Unit 4: Solving Linear Equations – GCSE Foundation Maths Rescue System

How to use this lesson

1. Complete the warm-up questions from memory
2. Study the worked examples carefully
3. Try the practice problems: Green → Amber → Red
4. Check your answers and complete the mastery checklist

GCSE Foundation Maths Revision Pathway

A structured 10-unit revision system to help Foundation students build from Grade 3 towards Grade 4/5.

1 Algebra & Sequences

2 Expanding & Factorising

3 Indices & Powers

📖 Unit 4 — Linear Equations (current)

5 Inequalities

6 Straight-Line Graphs

7 Non-Linear Graphs

8 Real-Life Graphs

9 Graph Applications

10 Coordinate Geometry

🔥

Warm-Up: Retrieval Practice

⏱ Quick Recall

These questions bridge Unit 2 concepts and prepare you for solving equations.

Q1 Expand: 2(x + 3)

Q2 Simplify: 5x − 2x

Q3 What is 12 ÷ 3?

Lesson Progress

✓ Warm-up — complete

→ Worked examples & key concepts

· Tiered practice (Green → Amber → Red)

· Mistake Detective & Examiner Lens

· Mastery checklist

💡

The Big Idea: The Balance Method

An equation is like a balanced scale — whatever you do to one side, you must do to the other.

3x + 5 = 14

−5 both sides

→

3x = 9

÷3 both sides

→

x = 3

The equality is preserved at every step. Use inverse operations to peel away layers and isolate x.

Whatever you do to one side, you must do to the other.

📖

Core Definitions

Equation

A statement showing two expressions are equal

Solve

Find the value of the variable

Inverse Operation

The opposite operation (add/subtract, multiply/divide)

Balance Method

Apply the same operation to both sides to isolate the variable

✏️

Worked Examples

Worked Example 01One-Step (Multiplication)

Solve: 4x = 20

Divide both sides by 4: 4x ÷ 4 = 20 ÷ 4

✓ Answer

x = 5

Worked Example 02One-Step (Addition)

Solve: x + 6 = 15

Subtract 6 from both sides: x + 6 − 6 = 15 − 6

✓ Answer

x = 9

Worked Example 03Two-Step (Positive Coefficient)

Solve: 3x + 7 = 22

Subtract 7: 3x = 15

Divide by 3: x = 5

✓ Answer

x = 5

Worked Example 04Two-Step (Negative Coefficient)KEY SKILL

Solve: −2x + 4 = 10

Subtract 4: −2x = 6

Divide by −2: x = −3

✓ Answer

x = −3

⚠️ 6 ÷ (−2) = −3. Dividing by a negative changes the sign.

Worked Example 05Fractional Solution

Solve: 3x = 10

Divide by 3: x = 10 ÷ 3

✓ Answer

x = 10/3

💡 Fractional answers are acceptable unless instructed otherwise.

Worked Example 06VerificationIMPORTANT

Check: x = 5 for 3x + 7 = 22

Substitute x = 5: 3(5) + 7 = 15 + 7 = 22 ✔

✓ Verified

x = 5 is correct

💡 Always check by substituting back into the original equation.

📐 Part 2 — Unknowns on Both Sides & Brackets

Now we tackle equations where x appears on both sides, or is inside a bracket.

Worked Example 07Unknowns on Both SidesKEY SKILL

Solve: 5x + 2 = 2x + 11

Subtract 2x from both sides: 3x + 2 = 11

Subtract 2 from both sides: 3x = 9

Divide both sides by 3: x = 3

✓

Check: 5(3)+2 = 17 and 2(3)+11 = 17 ✔

✓ Final Answer

x = 3

⚠️ Collect x-terms on the side with the larger coefficient — this avoids negatives.

Worked Example 08Unknowns on Both Sides (Harder)

Solve: 8x − 3 = 3x + 12

Subtract 3x from both sides: 5x − 3 = 12

Add 3 to both sides: 5x = 15

Divide by 5: x = 3

✓ Final Answer

x = 3

Worked Example 09Equations with BracketsNEW

Solve: 3(2x − 1) = 15

Expand the bracket first: 6x − 3 = 15

Add 3 to both sides: 6x = 18

Divide by 6: x = 3

✓ Final Answer

x = 3

💡 Remember Unit 2: expand brackets first, then solve as normal.

Worked Example 10Brackets on Both SidesEXAM STYLE

Solve: 4(x + 1) = 2(3x − 1)

Expand both brackets: 4x + 4 = 6x − 2

Subtract 4x from both sides: 4 = 2x − 2

Add 2 to both sides: 6 = 2x

Divide by 2: x = 3

✓

Check: 4(3+1) = 16 and 2(3×3−1) = 2(8) = 16 ✔

✓ Final Answer

x = 3

To isolate x, use the inverse: undo addition with subtraction, undo multiplication with division.

✅

Method Checklist

📋 Solving Linear Equations

✓

Identify the operation attached to the variable

✓

Undo addition/subtraction first

✓

Undo multiplication/division second

✓

Apply the same operation to both sides

✓

Show each step clearly

✓

Check solution by substitution

🧮

Arithmetic Support

Dividing by a negative changes the sign

−2x = 6 → x = −3

3x = 10 → x = 10/3 (fractions valid)

Substitute solution back to check

💪

Tiered Practice

🔓 How It Works

Tap any card to reveal the full working and answer.

