Mathematics · AQA 8300 §A3

Linear Equations

Spec: AQA 8300 §A3 ⭐⭐ 🕑 40 mins AQA · Edexcel · OCR Grade 9

Solve one-step, two-step and multi-step linear equations
Solve equations containing brackets and fractions
Form and solve equations from geometric and word problems
Check solutions by substitution back into the original equation
Solve equations with the unknown on both sides

🔑 Core Concepts

The Balance Method — Why It Works

An equation is a statement of equality: whatever is on the left equals whatever is on the right. Think of a perfectly balanced set of scales. The golden rule is: whatever you do to one side, you must do identically to the other side. This preserves equality throughout every step.

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DEFINITION — Linear Equation

An equation where the highest power of the unknown is 1, e.g. $3x + 5 = 14$. The word "linear" means the unknown appears only to the first power — no $x^2$, no $x^3$.

⚠️

IMPORTANT — The Balance Principle

Any operation (+, −, ×, ÷) applied to one side of an equation must be applied to every term on the other side. Failing to do this is the most common source of errors in exam solutions.

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EXAM TIP

Always write out every step — examiners award method marks for correct intermediate lines even if the final answer is wrong. A missing step means a missing mark.

One-Step and Two-Step Equations

A one-step equation requires a single inverse operation to isolate the unknown. A two-step equation involves two operations: typically undo addition/subtraction first, then multiplication/division (or the other way round depending on structure).

Two-Step Equation Template

$$ax + b = c \implies x = \frac{c - b}{a}$$

$a$ = coefficient of $x$ $b$ = constant added $c$ = right-hand side

✏️

WORKED EXAMPLE — Two-Step

Solve $5x - 3 = 17$.
Add 3 to both sides: $5x = 20$
Divide both sides by 5: $x = 4$
Check: $5(4) - 3 = 20 - 3 = 17$ ✓

✗

COMMON MISTAKE

Students sometimes divide before subtracting and write $\frac{5x-3}{5} = \frac{17}{5}$, then forget to divide the $-3$ term too. Always perform one clear step at a time.

Equations with Brackets — Expand First

When brackets appear, expand first, then collect like terms. Expansion means multiplying each term inside the bracket by the factor outside. Only after the equation is in standard $ax + b = c$ form should you apply the two-step method.

Expanding Brackets Rule

$$a(x + b) = ax + ab$$

$a$ = factor outside bracket $b$ = constant inside bracket

🧠

MEMORY TRICK — EXPAND THEN SOLVE

Say to yourself: "Brackets out, collect up, then solve." Never try to divide through a bracket that hasn't been expanded — this is a very common Grade 5 error.

✗

COMMON MISTAKE — Negative Outside Bracket

With $-3(2x - 5)$, the minus sign applies to both terms: $-6x + 15$, not $-6x - 15$. The sign of the second term flips because $-3 \times -5 = +15$.

Write equation with brackets

→

Expand brackets (distribute)

→

Collect like terms

→

Apply two-step method

→

Check by substitution

Unknowns on Both Sides

When the unknown appears on both sides of the equation, the strategy is to collect all $x$-terms on one side and all constants on the other. It is conventional (though not required) to move terms so the $x$-coefficient is positive — this reduces sign errors.

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DEFINITION — Collecting Terms

To collect $x$-terms: add or subtract the same multiple of $x$ from both sides. To collect constants: add or subtract the same constant from both sides. Do one step at a time.

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EXAM TIP — Which Side to Collect On?

Move $x$ to the side where its coefficient is larger. E.g. in $5x - 2 = 3x + 8$, the $5x$ side is larger, so subtract $3x$ from both sides: $2x - 2 = 8$. This keeps the coefficient positive and avoids dividing by a negative number.

Equations with Fractions — Multiply Through by LCM

Fractions in equations are best eliminated immediately. Find the Lowest Common Multiple (LCM) of all denominators in the equation, then multiply every single term on both sides by that LCM. This clears all fractions in one step and produces an ordinary integer equation.

