Solving Linear Equations — AQA GCSE Maths Revision Notes
What you'll learn
This topic covers solving linear equations: equations in which the unknown appears only to the first power. By the end of this guide you should be able to solve one-step and two-step equations, equations containing brackets, equations with the unknown on both sides, and equations containing fractions.
You should also be able to form an equation from a description in words, recognise when an equation has been rearranged incorrectly, and check a solution by substituting it back.
The organising idea is this: an equation describes a machine built around x. Numbers were done to x in a particular order, and the equation records the result. Solving means taking that machine apart, undoing each operation in the reverse of the order it was applied. Everything in this guide is that one idea, applied to progressively more complicated machines.
Key terms and definitions
Equation — a statement that two expressions are equal, containing at least one unknown. An equation can be solved; an expression cannot.
Linear — the unknown appears only to the power 1. No x², no 1/x, no √x.
Solution (or root) — the value of the unknown that makes both sides equal. A linear equation has exactly one.
Inverse operation — the operation that undoes another. Addition and subtraction are inverses; multiplication and division are inverses.
Coefficient — the number multiplying the unknown. In 5x − 3, the coefficient of x is 5.
Balance — the principle that both sides stay equal provided the same thing is done to each side.
Core concepts
The balance principle
The equals sign is a statement of fact: the two sides have the same value. That remains true as long as whatever is done to one side is done to the other side as well.
Add 4 to both sides, and they are still equal. Divide both sides by 3, and they are still equal. Add 4 to one side only, and the statement becomes false and every line after it is wrong.
This is why the working for an equation is written as a column, one line per operation, with the operation itself written at the side. It makes it visible that each step was applied to both halves.
Undoing in reverse order
To solve x + 7 = 12, one thing has been done to x: 7 was added. Subtract 7 from both sides and x = 5.
When two operations have been applied, they must be undone in reverse order — the same way you take off shoes and socks in the opposite order from putting them on.
In 3x + 4 = 19, the machine multiplied x by 3 and then added 4. So undo the addition first: 3x = 15. Then undo the multiplication: x = 5.
Undoing in the wrong order is the most common structural error in this topic. Dividing 3x + 4 = 19 by 3 first gives x + 4/3 = 19/3, which is correct but far harder to finish, and students who write x + 4 = 19/3 instead have lost the equation entirely — they divided only part of the left side.
The rule is therefore: numbers added or subtracted come off first; the coefficient comes off last.
Equations with brackets
There are two valid routes, and choosing the right one saves time.
Expanding is always safe. For 2(x + 3) = 16, expand to 2x + 6 = 16, subtract 6 to get 2x = 10, then divide to get x = 5.
Dividing first is quicker when the number outside divides the right-hand side exactly. Here 16 ÷ 2 = 8, so x + 3 = 8 and x = 5 immediately.
When expanding, the number outside multiplies every term inside. Writing 2(x + 3) = 2x + 3 is the error that loses most marks in this topic, and it is worth checking each bracket has been fully multiplied before continuing.
The unknown on both sides
When x appears on both sides, the first job is to gather it on one side. Subtract the smaller x-term from both sides, which keeps the remaining coefficient positive and avoids sign errors.
For 5x − 2 = 3x + 8, subtract 3x from both sides: 2x − 2 = 8. Now it is an ordinary two-step equation. Add 2: 2x = 10. Divide: x = 5.
If the x-terms are collected on the left when the larger is on the right, the working still succeeds but a negative coefficient appears — 5x − 2 = 7x + 8 becomes −2x = 10, so x = −5. That is perfectly correct; it simply invites a sign slip at the last division.
Equations with fractions
A fraction is a division, so the inverse is multiplication.
For x/4 = 3, multiply both sides by 4: x = 12.
Where the fraction has a whole expression on top, the multiplication clears the bar for the entire numerator. For (x + 1)/3 = 4, multiplying by 3 gives x + 1 = 12, so x = 11. The bracket matters: multiplying to get x + 1/3 = 12 would be treating the bar as though it covered only part of the top.
When several fractions appear, multiply every term by the lowest common denominator, which clears them all in one move. For x/2 + x/3 = 5, multiply throughout by 6: 3x + 2x = 30, so 5x = 30 and x = 6.
