S4 Mathematics - Term 1

Topic 2: Equations and Inequalities

Lesson 5: Solving Linear Equations Review

Duration: 40 minutes

Learning Outcomes

By the end of this lesson, you should be able to:

• Solve linear equations using inverse operations
• Verify solutions by substitution
• Apply systematic approaches to multi-step equations

Connection to Previous Lessons

Changing the subject:
y = 2x + 5 becomes x = (y - 5)/2

Solving an equation:
If y = 11, then x = (11 - 5)/2 = 3

Same skills, same inverse operations!

The Balance Principle

One-Step Equations

Example 1: 3x = 15

Divide both sides by 3:
x = 5

Check: 3(5) = 15 check!

One-Step Equations

Example 2: x - 7 = 12

Add 7 to both sides:
x = 19

Check: 19 - 7 = 12 check!

Two-Step Equations

Example: 2x + 3 = 11

Step 1: Subtract 3 from both sides
2x = 8

Step 2: Divide by 2
x = 4

Check: 2(4) + 3 = 8 + 3 = 11 check!

Two-Step Equations

Example: 4x - 5 = 15

Step 1: Add 5 to both sides
4x = 20

Step 2: Divide by 4
x = 5

Check: 4(5) - 5 = 20 - 5 = 15 check!

Pair Practice

Solve for x:

1. 5x = 35
2. x + 9 = 14
3. 3x + 2 = 17
4. 2x - 7 = 9

Answers

1. 5x = 35 then x = 7
2. x + 9 = 14 then x = 5
3. 3x + 2 = 17 then 3x = 15 then x = 5
4. 2x - 7 = 9 then 2x = 16 then x = 8

x on Both Sides

Example: 5x + 3 = 2x + 12

x on Both Sides - Steps

5x + 3 = 2x + 12

1. Subtract 2x: 3x + 3 = 12
2. Subtract 3: 3x = 9
3. Divide by 3: x = 3

Check: 5(3)+3 = 18; 2(3)+12 = 18 check!

Equations with Brackets

Example: 3(x + 2) = 15

Step 1: Expand the bracket
3x + 6 = 15

Step 2: Subtract 6
3x = 9

Step 3: Divide by 3
x = 3

Check: 3(3 + 2) = 3(5) = 15 check!

Equations with Fractions

Example: x/4 = 7

Multiply both sides by 4:
x = 28

Check: 28/4 = 7 check!

Complex Fractions

Example: (x + 3)/2 = 5

Step 1: Multiply by 2
x + 3 = 10

Step 2: Subtract 3
x = 7

Check: (7 + 3)/2 = 10/2 = 5 check!

Individual Practice

Solve for x:

1. 4x + 1 = 2x + 9
2. 2(x - 3) = 10
3. (x - 2)/3 = 4
4. 3x + 5 = x + 13

Answers

1. 4x + 1 = 2x + 9 then 2x = 8 then x = 4

2. 2(x - 3) = 10 then 2x - 6 = 10 then 2x = 16 then x = 8

3. (x - 2)/3 = 4 then x - 2 = 12 then x = 14

4. 3x + 5 = x + 13 then 2x = 8 then x = 4

The Solving Process

1. Simplify each side (expand brackets)
2. Collect x terms on one side
3. Collect constants on the other side
4. Isolate the variable
5. CHECK by substitution!

Exit Problems

Solve for x:

1. 2x + 5 = 17
2. 3(x - 1) = 12
3. 4x + 3 = x + 15

Answers

1. 2x + 5 = 17
   2x = 12, x = 6

2. 3(x - 1) = 12
   3x - 3 = 12, 3x = 15, x = 5

3. 4x + 3 = x + 15
   3x = 12, x = 4

Homework

Solve for x:

1. 3x + 7 = 22
2. 5x - 8 = 12
3. 2x + 4 = x + 10
4. 4(x + 2) = 20
5. (x + 5)/3 = 4
6. 5x - 3 = 2x + 9

Next Lesson

Lesson 6: Word Problems with Equations

We will:

• Translate word problems into equations
• Solve real-life problems using equations
• Practice setting up and solving

Credits

Created: December 2025
Based on: NCDC Lower Secondary Mathematics Syllabus (2019)

Source: National Curriculum Development Centre (NCDC), Uganda