What does the Equations and Inequalities strand of O-Level E-Maths cover?

This part of SEAB 4052 covers solving linear equations in one unknown (including those with fractions) and pairs of simultaneous linear equations by substitution and elimination; solving linear inequalities and representing the solution on a number line; solving quadratic equations by factorisation, by the quadratic formula and by completing the square; and forming linear and quadratic equations from worded problems and interpreting the solution in context. It is where algebraic manipulation is put to work to find unknowns.

When should I use the quadratic formula instead of factorising?

Try factorisation first because it is quick and exact when the quadratic factorises neatly. If the quadratic does not factorise with integers, use the quadratic formula, x equals minus b plus or minus the square root of b squared minus four a c, all over two a, which works for every quadratic. Completing the square is a third route that is especially useful when you also need the minimum or maximum value, or when the question explicitly asks for completed-square form.

Why does the inequality sign flip when you multiply or divide by a negative?

Multiplying or dividing both sides of an inequality by a negative number reverses the order of the two sides, so the inequality sign must be reversed to keep the statement true. For example, from minus two x is greater than six, dividing both sides by minus two gives x is less than minus three. Forgetting to reverse the sign is one of the most common and costly slips in this strand.

How do substitution and elimination differ for simultaneous equations?

Both solve a pair of linear equations for two unknowns. In substitution you rearrange one equation to make a variable the subject, then substitute that expression into the other equation. In elimination you add or subtract multiples of the equations so that one variable cancels. Elimination is usually faster when the coefficients line up easily; substitution is natural when one equation already has a variable as the subject.

How do I turn a worded problem into an equation?

Define the unknown clearly with a letter (for example, let the number be x), translate each relationship in the problem into algebra, form an equation, solve it, and then interpret the answer in the original context, including checking that it is sensible. Many marks are awarded for setting up the correct equation, so write down what your variable represents and state the final answer in words with units.

SingaporeMaths

O-Level E-Maths Equations and Inequalities: linear and simultaneous equations, linear inequalities, quadratic equations by factorisation, the formula and completing the square, and forming equations from words

An overview of the O-Level E-Maths Equations and Inequalities strand (SEAB 4052). Solving linear and simultaneous equations, handling linear inequalities and the sign-reversal rule, solving quadratics by factorisation, by the quadratic formula and by completing the square, and turning worded problems into equations, with links to every dot point.

Generated by Claude Opus 4.88 min readSEAB-4052Updated 2026-06-13

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

Jump to a section

What this strand is about
Linear equations and simultaneous equations
Linear inequalities
Quadratic equations: three methods
Word problems and modelling
How the strand is examined
Check your knowledge

What this strand is about

Equations and inequalities is where the algebraic manipulation of the Number and Algebra strand is put to work to find unknown quantities. The skills here, solving linear and quadratic equations and handling inequalities, appear throughout the syllabus: in coordinate geometry, in graph work, in mensuration and in real-world modelling. This overview ties the strand together and links to every dot point, each with worked answers and practice.

See the full set of dot points at /sg-o-level/mathematics/syllabus.

Linear equations and simultaneous equations

The strand begins with linear equations and simultaneous equations. A linear equation in one unknown is solved by doing the same operation to both sides until the variable is isolated, including clearing fractions by multiplying through by the common denominator. A pair of simultaneous linear equations is solved by substitution (making a variable the subject and substituting) or elimination (adding or subtracting to cancel a variable). The solution is the point where the two lines meet.

Linear inequalities

Linear inequalities are solved much like equations, with one crucial extra rule: when you multiply or divide both sides by a negative number, you must reverse the inequality sign. The solution is a range of values, shown on a number line (an open circle for a strict inequality, a filled circle when the endpoint is included), and you may be asked to list the integers that satisfy it.

Quadratic equations: three methods

Two dot points cover quadratics. Quadratic equations teaches solving by factorisation: rearrange into standard form $ax^2 + bx + c = 0$ , factorise, then apply the zero product property (if a product is zero, at least one factor is zero). Solving by the quadratic formula and completing the square handles quadratics that do not factorise neatly. The formula

x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

solves any quadratic, while completing the square both solves the equation and reveals the minimum or maximum value.