Clearing Fractions

$$\frac{x}{a} + \frac{x}{b} = c \xrightarrow{\times \text{lcm}(a,b)} \text{integer equation}$$

Multiply every term by LCM $\text{lcm}(a,b)$ = lowest common multiple of denominators

✏️

WORKED EXAMPLE — Clearing Fractions

Solve $\dfrac{x+1}{3} + \dfrac{x-2}{4} = 5$.
LCM of 3 and 4 is 12. Multiply every term by 12:
$4(x+1) + 3(x-2) = 60$
$4x + 4 + 3x - 6 = 60$
$7x - 2 = 60$
$7x = 62$
$x = \dfrac{62}{7}$ (exact fraction answer is acceptable in GCSE)

✗

COMMON MISTAKE — Forgetting to Multiply ALL Terms

In $\dfrac{x}{3} + 2 = 5$, multiplying by 3 gives $x + 6 = 15$, not $x + 2 = 15$. The integer $2$ must also be multiplied by 3 because it is a term in the equation.

Forming Equations from Geometry and Word Problems

Many exam questions require you to construct an equation from given information before solving it. This is often where Grade 9 marks are allocated — the ability to translate a geometric or contextual problem into algebraic language.

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DEFINITION — Forming an Equation

Identify the unknown, express all given quantities in terms of that unknown, then set two expressions equal to each other (or equal to a known total).

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EXAM TIP — Geometry Problems

For angle problems: use angle sum facts (triangle = 180°, straight line = 180°, quadrilateral = 360°). For perimeter: sum all sides = given perimeter. For area: use the area formula and set equal to the given area. Always define your unknown clearly at the start.

🧠

MEMORY TRICK — DFSS for Word Problems

Define the unknown → Form the equation → Solve the equation → State the answer in context. Never skip the last step — the question asks for a real-world answer, not just $x = \ldots$

Checking Solutions by Substitution

After solving any equation, substitute your answer back into the original equation (not a simplified version) and verify that both sides are equal. In an exam, a correct check proves your answer is right and can earn a final accuracy mark even if you made an arithmetic slip earlier.

🎯

EXAM TIP — Verification

Write the check explicitly: "Check: LHS = $\ldots$ = RHS ✓". Examiners look favourably on this and it immediately reveals any arithmetic errors before you move on.

🗼 Visual Notes

Linear
Equations

Types

One-step: $x + 5 = 9$
Two-step: $3x - 4 = 11$
With brackets: $2(x+3)=10$
Both sides: $5x+1=3x+9$

Fractions

Find LCM of denominators
Multiply every term by LCM
Clears all fractions at once
Answer may be a fraction

Geometry

Angles in triangle = 180°
Angles on straight line = 180°
Perimeter = sum of all sides
Area formula = given value

Method

Define unknown clearly
Balance: same to both sides
Collect like terms
Check by substitution

Grade 9 Skills

Brackets + fractions combined
Form equation from context
Prove a result using algebra
Equations leading to fractions

Checks

Substitute back into original
Verify LHS = RHS
Does answer make sense?
Show check working explicitly

Equation Types Comparison

Type	Example	Strategy	Steps
One-step	$x + 7 = 12$	Single inverse operation	1
Two-step	$3x - 4 = 11$	Undo +/−, then ÷	2
Brackets	$2(x + 3) = 14$	Expand, then two-step	3+
Both sides	$5x + 1 = 3x + 9$	Collect $x$, collect constants	3+
Fractions	$\frac{x+1}{3} = \frac{2x-1}{5}$	Multiply by LCM, then solve	4+
Forming (geometry)	Angles in triangle sum to 180°	Define variable, form, solve	4+

Which Strategy? — Decision Chain

Does the equation have fractions?

Yes →

Multiply every term by LCM of denominators

→

Does it have brackets?

Yes →

Expand all brackets

→

Collect $x$-terms (one side) & constants (other side)

→

Divide by coefficient of $x$

→

Check by substitution

✏️ Worked Examples

Grade 4–5 — Simple

Solve $4(2x - 3) = 20$

Expand the bracket

Multiply each term inside the bracket by 4:
$$4 \times 2x - 4 \times 3 = 20$$ $$8x - 12 = 20$$

Add 12 to both sides

$$8x - 12 + 12 = 20 + 12$$ $$8x = 32$$

Divide both sides by 8

$$x = \frac{32}{8} = 4$$

Check by substitution

LHS $= 4(2 \times 4 - 3) = 4(8 - 3) = 4 \times 5 = 20$ = RHS ✓

$x = 4$

Grade 6–7 — Medium

Solve $\dfrac{3x+1}{4} = \dfrac{x+5}{2}$

Identify the LCM of denominators

Denominators are 4 and 2. LCM(4, 2) = 4. Multiply every term by 4.