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Problems revealed

Tap problems to reveal working and answers

🟢 Green — Core Skills

🟢 Green · TYPE YOUR ANSWER

Solve: 3x = 18

✍️ Type Your Answer

Solve: 3x = 18

Divide both sides by 3: x = 18 ÷ 3 = 6

✓ x = 6

Green · Q1

Solve: 6x = 42

Divide both sides by 6: 6x ÷ 6 = 42 ÷ 6

✓ x = 7

Tap to reveal answer

Green · Q2

Solve: x − 9 = 5

Add 9 to both sides: x − 9 + 9 = 5 + 9

✓ x = 14

Tap to reveal answer

Green · Q3

Solve: 5x = −20

Divide both sides by 5: 5x ÷ 5 = −20 ÷ 5

✓ x = −4

Tap to reveal answer

🟡 Amber — Multi-Step

🟡 Amber · TYPE YOUR ANSWER

Solve: 5x − 3 = 12

Amber · Q4

Solve: 2x + 5 = 13

Step 1: Subtract 5 → 2x = 8 Step 2: Divide by 2 → x = 4

✓ x = 4

Tap to reveal answer

Amber · Q5

Solve: 4x − 7 = 9

Step 1: Add 7 → 4x = 16 Step 2: Divide by 4 → x = 4

✓ x = 4

Tap to reveal answer

Amber · Q6

Solve: −3x + 6 = 0

Step 1: Subtract 6 → −3x = −6 Step 2: Divide by −3 → x = 2

✓ x = 2

Tap to reveal answer

Amber · Q7

Solve: 5x = 12

Divide both sides by 5: x = 12 ÷ 5 Fractional answer is valid

✓ x = 12/5

Tap to reveal answer

🔴 Red — Exam Style

🔴 Red · TYPE YOUR ANSWER

Solve: 6x + 2 = 2x + 18

Red · 2 marks · Q8

Solve fully: 7x − 5 = 16

Step 1: Add 5 → 7x = 21 Step 2: Divide by 7 → x = 3

✓ x = 3

Tap to reveal answer

Red · 2 marks · Q9

Solve fully: −4x + 3 = 11

Step 1: Subtract 3 → −4x = 8 Step 2: Divide by −4 → x = −2

✓ x = −2

Tap to reveal answer

Red · 2 marks · Q10

Solve fully: 6x = 25

Divide by 6: x = 25 ÷ 6 Leave as fraction

✓ x = 25/6

Tap to reveal answer

🟡 Unknowns on Both Sides & Brackets

Amber · Q11

Solve: 4x + 1 = x + 10

Subtract x from both sides: 3x + 1 = 10 Subtract 1: 3x = 9 Divide by 3: x = 3

✓ x = 3

Tap to reveal answer

Amber · Q12

Solve: 2(x + 4) = 14

Expand: 2x + 8 = 14 Subtract 8: 2x = 6 Divide by 2: x = 3

✓ x = 3

Tap to reveal answer

Red · 3 marks · Q13

Solve: 7x − 5 = 3x + 11

Subtract 3x: 4x − 5 = 11 Add 5: 4x = 16 Divide by 4: x = 4 Check: 7(4)−5=23, 3(4)+11=23 ✓

✓ x = 4

Tap to reveal answer

Red · 3 marks · Q14

Solve: 5(2x − 3) = 25

Expand: 10x − 15 = 25 Add 15: 10x = 40 Divide by 10: x = 4

✓ x = 4

Tap to reveal answer

Red · 4 marks · Q15

Solve: 3(x + 2) = 2(2x − 1)

Expand both: 3x + 6 = 4x − 2 Subtract 3x: 6 = x − 2 Add 2: x = 8 Check: 3(10)=30, 2(15)=30 ✓

✓ x = 8

Tap to reveal answer

🔍

Mistake Detective

⚠️ These mistakes cost marks

Study each one carefully.

❌ Wrong

3x + 4 = 13 3x = 13 x = 13/3

The constant 4 was not subtracted from both sides!

✅ Correct

3x + 4 = 13 3x + 4 − 4 = 13 − 4 3x = 9 x = 3

❌ Wrong

5x + 2 = 2x + 11 5x = 2x + 9 5x − 2x = 9 3x = 9 ← correct here but…

Method works but the intermediate step is messy. Always subtract x-terms in one clean step.

✅ Correct

5x + 2 = 2x + 11 5x − 2x + 2 = 11 3x + 2 = 11 3x = 9, x = 3

❌ Wrong

2(3x − 1) = 10 2 × 3x − 1 = 10 6x − 1 = 10

Only multiplied the first term! The −1 must also be multiplied by 2.

✅ Correct

2(3x − 1) = 10 6x − 2 = 10 6x = 12, x = 2

🎓

Examiner Lens

🎓 To Gain Full Marks

Show each inverse operation clearly on a new line
Apply the operation to both sides (write “both sides”)
Box or underline your final answer
Substitute back to verify if unsure
Watch negative division carefully — show your working

🏆

Mastery Checklist

🏆 Unit 4 Mastery

✓

I can solve one-step equations confidently

✓

I can solve two-step equations with positive coefficients

✓

I can solve equations with negative coefficients (watching signs)

✓

I can recognise and write fractional solutions correctly

✓

I can check my solution by substituting back into the original equation

✓

I can solve equations with unknowns on both sides

✓

I can solve equations containing brackets

Nice work — you’ve completed Unit 4!

6 units remaining. Consistent practice is what turns Grade 3 into Grade 4/5. Keep going.

Ready for Unit 5?

If you can do these 3 things confidently, you're ready to move on:

I can solve one-step and two-step equations I can solve equations with brackets I can solve equations with unknowns on both sides

👨‍👩‍👧 For Parents

Your child has completed Unit 4 of a structured GCSE Foundation Maths revision programme covering Algebra and Graphs — key topics on every Foundation paper.

✔ AQA, Edexcel & OCR aligned

✔ 170+ practice questions

✔ Step-by-step explanations

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Unit 5: Linear Inequalities →

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