An inequality describes a region of possible values, not one final number. That difference changes the entire solution process. An equation may ask which value makes two expressions equal. An inequality asks which values keep one quantity greater than, less than, or bounded relative to another.
This seventeenth Secondary 4 Mathematics Learning Guide develops linear inequalities as a system of constraints. It belongs to the Secondary Mathematics Hub and S1–S4 Capability Map. The guide covers algebraic solving, sign reversal, number-line representation, combined constraints, modelling and verification at boundaries.
Current syllabus connection: inequality notation, simple linear inequalities and number-line interpretation sit within the Number and Algebra system of the 2026 O-Level Mathematics 4052 and 2027 SEC G3 Mathematics K310 syllabuses. Use the official 2026 O-Level syllabus page or 2027 SEC G3 syllabus page for cohort-specific assessment details.
Inequality symbols describe order
| Symbol | Meaning |
|---|---|
| x < 4 | x is less than 4 |
| x ≤ 4 | x is less than or equal to 4 |
| x > 4 | x is greater than 4 |
| x ≥ 4 | x is greater than or equal to 4 |
The equality bar matters. x < 4 excludes 4. x ≤ 4 includes it. On a number line, that inclusion or exclusion should be represented clearly using the convention required by the task.
The main solving rule is familiar—with one critical exception
You can add or subtract the same quantity on both sides of an inequality without changing its direction. You can also multiply or divide both sides by the same positive number without changing its direction.
But multiplying or dividing by a negative number reverses the inequality sign.
If a < b, then −a > −b.
The reason is numerical order. Since 2 < 5, multiplying both by −1 gives −2 > −5. The sign reversal is not a special trick invented for algebra; it preserves the true order after reflection across zero.
Worked Example 1 | Solve without sign reversal
Solve 3x + 5 < 20.
Subtract 5 from both sides:
3x < 15.
Divide by positive 3:
x < 5.
The answer is a range. Values such as 4, 0 and −10 satisfy the inequality; 5 and 6 do not.
Worked Example 2 | Reverse the sign once
Solve −4x + 7 ≥ 19.
Subtract 7:
−4x ≥ 12.
Divide by −4 and reverse the inequality:
x ≤ −3.
Check a value inside the solution, such as x = −4: −4(−4)+7 = 23, which is indeed at least 19. Check x = 0: the left side is 7, so 0 should not belong to the solution. The test supports the sign direction.
Boundary values tell you where the truth can change
For a linear inequality such as 2x − 1 > 7, the related equation 2x − 1 = 7 gives the boundary x = 4. Values on one side satisfy the inequality; values on the other side do not.
Testing the boundary and one nearby value can verify a solution quickly. For x > 4, the boundary 4 should not be included because the original sign is strict. At x = 5, 2x − 1 = 9, which is greater than 7.
Worked Example 3 | Brackets first, then inequality control
Solve 5 − 2(3x − 4) < 17.
Expand carefully:
5 − 6x + 8 < 17
13 − 6x < 17
−6x < 4.
Divide by −6 and reverse the sign:
x > −2/3.
A learner who expands −2(3x − 4) as −6x − 8 has already changed the expression before the inequality rule is even reached. Diagnosis should therefore separate bracket control from sign reversal.
Fractions: clear denominators with a known positive multiplier
Suppose we solve x/3 + 2 ≤ 7. Multiplying both sides by positive 3 preserves the direction. This gives x + 6 ≤ 21, so x ≤ 15.
If a multiplier contains an unknown whose sign is not known, do not casually multiply and assume the inequality direction stays fixed. At Secondary 4 level, the safest route in ordinary linear inequalities is usually to rearrange using known numerical multipliers.
Combined inequalities describe an interval
A statement such as 2 ≤ x < 7 means both conditions must be true at the same time: x is at least 2 and less than 7. It can be seen as the intersection of two simpler solution sets.
On a number line, the solution lies from 2 up to but not including 7. The left boundary is included; the right boundary is excluded.
Worked Example 4 | Solve a double inequality
Solve −5 < 2x + 1 ≤ 9.
Subtract 1 from all three parts:
−6 < 2x ≤ 8.
Divide all three parts by positive 2:
−3 < x ≤ 4.
The learner should not solve only the left half or right half and forget that both constraints must survive.
Integer solutions: the domain changes the final answer
If x is a real number and 2 < x ≤ 6, infinitely many values are allowed. If x is restricted to integers, the solutions are 3, 4, 5 and 6.
A question about numbers of people, books, tickets or whole objects may require an integer interpretation even when the algebra first produces a continuous interval.
Worked Example 5 | Maximum whole-number value
A van can carry at most 900 kg. Its fixed load is 420 kg, and each identical box has mass 32 kg. Find the maximum number of boxes.
