Secondary 3 Mathematics Classroom | Chapter 2: Linear Inequalities

SECONDARY 3 MATHEMATICS CLASSROOM · CHAPTER 2 · LINEAR INEQUALITIES · NUMBER LINES · G2/G3

Linear Inequalities: When the Answer Is a Region, Not a Point

An equation asks which value makes two expressions equal. An inequality asks which whole range of values makes one expression smaller, larger, or no greater/no less than another.

The older Secondary 3 E-Mathematics textbook places Linear Inequalities directly after Quadratic Equations and Functions, and that sequence remains useful. Chapter 1 trained students to solve for isolated roots. Chapter 2 changes the mathematical object: instead of one or two solution points, the solution is usually an interval on the number line. The algebra may look familiar, but the interpretation is different.

Classroom rule: solve as you would an equation → watch for multiplication or division by a negative number → reverse the inequality sign when required → combine simultaneous conditions → represent the final interval accurately → test a value inside and outside the region.

Current syllabus boundary. The current G3 syllabus includes solving linear inequalities in one variable, including simultaneous inequalities, and representing solutions on the number line. G2 students meet inequality ideas earlier, but the timing and depth of simultaneous inequalities can differ by route. Use the student’s school sequence as the controlling timetable.

Official reference: MOE G2 and G3 Mathematics Syllabuses.

Featured Answer: Why Does the Inequality Sign Reverse?

Because multiplying or dividing by a negative number reverses order on the number line. For example, 3<5. Multiplying both sides by −1 gives −3>−5. The values have reflected across zero, so the ordering reverses.

1. Inequality Symbols Carry Boundary Information

x<4 means values strictly less than 4.
x≤4 includes the boundary value 4.
x>4 means values strictly greater than 4.
x≥4 includes 4 and all greater values.

2. Equations and Inequalities Share Algebraic Moves

Solve x+7<12 by subtracting 7 from both sides: x<5.

The same balancing idea used in equations still applies because the same operation is performed on both sides.

3. Adding or Subtracting Does Not Reverse the Sign

If 2x−5≥9, add 5 to get 2x≥14, then divide by positive 2 to obtain x≥7.

4. Multiplying or Dividing by a Negative Reverses the Sign

Solve −3x>12. Dividing by −3 gives x<−4.

5. Teacher Model 1: Multi-Step Inequality

Solve 7−3x≥19.

−3x≥12.

Divide by −3 and reverse the sign: x≤−4.

6. Number Lines Show the Solution Set

For x<3, place an open point at 3 and shade to the left. For x≤3, use a closed point because the boundary is included.

7. Direction of Shading Comes From the Inequality

less than → shade left;
greater than → shade right.

Do not decide from the original appearance of the variable. Use the final solved statement.

8. Teacher Model 2: Fractional Coefficient

Solve (2x−1)/3<5.

2x−1<15 → 2x<16 → x<8.

9. Brackets Must Be Expanded Before Combining Terms

Solve 3(x−2)≤2x+5.

3x−6≤2x+5 → x≤11.

10. Simultaneous Inequalities Mean Both Conditions Must Hold

If x>2 and x≤7, the solution is 2<x≤7. This is the overlap of the two solution regions.

11. Teacher Model 3: Solve Two Conditions Separately

Find x satisfying x−3<7 and 4x−5>22.

First condition: x<10.

Second condition: 4x>27, so x>27/4.

Combined: 27/4<x<10.

12. Compound Notation Compresses an Overlap

−5<2x+1≤9 can be solved in one chain:

−6<2x≤8 → −3<x≤4.

13. Integer Solutions Are a Second Filtering Step

If −3<x≤4 and x is an integer, then x∈{−2,−1,0,1,2,3,4}.

14. Teacher Model 4: Boundary Inclusion Matters

For 1<x≤5, x=1 is excluded while x=5 is included. A number-line diagram must communicate that distinction with open and closed endpoints.

15. Inequalities Can Model Real Constraints

A lift has capacity 600 kg. If equipment already weighs 120 kg and each crate weighs 40 kg, the number n of crates must satisfy:

120+40n≤600.

40n≤480 → n≤12.

Since n is a non-negative whole number, the maximum is 12 crates.

16. Context Can Change a Continuous Range Into Discrete Values

An algebraic inequality may allow every real value in an interval, but counts of people, tickets or boxes usually require integer solutions.

17. Teacher Model 5: Budget Constraint

A student has at most $50. A fixed fee is $8 and each item costs $6. If n items are bought:

8+6n≤50 → 6n≤42 → n≤7.

Maximum number of whole items=7.

18. “At Least” and “At Most” Must Be Translated Precisely

at least 12 → x≥12;
at most 12 → x≤12;
more than 12 → x>12;
less than 12 → x<12.

19. Verification by Testing Values

If the solution is x<−4, test x=−5 in the original inequality and also test a value outside the region, such as x=0. The first should work and the second should fail.

20. A Boundary Test Checks Open or Closed Endpoints

For x≤7, substituting x=7 into the original inequality should satisfy it. For x<7, x=7 should fail.

