Solving Linear Equations

A linear equation contains a variable raised to the power of 1 (no $x^{2}$x2, $x^{3}$x3, etc.). Solving means finding the value of the unknown that makes the equation true. This lesson covers one-step, two-step, and multi-step equations, including those with unknowns on both sides, brackets, and fractions.

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Key Vocabulary

Term	Meaning
Equation	A mathematical statement with an equals sign
Solution	The value of the variable that makes the equation true
Inverse operation	The opposite operation (+ undone by −; $\times$× undone by $\div$÷)
Balancing	Doing the same thing to both sides of the equation

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One-Step Equations

Use a single inverse operation.

Worked Example 1

Solve: x + 7 = 12

x = 12 − 7 = 5

Worked Example 2

Solve: 3x = 21

$x = 21 \div 3 =$x=21÷3= 7

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Two-Step Equations

Apply inverse operations in reverse order (undo the last operation first, or work to isolate the variable).

Worked Example 3

Solve: 2x + 5 = 17

Step 1: 2x = 17 − 5 = 12

Step 2: $x = 12 \div 2 =$x=12÷2= 6

Worked Example 4

Solve: (x − 3)/4 = 5

Step 1: $x - 3 = 5 \times 4 = 20$x−3=5×4=20

Step 2: x = 20 + 3 = 23

Function Machine for Two-Step Equations

flowchart LR
    X[Input x] --> OP1[Multiply by 2]
    OP1 --> OP2[Add 5]
    OP2 --> OUT[Output 17]
    OUT -.reverse.-> INV2[Subtract 5]
    INV2 -.-> INV1[Divide by 2]
    INV1 -.-> X2[Recover x = 6]

The forward arrows show how 2x + 5 = 17 is built up from x. To solve, run the machine in reverse: undo the last operation first (subtract 5), then undo the previous one (divide by 2). This is the "balancing" method viewed as inverse operations.

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Equations with Unknowns on Both Sides

Collect the variable terms on one side and the constant terms on the other.

Worked Example 5

Solve: 5x + 2 = 3x + 14

Step 1: 5x − 3x = 14 − 2

Step 2: 2x = 12

Step 3: x = 6

Check: LHS = 5(6) + 2 = 32; RHS = 3(6) + 14 = 32 ✓

Worked Example 6

Solve: 7x − 3 = 2x + 22

5x = 25

x = 5

Worked Example 7

Solve: 3(x + 4) = 5x − 2

Step 1 (expand): 3x + 12 = 5x − 2

Step 2: 12 + 2 = 5x − 3x

Step 3: 14 = 2x

Step 4: x = 7

Check: LHS $= 3(7 + 4) = 3 \times 11 = 33$=3(7+4)=3×11=33; RHS = 5(7) − 2 = 35 − 2 = 33 ✓

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Equations with Brackets

Always expand the brackets first, then solve as normal.

Worked Example 8

Solve: 4(2x − 1) − 3(x + 5) = 7

Step 1: 8x − 4 − 3x − 15 = 7

Step 2: 5x − 19 = 7

Step 3: 5x = 26

Step 4: x = 26/5 = 5.2

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Equations with Fractions

Multiply every term by the lowest common denominator (LCD) to eliminate fractions.

Worked Example 9

Solve: x/3 + x/4 = 7

LCD of 3 and 4 is 12. Multiply every term by 12:

4x + 3x = 84

7x = 84

x = 12

Worked Example 10

Solve: (2x + 1)/5 = (x − 3)/2

Cross-multiply: 2(2x + 1) = 5(x − 3)

4x + 2 = 5x − 15

2 + 15 = 5x − 4x

x = 17

Check: LHS = (2(17) + 1)/5 = 35/5 = 7; RHS = (17 − 3)/2 = 14/2 = 7 ✓

Worked Example 11

Solve: (3x − 1)/6 − (x + 2)/4 = 1

LCD = 12. Multiply every term by 12:

2(3x − 1) − 3(x + 2) = 12

6x − 2 − 3x − 6 = 12

3x − 8 = 12

3x = 20

x = 20/3 (or 6⅔)

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Forming and Solving Equations

Many GCSE problems require you to set up the equation from a context, then solve it.

Worked Example 12

The perimeter of a rectangle is 52 cm. The length is 3 cm more than twice the width. Find the dimensions.

Let the width = w.

Length = 2w + 3.

Perimeter: 2(w + 2w + 3) = 52

2(3w + 3) = 52

6w + 6 = 52

6w = 46

$w = 46/6 = 23/3 \approx 7.67 cm$w=46/6=23/3≈7.67cm

Length $= 2(23/3) + 3 = 46/3 + 9/3 = 55/3 \approx 18.33 cm$=2(23/3)+3=46/3+9/3=55/3≈18.33cm

Check: Perimeter $= 2(23/3 + 55/3) = 2(78/3) = 2 \times 26 = 52$=2(23/3+55/3)=2(78/3)=2×26=52 ✓

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Common Mistakes and Misconceptions

Mistake	Correct approach
Subtracting from one side only	Always do the same to both sides
Forgetting to multiply ALL terms by the LCD	Every single term must be multiplied
Not expanding brackets before collecting terms	Expand first, then simplify
Writing x = 12 when you mean 2x = 12	Always complete the final step
Sign errors when moving terms across	If you subtract 3x from both sides, +3x becomes −3x (not +3x)

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Edexcel Exam Tips

• Edexcel GCSE papers love "form an equation" problems — especially using angles (angles in a triangle sum to $180°$180°) or perimeters.
• Always show your working line by line. On a 3-mark equation, you typically get: M1 for forming the equation, M1 for a correct algebraic step, A1 for the correct answer.
• If the question says "Give your answer as a fraction", do not convert to a decimal.
• On Paper 1 (non-calculator), equations with fractions are common — practise clearing fractions by multiplying by the LCD.
• If you have time, always check your answer by substituting back into the original equation.

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Practice Problems