An inequality does not ask for one answer. It asks for a region of possible answers. That changes the whole style of reasoning. Instead of finding a single value of x, the learner must preserve an order relationship, understand why multiplying or dividing by a negative reverses the inequality sign, and represent an entire solution set accurately.
This Secondary 3 Mathematics Learning Guide develops linear inequalities from meaning rather than from rules alone. It covers one-step and multi-step inequalities, double inequalities, number-line representation, restrictions from context, and problems where the answer is a range rather than a single number. It belongs to the Secondary Mathematics Hub.
The official 2027 SEC G3 Mathematics syllabus includes linear inequalities in one variable, simultaneous inequalities and number-line representation. This guide is G3-oriented; schools may distribute the topic differently across their Secondary 3 programme.
Use this route: diagnostic · why the rules work · simultaneous inequalities · word problems · practice · answers. All examples are original teaching examples.
Equation Versus Inequality
The equation x + 3 = 8 asks for the value that makes two quantities equal. The answer is x = 5. The inequality x + 3 < 8 asks for every value that keeps the left side smaller than the right. Its solution is x < 5.
Testing values makes the difference visible. If x = 4, then 4 + 3 = 7, which is below 8. If x = 0, the inequality is also true. If x = −100, it is still true. The solution is therefore not one point but an entire set of values.
The four common symbols are <, >, ≤ and ≥. The symbols ≤ and ≥ include equality. Thus x ≤ 5 includes 5 itself, while x < 5 does not.
A Five-Question Diagnostic
Try these without looking ahead. Solve x + 7 < 12. Solve 3x ≥ 18. Solve −2x < 10. Represent x ≥ 4 on a number line. Finally, write an inequality saying that a ticket price p dollars is at most $15.
The answers are x < 5; x ≥ 6; x > −5; a closed point at 4 with the line extending to the right; and p ≤ 15. If the third answer is x < −5, the main difficulty is probably the sign reversal after dividing by a negative. If the number-line point is open, the issue is interpreting ≥ rather than algebra.
These errors should not be repaired in the same way. Algebraic transformation, symbol meaning and graphical representation are separate capabilities even though they meet in one topic.
Why Adding the Same Quantity Preserves Order
If 3 < 7, then adding 5 gives 8 < 12. Both numbers move the same distance along the number line, so their order is unchanged. This is why adding or subtracting the same quantity from both sides of an inequality does not reverse the sign.
The same applies algebraically. From x + 4 < 11, subtract 4 from both sides to obtain x < 7. The operation changes the position of both quantities equally while preserving their relative order.
Why Multiplying by a Negative Reverses the Sign
Start with 2 < 5. Multiplying both sides by 3 gives 6 < 15; order is preserved. But multiplying by −3 gives −6 and −15. On the number line, −6 is greater than −15. Therefore the true statement is −6 > −15.
Multiplication by a negative reflects numbers across zero. Reflection reverses left and right, so the order reverses. That is the reason the inequality symbol flips when multiplying or dividing both sides by a negative number.
The rule is not “flip the sign whenever a negative number appears”. A negative number can appear without causing reversal. The reversal happens specifically when the transformation multiplies or divides both sides by a negative quantity.
Worked Example 1: A Multi-Step Inequality
Solve 5x − 7 ≤ 18. Add 7 to both sides to obtain 5x ≤ 25. Divide by positive 5, so the sign stays the same. Therefore x ≤ 5.
Check two values. At x = 5, the left side is 18, so equality is allowed. At x = 6, the left side is 23, which is not ≤ 18. The boundary behaves exactly as the final inequality predicts.
Worked Example 2: The Sign Reversal
Solve 7 − 3x > 19. Subtract 7 to obtain −3x > 12. Divide by −3 and reverse the inequality: x < −4.
Check x = −5: 7 − 3(−5) = 22, which is greater than 19. Check x = −3: 7 + 9 = 16, which is not greater than 19. This quick substitution exposes the wrong-sign answer immediately.
A useful habit is to predict direction before completing the arithmetic. Since making x more negative makes −3x larger, the solution should lie to the left of a boundary value. That qualitative check supports the algebraic rule.
Brackets and Fractions Do Not Change the Logic
The same principles apply when the inequality looks longer. Expand brackets carefully, collect like terms, preserve order under addition and subtraction, and inspect the sign of the quantity by which you finally divide.
For example, 2(3x − 1) ≥ 5x + 7 becomes 6x − 2 ≥ 5x + 7, hence x ≥ 9. No reversal occurs because the final coefficient of x is positive.
Worked Example 3: Variables on Both Sides
Solve 4 − 2x ≤ 3x + 19. Subtract 3x from both sides: 4 − 5x ≤ 19. Subtract 4: −5x ≤ 15. Divide by −5 and reverse the sign, giving x ≥ −3.
At x = −3, both sides equal 10, so the boundary is included. At x = −4, the left side is 12 while the right side is 7, so the original inequality fails. This verifies both the boundary and the direction.
Number-Line Representation
A number line communicates two things: the boundary and the direction of allowed values. An open point means the boundary is excluded, corresponding to < or >. A closed point means the boundary is included, corresponding to ≤ or ≥.
