Solving Linear Equations

SPEC A17 • ALGEBRA

⭐ High-priority topic, Foundation + Higher. Master the balance rule, equations with brackets, unknowns on both sides, and fractions. Guaranteed marks on every paper.

Why This Topic Matters

Solving linear equations is the single most reused skill in GCSE algebra. It shows up directly (e.g. "Solve 5(2x − 1) = 6x + 9", Jun 2022 P2H Q5) and indirectly — inside simultaneous equations, rearranging formulae, solving geometry problems, and any time you have to "form and solve an equation."

What you'll master

The balance rule — do the same thing to both sides.

Equations with brackets.

Equations with the unknown on both sides.

Equations with fractions — clear them first.

Finding approximate solutions using a graph.

Tier note

Basic equations are Foundation-tier content. Equations with the unknown on both sides are Additional Foundation + Higher. Brackets appear at every tier.

Key Terms

Equation

A mathematical statement with an equals sign, saying two expressions have the same value. Solving means finding the value(s) of the unknown that make it true.

Linear equation

An equation where the highest power of the unknown is 1 — no x², no √x, no 1/x. At GCSE, every linear equation in one unknown has exactly one solution.

Inverse operation

The operation that "undoes" another. Addition ↔ subtraction. Multiplication ↔ division. Squaring ↔ square root. We use inverses to isolate the unknown.

The Balance Rule

An equation is a balanced scale. Both sides have the same value. Whatever you do to one side, you must do to the other — otherwise the balance is broken.

The equation is balanced. Add 3 to the left → add 3 to the right. Divide the right by 4 → divide the left by 4. No exceptions.

The golden rule

Whatever you do to one side, do to the other.

The Order of Undoing (Reverse BIDMAS)

To isolate the unknown, apply inverse operations in reverse BIDMAS order — the opposite of how expressions are evaluated.

Evaluation order

Brackets → Indices → Division/Multiplication → Addition/Subtraction

Undoing (solving) order

Subtraction/Addition first → then Division/Multiplication → then powers → then brackets last (or expand brackets first if that's easier).

Equations with Brackets

Brackets in an equation can be dealt with in two ways: expand first (usually easier), or divide first (only if the RHS is a multiple of the outside factor).

Expand first

3(2x + 1) = 15

6x + 3 = 15

(− 3)

6x = 12

(÷ 6)

x = 2

Divide first

3(2x + 1) = 15

(÷ 3)

2x + 1 = 5

(− 1)

2x = 4

(÷ 2)

x = 2

Negative brackets — biggest trap

When expanding −3( x − 1), the sign flips on every term inside: −3(x − 1) = −3x+ 3 (not −3x − 3). A wrong sign here kills the rest of the solution.

Unknown on Both Sides

When the variable appears on both sides, collect the variable terms on one side, the constants on the other. A good habit: move the smaller x-coefficient so the remaining coefficient stays positive.

Why move the smaller one?

If you subtracted 3x instead, you'd get 4 = −2x + 10, which leads to negative arithmetic. Subtracting the smaller coefficient keeps working cleaner.

Equations with Fractions

The cleanest method: multiply every term by the lowest common multiple (LCM) of all denominators. This clears all fractions in one step — then you're left with a standard equation.

Simple example

x/3 = 4

(× 3 on both sides)

x = 12

Two denominators — use LCM

x/2 + x/3 = 5

(× 6, the LCM of 2 and 3)

3x + 2x = 30

5x = 30 → x = 6

Multiply every term

Including any plain integer on the other side. Missing one term is the #1 fraction-clearing error. If you're clearing with ×12, every term gets multiplied — no exceptions.

MCE Chain for Linear Equations

Every linear equation — however dressed up — follows the same process. Mark schemes split marks across the three stages.

MMethod. Look at the equation and decide the first move: expand brackets, clear fractions by multiplying by LCM, or collect variable terms. State what you'll do before you do it.

CCalculate. Apply inverse operations, one step per line. Each line has an equals sign. Keep the equation vertically aligned so the marker can follow.

EEvaluate. State x = value. Check by substituting back into the original equation — LHS must equal RHS. If it doesn't, find the slip.

Worked Example: Setup

Ex

Higher — 3 marks (Jun 2022 P2H Q5 authentic)

Solve 5(2x − 1) = 6x + 9.

M — Method

Two things here: brackets on the LHS, and the unknown on both sides.

Plan:

Expand the brackets on the left: 5(2x − 1) = 10x − 5.

Collect x-terms on one side (subtract 6x), constants on the other (add 5).

Divide to find x.

Worked Example: Calculate & Evaluate

C — Calculate

5(2x − 1) = 6x + 9

(expand brackets)

10x − 5 = 6x + 9

(subtract 6x from both sides)

4x − 5 = 9

(add 5 to both sides)

4x = 14

(divide both sides by 4)

x = 14/4 = 7/2 = 3.5

E — Evaluate (check)

LHS = 5(2(3.5) − 1) = 5(7 − 1) = 5 × 6 = 30

RHS = 6(3.5) + 9 = 21 + 9 = 30 ✓ Both sides equal 30, so the solution is verified.

