Algebra: Linear Equations, Quadratics and APs
Algebra is assessed in Mathematics Part I and rewards exact symbolic work: solving simultaneous equations, handling discriminants, and summing arithmetic progressions. Each mark comes from a stated rule applied in the right order, so write the rule before substituting.
Linear equations in two variables
Write both equations in the standard form a₁x + b₁y = c₁ and a₂x + b₂y = c₂ before using Cramer's rule: D = a₁b₂ − a₂b₁, Dx = c₁b₂ − c₂b₁, Dy = a₁c₂ − a₂c₁, then x = Dx/D and y = Dy/D, which needs D ≠ 0. The pair has a unique solution when a₁/a₂ ≠ b₁/b₂, infinitely many solutions when a₁/a₂ = b₁/b₂ = c₁/c₂, and no solution when a₁/a₂ = b₁/b₂ ≠ c₁/c₂.
Quadratic equations
For ax² + bx + c = 0 with a ≠ 0, the discriminant is Δ = b² − 4ac. If Δ > 0 the roots are real and distinct, if Δ = 0 they are real and equal, and if Δ < 0 there are no real roots. Roots satisfy α + β = −b/a and αβ = c/a, so x² − (α + β)x + αβ = 0 builds a quadratic from its roots.
Arithmetic progression
The nth term is tₙ = a + (n − 1)d, the sum is Sₙ = n/2[2a + (n − 1)d] = n/2(a + tₙ), and any single term can be recovered as tₙ = Sₙ − Sₙ₋₁.
Worked example
Solve 3x + 4y = 10 and 2x − 3y = 1. D = (3)(−3) − (4)(2) = −9 − 8 = −17; Dx = (10)(−3) − (4)(1) = −34; Dy = (3)(1) − (10)(2) = −17. Hence x = −34/−17 = 2 and y = −17/−17 = 1. Check: 3(2) + 4(1) = 10 and 2(2) − 3(1) = 1.
Traps that cost marks
Replacing the wrong column when forming Dx or Dy, and forgetting to move the constant to the right side first, are the commonest errors. With D = 0, do not stop at "no solution" - compare Dx and Dy. In an AP question, n must come out a positive integer; if it does not, recheck whether the question asks for a term or a sum, and whether d should be negative.