An equation asks when two quantities are equal. An inequality asks which values satisfy a comparison. That difference changes the shape of the answer. Instead of one value such as x = 4, an inequality may produce an entire range such as x < 4 or x ≥ 4.
This Secondary 2 Mathematics Learning Guide develops linear inequalities as a language of boundaries, permitted regions and constraints. It connects symbolic manipulation to number-line representation, checks the special effect of multiplying or dividing by negative numbers, and shows how inequalities appear naturally in capacity, budget and measurement problems.
Secondary Mathematics Hub: S1–S4 Capability Map · Secondary 2 Learning Guide, Batch 2, Guide 2. Companion guides cover quadratic functions, simultaneous linear equations, and quadratic equations.
Course boundary. The current MOE G2/G3 Mathematics syllabus includes Secondary Two simple linear inequalities such as ax + b ≤ c and ax + b < c, together with number-line representation. This guide develops those ideas and uses simple modelling contexts. More advanced systems of inequalities or quadratic inequalities belong to later or different courses unless specifically introduced by the school.
Navigate: Meaning · Solving · Negative multipliers · Number lines · Boundary values · Modelling · Practice and answers · Teaching and transfer.
1. Inequality symbols describe order, not approximate equality
The statement x < 5 means every permitted x is less than 5. The statement x ≤ 5 includes 5 itself. Likewise, x > 5 excludes 5 while x ≥ 5 includes it. The line beneath ≤ or ≥ records that equality at the boundary is allowed.
Do not read ≤ as approximately less than. It is an exact logical condition: either the value is less than the boundary or exactly equal to it. A value of 5.0001 does not satisfy x ≤ 5, however close it may look.
A solution is a set of values
For x + 3 < 8, the solution is x < 5. This does not mean x is 4. Any real number less than 5 works: 4, 0, −10 and 4.999 all satisfy the inequality. The output is a range rather than one isolated number.
To test a proposed value, substitute it into the original inequality. If x = 4, then 4 + 3 < 8 becomes 7 < 8, which is true. If x = 5, then 8 < 8 is false. This confirms that the boundary is excluded.
Worked example 1: translate words into a condition
A box can hold at most 18 kg. Its current contents weigh x kg. Write an inequality. At most means the value may be less than or equal to the limit, so x ≤ 18.
Compare this with weighs less than 18 kg, which would be x < 18. The single equality line changes whether exactly 18 kg is allowed. Reading the phrase precisely matters before any algebra begins.
2. Most equation moves still work because order is preserved
If a < b, then adding the same number to both sides preserves the order. For example, 3 < 7 remains true after adding 5: 8 < 12. Subtracting the same number also preserves order. This supports the familiar inverse-operation approach.
Likewise, multiplying or dividing both sides by the same positive number preserves the direction. If 2 < 5, then multiplying by 3 gives 6 < 15. The exceptional case is a negative multiplier, which reverses order and deserves its own section.
Worked example 2: one-step inequality
Solve x + 7 ≤ 12. Subtract 7 from both sides to obtain x ≤ 5. Check the boundary: x = 5 gives 12 ≤ 12, which is true. A smaller value such as x = 4 also works.
A larger value such as x = 6 gives 13 ≤ 12, which is false. Testing one value on each side of the boundary makes the final direction easier to trust.
Worked example 3: two-step inequality
Solve 3x + 4 > 19. Subtract 4 to get 3x > 15. Divide by positive 3 to obtain x > 5.
The boundary value 5 gives 19 > 19, which is false, so it is correctly excluded. Testing x = 6 gives 22 > 19, true. Testing x = 4 gives 16 > 19, false.
Keep the same transformation on the whole side
For 2(x + 3) ≤ 14, either divide by 2 first to obtain x + 3 ≤ 7, then x ≤ 4, or expand first to get 2x + 6 ≤ 14, then 2x ≤ 8, so x ≤ 4. Both routes preserve the same condition.
What is not valid is subtracting 3 from inside the bracket without applying an equivalent operation to the entire left side. Brackets represent grouped operations. Solve the relationship, not the appearance of individual symbols.
3. Multiplying or dividing by a negative reverses the inequality
Start with the true statement 2 < 5. Multiply both sides by −1. The values become −2 and −5, but −2 is greater than −5. Therefore the correct transformed statement is −2 > −5. The inequality direction reverses because multiplication by a negative reflects positions across zero on the number line.
This is not a special symbol trick. It follows from order. Numbers farther to the right are larger. Negation sends positive positions to matching negative positions on the opposite side, reversing left-right order.
Worked example 4: divide by a negative
Solve −4x ≤ 20. Divide by −4 and reverse the inequality: x ≥ −5. The boundary −5 gives 20 ≤ 20, true. A value greater than −5, such as x = 0, gives 0 ≤ 20, true.
If the sign were not reversed, the incorrect result x ≤ −5 would reject x = 0 even though it clearly satisfies the original inequality. Substitution is a strong way to expose this mistake.
