Algebra: revision notes and practice

What you will learn

• Distinguish expressions from equations and recognise like terms.
• Expand and factorise a single bracket correctly.
• Substitute positive or negative values into expressions.
• Solve and check linear equations, including unknowns on both sides.

Key rules

Keep equations balanced

Use the same operation on both sides. Undo addition or subtraction, then multiplication or division.

Expand every term

Multiply the number outside a bracket by every term inside. For example, k(x + b) = kx + kb.

Substitute carefully

Replace each letter with its value. Put negative values in brackets and calculate powers before multiplying.

Worked examples

Example 1: Solve 3x − 5 = 16.

1. Add 5 to both sides: 3x = 21.
2. Divide both sides by 3: x = 7.
3. Check: 3 × 7 − 5 = 16.

x = 7

Example 2: Expand −2(x − 4).

1. Multiply −2 by x to get −2x.
2. Multiply −2 by −4 to get +8.
3. Combine the two terms.

−2x + 8

A practice question

Solve 5x − 6 = -36. What is x?

1. Know what the symbols are saying

An expression, such as 3x + 5, describes a quantity but does not have an equals sign. An equation, such as 3x + 5 = 20, says that two quantities have the same value and can be solved for x. A formula is an equation that describes a relationship between quantities, for example distance = speed × time.

In 3x + 5, the terms are 3x and 5. The coefficient of x is 3. Writing 3x means 3 × x, not the two-digit number formed by putting 3 next to x. A variable can take different values, depending on the problem.

2. Collect only like terms

Like terms contain exactly the same variable part. You can combine 4x + 3x because both terms count copies of x; the result is 7x. However, x and x² are different kinds of terms, so 4x + 3x² cannot be combined into 7x².

Keep each sign attached to its term while collecting. For 5x + 2 − 3x + 7, the x terms give 2x and the constant terms give 9, producing 2x + 9. A useful written step groups 5x − 3x and 2 + 7 separately without changing any signs.

3. Expand a single bracket

Expanding reverses grouping: multiply the factor outside a bracket by every term inside it. Thus 4(x + 3) becomes 4x + 12. The 4 multiplies both x and 3. An area model can help: four equal strips, each of length x + 3, together have length 4x + 12.

Negative factors need extra care. For −3(2x − 5), multiply −3 by 2x to obtain −6x, then by −5 to obtain +15. The result is −6x + 15. If an expression has several brackets, expand each completely before collecting like terms.

a(b + c) = ab + ac

4. Take out a common factor

Factorising writes a sum as a product. In 6x + 9, both terms share a factor of 3, so the expression becomes 3(2x + 3). Divide each original term by the factor to find what belongs inside the bracket.

A common factor can include a variable. For 8x² − 12x, the greatest common factor is 4x, giving 4x(2x − 3). Expand your proposed answer to check it returns the original expression. Do not cancel an x from a sum unless it has first been taken out as a factor of every relevant term.

5. Substitute without losing negative signs

Substitution means replacing every occurrence of a letter with its stated value. If x = −2, then 3x² + 4x becomes 3(−2)² + 4(−2). Calculate the power first: 3 × 4 − 8 = 4. Writing brackets around a negative replacement protects the sign.

Read the expression carefully before entering it into a calculator. In 2x², only x is squared; in (2x)², the whole product is squared. With x = 3, these give 18 and 36 respectively. Fractions also act as grouping symbols, so work out a complete numerator and denominator before dividing.

6. Solve by keeping both sides equal

An equation behaves like a balance: performing the same valid operation on both sides keeps it true. To solve 4x − 7 = 21, add 7 to both sides to get 4x = 28, then divide both sides by 4 to get x = 7.

Undo operations in reverse order. In 3(x + 2) = 18, dividing both sides by 3 first gives x + 2 = 6; subtracting 2 gives x = 4. Expanding the bracket first also works. Choose the route with clear, manageable arithmetic rather than relying on a rule about moving terms.

7. Handle unknowns on both sides and fractions

For 5x + 2 = 2x + 17, subtract 2x from both sides to obtain 3x + 2 = 17. Subtract 2, then divide by 3, giving x = 5. Removing the smaller x term first often leaves a positive coefficient, but any correctly balanced sequence works.

For (x − 3)/4 = 5, multiply both sides by 4 to obtain x − 3 = 20, so x = 23. Multiplying a whole equation by a common denominator can clear several fractions. Apply that multiplier to every term on both sides, and never divide by an expression that might be zero without considering the restriction.

8. Check solutions and interpret formulae

Substitute a proposed solution into the original equation, before any rearrangement. For x = 5 in 5x + 2 = 2x + 17, both sides equal 27. Checking the original line can reveal a sign or arithmetic error that later working would otherwise hide.

To change the subject of a formula, use the same balancing principles. From v = u + at, subtract u to get v − u = at, then divide by a to get t = (v − u)/a, provided a ≠ 0. In a word problem, give the value a meaning and include any units or restrictions required by the situation.

Check your understanding

How to earn the marks

• Write one balanced operation per line when solving.
• Keep minus signs attached to terms and negative substituted values in brackets.
• Check factorisations by expanding and solutions by substitution.
• Show the equation you formed in a word problem before solving it.

A common mistake

A negative multiplied by a negative is positive. Also, (−3)² = 9, while −3² = −9.

Course context and official specifications

This guide covers selected skills. Some extensions are Higher content; choose your official tier separately from practice difficulty in the app. These are original questions and explanations.

• AQA GCSE Mathematics — A2, A4, A17

  Linear equations, expansion and substitution are available at both tiers. Difficulty changes the numbers and steps; it is separate from exam tier.

• Edexcel GCSE Mathematics — A2, A4, A17

  Linear equations, expansion and substitution are available at both tiers. Difficulty changes the numbers and steps; it is separate from exam tier.

• OCR GCSE Mathematics — 6.01, 6.02, 6.03

  Linear equations, expansion and substitution are available at both tiers. Difficulty changes the numbers and steps; it is separate from exam tier.

Keep your next step small

Try five questions, read the feedback, and revisit one thing you found difficult. The app tracks a 20-answer practice goal separately from accuracy.

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