Thus x < 3 is shown with an open point at 3 and the line extending left. The inequality x ≥ −2 has a closed point at −2 and extends right. The arrow or shaded direction matters as much as the point itself.
Do not decide direction from the visual appearance of the inequality symbol alone. Read it as a sentence: “x is less than 3” means values to the left of 3; “x is greater than or equal to −2” means values at −2 and to its right.
Simultaneous Inequalities Mean Both Conditions Must Hold
If x > 1 and x ≤ 5, the solution must satisfy both conditions. The common region is 1 < x ≤ 5. On a number line, it is the interval between 1 and 5, open at 1 and closed at 5.
The word “and” is an intersection idea: keep only values allowed by every condition. This connects naturally to the set language developed in the companion Set Language, Venn Diagrams and Counting guide.
Worked Example 4: Solve Two Conditions Separately, Then Intersect
Solve 2x + 1 > 5 and 3x − 4 ≤ 11. The first condition gives 2x > 4, so x > 2. The second gives 3x ≤ 15, so x ≤ 5.
Both must hold, so the answer is 2 < x ≤ 5. A value such as 4 satisfies both; a value such as 6 satisfies the first but not the second. This is why writing the two separate answers without combining them is incomplete.
Double Inequalities Can Be Manipulated Together
A statement such as −1 ≤ 2x + 3 < 9 says that the middle expression lies between two boundaries. One efficient route is to perform the same operation on all three parts.
Subtract 3 throughout: −4 ≤ 2x < 6. Divide throughout by positive 2: −2 ≤ x < 3. Because the divisor is positive, both inequality directions stay unchanged.
If dividing all three parts by a negative number, both inequality signs reverse. For example, 2 < −x ≤ 7 becomes, after division by −1, −2 > x ≥ −7. Rewriting in increasing order gives −7 ≤ x < −2.
When the Variable Is an Integer
An inequality may describe a continuous interval, while the context may permit only integer values. If 2.4 < n ≤ 6.8 and n is an integer, then n can be 3, 4, 5 or 6.
Do not round the boundaries as though they were approximate measurements. Instead, identify which integers actually lie inside the interval. Rounding 2.4 to 2 and 6.8 to 7 would incorrectly include values that do not satisfy the original condition.
Word Problems: Translate the Limiting Language
Words such as “at least”, “at most”, “more than”, “less than”, “no more than” and “minimum” determine the inequality. Translate the sentence before calculating.
| Phrase | Typical mathematical meaning |
|---|---|
| at least 20 | x ≥ 20 |
| more than 20 | x > 20 |
| at most 20 | x ≤ 20 |
| less than 20 | x < 20 |
| between 5 and 12 inclusive | 5 ≤ x ≤ 12 |
The word “inclusive” explicitly includes the stated boundaries. In ordinary contexts, phrases can sometimes be ambiguous, so use the exact language of the question rather than relying on a memorised word list without reading the sentence.
Worked Example 5: A Budget Constraint
A student has $42 for notebooks costing $5 each and a fixed delivery charge of $7. What is the greatest number of notebooks that can be bought? Let n be the number of notebooks, where n is a nonnegative integer.
The total cost is 5n + 7. It must not exceed 42, so 5n + 7 ≤ 42. Hence 5n ≤ 35 and n ≤ 7. Therefore the greatest possible integer is 7 notebooks.
The answer is not “n ≤ 7 notebooks” because the question asks for the greatest number. The inequality describes the full feasible set; the final sentence asks us to select its largest allowed integer.
Worked Example 6: A Range of Lengths
A rectangular strip has perimeter at most 50 cm. Its length is 4 cm more than its width. If the width is w cm, find the possible positive widths.
The length is w + 4. The perimeter condition gives 2w + 2(w + 4) ≤ 50. Thus 4w + 8 ≤ 50, so 4w ≤ 42 and w ≤ 10.5.
Because width must be positive, the complete contextual solution is 0 < w ≤ 10.5. The lower restriction came from geometry, not from the algebraic inequality alone.
Worked Example 7: Comparing Two Plans
Plan A costs $12 plus $3 per session. Plan B costs $24 plus $2 per session. For how many sessions is Plan A cheaper than Plan B? Let n be the number of sessions, with n a nonnegative integer.
Write the cost comparison: 12 + 3n < 24 + 2n. Subtract 2n and 12 to obtain n < 12. Therefore Plan A is cheaper for 0 to 11 sessions if zero sessions is meaningful in the context; for at least one session, it is cheaper for 1 to 11 sessions.
At n = 12 the costs are equal, so the strict inequality correctly excludes the boundary. This is why the difference between “cheaper than” and “no more expensive than” matters.
Four Common Errors and Their Repairs
Automatic sign flipping: the student reverses the inequality whenever a negative appears. Repair the idea that reversal is tied to multiplying or dividing both sides by a negative.
Boundary confusion: the algebra is correct but the open or closed point is wrong. Repair the meaning of strict versus inclusive symbols.
