Algebra & Equations for Matric Maths

Master Grade 12 Algebra & Equations — linear, quadratic, simultaneous and inequalities. CAPS-aligned examples and AI step-by-step explanations.

Algebra is the backbone of NSC Paper 1. Almost every question — Functions, Calculus, Sequences, Finance — depends on confident algebraic manipulation. CAPS Grade 12 requires you to solve linear equations, quadratic equations (factorising, quadratic formula, completing the square), simultaneous equations, inequalities, and equations involving surds and exponents. Examiners also test whether you can interpret your answers: rejecting a negative length, recognising that a square root cannot equal a negative number, or explaining why an equation has no real solutions. Clear, line-by-line working earns method marks even when an arithmetic slip costs you the final answer.

Key concepts

  • Solving quadratic equations by factorising, formula, and completing the square
  • Nature of roots — discriminant Δ = b² − 4ac
  • Solving simultaneous equations (linear-linear and linear-quadratic)
  • Quadratic and linear inequalities
  • Equations with surds and exponents
  • Rejecting invalid roots (from surd equations, zero denominators or context)
  • Making one variable the subject before substituting

Worked examples

Solve for x: 2x² − 5x − 3 = 0

Factorise: (2x + 1)(x − 3) = 0. So x = −½ or x = 3. Always check both roots satisfy the original equation.

For which values of k does x² + kx + 4 = 0 have equal roots?

Equal roots ⇒ Δ = 0. k² − 4(1)(4) = 0, so k² = 16 and k = ±4.

Solve simultaneously: y = 2x − 1 and x² + y² = 13

Substitute the linear equation into the quadratic: x² + (2x − 1)² = 13, so 5x² − 4x − 12 = 0 and (5x + 6)(x − 2) = 0. If x = 2 then y = 3; if x = −6/5 then y = −17/5. Check both pairs in x² + y² = 13.

Solve x² − x − 6 < 0

Factorise: (x − 3)(x + 2) < 0. The critical values are −2 and 3. The parabola opens upwards, so it is negative between its roots: −2 < x < 3.

Solve √(x + 7) = x − 5

Square both sides: x + 7 = x² − 10x + 25, so x² − 11x + 18 = 0 and (x − 9)(x − 2) = 0. Check: x = 9 gives √16 = 4 = 9 − 5 ✓. x = 2 gives √9 = 3 but 2 − 5 = −3 ✗. So x = 9 only.

Common mistakes

  • Dividing both sides by x (you lose the root x = 0).
  • Forgetting ± when taking square roots.
  • Flipping the inequality sign incorrectly when multiplying by a negative.
  • Squaring both sides of a surd equation without checking for extraneous roots.
  • Writing the solution of a quadratic inequality as two separate statements when it is one interval.
  • Checking a simultaneous-equation solution in only one of the two equations.

Exam tips

  • If the quadratic doesn't factorise quickly, jump to the formula — don't waste minutes.
  • For inequalities, sketch the parabola and read the sign from the graph.
  • State restrictions (e.g. x ≠ 0) before you start dividing.
  • With one linear and one quadratic equation, make x or y the subject of the linear equation first.
  • Give answers in the form asked for: exact, simplified or rounded to two decimal places.

In past papers

  • NSC Nov 2024 Paper 1 Q1 — solving quadratics and inequalities
  • NSC Nov 2023 Paper 1 Q1.4 — equations with surds

Frequently asked questions

What types of equations appear in Matric Paper 1?

Linear, quadratic, simultaneous, inequalities, surd equations and exponential/logarithmic equations. Question 1 of Paper 1 is almost always pure algebra.

Do I need to memorise the quadratic formula?

It is printed on the NSC information sheet, but you should still know it well: you need to identify a, b and c quickly, and the discriminant b² − 4ac inside it answers every nature-of-roots question.

How do I know when to reject a root?

Substitute each root into the original equation. Reject a root that makes a square root equal a negative number, makes a denominator zero, or does not fit the context, such as a negative length.

How do I solve a quadratic inequality quickly?

Make one side zero, factorise to find the critical values, then use a sign diagram or a rough sketch of the parabola to choose the interval where the expression has the required sign.

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