Solving Linear Equations

Linear equations are the foundation of algebra and appear on every GCSE Maths paper, regardless of exam board or tier. A linear equation contains an unknown — usually x — raised to the power of one, and your job is to find the value of that unknown. This skill feeds directly into harder topics such as simultaneous equations, inequalities, and forming equations from real-life contexts. On this page you will learn how to solve one-step, two-step, and multi-step linear equations, including those with brackets and fractions. By the end, you should feel confident tackling any linear equation the examiner can throw at you.

What Is a Linear Equation?

A linear equation is an equation where the highest power of the unknown is 1. The graph of a linear equation is always a straight line — hence the name.

Examples: 3x + 5 = 20, 2(x − 4) = 10, (x + 3)/2 = 7.

The golden rule of equation solving is: whatever you do to one side, you must do to the other side. This keeps the equation balanced.

If a + b = c, then a = c − b

If ax = b, then x = b ÷ a

The aim is always to isolate the unknown on one side of the equals sign.

Step-by-Step Method

One-step and two-step equations

Look at what is happening to x. For instance, in 3x + 5 = 20, x is multiplied by 3 and then 5 is added.
Undo operations in reverse order (think of "peeling off layers"). Subtract 5 first: 3x = 15. Then divide by 3: x = 5.
Check by substituting back: 3(5) + 5 = 20 ✓.

Equations with brackets

Expand the brackets first. For 4(2x − 1) = 28, expand to get 8x − 4 = 28.
Solve as a two-step equation. Add 4: 8x = 32. Divide by 8: x = 4.

Equations with the unknown on both sides

Collect the x terms on the side where there are more of them. For 5x + 3 = 2x + 18, subtract 2x from both sides: 3x + 3 = 18.
Then solve normally. Subtract 3: 3x = 15. Divide by 3: x = 5.

Equations with fractions

Multiply every term by the lowest common denominator (LCD) to clear the fractions.
Solve the resulting equation as normal.

Worked Example 1 — Foundation Level

Question: Solve 7x − 3 = 25.

Working:

Step 1: Add 3 to both sides. 7x = 28

Step 2: Divide both sides by 7. x = 4

Check: 7(4) − 3 = 28 − 3 = 25 ✓

Answer: x = 4

Worked Example 2 — Higher Level

Question: Solve (3x + 1)/4 − (x − 2)/3 = 2.

Working:

Step 1: Find the LCD of 4 and 3, which is 12. Multiply every term by 12.

12 × (3x + 1)/4 − 12 × (x − 2)/3 = 12 × 2

3(3x + 1) − 4(x − 2) = 24

Step 2: Expand brackets.

9x + 3 − 4x + 8 = 24

Step 3: Simplify the left side.

5x + 11 = 24

Step 4: Subtract 11.

5x = 13

Step 5: Divide by 5.

x = 13/5 = 2.6

Check: (3(2.6) + 1)/4 − (2.6 − 2)/3 = (8.8)/4 − (0.6)/3 = 2.2 − 0.2 = 2 ✓

Answer: x = 2.6

Common Mistakes

Forgetting to apply an operation to both sides. If you subtract 5 from the left, you must subtract 5 from the right too. The equals sign means both sides are balanced.
Sign errors when expanding brackets. In −2(x − 3), the answer is −2x + 6, not −2x − 6. Remember: a negative times a negative gives a positive.
Dividing only one term by the coefficient. In 3x + 6 = 21, some students divide the 6 by 3 as well. You should subtract 6 first, then divide.
Leaving fractions unfinished. If x = 15/4, write it as 3.75 or 3¾ unless the question asks for a specific form. On the calculator paper, decimals are usually fine.
Not checking the answer. A quick substitution back into the original equation catches most errors and only takes a few seconds.

Exam Tips

Show every step of working. Even if you can solve it in your head, the method marks require written steps. On AQA and Edexcel, each operation on both sides is typically worth one mark.
If fractions appear, clear them immediately. Multiply through by the LCD at the start — this makes the rest of the equation far simpler.
Circle or underline your final answer. Examiners scan quickly; make it easy for them to find x = ... in your working.
Practise equations with x on both sides. These are the most common mid-difficulty questions on Foundation papers and appear early on Higher papers.

Practice Questions

Q1 (Foundation): Solve 4x + 9 = 37.

Answer: x = 7

Q2 (Foundation/Higher): Solve 3(2x − 5) = 4x + 7.

Answer: x = 11

Q3 (Higher): Solve (5x − 1)/3 = (2x + 4)/2.