What does the Equations and Inequalities strand of N(A)-Level Mathematics cover?

This strand of SEAB 4045 (Mathematics Syllabus A) covers solving linear equations by the balance method, including forming them from word problems; solving linear inequalities and showing the solution range on a number line; solving quadratic equations by factorising or by the quadratic formula; and solving a pair of simultaneous linear equations by elimination or substitution. These techniques underpin the graphs strand, where the same solutions appear as intercepts and intersection points.

What is the balance method for solving equations?

The balance method keeps an equation balanced by doing the same inverse operation to both sides until the unknown is alone. Expand any brackets and clear fractions first, then collect the unknown terms on one side and the numbers on the other, and finally divide to isolate the unknown. Always check your answer by substituting it back into the original equation.

When does an inequality sign flip?

The inequality sign reverses only when you multiply or divide both sides by a negative number. For example, from $-2x > 6$ you divide both sides by $-2$ and flip the sign to get $x < -3$. Adding or subtracting a number, or multiplying or dividing by a positive number, leaves the sign unchanged.

What is the difference between elimination and substitution?

Both solve a pair of simultaneous linear equations. Elimination matches the coefficient of one unknown, then adds or subtracts the equations to remove it. Substitution makes one unknown the subject of one equation and puts that expression into the other. Use elimination when coefficients line up easily, and substitution when one unknown already stands alone; always substitute back to find the second unknown and check the pair.

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N(A)-Level Mathematics Equations and Inequalities: linear equations, linear inequalities, quadratic equations, and simultaneous equations

Q: How do I choose between factorising and the quadratic formula?

Try factorising first, because it is faster: make one side zero, split the quadratic into two brackets, and set each bracket to zero. If the quadratic does not factorise neatly, use the quadratic formula $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, which works for every quadratic. The $\pm$ gives the two roots, so a quadratic usually has two solutions.

Q: What is the difference between elimination and substitution?

Both solve a pair of simultaneous linear equations. Elimination matches the coefficient of one unknown, then adds or subtracts the equations to remove it. Substitution makes one unknown the subject of one equation and puts that expression into the other. Use elimination when coefficients line up easily, and substitution when one unknown already stands alone; always substitute back to find the second unknown and check the pair.

An overview of the N(A)-Level Mathematics Equations and Inequalities strand (SEAB 4045). Solving linear equations by the balance method, linear inequalities and their number-line solutions, quadratic equations by factorising and by formula, and simultaneous linear equations by elimination and substitution, with links to every dot point.

Generated by Claude Opus 4.88 min readSEAB-4045Updated 2026-06-13

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

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Why equations and inequalities matter
Linear equations: the balance method
Linear inequalities and the number line
Quadratic equations: two roots
Simultaneous linear equations
Check your knowledge

Why equations and inequalities matter

Solving equations is the engine room of N(A)-Level Mathematics (SEAB 4045, Mathematics Syllabus A). Almost every applied question, from finding an unknown length to working out where two graphs meet, ends with an equation to solve. This strand builds that skill in stages: from the straight-line balance of a linear equation, through the range answers of an inequality, up to the two-root world of quadratics and the paired logic of simultaneous equations. This overview links to every dot point in the module, each with its own worked answers and practice.

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

Linear equations: the balance method

Linear equations are solved by the balance method: whatever you do to one side, do to the other. Expand brackets and clear fractions first, then gather the unknown terms on one side and the numbers on the other, and divide to leave the unknown alone. Forming an equation from a word problem is half the skill: let a letter stand for the unknown, translate each fact into algebra, then solve. Always check by substituting your answer back into the original equation.

Linear inequalities and the number line

Linear inequalities follow exactly the same balance steps as equations, with one extra rule: multiplying or dividing both sides by a negative number reverses the inequality symbol. The answer is a range rather than a single value, shown on a number line with a filled circle for $\leq$ or $\geq$ (the endpoint is included), an open circle for a strict inequality (the endpoint is excluded), and an arrow pointing toward the valid values.

