Mixed Algebraic Problems — Solutions

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1. Mixed algebra techniques (set 1)

   1. 4(3x − 2) + 5x: 12x − 8 + 5x = 17x − 8
   2. (x + 6)(x − 2): x² + 4x − 12
   3. Factorise 12x² − 8x: 4x(3x − 2)
   4. Factorise x² − 49: (x + 7)(x − 7)
   5. Solve 5x + 3 = 28: 5x = 25 → x = 5
   6. Solve 3(2x − 1) = 21: 2x − 1 = 7 → 2x = 8 → x = 4
   7. x² − 5x + 6 = 0: (x − 2)(x − 3) = 0 → x = 2 or x = 3
   8. 2x² − 3x + 1, x = −2: 2(4) − 3(−2) + 1 = 8 + 6 + 1 = 15

2. Mixed algebra techniques (set 2)

   1. (2x + 3)(3x − 5): 6x² − 10x + 9x − 15 = 6x² − x − 15
   2. Factorise x² − 3x − 10: (x − 5)(x + 2)
   3. Solve (2x + 1) ÷ 3 = 5: 2x + 1 = 15 → 2x = 14 → x = 7
   4. Make y the subject of 3x + 2y = 12: 2y = 12 − 3x → y = (12 − 3x) ÷ 2
   5. (x + 4)² − (x − 3)²: x² + 8x + 16 − (x² − 6x + 9) = 14x + 7
   6. Factorise 4x² − 36: 4(x² − 9) = 4(x + 3)(x − 3)
   7. Solve x² + x − 12 = 0: (x + 4)(x − 3) = 0 → x = −4 or x = 3
   8. A = πr², r = 4: A = 16π ≈ 50.3 cm²

3. Connecting techniques

   1. x² + 7x + 12: (x + 3)(x + 4), so possible dimensions are (x + 3) and (x + 4)
   2. Expand and apply to x² + 9x + 20: (x + a)(x + b) = x² + (a+b)x + ab. For x² + 9x + 20: a + b = 9, ab = 20 → a = 4, b = 5. So (x + 4)(x + 5).
   3. y = 3x + 2 → x = (y − 2) ÷ 3; y = 11: x = (11 − 2) ÷ 3 = 9 ÷ 3 = 3
   4. (x + 5)² − (x + 2)² = 21: x² + 10x + 25 − (x² + 4x + 4) = 6x + 21. Set 6x + 21 = 21 → 6x = 0 → x = 0

4. Justify and explain

   1. (x + 3)² error: The student forgot the middle term. Correct: (x + 3)² = x² + 6x + 9
   2. x² − 4x = 0 error: x(x − 4) = 0 gives x = 0 or x = 4. The student ignored x = 0 (the null factor law applies to both factors).
   3. A = ½bh error: Cannot subtract ½b from A. Correct: 2A = bh → h = 2A ÷ b
   4. (2x + 3)(2x − 3) difference of squares: (2x)² − 3² = 4x² − 9. When x = 5: 4(25) − 9 = 100 − 9 = 91

5. Multi-step problem solving

   1. Rectangular garden:
      1. Length: l = 2w + 3
      2. Equation: 2(2w + 3 + w) = 42 → 2(3w + 3) = 42 → 6w + 6 = 42 → 6w = 36 → w = 6 m
      3. Length and area: l = 2(6) + 3 = 15 m. Area = 15 × 6 = 90 m²
   2. Consecutive odd integers:
      1. Equation: n(n + 2) = 143
      2. Expand: n² + 2n − 143 = 0
      3. Solve: (n + 13)(n − 11) = 0 → n = 11 (taking positive). The integers are 11 and 13.
   3. Ball h = −5t² + 15t:
      1. Factorise: h = 5t(3 − t) or −5t(t − 3)
      2. h = 0: 5t(3 − t) = 0 → t = 0 or t = 3. Ball hits the ground at t = 3 s.
      3. t = 1.5: h = −5(2.25) + 15(1.5) = −11.25 + 22.5 = 11.25 m. This is the midpoint of the flight, suggesting the ball reaches maximum height at t = 1.5 s (axis of symmetry).

