Linear equations

Solve equations in one variable by keeping both sides balanced, including brackets, fractions and special cases.

Also known as: solving equations, one-variable equations.

SAT · Algebra GCSE Maths · Algebra IGCSE Maths · Algebra and graphs

Key ideas

An equation is a balance: whatever you do to one side, do to the other.
Undo operations in reverse order: brackets and fractions first, then collect terms.
Collect the variable on the side where its coefficient stays positive to avoid sign errors.
If the variable disappears, the equation has either no solution or infinitely many solutions.

The balance idea

A linear equation has the variable only to the first power, for example $3x + 7 = 22$ . Solving means finding the value of $x$ that makes both sides equal.

Think of the equals sign as a balance. Subtracting $7$ from both sides keeps it level:

$3x + 7 - 7 = 22 - 7 \quad\Rightarrow\quad 3x = 15$

Then divide both sides by $3$ to get $x = 5$ . Always check by substituting back: $3(5) + 7 = 22$ ✓.

Brackets, fractions and unknowns on both sides

Brackets: expand first. $4(x - 2) = 2x + 6$ becomes $4x - 8 = 2x + 6$ .
Unknowns on both sides: subtract the smaller $x$ -term from both sides: $2x - 8 = 6$ , so $x = 7$ .
Fractions: multiply every term by the lowest common denominator. For $\frac{x}{3} + \frac{x}{4} = 7$ , multiply by $12$ : $4x + 3x = 84$ , so $x = 12$ .

No solution or infinitely many

Sometimes the variable cancels out completely:

$2(x + 3) = 2x + 5$ becomes $6 = 5$ , which is false, so there is no solution.
$2(x + 3) = 2x + 6$ becomes $6 = 6$ , which is always true, so there are infinitely many solutions.

Exam questions often ask for the value of a constant that creates one of these cases. Make the $x$ -coefficients equal, then compare the constants.

Worked examples

Example 1 · Unknowns on both sides

Solve $5x - 4 = 2x + 11$ .

Step 1.Subtract $2x$ from both sides: $3x - 4 = 11$ .
Step 2.Add $4$ to both sides: $3x = 15$ .
Step 3.Divide by $3$ : $x = 5$ .
Step 4.Check: $5(5) - 4 = 21$ and $2(5) + 11 = 21$ ✓.

Answer: $x = 5$

Example 2 · Equation with fractions

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

Step 1.Multiply both sides by $6$ (the LCD): $2(2x + 1) = 3(x - 2)$ .
Step 2.Expand: $4x + 2 = 3x - 6$ .
Step 3.Subtract $3x$ and $2$ from both sides: $x = -8$ .

Answer: $x = -8$

Common mistakes

Multiplying only the first term of a bracket: $3(x + 4) \neq 3x + 4$ .
Forgetting to multiply every term (including whole numbers) when clearing fractions.
Sign errors when moving terms: subtracting $-3$ means adding $3$ .

Try it yourself

Original practice questions. Your answer is checked instantly.

Question 1Linear equationsEasier

What value of $x$ satisfies $3x + 7 = 22$ ?

Question 2Linear equationsEasier

If $5x - 4 = 2x + 11$ , what is the value of $x$ ?