/2
04 · Both conditions
3x − 1 > 2 and 12 − 2x ≥ 4
Hint
Solve each condition and intersect the intervals.
Worked solution
x > 1 and x ≤ 4
1 < x ≤ 4
Interval: (1, 4].
05 · Either condition
2x + 1 < −5 or 4 − x ≤ 0
Hint
The result is a union, not an overlap.
Worked solution
x < −3 or x ≥ 4
Interval notation: (−∞, −3) ∪ [4, ∞).
06 · Find the error
A student changes −4x < 8 into x < −2. Correct the answer and disprove the proposed result with a test value.
Hint
Dividing by a negative number reverses the direction.
Worked solution
The correct answer is x > −2. For example x = 0 satisfies 0 < 8, but is excluded by the student’s proposed interval. The boundary −2 gives equality and is excluded.
07 / Recap
Order and logic both matter.
- Reverse the direction when multiplying or dividing by a negative number.
- Simplify fully: powers may cancel.
- Distinguish strict and inclusive endpoints.
- “And” is intersection; “or” is union.
- Use the given domain: real intervals and integer lists are different answers.
Section 1 of 7 · The rules