Mark breakdown & "follow-through"

M1 for correct expansion (10x − 5). M1 for collecting x-terms and constant terms correctly (4x = 14). A1 for x = 3.5 (or 7/2, 14/4, 3½ — any equivalent form).

The mark scheme uses A1ft (follow-through) — if you make a small arithmetic slip early but your method stays consistent, the answer mark can still be awarded for the correct answer to your equation. One more reason to always show working.

Approximate Solution Using a Graph

The spec requires you to know the graphical method too. Plot both sides of the equation as straight lines — the x-coordinate of their intersection is the solution.

To solve x + 3 = 2x + 1, plot y = x + 3 and y = 2x + 1. Their intersection has x = 2 — the solution.

Why "approximate"?

Unless the intersection lands on integer coordinates, you can only read the x-value to the accuracy of the grid. Algebra gives the exact answer; graphs give a visual check or an estimate when algebra is hard.

Weak vs Strong: Distributing Negatives

When a negative factor multiplies a bracket, every term inside changes sign. Getting this wrong is the most common linear-equation error.

Weak student

Writes −3(x − 1) = −3x− 3.

Kept the "−" on the inside and just slapped it onto both terms. Wrong.

Strong student

Writes −3(x − 1) = −3x+ 3.

The −3 multiplies x (→ −3x) and multiplies −1 (→ +3). Both signs handled.

Quick check

After expanding a bracket with a negative factor, there should be one sign change inside — not zero, not two. If every sign stayed the same, you forgot to distribute the minus.

Always Check Your Solution

A linear equation has exactly one solution — and checking takes 30 seconds. Substitute your answer back into the original equation. Both sides must evaluate to the same number.

Example check — 5(2x − 1) = 6x + 9 with x = 3.5

LHS = 5(2 × 3.5 − 1) = 5(6) = 30

RHS = 6 × 3.5 + 9 = 21 + 9 = 30

30 = 30 ✓

Why this matters in exams

Method marks can still be earned even if the final answer is wrong — but only if you've shown your working. Checking also catches arithmetic slips before you move on. A 30-second check can rescue a 3-mark question.

Examiner Alerts

Sign flip on negatives

Expanding −a(b − c) gives −ab + ac. Every minus sign inside a negative bracket becomes a plus. Mis-distributing costs the method mark even with correct arithmetic afterwards.

Multiply every term when clearing fractions

When you multiply through by the LCM, every term on both sides gets multiplied — including any plain integers. Miss one and the equation becomes unbalanced.

Equals sign on every line

Each line of algebra should be a complete equation with an "=" sign. This helps the examiner trace your method and makes it easier to earn the M marks. Lines like "− 3" or "÷ 4" scribbled alongside working aren't full equations — write the resulting equation each time.

Keep fractions exact

If x = 14/4, leave it as 7/2 or 3.5 — whichever the question prefers. Don't round to 3 and call it done. "14/4" alone (unsimplified) is usually accepted on the mark scheme, but 7/2 or 3.5 is cleaner.

Practice Question 1

Brackets. Solve 6y − 9 = 2(y − 8). [3 marks]

M

Expand the bracket on the right. Then collect y-terms on one side.

C

6y − 9 = 2y − 16 (expand). 6y − 2y = −16 + 9 → 4y = −7. y = −7/4 = −1.75.

E

Check: LHS = 6(−7/4) − 9 = −42/4 − 36/4 = −78/4 = −19.5. RHS = 2(−7/4 − 8) = 2(−39/4) = −39/2 = −19.5 ✓. y = −1.75 (or −7/4).

Practice Question 2

Unknown on both sides. Solve 5w − 17 = 2w + 4. [3 marks]

M

Subtract the smaller w-coefficient (2w). Add 17 to the other side. Divide.

C

5w − 2w = 4 + 17 → 3w = 21 → w = 7.

E

Check: LHS = 5(7) − 17 = 18. RHS = 2(7) + 4 = 18 ✓. w = 7.

Practice Question 3

Fractions. Solve (x + 2)/3 − (x − 1)/4 = 5. [3 marks]

M

LCM of 3 and 4 is 12. Multiply every term by 12 to clear fractions.

C

4(x + 2) − 3(x − 1) = 60

4x + 8 − 3x + 3 = 60 (careful with −3(x−1) = −3x + 3)

x + 11 = 60 → x = 49.

E

Check: (49 + 2)/3 − (49 − 1)/4 = 51/3 − 48/4 = 17 − 12 = 5 ✓. x = 49.

Summary

The Balance Rule

Whatever you do to one side, do to the other.

Strategy by Equation Type

TypeFirst move

Simple xApply inverse operations (reverse BIDMAS)

BracketsExpand first (or divide if RHS is a multiple of the factor)

Unknown both sidesSubtract the smaller x -coefficient from both sides

FractionsMultiply every term by the LCM of denominators

Approximate / graphicalPlot y = LHS and y = RHS; read x -coordinate of intersection

Golden habits

Equals sign on every line

One operation per line

Keep fractions exact

Always check with substitution

Top traps

−3(x−1) = −3x+ 3

Multiply every term by LCM

Subtract smaller coefficient to keep signs positive

Top Exam Tip

Linear equations reward neat working more than clever shortcuts. One step per line, equals sign on every line, check at the end. Method marks are generous but only if the method is visible.