Algebraic Fractions
Prepared by Miss Clarissa Ng
www.clartutors.com
www.clartutors.com
Q1. Simplify completely by factorising the numerator and denominator:
9x² − 166x² − 8xQ2. Simplify completely by factorising the numerator and denominator:
x² + x − 123x² − 7x − 6Q3. Express as a single fraction in its simplest form:
23x − 2 + 5x + 4Q4. Express as a single fraction in its simplest form:
3xx² − 25 − 2x − 5Q5. Express as a single fraction in its simplest form:
4x² − 2x − 8 − 3x² + 5x + 6Q6. Express as a single fraction in its simplest form:
(a) 1x² − 9 + 2x + 3
(b) 3x² + x − 1x + 1
Q7. Express as a single fraction in its simplest form:
(a) x + 2x² − 4 + x − 1x² − 2x
(b) 4x² − 6x + 9 − 3x − 3
Q8. Express as a single fraction in its simplest form:
(a) 2xx² − 5x + 3x − 5
(b) 1x² + x − 1x² − 1
Q9. Simplify completely:
x² − 3x2x + 10 × x² + 8x + 15x² − 9Q10. Simplify completely:
2x² − 7x − 49x² − 1 ÷ 2x + 16x − 2Q11. Simplify completely:
x² − 4y²3x² + 5xy − 2y² × 6x − 2y4x + 8yQ12. Simplify completely:
(a) x² − 9x + 1 ÷ x − 3x² + x
(b) 2x + 6x² − 4 ÷ x + 3x − 2
Q13. Simplify completely:
(a) x² + xx² − 1 ÷ x + 1x − 1
(b) 3x² − 6xx² − 9 ÷ 2xx + 3
Q14. Simplify completely:
(a) x² − 4y²x + 2y ÷ x − 2yx + 2y
(b) 4x² − 12xyx² − 9 ÷ 2xx + 3
| Q1 | (3x − 4)(3x + 4) ÷ 2x(3x − 4) = 3x + 42x |
| Q2 | (x + 4)(x − 3) ÷ (3x + 2)(x − 3) = x + 43x + 2 |
| Q3 | common denominator (3x − 2)(x + 4): [2(x + 4) + 5(3x − 2)] ÷ [(3x − 2)(x + 4)] = 17x − 2(3x − 2)(x + 4) |
| Q4 | x² − 25 = (x − 5)(x + 5): [3x − 2(x + 5)] ÷ [(x − 5)(x + 5)] = x − 10x² − 25 |
| Q5 | (x − 4)(x + 2) and (x + 3)(x + 2) → LCM = (x − 4)(x + 3)(x + 2): [4(x + 3) − 3(x − 4)] ÷ [(x − 4)(x + 3)(x + 2)] = x + 24(x − 4)(x + 3)(x + 2) |
| Q6(a) | x² − 9 = (x − 3)(x + 3): [1 + 2(x − 3)] ÷ [(x − 3)(x + 3)] = 2x − 5x² − 9 |
| Q6(b) | x² + x = x(x + 1): [3 − x] ÷ [x(x + 1)] = 3 − xx(x + 1) |
| Q7(a) | (x − 2)(x + 2) and x(x − 2) → LCM = x(x − 2)(x + 2): [x(x + 2) + (x − 1)(x + 2)] ÷ [x(x − 2)(x + 2)] = (2x² + 3x − 2) → 2x − 1x(x − 2) (cancel x + 2) |
| Q7(b) | x² − 6x + 9 = (x − 3)²: [4 − 3(x − 3)] ÷ [(x − 3)²] = 13 − 3x(x − 3)² |
| Q8(a) | x² − 5x = x(x − 5): [2x + 3x] ÷ [x(x − 5)] = 5x − 5 (cancel x) |
| Q8(b) | x(x + 1) and (x − 1)(x + 1) → LCM = x(x + 1)(x − 1): [(x − 1) − x] ÷ [x(x + 1)(x − 1)] = −1x(x² − 1) |
| Q9 | x(x − 3) ÷ [2(x + 5)] × (x + 3)(x + 5) ÷ [(x − 3)(x + 3)] — cancel (x − 3), (x + 5), (x + 3) → x2 |
| Q10 | (2x + 1)(x − 4) ÷ [(3x − 1)(3x + 1)] × [2(3x − 1) ÷ (2x + 1)] — cancel (2x + 1), (3x − 1) → 2(x − 4)3x + 1 |
| Q11 | x² − 4y² = (x − 2y)(x + 2y); 3x² + 5xy − 2y² = (3x − y)(x + 2y); 6x − 2y = 2(3x − y); 4x + 8y = 4(x + 2y) — cancel (3x − y), one (x + 2y) → x − 2y2(x + 2y) |
| Q12(a) | x² − 9 = (x − 3)(x + 3); x² + x = x(x + 1): [(x − 3)(x + 3) ÷ (x + 1)] × [x(x + 1) ÷ (x − 3)] — cancel (x − 3), (x + 1) → x(x + 3) |
| Q12(b) | 2x + 6 = 2(x + 3); x² − 4 = (x − 2)(x + 2): [2(x + 3) ÷ ((x − 2)(x + 2))] × [(x − 2) ÷ (x + 3)] — cancel (x + 3), (x − 2) → 2x + 2 |
| Q13(a) | x² + x = x(x + 1); x² − 1 = (x − 1)(x + 1): [x(x + 1) ÷ ((x − 1)(x + 1))] × [(x − 1) ÷ (x + 1)] — cancel (x + 1), (x − 1) → xx + 1 |
| Q13(b) | 3x² − 6x = 3x(x − 2); x² − 9 = (x − 3)(x + 3): [3x(x − 2) ÷ ((x − 3)(x + 3))] × [(x + 3) ÷ 2x] — cancel x, (x + 3) → 3(x − 2)2(x − 3) |
| Q14(a) | x² − 4y² = (x − 2y)(x + 2y): [(x − 2y)(x + 2y) ÷ (x + 2y)] × [(x + 2y) ÷ (x − 2y)] — cancel (x − 2y), one (x + 2y) → x + 2y |
| Q14(b) | 4x² − 12xy = 4x(x − 3y); x² − 9 = (x − 3)(x + 3): [4x(x − 3y) ÷ ((x − 3)(x + 3))] × [(x + 3) ÷ 2x] — cancel x, (x + 3) → 2(x − 3y)x − 3 |