A21 - Forming and solving equations | AQA GCSE Mathematics

An equation starts with a relationship: two costs are equal, the sides add to a perimeter, or two purchases have known totals. You will turn words and procedures into algebra, form one or two equations, and interpret the answers in the original problem. Choosing what your letters mean is the first useful step.

Turning words and procedures into algebra

A school trip costs £35 for a booking plus £3 for each student. The total is £89. How many students are going? You could try numbers, but one equation will describe the whole situation and give an answer you can check.

You already know

An expression has no equals sign, such as $3n+35$ . An equation states that two expressions have equal values. You can solve linear and simultaneous linear equations by keeping both sides balanced.

Let $n$ be the number of students. The student charge is $3n$ pounds, and the booking is added once. The total cost, $C$ pounds, is $C=35+3n$ . This is a formula: it works for different group sizes. The trip with a total of £89 gives the equation $35+3n=89$ .

An algebraic model

An expression, formula or equation representing a stated relationship. Define each letter, including its units where needed. For example, $C=35+3n$ models the trip's cost in pounds. $C=35n+3$ does not model that pricing rule: it charges the booking fee for every student.

Read a procedure in its stated order. “Multiply a number by $5$ , then add $4$ ” gives $5n+4$ . “Add $4$ , then multiply by $5$ ” gives $5(n+4)$ . The bracket keeps the first result together. These translation methods are not on the formulae sheet: learn them; a context supplies its own relationship.

Worked example 1: writing each stage

Question. A machine multiplies its input $n$ by $5$ , adds $4$ , then divides by $3$ . Write an expression for its output. Non-calculator.

Plan. Write the result after each instruction, grouping the whole quantity being divided.

\begin{aligned} n&\longrightarrow5n && \text{multiply by }5\\ 5n&\longrightarrow5n+4 && \text{add }4\\ 5n+4&\longrightarrow\frac{5n+4}{3} && \text{divide the whole result} \end{aligned}

Why this step? The division is applied after the addition, so the fraction line must cover both terms.

Answer. $\dfrac{5n+4}{3}$ .

Check. With input $n=2$ , the instructions give $10$ , then $14$ , then $14/3$ . The expression gives the same result.

Change only the order of the first two instructions. The bracket now belongs before the multiplication.

Worked example 2: changing the order

Question. A machine adds $4$ to its input $n$ , multiplies by $5$ , then divides by $3$ . Write an expression for its output. Non-calculator.

Plan. Keep the sum $n+4$ together when multiplying.

\begin{aligned} n&\longrightarrow n+4 && \text{add }4\\ n+4&\longrightarrow5(n+4) && \text{multiply the whole sum}\\ 5(n+4)&\longrightarrow\frac{5(n+4)}{3} && \text{divide by }3 \end{aligned}

Why this step? Multiplying the result of an addition multiplies every part of that result, not just $n$ .

Answer. $\dfrac{5(n+4)}{3}$ .

Check. At $n=2$ , the instructions give $6$ , then $30$ , then $10$ . The expression gives $30/3=10$ .

This also explains a number trick. Add $4$ to any number, multiply by $5$ , then subtract five times your original number. The result is always $20$ , because $5(n+4)-5n=20$ . The “guess” at the end needs no knowledge of your starting number.

Watch out

Wrong: “ $4$ less than $n$ ” becomes $4-n$ . Right: start with $n$ and subtract $4$ , giving $n-4$ . Check a phrase with a concrete input before writing its algebra.

Follow the order with different instructions now.

Forming one equation from a context

Choose a letter for the unknown, write expressions for the quantities, then use the statement that connects them. A known total gives an equals sign. After solving, say what the value means: a bare number may be the letter's value rather than the answer requested.

In the opening problem, $n$ counts students, so it must be a non-negative whole number. The total cost has the same units on both sides of the equation: pounds. The model also assumes the stated price applies to every student.

Worked example 3: a fixed charge and a charge each

Question. A trip costs £35 for a booking plus £3 per student. The total is £89. Form an equation and work out the number of students. Non-calculator.

Plan. Let $n$ count students and equate the two descriptions of the total cost.

\begin{aligned} 35+3n&=89 && \text{fixed charge plus student charges}\\ 3n&=54 && \text{subtract }35\\ n&=18 && \text{divide by }3 \end{aligned}

Why this step? Only the per-student charge is multiplied by $n$ . The booking appears once.

Answer. $18$ students are going.

Check. $35+3(18)=35+54=89$ pounds, as required.

Change only the total cost to £95. Keeping the model but changing its total helps separate the relationship from the arithmetic.

Worked example 4: interpreting a changed total

Question. The same pricing rule gives a total of £95 for another trip. Form an equation and work out the number of students. Non-calculator.

Plan. Keep $35+3n$ for the charges and set it equal to the new total.

\begin{aligned} 35+3n&=95 && \text{new known total}\\ 3n&=60 && \text{subtract the fixed charge}\\ n&=20 && \text{divide by the price each} \end{aligned}

Why this step? Subtracting the fixed charge leaves exactly the amount paid for students.

Answer. $20$ students.

Check. The total rose by £6, enough for $2$ more students at £3 each: $18+2=20$ .

Some contexts put expressions on both sides. If one person's age is $x$ and another's is $x+4$ , “twice the younger age equals the older age plus $7$ ” gives $2x=(x+4)+7$ . Keep the meanings attached to the expressions while solving.

Exam tip

When asked to form an equation, show the equation before solving it. Define your letter and finish with the quantity requested, including a unit or description.

Try a different fixed charge and rate.

Forming equations in geometry

Geometry gives relationships too. A perimeter is the total length around a shape. An angle problem may tell you that some angles make a straight line or a full turn. Use the relationship, rather than measuring a sketch that is not drawn accurately.

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A21 - Forming and solving equations

Turning words and procedures into algebra

Forming one equation from a context

Forming equations in geometry

Continue this lesson.

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