21. Misconception Clinic: Reverse the Sign Every Time You Move a Term

Repair: the sign reverses only when multiplying or dividing the whole inequality by a negative quantity, not when adding or subtracting.

22. Misconception Clinic: Forget to Reverse After Dividing by −3

Repair: explicitly mark the negative divisor before the final step.

23. Misconception Clinic: Open and Closed Points Are Decorative

Repair: endpoint style records whether equality is included.

24. Misconception Clinic: “And” Means Combine All Values From Both Sets

Repair: simultaneous inequalities joined by “and” require the overlap where both conditions hold.

25. Misconception Clinic: Every Real Solution Is Allowed in Context

Repair: context may require whole numbers, non-negative values or another stated domain.

26. Guided Practice A: Single Inequalities

2x+5<17.
9−4x≤21.
5(x−1)>3x+7.
(3x+2)/4≥5.

Solutions

x<6. −4x≤12, so x≥−3. 5x−5>3x+7, so x>6. 3x+2≥20, so x≥6.

27. Guided Practice B: Simultaneous Inequalities

Solve x+2>5 and 2x≤14.
Solve −7<3x−1≤11.
List integer values satisfying −2<x≤3.

Solutions

3<x≤7. −6<3x≤12, so −2<x≤4. Integers: −1,0,1,2,3.

28. Guided Practice C: Context

A hall can hold at most 240 people. There are already 36 organisers inside. Each bus brings 28 students. Find the greatest possible number of full buses.

Worked solution

36+28b≤240 → 28b≤204 → b≤7.285… Since b is a whole number, greatest possible b=7.

29. Challenge Practice: Detect the Sign Error

A learner solves 5−2x>11 as x>−3. Diagnose the error.

Answer

5−2x>11 gives −2x>6. Dividing by −2 requires reversal, so the correct solution is x<−3.

30. Assessment Method: Mark the Reversal Step Explicitly

When dividing by a negative number, write a short note or circle the divisor. This turns a common hidden error into a visible decision.

31. Assessment Method: Draw the Number Line After Solving

Do not sketch from memory before the final inequality is stable. Solve first, then transfer the final statement faithfully to the line.

32. Assessment Method: Test One Inside and One Outside

This is a fast way to detect an accidentally reversed region.

33. Oral Classroom Check

How is an inequality different from an equation?
When does the inequality sign reverse?
Why does multiplying by −1 reverse order?
What does an open endpoint mean?
What does a closed endpoint mean?
What does “and” mean for two inequality conditions?
How do you solve a compound inequality such as −5<2x+1≤9?
Why might an answer need to be restricted to integers?
How can you verify the direction of a solution interval?
What is the difference between “at most” and “less than”?

34. Exit Ticket

Solve x−6<9.
Solve −4x≥20.
Solve 3(x+2)<2x+11.
Solve (x−1)/2≤4.
Solve x>−1 and x≤6.
Solve −9<2x+1≤7.
List integers satisfying 1<x≤5.
Represent x<3 on a number line.
Translate “at least 20” into an inequality.
Explain why the sign reverses when dividing by a negative number.

Exit-ticket solutions

x<15. x≤−5. 3x+6<2x+11, so x<5. x−1≤8, so x≤9. −1<x≤6. −10<2x≤6, so −5<x≤3. Integers 2,3,4,5. Open point at 3, shade left. x≥20. Multiplication/division by a negative reflects order across zero, so the inequality direction reverses.

35. The Seven-Day Return Cycle

Day 0: solve and graph single inequalities.
Day 1: negative-coefficient reversal and endpoint control.
Day 3: simultaneous inequalities and integer filtering.
Day 7: changed context problem with an “at most/at least” translation.

36. The Full Inequality Routine

translate condition → simplify → isolate variable → reverse sign if dividing/multiplying by a negative → combine conditions → apply domain → draw number line → test values.

37. Connect Back to Chapter 1

Return to Secondary 3 Chapter 1: Quadratic Equations and Functions when equation balancing or algebraic sign control is unstable. Chapter 1 solves equality; Chapter 2 extends the same algebra into ranges.

38. Specialist Companions

39. Why This Chapter Matters for Chapter 3

Linear inequalities demand exact algebraic order control. Chapter 3 shifts to multiplicative structure through indices and standard form. The same discipline continues: operations have rules, signs matter, and an answer must be interpreted structurally rather than copied from a calculator.

40. Ready for Chapter 3?

solve one-variable linear inequalities;
reverse the sign correctly with negative multiplication/division;
represent strict and inclusive boundaries correctly;
solve simultaneous inequalities;
filter integer solutions when context requires them;
translate at least/at most/more than/less than accurately;
verify an interval by testing values.

If one item is weak, return to the smallest section that owns it and solve a changed example. When the route is stable, continue to Chapter 3: Indices and Standard Form.

Small Group Tutorials

Secondary 3 Mathematics Classroom | Chapter 2: Linear Inequalities | G2/G3