Clear fractions

$$4 \times \frac{3x+1}{4} = 4 \times \frac{x+5}{2}$$ $$3x + 1 = 2(x + 5)$$

Expand the bracket on the right

$$3x + 1 = 2x + 10$$

Collect unknowns on the left, constants on the right

Subtract $2x$ from both sides: $x + 1 = 10$
Subtract 1 from both sides: $x = 9$

Check by substitution into the original

LHS $= \dfrac{3(9)+1}{4} = \dfrac{28}{4} = 7$
RHS $= \dfrac{9+5}{2} = \dfrac{14}{2} = 7$ ✓

$x = 9$

Grade 9 — Multi-Step with Forming

The diagram shows a rectangle. The length is $(3x + 5)$ cm and the width is $(x + 2)$ cm. The perimeter of the rectangle is 58 cm. Find the area of the rectangle.

Form the equation using the perimeter formula

Perimeter of a rectangle $= 2(\text{length} + \text{width})$
$$2\bigl[(3x+5) + (x+2)\bigr] = 58$$

Simplify inside the brackets

$$2(4x + 7) = 58$$

Divide both sides by 2

$$4x + 7 = 29$$

Solve for $x$

$$4x = 22 \implies x = 5.5$$

Find the dimensions

Length $= 3(5.5) + 5 = 16.5 + 5 = 21.5$ cm
Width $= 5.5 + 2 = 7.5$ cm

Calculate the area

$$\text{Area} = 21.5 \times 7.5 = 161.25 \text{ cm}^2$$

Verify the perimeter

$2(21.5 + 7.5) = 2 \times 29 = 58$ cm ✓

$x = 5.5$, Area $= 161.25 \text{ cm}^2$

❓ Exam Questions

Q1 1 mark

Solve $7x - 4 = 24$.

Mark Scheme:
$7x = 28$ (B1 for correct first step, or)
$x = 4$ [1 mark]

Check: $7(4) - 4 = 28 - 4 = 24$ ✓

Q2 2 marks

Solve $3(2x - 5) = 2x + 7$.

Mark Scheme:
Expand: $6x - 15 = 2x + 7$ (M1)
Collect: $4x = 22 \implies x = 5.5$ (A1)

Check: LHS $= 3(11-5) = 3 \times 6 = 18$; RHS $= 11 + 7 = 18$ ✓

Q3 3 marks

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

Mark Scheme:
LCM of 5 and 3 is 15. Multiply every term by 15: (M1)
$3(2x+3) - 5(x-1) = 15$
$6x + 9 - 5x + 5 = 15$ (M1 for correct expansion of both brackets)
$x + 14 = 15$
$x = 1$ (A1)

Check: $\dfrac{5}{5} - \dfrac{0}{3} = 1 - 0 = 1$ ✓

Q4 4 marks

The angles of a triangle are $(2x + 10)°$, $(3x - 20)°$ and $(x + 30)°$. Find the value of $x$ and state whether the triangle is equilateral, isosceles or scalene.

Mark Scheme:
Angles in a triangle sum to 180°: (M1 for using angle sum)
$(2x+10) + (3x-20) + (x+30) = 180$
$6x + 20 = 180$ (M1 for correct collection)
$6x = 160$
$x = 26\tfrac{2}{3}°$ (or $\frac{80}{3}$) (A1)

Angles: $2(80/3)+10 = 63\tfrac{1}{3}°$; $3(80/3)-20 = 60°$; $(80/3)+30 = 56\tfrac{2}{3}°$
All three angles are different, so the triangle is scalene. (A1)

Q5 4 marks

A rectangle has length $(5x - 3)$ cm and width $(2x + 1)$ cm. Its perimeter equals its area (numerically). Form and solve an equation to find $x$. Give your answer to 2 decimal places.

Mark Scheme:
Perimeter $= 2[(5x-3)+(2x+1)] = 2(7x-2) = 14x - 4$ (M1)
Area $= (5x-3)(2x+1) = 10x^2 + 5x - 6x - 3 = 10x^2 - x - 3$
Setting equal: $14x - 4 = 10x^2 - x - 3$ (M1 for forming equation)
$10x^2 - 15x + 1 = 0$
Using the quadratic formula: $x = \dfrac{15 \pm \sqrt{225 - 40}}{20} = \dfrac{15 \pm \sqrt{185}}{20}$ (M1)
$x \approx \dfrac{15 + 13.60}{20} \approx 1.43$ or $x \approx \dfrac{15 - 13.60}{20} \approx 0.07$ (reject as width becomes negative) (A1)
$x \approx 1.43$

Q6 6 marks

Prove that if the three consecutive integers are $n-1$, $n$ and $n+1$, then the sum of the largest two is always 3 more than twice the smallest. Then use this to form and solve an equation: if the sum of the three consecutive integers is 51, find the integers.

