A17, A21AlgebraFoundation & HigherGrades 2–6

Linear Equations

Q: Can the answer be negative or a fraction?

Yes. Equations often have answers like x = -3 or x = 52. Only reject such an answer if the context makes it impossible — a length cannot be negative.

Q: What is the difference between an equation and an identity?

An equation such as 2x + 1 = 7 is true for particular values (x = 3). An identity such as 2(x + 1) 2x + 2 is true for every value of x.

Solving an equation means finding the value of the letter that makes both sides equal. The whole topic rests on one rule — whatever you do to one side, do to the other — and it earns marks far beyond this lesson, because equations sit inside angle, perimeter, sequence and probability questions too.

45 min Video by Cognito AQA GCSE Maths

Before you start

Use negative numbers and expand a single bracket.

Try this first. If you need the answer, revisit the prerequisite before continuing.

Starting check: Expand 3(x − 2). Why is the constant negative?

Show answer and reasoning

3x − 6. Multiplication by 3 applies to both terms, including −2.

The big picture

Think of an equation as a set of scales in balance. Adding, subtracting, multiplying or dividing both sides by the same thing keeps it balanced while making it simpler, until the letter stands alone.

Undo the operations in the reverse order to the one they were done in. In $3x + 5$ , the $x$ was multiplied by 3 and then had 5 added — so subtract 5 first, then divide by 3. On AQA papers, the harder marks are usually for forming the equation from words or a diagram, not for solving it.

GCSE Maths - How to Solve Algebraic Equations (Part 1 of 3) (2026/27 exams)Curated by Kepler, not produced by us. Watch it, then read the notes below.

Open on YouTube

Watch, pause, try

Video by Cognito. This is an introductory equation video. Brackets and unknowns on both sides need the additional written examples.

Viewing focus: Describe each operation performed on both sides, instead of treating terms as moving by magic.

After watching, attempt these questions before opening the answers. You can also use the written lesson below.

After watching 1: Solve 4x + 7 = 31 and check your answer.

Show answer and reasoning

Subtract 7 from both sides: 4x = 24. Divide both sides by 4: x = 6. Check: 4 × 6 + 7 = 31.

After watching 2: Why does subtracting 7 from just one side fail to preserve an equation?

Show answer and reasoning

Equality says the two expressions have the same value. Changing only one side generally makes their values different.

KEPLER VISUAL LAB

Explore the idea

Predict what will change, then use the controls to test your explanation.

Start with three equal unknown amounts and 6 on the left, balanced by 18 on the right.

On a narrow screen, scroll the diagram sideways to read every label.

What you'll learn

Solve one- and two-step equations using inverse operations
Solve equations with brackets and with the unknown on both sides
Clear fractions by multiplying every term by a common denominator
Form and solve equations from worded problems and diagrams

Keeping it balanced

Use inverse operations to get the letter on its own: the inverse of $+$ is $-$ , and the inverse of $\times$ is $\div$ . Apply each one to both sides, and write each step on a new line.

Deal with any adding or subtracting first, then the multiplying or dividing. Finally, check by substituting your answer back into the original equation.

Both sides stay equal at every step:

3x + 5 = 20 \Rightarrow 3x = 15 \Rightarrow x = 5

Example 1Solve

3x + 5 = 20

1Subtract 5 from both sides:

3x = 15

2Divide both sides by 3:

x = 5

3Check:

3(5) + 5 = 20

✓

Answer

x = 5

Your turnSame method, new numbers — try it before moving on.

Solve $4x - 7 = 25$ .

Your answer

Show hint

Undo the $-7$ first by adding 7 to both sides.

Show worked solution

Solution

1Add 7 to both sides: $4x = 32$ .
2Divide by 4: $x = 8$ .

Answer: $x = 8$

Example 2Solve

\dfrac{x}{4} - 3 = 2

1Add 3 to both sides:

\dfrac{x}{4} = 5

2Multiply both sides by 4:

x = 20

Answer

x = 20

Your turnSame method, new numbers — try it before moving on.

Solve $\dfrac{x}{6} + 4 = 9$ .

Your answer

Show worked solution

Solution

1Subtract 4: $\dfrac{x}{6} = 5$ .
2Multiply by 6: $x = 30$ .

Answer: $x = 30$

Tip — Answers do not have to be whole numbers. $2x = 7$ gives $x = 3.5$ (or $\frac{7}{2}$ ) — do not round unless the question tells you to.

Brackets and unknowns on both sides

Expand any brackets first. Then, if the letter appears on both sides, subtract the smaller letter term from both sides so the letter is only on one side.

Collecting the letters on the side with the larger coefficient keeps the coefficient positive, which avoids sign errors.