Let n be the number of boxes. The total mass constraint is:
420 + 32n ≤ 900.
Then 32n ≤ 480, so n ≤ 15. Since n is a non-negative integer, the maximum is 15 boxes.
Check: 420 + 32(15) = 900 exactly. Sixteen boxes would give 932 kg, exceeding the limit.
Worked Example 6 | Minimum whole-number value
A club needs to collect at least $1200. It already has $275 and each ticket sold brings in $18. How many additional tickets must be sold?
Let t be the number of tickets:
275 + 18t ≥ 1200.
18t ≥ 925, so t ≥ 51.388… . Because partial tickets cannot be sold, the minimum integer is 52 tickets.
Rounding to the nearest whole number would give 51 and fail the condition. Inequality context determines whether to round up or down.
Inequalities and percentages
A requirement such as “the final amount must be at least 80% of the original” can be translated before solving. If the original amount is P and a fee x is deducted, then P − x ≥ 0.8P.
Rearranging gives x ≤ 0.2P. The inequality has exposed the maximum allowable fee relative to the starting value.
Worked Example 7 | A budget constraint
A student has $75. A notebook costs $8 and each pen costs $2.50. If the student must buy the notebook, find the maximum number of pens that can be purchased without exceeding the budget.
Let p be the number of pens:
8 + 2.5p ≤ 75.
2.5p ≤ 67, so p ≤ 26.8. Since p is a whole number, the maximum is 26 pens.
Twenty-six pens cost $65, giving a total of $73. Twenty-seven pens would make the total $75.50, so the answer is verified by the boundary.
A number line is part of the answer
When a number-line representation is required, include the boundary correctly and show the direction of the solution. The number line is not merely an illustration; it communicates the set of allowed values.
For x ≥ 2, the point 2 is included and the solution extends to the right. For x < −1, the point −1 is excluded and the solution extends to the left.
Common failure modes
| Error | Cause | Repair |
|---|---|---|
| Sign not reversed after dividing by a negative | Equation rules transferred without order reasoning | Test a simple numerical example such as 2<5 |
| Boundary included when sign is strict | < confused with ≤ | Substitute the boundary into the original inequality |
| One half of a double inequality is lost | Combined constraint not preserved | Operate on all three parts together |
| Continuous answer used for a count | Domain of variable ignored | State whether x is real, integer or non-negative |
| Rounds 51.388 down in a minimum problem | Ordinary rounding used instead of constraint checking | Test neighbouring integers against the original condition |
| Number line points the wrong direction | Solution meaning not checked | Test one simple value from each side of boundary |
Verification: test the boundary and one interior point
- Substitute the boundary into the original inequality.
- Check whether equality should be allowed.
- Choose one value inside the proposed solution and test it.
- Choose one value outside and confirm that it fails.
- If the variable is a count, check neighbouring integers.
- Ensure the number line matches the algebraic statement.
Independent practice
- Solve 4x − 7 > 13.
- Solve 9 − 3x ≤ 21.
- Solve −4 < 2x + 6 ≤ 14.
- Find the integer solutions to −2 ≤ x < 4.
- A lift can carry at most 600 kg. Five people have total mass 410 kg. Each identical crate has mass 28 kg. Find the maximum number of crates.
- A charity needs at least $950. It has $215 and receives $24 for each item sold. Find the minimum number of items it must sell.
Explained answers
1. 4x > 20, so x > 5.
2. −3x ≤ 12. Divide by −3 and reverse the sign: x ≥ −4.
3. Subtract 6 to get −10 < 2x ≤ 8. Divide by 2: −5 < x ≤ 4.
4. The integers are −2, −1, 0, 1, 2, 3.
5. 410 + 28c ≤ 600 gives 28c ≤ 190, so c ≤ 6.785… . Maximum whole number: 6 crates.
6. 215 + 24n ≥ 950 gives n ≥ 30.625, so the minimum whole number is 31 items.
Teaching sequence: build the meaning before the manipulation
Start with number-line comparisons and ask learners to explain why multiplying by −1 reverses order. Then solve ordinary linear inequalities before introducing double inequalities and context problems.
Next, use counts, budgets and capacity limits so the learner has to convert a continuous algebraic bound into a valid integer decision. Finish with mixed questions in which the word “maximum”, “minimum”, “at most” or “at least” must be translated before any algebra begins.
Connect this guide to Ratio, Percentage, Rates and Financial Mathematics for applied constraints and to Error Analysis, Corrections and Full-Paper Recovery when sign or boundary mistakes recur.
Final thought
An inequality is a mathematical boundary. Strong solutions preserve that boundary through algebra, represent the permitted region clearly and return the result to the practical meaning of the variable.
Solve the boundary. Then decide which side of it the problem allows.
Return to the Secondary Mathematics Hub.