Forming an equation from words
Exam questions frequently describe a situation and ask you to solve it. Translate one phrase at a time, in the order given.
"I think of a number, multiply it by 4 and add 5 to get 29" becomes 4x + 5 = 29. Note that the order of the words is the order the machine was built, which is exactly the order you will undo in reverse: subtract 5, then divide by 4, giving x = 6.
Checking
Substituting the answer back into the original equation confirms it completely. For x = 5 in 5x − 2 = 3x + 8: the left is 25 − 2 = 23 and the right is 15 + 8 = 23. ✓
Use the original equation, not a rearranged line — checking against a line that already contains an error will confirm the error.
Worked examples
Example 1: Two operations, undone in reverse
Solve 4x − 5 = 23.
The machine multiplied by 4, then subtracted 5. Undo the subtraction first by adding 5 to both sides: 4x = 28.
Now undo the multiplication by dividing both sides by 4: x = 7.
Check: 4 × 7 − 5 = 28 − 5 = 23. ✓
Example 2: Brackets and the unknown on both sides
Solve 3(2x − 1) = 4x + 9.
Expand the bracket first, multiplying both terms inside by 3: 6x − 3 = 4x + 9.
Subtract the smaller x-term, 4x, from both sides: 2x − 3 = 9.
Add 3 to both sides: 2x = 12. Divide by 2: x = 6.
Check in the original: the left is 3(12 − 1) = 3 × 11 = 33, and the right is 24 + 9 = 33. ✓
Example 3: A fraction with an expression on top
Solve (2x + 1)/5 = 3.
The bar divides the whole of 2x + 1 by 5, so multiply both sides by 5: 2x + 1 = 15.
Subtract 1: 2x = 14. Divide by 2: x = 7.
Check: (14 + 1)/5 = 15/5 = 3. ✓ A student who wrote 2x + 1/5 = 15 would have multiplied only the first term, and this check would have caught it.
Common mistakes and how to avoid them
Changing only one side. Write the operation at the side of the line — "− 5" or "÷ 4" — and apply it to both halves before moving on.
Undoing in the wrong order. Numbers added or subtracted come off before the coefficient. Reversing this makes the arithmetic much harder even when it is legitimate.
Expanding a bracket incompletely. In 2(x + 3), both the x and the 3 are multiplied by 2. Check each bracket has as many products as it had terms.
Losing a sign across the equals sign. A term that moves changes sign: +6 on the left becomes −6 on the right. If you prefer not to think in terms of moving, simply subtract 6 from both sides and the sign takes care of itself.
Treating a fraction bar as covering only one term. The bar acts as a bracket around everything above it.
Stopping at 3x = 15. The question asked for x, not for 3x. Complete the final division.
Checking against a rearranged line. Substitute into the original equation, so an earlier error cannot hide.
Exam technique for "Solving Linear Equations"
Set the working out vertically, one operation per line, with the operation written beside it. Method marks in multi-step equation questions are awarded for the correct sequence of steps, and a visible sequence can earn credit even when the final arithmetic slips.
When an equation has brackets, look at the right-hand side before expanding. If the number outside the bracket divides it exactly, dividing first is faster and removes a chance to make a multiplication error.
Collect the unknowns on whichever side leaves a positive coefficient. It costs nothing and removes the commonest sign error.
Substitute the answer back into the original equation and write the check down. It takes one line and confirms the whole solution.
If the answer is a fraction or a negative number, do not assume it is wrong. Linear equations at GCSE frequently have solutions such as x = −3 or x = 2.5.
Quick revision summary
An equation is a machine built around x. Solve it by undoing the operations in the reverse of the order they were applied, doing the same thing to both sides every time.
Numbers added or subtracted come off first; the coefficient comes off last. So 3x + 4 = 19 becomes 3x = 15, then x = 5.
For brackets, either expand fully — multiplying every term inside — or, when the outside number divides the other side exactly, divide first.
With the unknown on both sides, subtract the smaller x-term from both sides so the remaining coefficient is positive, then finish as an ordinary two-step equation.
A fraction bar acts as a bracket: multiplying by the denominator clears the whole of the numerator, so (x + 1)/3 = 4 gives x + 1 = 12. With several fractions, multiply every term by the lowest common denominator.
Form an equation from words by translating in the order the words appear, then solve in reverse order.
Always check by substituting into the original equation.