Example 1Solve

5(x - 2) = 3x + 8

1Expand:

5x - 10 = 3x + 8

2Subtract

3x

from both sides:

2x - 10 = 8

3Add 10:

2x = 18

4Divide by 2:

x = 9

Answer

x = 9

Your turnSame method, new numbers — try it before moving on.

Solve $3(2x + 1) = 4x + 15$ .

Your answer

Show hint

Expand to get $6x + 3$ , then subtract $4x$ from both sides.

Show worked solution

Solution

1Expand: $6x + 3 = 4x + 15$ .
2Subtract $4x$ : $2x + 3 = 15$ .
3Subtract 3: $2x = 12$ , so $x = 6$ .

Answer: $x = 6$

Example 2Solve

7 - 2x = 3x - 13

1Add

2x

to both sides (the right has the larger

x

term):

7 = 5x - 13

2Add 13:

20 = 5x

3Divide by 5:

x = 4

Answer

x = 4

Your turnSame method, new numbers — try it before moving on.

Solve $12 - x = 2x - 9$ .

Your answer

Show worked solution

Solution

1Add $x$ : $12 = 3x - 9$ .
2Add 9: $21 = 3x$ .
3Divide by 3: $x = 7$ .

Answer: $x = 7$

Why "change the sign" works

Subtracting $3x$ from both sides is exactly what people mean by "move $3x$ over and change its sign". Saying it as an operation on both sides keeps the logic honest — and stops the classic slip of changing a sign that should not change.

Equations with fractions

Multiply every term on both sides by the lowest common multiple (LCM) of the denominators. All the fractions cancel in one step.

Put brackets round any numerator with more than one term before you multiply, so a minus sign in front applies to the whole numerator.

Example 1Solve

\dfrac{3x - 1}{2} = \dfrac{x + 5}{3}

1The LCM of 2 and 3 is 6. Multiply both sides by 6:

3(3x - 1) = 2(x + 5)

2Expand:

9x - 3 = 2x + 10

3Subtract

2x

and add 3:

7x = 13

Answer

x = \dfrac{13}{7}

Your turnSame method, new numbers — try it before moving on.

Solve $\dfrac{x}{2} + \dfrac{x}{3} = 10$ .

Your answer

Show hint

Multiply every term — including the 10 — by 6.

Show worked solution

Solution

1Multiply every term by 6: $3x + 2x = 60$ .
2 $5x = 60$ , so $x = 12$ .

Answer: $x = 12$

Example 2Solve

\dfrac{x + 4}{3} - \dfrac{x - 2}{5} = 2

1Multiply every term by 15:

5(x + 4) - 3(x - 2) = 30

2Expand carefully:

5x + 20 - 3x + 6 = 30

3Simplify:

2x + 26 = 30

, so

2x = 4

Answer

x = 2

Your turnSame method, new numbers — try it before moving on.

Solve $\dfrac{2x + 1}{3} - \dfrac{x - 1}{2} = 2$ .

Your answer

Show hint

Multiply by 6. Watch the sign when you expand $-3(x - 1)$ .

Show worked solution

Solution

1Multiply every term by 6: $2(2x + 1) - 3(x - 1) = 12$ .
2Expand: $4x + 2 - 3x + 3 = 12$ .
3Simplify: $x + 5 = 12$ , so $x = 7$ .

Answer: $x = 7$

Tip — $-3(x - 2)$ expands to $-3x + 6$ , not $-3x - 6$ . A negative times a negative is positive.

Forming equations

Choose a letter for the unknown and say what it stands for. Write every other quantity in terms of it, then use a fact from the question — a total, a perimeter, an angle sum, "the same as" — to write an equation.

Solve it, then answer the question that was actually asked. If it asks for the longest side, finding $x$ is only part of the way there.

The perimeter is 32 cm:

(x + 3) + 2x + (2x - 1) = 32

Example 1The triangle above has a perimeter of 32 cm. Find the length of its longest side.

1Add the sides:

x + 3 + 2x + 2x - 1 = 32

2Simplify:

5x + 2 = 32

, so

5x = 30

and

x = 6

3Sides:

6 + 3 = 9

2(6) = 12

2(6) - 1 = 11

Answer

The longest side is 12 cm.

Your turnSame method, new numbers — try it before moving on.

I think of a number, multiply it by 4 and subtract 9. The result is the same as doubling the number and adding 13. What is the number?

Your answer

Show hint

Let the number be $n$ . "The same as" is the equals sign.

Show worked solution

Solution

1 $4n - 9 = 2n + 13$
2Subtract $2n$ : $2n - 9 = 13$ .
3Add 9: $2n = 22$ , so $n = 11$ .

Answer: $11$

Tip — Check your answer in the story, not just the equation: $9 + 12 + 11 = 32$ ✓.