Quadratic equations: two roots

Quadratic equations of the form $ax^2 + bx + c = 0$ are solved by first making one side zero. Then either factorise into two brackets and apply the zero product rule (if a product is zero, at least one factor is zero), or, when the quadratic does not factorise neatly, use the quadratic formula

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

The

\pm

produces two roots, so give both solutions and check each by substitution.

Simultaneous linear equations

Simultaneous linear equations are two equations in two unknowns that share one common solution. Solve them by elimination (match a coefficient, then add or subtract to remove one unknown) or by substitution (make one unknown the subject and put it into the other equation). Solve for the first unknown, substitute back for the second, then check the pair in the equation you did not use. When forming the pair from a word problem, write one equation per fact.

Check your knowledge

A mix of equation and inequality questions covering the strand. Attempt them, then check the solutions.

Solve $3(x - 2) = 2x + 5$ . (2 marks)
Solve the inequality $5 - 2x \geq 11$ and show the solution on a number line. (2 marks)
Solve $x^2 - 7x + 12 = 0$ . (2 marks)
Solve $x^2 + 3x - 1 = 0$ , giving your answers to two decimal places. (3 marks)
Solve the simultaneous equations $2x + y = 7$ and $x - y = 2$ . (3 marks)

Solutions

The five questions cover the balance method, a linear inequality, factorising a quadratic, the quadratic formula, and solving simultaneous equations by elimination.

Step 1: Solve $3(x - 2) = 2x + 5$ by the balance method (Q1)

Expand the bracket first, then gather the $x$ terms on one side and the numbers on the other. Check by substituting back:

3x - 6 = 2x + 5 \implies 3x - 2x = 5 + 6 \implies x = 11.

Check: $3(11 - 2) = 27$ and $2(11) + 5 = 27$ . Both sides agree.

Final answer (Q1): $x = 11$ .

Step 2: Solve $5 - 2x \geq 11$ and show it on a number line (Q2)

Apply the same balance steps as an equation. When dividing by a negative number, the inequality sign must reverse to keep the statement true:

-2x \geq 6 \implies x \leq -3.

On the number line, place a filled circle at $-3$ (because $-3$ is included) with an arrow extending to the left toward smaller values.

Final answer (Q2): $x \leq -3$ , shown with a filled circle at $-3$ and an arrow to the left.

Step 3: Solve $x^2 - 7x + 12 = 0$ by factorising (Q3)

Look for two numbers that multiply to $12$ and add to $-7$ : these are $-3$ and $-4$ . Write the factorised form and apply the zero product rule:

(x - 3)(x - 4) = 0 \implies x = 3 \text{ or } x = 4.

Final answer (Q3): $x = 3$ or $x = 4$ .

Step 4: Solve $x^2 + 3x - 1 = 0$ using the quadratic formula (Q4)

Identify $a = 1$ , $b = 3$ , $c = -1$ and substitute into $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ :

x = \frac{-3 \pm \sqrt{9 + 4}}{2} = \frac{-3 \pm \sqrt{13}}{2}.

Evaluating each root to two decimal places gives $x \approx 0.30$ or $x \approx -3.30$ .

Final answer (Q4): $x = 0.30$ or $x = -3.30$ (to two decimal places).

Step 5: Solve $2x + y = 7$ and $x - y = 2$ simultaneously (Q5)

Adding the two equations eliminates $y$ because the $+y$ and $-y$ cancel, giving a single equation in $x$ :

(2x + y) + (x - y) = 7 + 2 \implies 3x = 9 \implies x = 3.

Substitute $x = 3$ into the simpler equation $x - y = 2$ to find $y$ :

3 - y = 2 \implies y = 1.

Final answer (Q5): $x = 3$ , $y = 1$ .

Sources & how we know this

Singapore-Cambridge GCE N(A)-Level Mathematics (Syllabus A, 4045) — Singapore Examinations and Assessment Board (2026)