Worked example 5: isolate the variable carefully
Solve 7 − 3x > 16. Subtract 7 to obtain −3x > 9. Divide by −3 and reverse the direction: x < −3.
Check x = −4: 7 − 3(−4) = 19, and 19 > 16 is true. Check x = −2: 13 > 16 is false. The values support the direction x < −3.
An alternative route can avoid a negative final division
For 7 − 3x > 16, you may subtract 16 and add 3x to both sides, obtaining 3x > 9? That would be wrong because 7 − 16 = −9, giving −9 > 3x, which is equivalent to 3x < −9 and therefore x < −3.
This example shows why algebraic bookkeeping matters. Changing which side the variable appears on does not automatically remove the need to track direction. Write each equivalent line clearly rather than mentally moving terms across the sign.
4. A number line is a picture of the solution set
For x < 3, mark an open circle at 3 and shade or draw the ray to the left. The open circle means 3 is excluded. For x ≤ 3, use a filled circle at 3 because the boundary is included.
For x > 3 or x ≥ 3, the ray extends to the right because larger numbers lie to the right on the standard number line. The diagram should match the symbolic statement exactly; it is not decorative.
Worked example 6: read a number line back into symbols
A number line shows a filled circle at −2 and a ray to the right. The solution is x ≥ −2. The filled point includes −2; the direction to the right includes all larger values.
An open circle at the same location would instead mean x > −2. The difference may look visually small, but mathematically it decides whether one boundary value belongs to the solution.
Do not confuse arrow direction with the inequality symbol shape
A common shortcut says the inequality sign points toward the smaller number. That can help when comparing two fixed numbers, but it is risky when graphing a variable range. The more reliable method is to read the meaning: x < 4 means values smaller than 4, which lie to the left.
Likewise x ≥ −3 means values at least −3, so start at −3 and move right. Translate the words, then draw the set.
5. The boundary value deserves a deliberate test
For 2x + 1 ≤ 9, solving gives x ≤ 4. The value x = 4 is the boundary because it makes the two sides equal. The original symbol allows equality, so the boundary belongs to the solution.
For 2x + 1 < 9, the same algebra gives x < 4, but now x = 4 is excluded. The numerical boundary is unchanged; only membership changes. This is why copying the final symbol from memory can alter the solution set.
Integer answers may need a final interpretation
Suppose a model gives n ≤ 7.6, where n is the number of whole boxes. Then the greatest permitted whole-number value is 7. The symbolic real-number solution and the contextual count are related but not identical.
If instead n ≥ 7.6 and n counts whole boxes, the least permitted value is 8. Do not round according to ordinary decimal rules without considering the inequality direction. The question asks for a permitted integer, not the nearest integer.
Worked example 7: capacity and whole-number interpretation
A container already holds 5 kg and can hold at most 29 kg. Identical packets each weigh 3 kg. What is the greatest number n of packets that can be added? The condition is 5 + 3n ≤ 29.
Subtract 5: 3n ≤ 24. Divide by 3: n ≤ 8. Since n is a non-negative whole number, the greatest possible number is 8 packets. Check: 5 + 3(8) = 29, exactly at the permitted capacity.
6. Inequalities model limits, requirements and permitted regions
Words such as at most, no more than, at least, minimum, maximum, below and exceeds often signal inequality relationships. Their meaning must be translated carefully. At least 12 means x ≥ 12. More than 12 means x > 12. No more than 12 means x ≤ 12.
Context can also impose additional conditions. A length must usually be positive. A count may need to be a whole number. A time interval may have a stated upper limit. Solve the algebraic inequality, then intersect it with the contextual restrictions.
Worked example 8: budget boundary
A fictional activity has a fixed charge of 12 dollars and an additional 4 dollars per participant. The budget is at most 80 dollars. Find the greatest number of participants. Let n be the number of participants. Then 12 + 4n ≤ 80.
Solving gives 4n ≤ 68, so n ≤ 17. The greatest non-negative whole-number value is 17 participants. The total is 80 dollars exactly, which is allowed because the phrase at most includes equality.
Worked example 9: a minimum requirement
A learner needs at least 120 total practice points. They already have 72 points and gain 8 points per completed task. Find the minimum number of additional tasks. The condition is 72 + 8n ≥ 120.
Then 8n ≥ 48, so n ≥ 6. The minimum whole number is 6 tasks. Five tasks would give 112, which fails the requirement; six gives exactly 120.
Worked example 10: when the algebra gives a non-integer boundary
A fictional shuttle can carry at most 320 kg of equipment. A fixed frame weighs 74 kg and each crate weighs 18 kg. What is the greatest whole number of crates? The inequality is 74 + 18n ≤ 320.
This gives 18n ≤ 246, so n ≤ 13.666… . The greatest permitted whole number is 13, not 14. Check: 74 + 18(13) = 308 kg, while 14 crates would give 326 kg and exceed capacity.
