Solving Linear Equations Algebraically Revision Notes | GCSE AQA Higher Maths

1. Why Solving Equations Is the Engine of Algebra

Every time you see an unknown in a maths problem — a missing length, an unknown age, an unpriced item — you are looking at a situation that needs an equation. Solving that equation is how you extract the answer from the algebra.

Linear equations are the most common type you will meet at GCSE. They appear on both questions without a calculator and calculator-allowed exam questions , and they underpin almost every other algebraic skill: rearranging formulae, solving simultaneous equations, and working with sequences all rely on the same core method you will practise here.

The good news is that there is one consistent strategy: whatever you do to one side, you do to the other. Once that idea clicks, every linear equation — no matter how it is dressed up — follows the same logical path.

2. The Core Toolkit: Definitions and the Balancing Method

Think of an equation as a perfectly balanced set of scales. The equals sign is the pivot. Whatever weight you add or remove from one side, you must do the same to the other — otherwise the scales tip and the equation is no longer true.

The goal is always to get the unknown on its own on one side. You do this by working backwards through the operations that have been applied to it, using inverse operations at each step.

Before moving to harder examples, start with the simplest case: a one-step equation. If $x + 5 = 12$, the only operation applied to $x$ is adding 5. The inverse is subtracting 5 from both sides, giving $x = 7$. That single idea scales to every equation in this lesson.

Worked example: Solving a Two-Step Linear Equation

Solve $3x - 7 = 14$. (Without a Calculator)

1. $3x - 7 + 7 = 14 + 7$ $3x = 21$ — The inverse of subtracting 7 is adding 7. This removes the constant from the left-hand side. [M1]
2. $\frac{3x}{3} = \frac{21}{3}$ $x = 7$ — The inverse of multiplying by 3 is dividing by 3. The unknown is now isolated. [A1]
3. $3(7) - 7 = 21 - 7 = 14$ ✓ — Substituting $x = 7$ into the original equation gives 14, which matches the right-hand side. The solution is confirmed.

Answer: $x = 7$

Notice the order of inverse operations: When solving, undo operations in reverse order to how they were applied to $x$. Since multiplication was applied before addition in $ax + b$, you undo addition first, then multiplication.

3. Unknowns on Both Sides and Equations with Brackets

Once you are confident with two-step equations, the next challenge is equations where the unknown appears on both sides. The strategy is to collect all the $x$ terms on one side first, then treat it as a standard two-step equation.

To collect $x$ terms, subtract the smaller $x$ term from both sides. This keeps the coefficient of $x$ positive, which reduces the chance of sign errors later.

Worked example: Solving an Equation with Unknowns on Both Sides

Solve $7x - 4 = 3x + 12$. (Without a Calculator)

1. $7x - 4 - 3x = 3x + 12 - 3x$ $4x - 4 = 12$ — Collecting $x$ terms on the left. Subtracting the smaller $x$ term keeps the coefficient positive. [M1]
2. $4x - 4 + 4 = 12 + 4$ $4x = 16$ — Removing the constant from the left-hand side using the inverse operation.
3. $\frac{4x}{4} = \frac{16}{4}$ $x = 4$ — Isolating $x$ by dividing by its coefficient. [A1]
4. Left: $7(4) - 4 = 28 - 4 = 24$ Right: $3(4) + 12 = 12 + 12 = 24$ ✓ — Both sides equal 24, confirming $x = 4$ is correct.

Answer: $x = 4$

Brackets add one extra step at the start: expand them before doing anything else. Use the distributive law — multiply every term inside the bracket by the term outside.

Worked example: Solving an Equation with Brackets

Solve $4(2x - 3) = 5x + 9$. (Without a Calculator)

1. $4 \times 2x - 4 \times 3 = 5x + 9$ $8x - 12 = 5x + 9$ — Multiply every term inside the bracket by 4. Do not forget to multiply the constant term as well. [M1]
2. $8x - 12 - 5x = 5x + 9 - 5x$ $3x - 12 = 9$ — Collecting $x$ terms on the left-hand side.
3. $3x - 12 + 12 = 9 + 12$ $3x = 21$ — Removing the constant from the left-hand side.
4. $\frac{3x}{3} = \frac{21}{3}$ $x = 7$ — Isolating $x$. [A1]
5. Left: $4(2(5.6) - 1) = 4(10.2) = 40.8$ Right: $5(5.6) + 9 = 28 + 9 = 40.8$ ✓ — Both sides equal 40.8. The solution is confirmed.

Answer: $x = 7$

This is where it gets interesting — what if there are two sets of brackets? Expand both, collect like terms, then proceed as normal. The method never changes; it just takes an extra line of working.

This section contains a different exercise. Please refer to your workbook or teacher for proper contents.

4. Equations Involving Fractions

Fractions in equations look alarming, but there is a clean trick: multiply every term by the denominator (or the lowest common multiple of all denominators if there are several) to clear the fractions entirely. After that, you have a standard equation with no fractions at all.

This approach works because multiplying both sides by the same number preserves the equality — it is the balancing method applied to fractions. Once the fractions are gone, proceed with the usual steps.

Worked example: Solving an Equation with a Fractional Coefficient

Solve $\frac{x}{4} + 3 = 7$. (Without a Calculator)

1. $\frac{x}{4} + 3 - 3 = 7 - 3$ $\frac{x}{4} = 4$ — Clear the constant first to isolate the fractional term. [M1]
2. $\frac{x}{4} \times 4 = 4 \times 4$ $x = 16$ — Multiplying by 4 is the inverse of dividing by 4. This clears the fraction. [A1]
3. $\frac{16}{4} + 3 = 4 + 3 = 7$ ✓ — Substituting $x = 16$ gives 7, which matches the right-hand side.

Answer: $x = 16$

When the equation has a fraction on both sides, such as $\frac{2x+1}{3} = \frac{x-2}{2}$, multiply every term by the LCM of the denominators (here, 6) to clear both fractions at once.

5. Deliberate Practice: Try These Yourself

The only way to build fluency with equation solving is to practise the method repeatedly until each step becomes automatic. Work through these problems before checking the solutions — even attempting a question and getting it wrong is more valuable than reading the answer passively.

For each question, write out every step on a separate line. Examiners award method marks for correct working even if the final answer contains an arithmetic slip, so never skip steps.

The graph above shows $y = 3x - 7$ and the horizontal line $y = 14$. The $x$-coordinate of their intersection is the solution to $3x - 7 = 14$. You can read off $x = 7$ from the graph — which matches the algebraic solution from Section 2. This is a useful sanity check: if your algebraic answer does not match the graph's intersection, something has gone wrong in your working.

6. Exam Traps, Common Mistakes, and Mark-Winning Technique

Most marks lost on linear equation questions are not lost because students do not know the method — they are lost because of predictable, avoidable errors. Knowing what these errors are puts you ahead of the majority of candidates.

The single most common error is failing to expand a bracket fully. When you see $-2(x - 3)$, the minus sign must be distributed to both terms: $-2x + 6$, not $-2x - 6$. A sign error here corrupts every subsequent step.

The second most common error is moving a term to the other side without changing its sign. If $3x - 4 = 11$, adding 4 to both sides gives $3x = 15$, not $3x = 7$. Think of it as the term changing sign when it crosses the equals sign.

In questions without a calculator, you must show all arithmetic working explicitly. Writing $3x = 21$ and then $x = 7$ without showing the division step may cost you a method mark if the examiner cannot follow your reasoning. One line per operation is the safest habit.