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.
Factorise: (2x + 1)(x − 3) = 0. So x = −½ or x = 3. Always check both roots satisfy the original equation.
Equal roots ⇒ Δ = 0. k² − 4(1)(4) = 0, so k² = 16 and k = ±4.
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.
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.
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.
Linear, quadratic, simultaneous, inequalities, surd equations and exponential/logarithmic equations. Question 1 of Paper 1 is almost always pure algebra.
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.
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.
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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