Failure to intersect: two simultaneous inequalities are solved correctly but left as separate statements. Repair the idea that “and” requires values satisfying both conditions.
Ignoring the context domain: a mathematically valid negative value is accepted for a length or number of people. Repair the return from algebra to the original meaning.
Inequalities and Graphs
The number line is the simplest graph of a one-variable solution set. In later work, inequalities can also describe regions on coordinate axes. Even when the course remains in one variable, thinking in terms of a region rather than a single point prepares the learner for a broader idea of constraints.
The companion Quadratic, Power and Exponential Graphs guide develops how equations produce curves, intersections and tangent gradients. An inequality instead asks which side of a boundary remains allowed.
Independent Practice
Solve each inequality over the real numbers unless the question specifies integers. Show the transformation that determines whether the inequality sign changes.
1. x − 4 > 9.
2. 6x ≤ 30.
3. −4x ≥ 20.
4. 3x + 7 < 2x + 15.
5. 5 − 2x ≥ 13.
6. 4(2x − 1) < 3x + 16.
7. Solve x > −2 and x ≤ 4.
8. Solve 2x − 1 ≥ 5 and 3x + 2 < 20.
9. Solve −3 ≤ 2x + 1 < 9.
10. Find the integer values satisfying −1.2 < n ≤ 4.6.
11. A taxi fare is modelled as $4 plus $2.50 per kilometre. A passenger wants the fare to be at most $24. Find the maximum whole number of kilometres permitted by this simplified model.
12. A rectangle has width w cm and length w + 5 cm. Its perimeter must be less than 46 cm. Find the possible positive widths.
13. A club needs at least $300. It already has $84 and earns $18 from each registration. Find the minimum whole number of registrations needed.
14. Explain why solving −2x < 6 as x < −3 is wrong.
Explained Answers
1. Add 4 to obtain x > 13.
2. Divide by positive 6 to obtain x ≤ 5.
3. Divide by −4 and reverse the sign: x ≤ −5.
4. Subtract 2x and then 7: x < 8.
5. Subtract 5: −2x ≥ 8. Divide by −2 and reverse the sign: x ≤ −4.
6. Expand to 8x − 4 < 3x + 16. Then 5x < 20, so x < 4.
7. The common region is −2 < x ≤ 4.
8. The first condition gives x ≥ 3. The second gives x < 6. Therefore 3 ≤ x < 6.
9. Subtract 1 throughout to obtain −4 ≤ 2x < 8, then divide by 2: −2 ≤ x < 4.
10. The integers strictly greater than −1.2 and no greater than 4.6 are −1, 0, 1, 2, 3 and 4.
11. 4 + 2.5d ≤ 24 gives 2.5d ≤ 20 and d ≤ 8. The maximum whole number is 8 km under the stated simplified model.
12. 2w + 2(w + 5) < 46 gives 4w + 10 < 46, so w < 9. Together with positivity: 0 < w < 9.
13. 84 + 18n ≥ 300 gives 18n ≥ 216 and n ≥ 12. The minimum whole number is 12 registrations.
14. Dividing −2x < 6 by −2 requires reversing the inequality. The correct answer is x > −3. Checking x = 0 confirms it: 0 < 6 is true, so zero must lie in the solution set.
A Reliable Checking Routine
Choose one value clearly inside the proposed solution region and one value clearly outside it. Substitute both into the original inequality. Then check the boundary separately if equality might be included. This three-point test is simple and catches many direction and endpoint errors.
For a contextual problem, also check units and discreteness. A solution such as n ≥ 11.2 may need the integer conclusion n ≥ 12 if n counts people. A width interval must include the physical requirement that length is positive.
Teacher and Parent Prompts
Ask “Does the answer represent one value or many?” before correcting algebra. Ask “What operation changed the order?” when the sign reversal is uncertain. For simultaneous inequalities, ask the learner to point to the overlap on a number line before compressing it into symbolic notation.
For extension, change the context after the algebra is correct. Ask for the maximum affordable number, the minimum required number, or the full feasible range. This forces the learner to interpret the same inequality in different ways.
Questions Students Often Ask
Why does the sign reverse only for negative multiplication or division? Because multiplication by a negative reflects the number line and reverses order.
Can an inequality answer be a decimal? Yes. The boundary can be any real number. If the variable must be an integer, select only the integer values inside the resulting interval.
How do I know whether the circle on a number line is open or closed? Strict symbols < and > exclude the boundary, so use an open point. Inclusive symbols ≤ and ≥ include it, so use a closed point.
Why do word problems sometimes need another restriction? The algebra describes the mathematical relation. The context can further restrict the domain—for example, lengths must be positive and counts may need to be whole numbers.
Continue the Secondary 3 Learning Route
Continue with Set Language, Venn Diagrams and Counting to make intersection and union explicit; Matrices, Operations and Information Representation for structured arrays; and Quadratic, Power and Exponential Graphs for graphical relationships.
Inequality work is successful when the student can preserve order, represent a region and return that region to the meaning of the question. Return to the Secondary Mathematics Hub for the full S1–S4 map.
