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Secondary 3 Elementary Mathematics Algebra Functions Quiz

Free Sec 3 E Maths Algebra Functions quiz, Gemma31B Exam version, with questions, answers, and O Level-style practice for Singapore students.

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Secondary 3 Elementary Mathematics From Real Exams Generated by Gemma 4 31B Updated 2026-09-16
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Secondary 3 Elementary Mathematics Quiz - Algebra Functions (Answers)

  1. (3x−7)(3x+7)(3x - 7)(3x + 7) [2]
  2. (3a−b)(x−2y)(3a - b)(x - 2y) [2]
  3. 4(x+1)−2(x−3)(x−3)(x+1)=4x+4−2x+6(x−3)(x+1)=2x+10(x−3)(x+1)\frac{4(x+1) - 2(x-3)}{(x-3)(x+1)} = \frac{4x+4-2x+6}{(x-3)(x+1)} = \frac{2x+10}{(x-3)(x+1)} [3]
  4. (2x+1)(x−3)=0⇒x=−0.5,x=3(2x + 1)(x - 3) = 0 \Rightarrow x = -0.5, x = 3 [2]
  5. x=−8±64−442=−8±202≈−1.76,−6.24x = \frac{-8 \pm \sqrt{64 - 44}}{2} = \frac{-8 \pm \sqrt{20}}{2} \approx -1.76, -6.24 [3]
  6. 3x−12<5x+2⇒−14<2x⇒x>−73x - 12 < 5x + 2 \Rightarrow -14 < 2x \Rightarrow x > -7 [2]
  7. 2x≤6⇒x≤32x \le 6 \Rightarrow x \le 3 AND 3x>−6⇒x>−23x > -6 \Rightarrow x > -2. Solution: −2<x≤3-2 < x \le 3 [3]
  8. 3x(x+1)=5(x−2)⇒3x2+3x=5x−10⇒3x2−2x+10=03x(x+1) = 5(x-2) \Rightarrow 3x^2 + 3x = 5x - 10 \Rightarrow 3x^2 - 2x + 10 = 0. Discriminant D=4−120=−116D = 4 - 120 = -116. No real solutions. [3]
  9. (a) (x−3)2−19(x-3)^2 - 19 [2] (b) (3,−19)(3, -19) [1]
  10. (a) (−5,0)(-5, 0) and (3,0)(3, 0) [2] (b) x=−5+32⇒x=−1x = \frac{-5+3}{2} \Rightarrow x = -1 [2]
  11. Part 1: x+3<2x−1⇒x>4x+3 < 2x-1 \Rightarrow x > 4. Part 2: 2x−1≤4x+103⇒6x−3≤4x+10⇒2x≤13⇒x≤6.52x-1 \le \frac{4x+10}{3} \Rightarrow 6x-3 \le 4x+10 \Rightarrow 2x \le 13 \Rightarrow x \le 6.5. Solution: 4<x≤6.54 < x \le 6.5 [4]
  12. (a) (11−y)(11+y)(11-y)(11+y) [2] (b) y=11,y=−11y = 11, y = -11 [2]
  13. 2x−(x+3)x+3=4x+3⇒x−3=4⇒x=7\frac{2x - (x+3)}{x+3} = \frac{4}{x+3} \Rightarrow x-3 = 4 \Rightarrow x = 7 [3]
  14. (a) x=0⇒y=22−5=−1x=0 \Rightarrow y = 2^2 - 5 = -1. Point (0,−1)(0, -1) [2] (b) (x+2)2=5⇒x+2=±5⇒x=−2±5(x+2)^2 = 5 \Rightarrow x+2 = \pm\sqrt{5} \Rightarrow x = -2 \pm \sqrt{5} [3]
  15. x=−10±100−(−24)6=−10±1246x = \frac{-10 \pm \sqrt{100 - (-24)}}{6} = \frac{-10 \pm \sqrt{124}}{6}. x1≈−3.52,x2≈0.188x_1 \approx -3.52, x_2 \approx 0.188 [4]
  16. (2x+3)(x−1)=54⇒2x2+x−3=54⇒2x2+x−57=0(2x+3)(x-1) = 54 \Rightarrow 2x^2 + x - 3 = 54 \Rightarrow 2x^2 + x - 57 = 0. (2x+19)(x−3)=0(2x+19)(x-3) = 0. Since xx must be positive, x=3x = 3 [4]
  17. 3(3x−4)+(x−2)(2x+1)(2x+1)(3x−4)=9x−12+2x2+x−4x−2(2x+1)(3x−4)=2x2+6x−14(2x+1)(3x−4)\frac{3(3x-4) + (x-2)(2x+1)}{(2x+1)(3x-4)} = \frac{9x-12 + 2x^2+x-4x-2}{(2x+1)(3x-4)} = \frac{2x^2+6x-14}{(2x+1)(3x-4)} [4]
  18. Part 1: 2x−5≤3x+3⇒x≥−82x-5 \le 3x+3 \Rightarrow x \ge -8. Part 2: x+1<x+72⇒2x+2<x+7⇒x<5x+1 < \frac{x+7}{2} \Rightarrow 2x+2 < x+7 \Rightarrow x < 5. Solution: −8≤x<5-8 \le x < 5 [4]
  19. (a) (1,9)(1, 9) [1] (b) 0=−(x−1)2+9⇒(x−1)2=9⇒x−1=±3⇒x=4,x=−20 = -(x-1)^2 + 9 \Rightarrow (x-1)^2 = 9 \Rightarrow x-1 = \pm 3 \Rightarrow x = 4, x = -2. Points: (4,0),(−2,0)(4, 0), (-2, 0) [3]
  20. x(x+1)+(x−3)(x−2)x(x−2)=3⇒x2+x+x2−5x+6x2−2x=3⇒2x2−4x+6=3x2−6x⇒x2−2x−6=0\frac{x(x+1) + (x-3)(x-2)}{x(x-2)} = 3 \Rightarrow \frac{x^2+x + x^2-5x+6}{x^2-2x} = 3 \Rightarrow 2x^2-4x+6 = 3x^2-6x \Rightarrow x^2-2x-6 = 0. x=2±4+242=2±282=1±7x = \frac{2 \pm \sqrt{4+24}}{2} = \frac{2 \pm \sqrt{28}}{2} = 1 \pm \sqrt{7} [4]