📘 GCSE Foundation Maths Rescue Pack

Unit 4: Solving Linear Equations

Grade 4/5 Focus · Master the balance method to isolate x and find the solution

Balance Method One-Step Equations Two-Step Equations Negative Coefficients Unknowns Both Sides Brackets Verification
AQA 8300 Edexcel 1MA1 OCR J560
🎯 Unit Progress
0 / 5 skills
1. Algebra FREE 2. Expanding 3. Indices 4. Equations 5. Inequalities 6. Graphs 7. Non-Linear 8. Real-Life 9. Applications 10. Coordinate
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 FREE
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.

The Balance Method: Do the Same to Both Sides 3x + 5 14 − 5 from both sides 3x = 9
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
1
Divide both sides by 4: 4x ÷ 4 = 20 ÷ 4
✓ Answer
x = 5
Worked Example 02One-Step (Addition)
Solve: x + 6 = 15
1
Subtract 6 from both sides: x + 6 − 6 = 15 − 6
✓ Answer
x = 9
Worked Example 03Two-Step (Positive Coefficient)
Solve: 3x + 7 = 22
1
Subtract 7: 3x = 15
2
Divide by 3: x = 5
✓ Answer
x = 5
Worked Example 04Two-Step (Negative Coefficient)KEY SKILL
Solve: −2x + 4 = 10
1
Subtract 4: −2x = 6
2
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
1
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
1
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
1
Subtract 2x from both sides: 3x + 2 = 11
2
Subtract 2 from both sides: 3x = 9
3
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
1
Subtract 3x from both sides: 5x − 3 = 12
2
Add 3 to both sides: 5x = 15
3
Divide by 5: x = 3
✓ Final Answer
x = 3
Worked Example 09Equations with BracketsNEW
Solve:   3(2x − 1) = 15
1
Expand the bracket first: 6x − 3 = 15
2
Add 3 to both sides: 6x = 18
3
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)
1
Expand both brackets: 4x + 4 = 6x − 2
2
Subtract 4x from both sides: 4 = 2x − 2
3
Add 2 to both sides: 6 = 2x
4
Divide by 2: x = 3
✓
Check: 4(3+1) = 16 and 2(3×3−1) = 2(8) = 16 ✔
✓ Final Answer
x = 3
Inverse Operations — Use the Opposite to Undo + add − subtract × multiply ÷ divide ⇆ ⇆
To isolate x, use the inverse: undo addition with subtraction, undo multiplication with division.
✅
Method Checklist
📋 Solving Linear Equations
🧮
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.
0 / 16
Problems revealed
Tap problems to reveal working and answers
🟢 Green — Core Skills
🟢 Green · TYPE YOUR ANSWER
Solve: 3x = 18
>
🔍
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
🏆
Mastery Checklist
🏆 Unit 4 Mastery
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