What is not checking against the original wording?

Substituting back into the words, not just the equation, catches misread relationships.

SingaporeMathsSyllabus dot point

How do we turn a worded situation into equations and solve it?

Formulate linear and quadratic equations from worded problems, solve them, and interpret the solution in context

A focused answer to the O-Level E-Maths outcome on forming and solving equations from words. Defining variables, translating relationships into equations, solving, and checking that the answer makes sense in context.

Generated by Claude Opus 4.88 min answerUpdated 2026-06-06

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
The answer
Examples in context
Try this

What this dot point is asking

SEAB wants you to read a worded problem, define variables, translate the relationships into linear or quadratic equations, solve them, and interpret the answer in context, including rejecting solutions that do not make physical sense. This ties together every equation-solving skill in the strand.

The answer

Define the variable clearly

Begin by stating what your letter represents, with units, for example let the width be $x$ cm. A clear definition keeps the algebra meaningful and earns method marks even before you solve.

Translate the words into an equation

Convert each piece of information into a symbol relationship. Common cues are sum (add), product (multiply), is or equals (an equals sign), more than, twice, consecutive, and area or perimeter formulas. If two unknowns appear, look for two pieces of information to form simultaneous equations.

Choose the right kind of equation

A relationship involving only first powers gives a linear equation; an area, a product of two unknowns, or a squared quantity gives a quadratic. Recognising which arises tells you which solving method to reach for.

Solve and interpret in context

Solve the equation, then read the answer back into the situation. A length, age, or number of objects cannot be negative, so reject any solution that breaks the context, and round only if the situation requires whole numbers.

Worked example

A father is $30$ years older than his son. In $5$ years the father will be three times as old as his son. Find their present ages.

Step 1: Define the variables

Let the son's present age be $x$ years, so the father's present age is $(x + 30)$ years.

Step 2: Translate the future condition

In $5$ years the son is $(x + 5)$ and the father is $(x + 35)$ . The father is then three times as old: $x + 35 = 3(x + 5)$ .

Step 3: Solve the equation

Expand: $x + 35 = 3x + 15$ . Collect: $35 - 15 = 3x - x$ , so $20 = 2x$ and $x = 10$ .

Step 4: Interpret in context

The son is $10$ years old now and the father is $40$ . Check: in $5$ years they are $15$ and $45$ , and $45 = 3 \times 15$ , which is correct.

Examples in context

Example 1. Mixing problems. Combining two quantities at different prices or concentrations, such as blending coffee beans, gives equations relating amounts and totals. Defining the amount of one blend as a variable turns the situation into a solvable linear equation.

Example 2. Number puzzles. Many problems describe relationships between unknown numbers, for example a number whose square exceeds itself by $12$ . This becomes $x^2 = x + 12$ , a quadratic whose roots are the candidate numbers, with the context deciding which to keep.

Try this

Q1. A number multiplied by $4$ and then increased by $7$ gives $31$ . Form and solve an equation. [2 marks]

Cue. $4x + 7 = 31$ , so $4x = 24$ and $x = 6$ .

Q2. Two consecutive even numbers have a sum of $54$ . Find them. [2 marks]

Cue. Let them be $n$ and $n + 2$ : $2n + 2 = 54$ , so $n = 26$ , giving $26$ and $28$ .

Q3. A square has area $5\ \text{cm}^2$ more than its perimeter (treating both as numbers). If the side is $x$ , form the equation. [2 marks]

Cue. Area is $x^2$ and perimeter is $4x$ , so $x^2 = 4x + 5$ , that is $x^2 - 4x - 5 = 0$ .

Exam-style practice questions

Practice questions written in the style of SEAB exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Original4 marksThe sum of three consecutive integers is

72

. By forming an equation, find the three integers.

Show worked answer →

Let the smallest integer be $n$ . The three consecutive integers are $n$ , $n + 1$ and $n + 2$ .

Form the equation: $n + (n + 1) + (n + 2) = 72$ , so $3n + 3 = 72$ .

Solve: $3n = 69$ , giving $n = 23$ .

The three integers are $23$ , $24$ and $25$ .

Markers reward defining the variable, forming a correct equation, solving it, and stating all three integers.

Original5 marksA rectangle is

5\ \text{cm}

longer than it is wide and has an area of

84\ \text{cm}^2

. By forming an equation, find its width.

Show worked answer →

Let the width be $x$ cm, so the length is $(x + 5)$ cm.

Area gives the equation: $x(x + 5) = 84$ , so $x^2 + 5x - 84 = 0$ .

Factorise: two numbers multiplying to $-84$ and adding to $5$ are $12$ and $-7$ , so $(x + 12)(x - 7) = 0$ .

This gives $x = -12$ or $x = 7$ . A width cannot be negative, so $x = 7\ \text{cm}$ .

Markers reward defining the variable, forming the quadratic, solving it, and rejecting the negative root with a reason.