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School SEE Compulsory Mathematics

Algebraic Fraction

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Simplify single fraction → cancel common factors using identities (x²−y², x³±y³). Add/subtract → same denominator: add numerators; different denominator: take LCM. Complex fractions → numerator becomes x²+y², 2xy, or 0; common results: x−y, a/(a²−b²), or 0. Value questions → cross-multiply → compare coefficients;

1. Reducing to Lowest Terms (Most important)

  • Cancel common factors only
  • Always write identities ready:
    • x² – y² = (x–y)(x+y)
    • x² + 2xy + y² = (x+y)²
    • x³ + y³ = (x+y)(x² – xy + y²)
    • x³ – y³ = (x–y)(x² + xy + y²)
    • a² + 2ab + b² = (a+b)²

Examples

  1. (x² – 16)/(x² – 8x + 16) = (x–4)(x+4)/(x–4)² = (x+4)/(x–4)
  2. (a³ – 8)/(a² – 4) = (a–2)(a² + 2a + 4)/(a–2)(a+2) = (a² + 2a + 4)/(a+2)

2. Addition & Subtraction – Golden Rule

Same denominator → add/subtract numerator Different denominator → LCM of denominator

Trick: If denominator is (x–y)(x+y), numerator becomes something like x² + y² or 2xy

3. Complex Fractions – 90% SEE questions are these types

 

TypeTypical AnswerExample1/(a–b) – b/(a² – b²)a/(a² – b²)Very commonx/(x–y) + x/(x+y) + 2xy/(x² + y²)2x²/(x² – y²)Direct SEE1/(2(x–y)) – 1/(2(x+y)) – y/(x² – y²)0Favourite questionx²/(x+y) – y²/(x+y)x – yEasiesta/(a+b) + b/(a–b) + 2ab/(a² – b²)2a²/(a² – b²)Common

4. Find Value of a, b, c Questions (Cross multiply method)

Method: Bring to one side = 0 → cancel → compare coefficients

Example a/(x+1) + 2/(x+3) = (x+8)/(x² + 4x + 3) → a(x+3) + 2(x+1) = x+8 → ax + 3a + 2x + 2 = x + 8 → (a+2)x + (3a+2) = x + 8 → a+2 = 1 → a = –1 → 3a+2 = 8 → a = 2 (wrong) → wait, correct a = –1 only

5. Top 10 SEE Questions (Must know)

  1. x²/(x+y) – y²/(x+y) = x – y
  2. 1/(a–b) – b/(a² + b²) = a/(a² + b²)
  3. x/(x–y) + x/(x+y) + 2xy/(x² + y²) = 2x²/(x² – y²)
  4. 1/(2(x–y)) – 1/(2(x+y)) – y/(x² – y²) = 0
  5. (a² + b²)/(a² – b²) – (a – b)/(a + b) = 2b²/(a² – b²)
  6. a/(a+b) + b/(a–b) + 2ab/(a² – b²) = 2a²/(a² – b²)
  7. (x² – y²)/(x+y) + y = x
  8. 1/(x+1) + 1/(x+2) + 1/(x+3) = something → LCM = (x+1)(x+2)(x+3)
  9. If a/(2x+1) + b/(x–3) = (x+5)/(2x² – 5x – 3) → a = 3, b = –1
  10. (a³ + b³)/(a² – ab + b²) + a = 2a (very rare but important)

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