Mark Scheme:
Part (a) — Proof:
Sum of largest two: $n + (n+1) = 2n + 1$ (M1)
Twice the smallest $+ 3$: $2(n-1) + 3 = 2n - 2 + 3 = 2n + 1$ (M1)
Since both expressions equal $2n + 1$, the result is proved. (A1 for conclusion)

Part (b) — Form and solve:
$(n-1) + n + (n+1) = 51$ (M1 for forming equation)
$3n = 51$ (M1 for correct simplification)
$n = 17$
Integers are $16, 17, 18$ (A1 for all three correct)

⭐ Grade 9 Model Answers

Full Annotated Solution — Q6 (6 marks)

⚠️

IMPORTANT — Why This Is Grade 9

This question combines algebraic proof (requiring generalisation and rigorous logical argument) with forming and solving a linear equation. You must demonstrate both that a result is always true (using algebra with no specific numbers) and then apply equation-solving skills. The proof component is assessed at Grade 8–9 only.

Grade 9 — Full Annotated Model Answer

Q6 full solution with examiner commentary

Set up general form — earning the first method mark

Write the three consecutive integers as $n-1$, $n$, $n+1$. This is the key insight: using a general variable means we can prove the result for all sets of consecutive integers, not just one example.

Examiner note: Using a specific example (e.g. 4, 5, 6) earns zero marks for the proof. Only an algebraic general argument scores.

Express "sum of the largest two" algebraically

The two largest integers are $n$ and $n+1$.
$$\text{Sum of largest two} = n + (n+1) = 2n + 1$$ Examiner note: This is the first expression to simplify. Write it clearly; do not skip the intermediate step.

Express "3 more than twice the smallest" algebraically

The smallest integer is $n - 1$.
$$2(n-1) + 3 = 2n - 2 + 3 = 2n + 1$$ Examiner note: Expanding the bracket correctly and obtaining $2n+1$ earns the second method mark. Ensure you write "= $2n+1$" explicitly.

Write the conclusion of the proof

Both expressions simplify to $2n+1$.
Therefore, the sum of the largest two is always equal to 3 more than twice the smallest. QED.

Examiner note: Always write a concluding sentence. Without it, the accuracy mark for the proof may be withheld. The phrase "for all values of $n$" reinforces generality.

Form the equation for Part (b)

$$(n-1) + n + (n+1) = 51$$ Simplify the left-hand side by collecting like terms:
$$3n = 51$$ Examiner note: The $-1$ and $+1$ cancel, leaving $3n$. This is a neat result that always happens with consecutive integers centred on $n$ — a nice observation worth mentioning in a "show that" context.

Solve and state the final answer in context

$$n = 17$$ The three consecutive integers are $\mathbf{16, 17, 18}$.

Examiner note: The question asks for the integers, not just $n$. Stating only $n=17$ earns no final accuracy mark. Always answer what was asked.

Proof: Both sides $= 2n+1$ QED. Integers: $16, 17, 18$.

📋 Revision Sheet

Key Definitions

Term	Meaning
Linear equation	Equation where the unknown is to power 1 only
Balance method	Apply identical operations to both sides
LCM	Lowest Common Multiple — used to clear fractions
Expand	Multiply out brackets
Collect like terms	Group terms with the same variable together
Substitution check	Replace $x$ with your answer to verify LHS = RHS

Essential Formulae

Two-step template: $$ax + b = c \implies x = \frac{c-b}{a}$$

Clearing fractions: multiply every term by LCM of all denominators

Bracket expansion: $a(b + c) = ab + ac$

Perimeter of rectangle: $P = 2(l + w)$

Angles in triangle: $a + b + c = 180°$

Angles on straight line: $= 180°$

Memory Hooks

EXPAND THEN SOLVE — never divide into an unexpanded bracket
LCM clears all fractions in one step — multiply every single term
DFSS — Define, Form, Solve, State (word problems)
Move $x$ to the bigger side — keeps coefficient positive
Check = free marks — always substitute back
Proof needs algebra — examples do not prove; they illustrate

Exam Tips

Show every step — method marks are available even if the answer is wrong
If the equation has fractions, eliminate them first before anything else
Define your unknown with units: "Let $x$ = width in cm"
For geometry questions, write down the angle/perimeter/area rule you're using
Answers can be fractions — leave as an exact fraction unless asked to round
For proofs, use algebra throughout — never substitute a specific number
Write "Check:" line — this demonstrates rigour and can recover an accuracy mark

🔄 Flashcards

Click any card to reveal the answer. Use these to test yourself without looking at your notes.

