Creating Algebraic Expressions from Situations Revision Notes | GCSE AQA Higher Maths

1. From Words to Algebra

Every algebraic expression started life as a sentence. Someone described a situation in words, then translated it into maths. That translation skill — turning a real situation into an expression or formula — is exactly what this lesson is about.

Think about buying items at a market stall. If apples cost 30p each and you buy $n$ apples, the total cost in pence is $30n$. You have taken a real situation, chosen a letter for the unknown quantity, and written an expression. That is the whole process.

The key skill is reading the language of a situation carefully. Certain words signal certain operations, and recognising those signals is what makes the translation reliable.

• "More than", "increased by", "sum of" → addition ($+$)
• "Less than", "decreased by", "difference" → subtraction ($-$)
• "Times", "product of", "multiplied by", "groups of" → multiplication ($\times$, written as a coefficient)
• "Shared equally", "divided by", "per", "each" → division ($\div$, written as a fraction)
• "Squared", "the square of" → power of 2 ($x^2$)

Keeping this list in mind as you read a question is the first step. The second step is choosing a sensible letter for the unknown — then writing the expression carefully.

2. Building Expressions Step by Step

Now that you know the language signals, here is a reliable method for translating any situation into an expression. Work through it on every question until it becomes automatic.

Worked example: Translating a Situation into an Expression

Mohammed thinks of a number. He multiplies it by 4 and then subtracts 7. Write an expression for Mohammed's result.

1. Let $n$ = the number Mohammed thinks of. — Choose a letter. $n$ is a natural choice for 'a number'. State what it represents — this earns a method mark in an exam.
2. Multiplies by 4 → $4n$ — "Multiplies by 4" means 4 lots of $n$. Write the coefficient directly in front of the letter, with no multiplication sign.
3. Subtracts 7 → $4n - 7$ — "Subtracts 7" means take 7 away from the result of step 2. Append $- 7$ to the expression.
4. $4n - 7$ — This is the complete expression. There is no equals sign because we have not been told the result — it is an expression, not an equation. [M1 for correct expression]

Answer: $4n - 7$

Notice the order matters. "Multiplies by 4 then subtracts 7" gives $4n - 7$. If the question said "subtracts 7 then multiplies by 4", the expression would be $4(n - 7)$. Read the order of operations in the question carefully.

3. Expressions with More Than One Variable

Some situations involve more than one unknown quantity. When that happens, you need a different letter for each unknown. The process is the same — you are still translating language into algebra — but now you have two (or more) letters to manage.

Worked example: Expression with Two Variables

Priya buys $a$ apples at 40p each and $b$ bananas at 25p each. Write an expression for the total cost in pence.

1. Let $a$ = number of apples, $b$ = number of bananas. — Two different quantities, so two different letters. State both clearly.
2. $40 \times a = 40a$ pence — 40p per apple, $a$ apples. Multiply the price by the quantity.
3. $25 \times b = 25b$ pence — 25p per banana, $b$ bananas. Same structure as step 2.
4. $40a + 25b$ pence — Add the two costs together. The expression has two terms, one for each item. [M1 for correct expression]

Answer: $40a + 25b$ pence

Each term in the expression corresponds to one part of the situation. This one-to-one mapping between the real context and the algebra is what makes expressions so powerful — and it is also what the examiner is checking.

4. Writing Formulae from Described Relationships

A formula goes one step further than an expression: it has an equals sign, connecting an output quantity to the variables that determine it. When a question describes a rule or procedure that produces a result, you are being asked to write a formula.

Worked example: Writing a Formula from a Description

A taxi company charges a fixed fee of £3 plus £2 for every kilometre travelled. Write a formula for the total charge $C$ (in pounds) for a journey of $d$ kilometres.

1. Total charge $= C$ — The formula will have $C$ on the left-hand side. The question tells you to use $C$.
2. Fixed fee $= £3$ — This amount is added regardless of distance. It is a constant in the formula.
3. £2 per kilometre for $d$ kilometres $= 2d$ — "£2 for every kilometre" means multiply the rate by the distance. This gives the term $2d$.
4. $C = 3 + 2d$ — Fixed fee plus variable cost. The formula can also be written $C = 2d + 3$. Both are correct. [M1 for correct formula structure, A1 for correct coefficients]
5. If $d = 5$: $C = 3 + 2 \times 5 = 3 + 10 = £13$ — Substituting a value and checking it makes sense in context is a reliable way to verify your formula. A 5 km journey costing £13 is plausible.

Answer: $C = 3 + 2d$

The checking step in the worked example above is worth doing in any exam question where you have time. If the formula gives a nonsensical answer for a simple input, something has gone wrong in the translation.

5. Perimeter and Area: Geometry Meets Algebra

Geometry problems are a common source of algebraic expression questions at GCSE. You are often given a shape with some sides labelled in terms of a variable, and asked to write an expression for the perimeter or area.

Worked example: Writing an Expression for Perimeter

A rectangle has length $(3x + 2)$ cm and width $(x - 1)$ cm. Write an expression for the perimeter of the rectangle. Simplify your answer.

1. $P = 2l + 2w$ — Perimeter of a rectangle is twice the length plus twice the width. Both pairs of opposite sides are equal.
2. $P = 2(3x + 2) + 2(x - 1)$ — Replace $l$ with $(3x + 2)$ and $w$ with $(x - 1)$. Keep brackets — you will expand them next.
3. $P = 6x + 4 + 2x - 2$ — $2 \times (3x + 2) = 6x + 4$ and $2 \times (x - 1) = 2x - 2$. Multiply every term inside each bracket by 2.
4. $P = (6x + 2x) + (4 - 2) = 8x + 2$ — Group the $x$ terms together and the number terms together. [M1 for correct expansion, A1 for simplified expression]

Answer: $P = (8x + 2)$ cm

Area questions follow the same principle. Write the area formula, substitute the algebraic expressions for the dimensions, then expand and simplify. The result is an algebraic expression for the area in terms of the variable.

6. Practice: Translating a Range of Situations

The best way to consolidate this skill is to practise translating a variety of situations. The problems below cover the main types you will meet in an Higher-tier exam. Work through each one before checking the solution.

7. Exam Technique and Common Mistakes

Translating situations into algebra is a skill where marks are lost in predictable ways. Knowing the traps in advance means you can sidestep them under exam conditions.