Think like an examiner

Show each operation as a new line. A correct first step earns a method mark even if a later slip loses the answer.
Leave answers as exact fractions like $\frac{13}{7}$ unless the question asks for a decimal.
In worded or diagram questions, a final answer of " $x = 6$ " often scores nothing if the question asked for a length or an angle.

The method

\text{same operation on both sides}

The balance rule — every step, every term.

ax + b = c \Rightarrow x = \dfrac{c - b}{a}

Undo

+b

first, then

\times a

\times \text{LCM of denominators}

Clears every fraction in one step.

Watch out for these

Dividing only one term:

4x + 8 = 20 \Rightarrow x + 8 = 5

Subtract 8 first (

4x = 12

), or divide every term by 4 (

x + 2 = 5

Expanding

-3(x - 2)

-3x - 6

Negative × negative is positive:

-3x + 6

Multiplying only the fractions by the LCM and forgetting the whole number on the other side.

Every term on both sides gets multiplied:

\frac{x}{2} + \frac{x}{3} = 10

becomes

3x + 2x = 60

From

5x = 30

, writing

x = 150

x = 25

5x

means

5 \times x

, so divide by 5:

x = 6

Stretch yourself

A rectangle has length $(3x - 2)$ cm and width $(x + 1)$ cm. A square has side $(x + 3)$ cm. The rectangle and the square have the same perimeter. Which has the larger area, and by how much?

Hint — Form an equation from the perimeters, solve for $x$ , then work out both areas.

Reveal solution

SolutionRectangle perimeter:

2(3x - 2) + 2(x + 1) = 8x - 2

. Square perimeter:

4(x + 3) = 4x + 12

. So

8x - 2 = 4x + 12

, giving

x = 3.5

. The rectangle is

8.5 \times 4.5 = 38.25

cm² and the square is

6.5 \times 6.5 = 42.25

cm². The square is larger, by

4

cm².

\blacksquare

Questions students ask

Which side should I collect the

x

terms on?

Either side gives the right answer. Collecting on the side with the larger

x

coefficient keeps it positive, so there is less chance of a sign error.

Can the answer be negative or a fraction?

Yes. Equations often have answers like

x = -3

x = \frac{5}{2}

. Only reject such an answer if the context makes it impossible — a length cannot be negative.

What is the difference between an equation and an identity?

An equation such as

2x + 1 = 7

is true for particular values (

x = 3

). An identity such as

2(x + 1) \equiv 2x + 2

is true for every value of

x

Key takeaways

Keep the balance: do the same operation to both sides.
Undo operations in reverse order — addition and subtraction before multiplication and division.
Expand brackets, then collect the letter terms on one side.
Clear fractions by multiplying every term by the LCM of the denominators.
When forming equations, define the letter, use a fact from the question, and answer what was asked.

Exam-style questions

Practise it the way it is examined

4 original questions · 15 marks, written to match the style and mark allocation of the real papers. Work on paper, then open the mark scheme and tick the marks you earned — method marks count even if the final answer slips.

Question 1Grade 33 marks

Solve $4(x - 3) = 2x + 7$ .[3]
Show mark scheme
- M1Expands the bracket correctly: $4x - 12$
- M1Collects terms correctly, e.g. $2x = 19$
- A1 $x = 9.5$ or $\frac{19}{2}$
Answer: $x = 9.5$

Question 2Grade 45 marks

A rectangle has length $(3x + 2)$ cm and width $(x + 4)$ cm. Its perimeter is 44 cm.

(a)Work out the value of $x$ .[3]
Show mark scheme (a)
- M1A correct expression for the perimeter, e.g. $2(3x + 2) + 2(x + 4)$ or $8x + 12$
- M1Sets it equal to 44 and rearranges: $8x = 32$
- A1 $x = 4$
Answer: $x = 4$
(b)Work out the area of the rectangle.[2]
Show mark scheme (b)
- M1Substitutes their $x$ to find both lengths: 14 and 8
- A1112 cm² (follow through from their $x$ )
Answer: $112$ cm²

Question 3Grade 54 marks

The angles of a quadrilateral, in degrees, are $x$ , $2x + 10$ , $3x - 30$ and $80$ .

Work out the size of the largest angle.[4]
Show mark scheme
- M1Forms an equation using the angle sum: $x + 2x + 10 + 3x - 30 + 80 = 360$
- M1Simplifies to $6x + 60 = 360$ or $6x = 300$
- A1 $x = 50$
- B1States the largest angle is $120^\circ$
Answer: $120^\circ$

Examiner note

Stopping at $x = 50$ loses the final mark — the question asks for an angle, and the angles are 50°, 110°, 120° and 80°.