Word problems and modelling

Word problems and modelling is the application: define a variable, translate the words into a linear or quadratic equation, solve, and interpret the answer in context. SEAB sets these in everyday and financial settings, so checking that the answer is sensible (a length cannot be negative, for instance) is part of the marks.

How the strand is examined

Choose an efficient method. Factorisation before the formula; elimination or substitution depending on the coefficients. The right choice saves time and reduces error.
Reverse the sign for inequalities. Multiplying or dividing by a negative flips the inequality; this is a frequent source of lost marks.
Define variables and interpret answers. In word problems, state what the letter stands for and give the final answer in context with units, rejecting any solution that is not sensible.

Check your knowledge

Attempt these, then check the solutions.

Solve $\dfrac{2x - 1}{3} = 5$ . (2 marks)
Solve the simultaneous equations $3x + 2y = 12$ and $x - y = 1$ . (3 marks)
Solve the inequality $5 - 2x > 11$ and list the positive integer solutions, if any. (2 marks)
Solve $x^2 - 5x + 6 = 0$ by factorisation. (2 marks)
Solve $x^2 + 4x - 2 = 0$ , giving your answers to 2 decimal places. (3 marks)

Solutions

Step 1: Clear the fraction by multiplying both sides by the denominator (Q1)

Fractions in equations are removed most cleanly by multiplying every term by the denominator, here $3$ . This leaves a straightforward linear equation that is solved by inverse operations.

\frac{2x - 1}{3} = 5 \implies 2x - 1 = 15 \implies 2x = 16 \implies x = 8.

Final answer: $x = 8$

Step 2: Use substitution to reduce two equations to one (Q2)

The second equation $x - y = 1$ can be rearranged as $x = y + 1$ , which is already solved for $x$ . Substituting this expression into the first equation replaces $x$ entirely and produces a single equation in $y$ that can be solved directly.

From $x - y = 1$ : $x = y + 1$ . Substituting into $3x + 2y = 12$ :

3(y + 1) + 2y = 12 \implies 3y + 3 + 2y = 12 \implies 5y = 9 \implies y = 1.8.

Then $x = 1.8 + 1 = 2.8$ . Check: $3(2.8) + 2(1.8) = 8.4 + 3.6 = 12$ . Correct.

Final answer: $x = 2.8$ , $y = 1.8$

Step 3: Solve the inequality, reversing the sign when dividing by a negative (Q3)

Inequalities are solved like equations except for one critical rule: dividing or multiplying by a negative number reverses the direction of the inequality sign. Here the coefficient of $x$ is $-2$ , so the sign reverses when dividing.

5 - 2x > 11 \implies -2x > 6 \implies x < -3.

All solutions are less than $-3$ , so there are no positive integer solutions.

Final answer: $x < -3$ ; no positive integer solutions

Step 4: Factorise by finding two numbers with the right product and sum (Q4)

For $x^2 - 5x + 6 = 0$ , you need two numbers that multiply to $+6$ (the constant term) and add to $-5$ (the coefficient of $x$ ). The pair $-2$ and $-3$ satisfies both conditions. Once factorised, the zero-product property gives the solutions directly.

x^2 - 5x + 6 = (x - 2)(x - 3) = 0 \implies x = 2 \text{ or } x = 3.

Final answer: $x = 2$ or $x = 3$

Step 5: Apply the quadratic formula when the equation does not factorise neatly (Q5)

For $x^2 + 4x - 2 = 0$ the discriminant is $b^2 - 4ac = 16 + 8 = 24$ , which is not a perfect square, so factorisation with integers is not possible. The quadratic formula works for any quadratic regardless of whether it factorises.

x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-4 \pm \sqrt{24}}{2}.

x = \frac{-4 + \sqrt{24}}{2} \approx 0.45 \quad \text{or} \quad x = \frac{-4 - \sqrt{24}}{2} \approx -4.45.

Final answer: $x = 0.45$ or $x = -4.45$ (to 2 decimal places)

Sources & how we know this

Singapore-Cambridge GCE O-Level Mathematics (Syllabus 4052) — Singapore Examinations and Assessment Board (2026)