The graph of a linear inequality has a boundary interpretation
On a one-dimensional number line, solving 3x + 2 ≤ 14 gives x ≤ 4. The equality 3x + 2 = 14 identifies the boundary x = 4. The inequality selects one side of that boundary.
This idea prepares students for later coordinate inequalities. Even before shading regions in two dimensions, the central logic is already present: find the boundary, determine whether it is included, then determine which side satisfies the condition.
7. Mixed practice: solve, represent, interpret
For each question, keep the original inequality visible until the final check. When you divide or multiply by a negative, write the reversed sign explicitly rather than relying on memory. If the variable is a count, interpret the real-number result before giving the contextual answer.
Questions 1–6. 1. Solve x + 9 < 15. 2. Solve 4x ≥ 28. 3. Solve 3x − 5 ≤ 16. 4. Solve 2(x + 4) > 18. 5. Solve −5x < 20. 6. Solve 8 − 2x ≥ 14.
Questions 7–12. 7. Represent x < 3 on a number line. 8. Represent x ≥ −4 on a number line. 9. A number line has an open circle at 6 and a ray left. Write the inequality. 10. A filled circle is at −1 with a ray right. Write the inequality. 11. Test whether x = 5 satisfies 2x + 1 ≤ 11. 12. Test whether x = 5 satisfies 2x + 1 < 11.
Questions 13–18. 13. A total may not exceed 90. A fixed amount is 18 and each item adds 6. Find the greatest whole number of items. 14. A score must be at least 150. A learner has 94 and gains 7 per task. Find the minimum whole number of tasks. 15. Explain why dividing −3x > 12 by −3 gives x < −4. 16. A model gives n ≤ 5.8 where n counts people. Find the greatest permitted n. 17. A model gives n ≥ 5.8 where n counts people. Find the least permitted n. 18. Translate no fewer than 20 into an inequality.
Explained answers: questions 1–6
1. x < 6. 2. x ≥ 7. 3. 3x ≤ 21, so x ≤ 7. 4. x + 4 > 9, so x > 5. 5. Divide by −5 and reverse: x > −4. 6. −2x ≥ 6, so x ≤ −3.
Explained answers: questions 7–12
7. Open circle at 3, ray left. 8. Filled circle at −4, ray right. 9. x < 6. 10. x ≥ −1. 11. Yes, because 11 ≤ 11 is true. 12. No, because 11 < 11 is false.
Explained answers: questions 13–18
13. 18 + 6n ≤ 90 gives n ≤ 12, so 12 items. 14. 94 + 7n ≥ 150 gives n ≥ 8, so 8 tasks. 15. Multiplication by a negative reverses number-line order; the transformed inequality must reverse direction to remain equivalent.
16. 5. 17. 6. 18. x ≥ 20. The phrase no fewer than includes 20 itself.
8. Teach inequalities as boundaries, not altered equations
A useful introduction compares x = 4, x < 4 and x ≤ 4 on the same number line. Ask how many values each statement permits and whether 4 itself belongs. This makes the change from equations to solution sets visible before algebraic manipulation is added.
Next use small true numerical inequalities such as 2 < 5 and apply the same operation to both sides. Positive multiplication preserves order; negative multiplication reverses it. Let the student observe the number-line positions before formalising the sign-reversal rule.
Diagnose the first failure
If the algebra is correct but the number-line circle is wrong, repair boundary inclusion. If the number line is correct but the symbolic inequality points the wrong way after dividing by a negative, repair order reversal. If the symbolic solution is right but the final whole-number answer is impossible, repair contextual interpretation.
A single mark labelled inequality error is too broad. The learner may understand comparison perfectly and only have an integer-rounding issue, or may manipulate confidently while misunderstanding the solution set. The repair should match the first weak link.
Questions parents can ask
Does the boundary value count? Which direction on the number line contains the permitted values? Did you multiply or divide by a negative anywhere? If the answer is a number of objects, which whole numbers are actually possible? These questions support reasoning without supplying the solution.
9. The transfer test: inequality as a decision rule
A fictional workshop can use at most 200 minutes. Setup takes 32 minutes and each repeated operation takes 14 minutes. Let n be the number of complete operations. The condition is 32 + 14n ≤ 200.
Then 14n ≤ 168, so n ≤ 12. The greatest whole number is 12. Check the boundary: 32 + 14(12) = 200. Thirteen operations would require 214 minutes and violate the condition.
Now change the wording from at most 200 minutes to less than 200 minutes. The algebraic boundary remains n < 12, so the greatest whole number becomes 11. One word changed the inclusion of the boundary and therefore changed the practical decision.
Connect back to Equations and Equality when preserving equivalent transformations needs repair. Use Linear Graphs, Coordinates and Relationships to see how boundaries and linear models appear graphically. Continue to Simultaneous Linear Equations when two equality constraints must be satisfied at once.
Find the boundary. Decide whether it belongs. Preserve the order. Interpret the permitted values.