Question 4Grade 63 marks

Solve $\dfrac{2x + 3}{4} - \dfrac{x - 1}{3} = 2$ .[3]
Show mark scheme
- M1Multiplies every term by 12 (or uses a common denominator): $3(2x + 3) - 4(x - 1) = 24$
- M1Expands correctly, including the sign: $6x + 9 - 4x + 4$
- A1 $x = 5.5$ or $\frac{11}{2}$
Answer: $x = 5.5$

Unlimited practice

Skill drill

A new question every time, marked instantly, with a full worked solution. Aim for a streak of five, then move up a level.

0 / 0 first-time correct Streak 0Best at this level: 0

Generating a question…

Complete the method

Some of the setup is done. Finish it on paper, then compare your reasoning.

Your turn: Finish solving 5x − 4 = 2x + 11: subtract 2x from both sides to get 3x − 4 = 11. What comes next?

Show a hint

Undo the subtraction before undoing the multiplication.

Show answer and reasoning

Add 4 to both sides: 3x = 15. Divide by 3: x = 5. Both original sides equal 21.

Spot and repair the mistake

Identify the first incorrect step and explain how to fix it.

Diagnose it: A student solves 2(x + 5) = 18 as 2x + 5 = 18, then x = 6.5. Locate the first error.

Show answer and reasoning

The bracket was expanded incorrectly: 2(x + 5) = 2x + 10. Then 2x = 8, so x = 4. Substitution gives 2(4 + 5) = 18.

Apply it in an exam

Write a full response before revealing the solution. These are Kepler-authored exam-style questions; the suggested marks are a self-check guide.

Question 1 · 4 marks: A taxi charges £3 plus £2.40 per kilometre. Another charges £6 plus £1.80 per kilometre. For what journey length are their fares equal?

Show answer and reasoning

Let d be the distance in kilometres. 3 + 2.4d = 6 + 1.8d gives 0.6d = 3, so d = 5 km. Each fare is £15.

Form expressions for both fares using the same distance.
Equate the fares.
Solve to obtain 5 km.
Check the common fare is £15.

Explore further

AQA 8300 specification (opens in a new tab)
Check the assessed content, tier requirements and official course documents.
Corbettmaths: solving equations practice (opens in a new tab)
Practise inverse operations and equations with the unknown on both sides.

Recall and check later

Return with the notes closed. Answer from memory, check the reasoning, then practise anything you missed.

Recall 1: Tomorrow: solve 7 − 2x = 15.

Show answer and reasoning

Subtract 7: −2x = 8, then divide by −2: x = −4.

Recall 2: In a week: solve 3(x − 1) = x + 9 and verify.

Show answer and reasoning

3x − 3 = x + 9, so 2x = 12 and x = 6. Both original sides equal 15.

Test yourself

Answer a few questions, then review anything you missed.

Prefer paper? Open the written self-check

Answer each question before opening its answer. This self-check does not record a quiz score.

Solve $x + 7 = 12$ .
- $x = 5$
- $x = 19$
- $x = 7$
- $x = \frac{12}{7}$
Show answer and explanation
Answer: $x = 5$
Subtract 7 from both sides.
Solve $3x - 4 = 11$ .
- $x = 5$
- $x = \frac{7}{3}$
- $x = 45$
- $x = 3$
Show answer and explanation
Answer: $x = 5$
Add 4: $3x = 15$ , then divide by 3.
Solve $\dfrac{x}{5} = 7$ .
- $x = 35$
- $x = 1.4$
- $x = 12$
- $x = 2$
Show answer and explanation
Answer: $x = 35$
Multiply both sides by 5.
Solve $2(x + 5) = 16$ .
- $x = 3$
- $x = 5.5$
- $x = 13$
- $x = 8$
Show answer and explanation
Answer: $x = 3$
$2x + 10 = 16$ , so $2x = 6$ .
Solve $6x + 1 = 2x + 13$ .
- $x = 3$
- $x = 1.75$
- $x = 4$
- $x = 7$
Show answer and explanation
Answer: $x = 3$
Subtract $2x$ : $4x + 1 = 13$ , so $4x = 12$ .
Solve $9 - 2x = 15$ .
- $x = -3$
- $x = 3$
- $x = 12$
- $x = -12$
Show answer and explanation
Answer: $x = -3$
Subtract 9: $-2x = 6$ , then divide by $-2$ .
Solve $\dfrac{x + 1}{2} = \dfrac{x + 4}{3}$ .
- $x = 5$
- $x = 3$
- $x = 1$
- $x = 7$
Show answer and explanation
Answer: $x = 5$
$3(x + 1) = 2(x + 4)$ , so $3x + 3 = 2x + 8$ .
Three consecutive whole numbers add up to 72. What is the largest?
- $25$
- $24$
- $23$
- $26$
Show answer and explanation
Answer: $25$
$n + (n + 1) + (n + 2) = 72$ , so $3n + 3 = 72$ and $n = 23$ . The numbers are 23, 24 and 25.

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