Solving Linear Equations

A linear equation is an equation where the highest power of the unknown is 1. Solving a linear equation means finding the value of the unknown that makes the equation true. This is one of the most frequently tested skills in the AQA GCSE Mathematics exam and appears on both Foundation and Higher tier papers.

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The Golden Rule

Whatever you do to one side of an equation, you must do the same to the other side. Think of the equals sign as a balance — both sides must remain equal at all times.

flowchart LR
    A[Equation] --> B[Identify what is being done to the unknown]
    B --> C[Undo operations in reverse order]
    C --> D[Simplify each step]
    D --> E[Isolate the unknown on one side]
    E --> F[Check by substituting back]

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One-Step Equations

These require just one operation to isolate the unknown.

Worked Example 1

Solve: x + 7 = 12

Subtract 7 from both sides: x = 12 - 7 x = 5

Worked Example 2

Solve: 3x = 18

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

Worked Example 3

Solve: x / 4 = 5

Multiply both sides by 4: x = 5 x 4 x = 20

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Two-Step Equations

These require two operations to solve. Always deal with addition/subtraction first, then multiplication/division.

Worked Example 4

Solve: 2x + 3 = 11

Step 1: Subtract 3 from both sides: 2x = 11 - 3 2x = 8

Step 2: Divide both sides by 2: x = 8 / 2 x = 4

Check: 2(4) + 3 = 8 + 3 = 11. Correct.

Worked Example 5

Solve: 5x - 7 = 23

Step 1: Add 7 to both sides: 5x = 23 + 7 5x = 30

Step 2: Divide by 5: x = 30 / 5 x = 6

Exam Tip: Always check your answer by substituting it back into the original equation. This takes only a few seconds and can save you marks if you have made an arithmetic error.

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Equations with Unknowns on Both Sides

When the unknown appears on both sides of the equation, you must first collect all the unknown terms on one side and all the number terms on the other.

Worked Example 6

Solve: 5x + 2 = 3x + 10

Step 1: Subtract 3x from both sides: 5x - 3x + 2 = 10 2x + 2 = 10

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

Step 3: Divide by 2: x = 4

Check: LHS = 5(4) + 2 = 22. RHS = 3(4) + 10 = 22. Correct.

Worked Example 7

Solve: 7x - 3 = 2x + 17

Step 1: Subtract 2x from both sides: 5x - 3 = 17

Step 2: Add 3 to both sides: 5x = 20

Step 3: Divide by 5: x = 4

Exam Tip: When there are unknowns on both sides, always move the smaller unknown term to the other side. This avoids dealing with negative coefficients and reduces the chance of sign errors.

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Equations with Brackets

If an equation contains brackets, expand the brackets first and then solve as normal.

Worked Example 8

Solve: 3(x + 4) = 21

Expand: 3x + 12 = 21

Subtract 12: 3x = 9

Divide by 3: x = 3

Worked Example 9

Solve: 2(3x - 1) = 4(x + 3)

Expand both sides: 6x - 2 = 4x + 12

Subtract 4x from both sides: 2x - 2 = 12

Add 2 to both sides: 2x = 14

Divide by 2: x = 7

Worked Example 10

Solve: 5(2x + 1) - 3(x - 2) = 25

Expand: 10x + 5 - 3x + 6 = 25

Collect like terms: 7x + 11 = 25

Subtract 11: 7x = 14

Divide by 7: x = 2

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Equations with Fractions

To solve equations with fractions, multiply every term by the lowest common denominator (LCD) to eliminate the fractions.

Worked Example 11

Solve: x/3 + x/4 = 7

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

12 x (x/3) + 12 x (x/4) = 12 x 7

4x + 3x = 84

7x = 84

x = 12

Worked Example 12

Solve: (2x + 1)/5 = 3

Multiply both sides by 5: 2x + 1 = 15

Subtract 1: 2x = 14

Divide by 2: x = 7

Exam Tip: When an equation has fractions, start by multiplying every term by the LCD. This clears all the fractions in one step and makes the equation much easier to solve.

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Forming and Solving Equations from Context

In exam questions, you are often given a word problem or a geometric diagram and asked to form an equation and then solve it.

Worked Example 13

A rectangle has a length of (3x + 2) cm and a width of (x + 1) cm. The perimeter is 46 cm. Find the value of x.

Perimeter = 2 x length + 2 x width

46 = 2(3x + 2) + 2(x + 1)

46 = 6x + 4 + 2x + 2

46 = 8x + 6

40 = 8x

x = 5

The length is 3(5) + 2 = 17 cm and the width is 5 + 1 = 6 cm.

Check: Perimeter = 2(17) + 2(6) = 34 + 12 = 46 cm. Correct.

Worked Example 14

The angles of a triangle are (2x + 10) degrees, (3x + 20) degrees, and (x + 30) degrees. Find the value of x and each angle.

Angles in a triangle add up to 180 degrees:

(2x + 10) + (3x + 20) + (x + 30) = 180

6x + 60 = 180

6x = 120

x = 20

Angles: 2(20) + 10 = 50 degrees, 3(20) + 20 = 80 degrees, 20 + 30 = 50 degrees.

Check: 50 + 80 + 50 = 180 degrees. Correct.