✗ Common Mistakes

✗

MISTAKE 1 — Not multiplying ALL terms when clearing fractions

What students do: In $\dfrac{x}{3} + 5 = 9$, they multiply only the fraction by 3, getting $x + 5 = 27$.
Why marks are lost: The equation is now wrong, and all subsequent working is incorrect — up to 3 marks lost.
How to avoid it: Write "multiply every term by 3:" and then explicitly multiply each term: $\dfrac{x}{3} \times 3 + 5 \times 3 = 9 \times 3 \implies x + 15 = 27$.

✗

MISTAKE 2 — Sign errors when expanding brackets with a negative factor

What students do: Expand $-2(3x - 4)$ as $-6x - 8$ instead of $-6x + 8$.
Why marks are lost: Incorrect expansion leads to a wrong answer; the method mark for expansion is also lost.
How to avoid it: Multiply each term separately and state the sign: $-2 \times 3x = -6x$; $-2 \times (-4) = +8$. Negative times negative is positive.

✗

MISTAKE 3 — Dividing by the bracket instead of expanding

What students do: Solve $3(x + 4) = 21$ by writing $x + 4 = \frac{21}{3} = 7$ (correct!) but then try the same trick on $3(x+4) = 2x + 9$ by writing $x + 4 = \frac{2x+9}{3}$, creating a more complicated equation.
Why marks are lost: This approach is not wrong per se, but students misapply it and lose marks.
How to avoid it: Always expand first — it is the safe, reliable method. Only divide through when one side has no variable terms.

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MISTAKE 4 — Not checking the answer by substitution

What students do: Solve the equation but do not substitute back in to verify.
Why marks are lost: If the answer is wrong due to an arithmetic error, the check would have caught it. In some mark schemes, a correct check can earn back an accuracy mark.
How to avoid it: Make checking automatic — always write "Check: LHS = ... = RHS ✓" as the final line of every solution.

✗

MISTAKE 5 — Using a specific example as a proof

What students do: To prove a result about consecutive integers, they use 4, 5, 6 and show the result holds. They write "therefore it always works."
Why marks are lost: A single example is not a proof. The entire proof component (usually 3 marks) is lost. One example only shows the result holds for those specific numbers.
How to avoid it: Use algebra. Let consecutive integers be $n-1$, $n$, $n+1$ and prove the result generally using these expressions.

✗

MISTAKE 6 — Answering with $x =$ only, ignoring what was asked

What students do: In a word problem or geometric context, they find $x = 5$ but the question asked for the length of a side, the area, or the actual angles — so the answer is incomplete.
Why marks are lost: The final accuracy mark typically requires the quantity asked for in context, not just $x$. This is an avoidable loss of 1 mark.
How to avoid it: Re-read the question after finding $x$. Ask: "Did the question ask for $x$ or for something else?" Then calculate and state that quantity with correct units.

✅ Final Checklist

Click each item when you are confident with it. Your progress is saved automatically.

I can solve one-step equations using a single inverse operation
I can solve two-step equations by undoing addition/subtraction, then division
I can expand a bracket correctly, including when the factor outside is negative
I can solve equations with brackets by expanding first then solving
I can solve equations with unknowns on both sides by collecting like terms
I can find the LCM of two or more denominators
I can clear fractions from an equation by multiplying every term by the LCM
I can solve equations that have both brackets and fractions
I can form an equation from a geometric problem (perimeter, area, angles)
I can form an equation from a word problem and solve it, stating the answer in context
I always check my answer by substituting back into the original equation
I understand that a single numerical example does not constitute an algebraic proof
I can construct an algebraic proof using general variable expressions
I know to state the final answer in the units and context the question requires
I can show all working clearly and earn method marks even if the final answer is wrong

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