6. Identify technique and solve (set 3)

   1. (3x − 4)² + 2(x + 1): Expanding. 9x² − 24x + 16 + 2x + 2 = 9x² − 22x + 18
   2. Factorise 2x³ − 8x: HCF. 2x(x² − 4) = 2x(x + 2)(x − 2)
   3. Solve 3x ÷ (x − 2) = 6: Multiply both sides by (x − 2): 3x = 6(x − 2) → 3x = 6x − 12 → −3x = −12 → x = 4
   4. V = &frac43;πr³, make r then find r when V = 113.1: r³ = 3V ÷ (4π) = 3(113.1) ÷ (4 × 3.14) = 339.3 ÷ 12.56 ≈ 27.0 → r = ³√27 = 3.0 cm

7. Error detection and correction

   1. (x − 5)² error: Student forgot the middle term. Correct: (x − 5)² = x² − 10x + 25
   2. x² + 5x + 6 error: Need a + b = 5 and ab = 6, so a = 2, b = 3. Correct: (x + 2)(x + 3)
   3. 2x² = 50 error: Missing x = −5. x² = 25 gives x = ±5. Both solutions are valid.
   4. s = ut + ½at², make u error: Student did not divide by t. Correct: s − ½at² = ut → u = (s − ½at²) ÷ t

8. Choose and chain techniques

   1. Side length of square area = x² + 10x + 25:
      1. Factorise: (x + 5)², so side = (x + 5) cm
      2. x = 3: Side = 8 cm; perimeter = 4 × 8 = 32 cm
   2. Area = x² + x − 6, width = (x − 2):
      1. Factorise: (x + 3)(x − 2), so length = (x + 3) m
      2. x = 5: Area = 25 + 5 − 6 = 24 m²; dimensions: width = 3 m, length = 8 m; 3 × 8 = 24 ✓
   3. (x + 3)² − 9 = x(x + 6): (x + 3)² − 9 = x² + 6x + 9 − 9 = x² + 6x = x(x + 6) ✓. Setting = 0: x = 0 or x = −6.

9. Real-world multi-step

   1. Picture frame:
      1. Total dimensions: (x + 6) wide × (x + 10) tall: Total area = (x + 6)(x + 10)
      2. Expand: x² + 10x + 6x + 60 = x² + 16x + 60
      3. x = 10: 100 + 160 + 60 = 320 cm²
   2. n + n² = 42:
      1. Equation: n² + n = 42
      2. Rearrange and factorise: n² + n − 42 = 0 → (n + 7)(n − 6) = 0
      3. Solutions: n = 6 or n = −7
   3. h = 80 − 5t²:
      1. h = 0: 80 − 5t² = 0 → 5t² = 80 → t² = 16 → t = 4 s
      2. Make t the subject: 5t² = 80 − h → t² = (80 − h) ÷ 5 → t = √((80 − h) ÷ 5)
      3. h = 35: t = √((80 − 35) ÷ 5) = √(45 ÷ 5) = √9 = 3 s

10. Strategy selection and justification

    1. (2x + 5)² − (2x − 5)²: DOPS: (2x + 5 + 2x − 5)(2x + 5 − 2x + 5) = (4x)(10) = 40x. When x = 3: 40 × 3 = 120.
    2. 3x² − 27: Factorise: 3(x² − 9) = 3(x + 3)(x − 3). Setting = 0: x = ±3.
    3. (x + 2) ÷ 4 = (2x − 1) ÷ 3: Cross-multiply: 3(x + 2) = 4(2x − 1) → 3x + 6 = 8x − 4 → −5x = −10 → x = 2
    4. A = π(R² − r²):
       1. Factorise: R² − r² = (R + r)(R − r)
       2. Make R the subject: A ÷ π = R² − r² → R² = A ÷ π + r² → R = √(A ÷ π + r²)
       3. A = 75.4, r = 3: R = √(75.4 ÷ 3.14 + 9) = √(24 + 9) = √33 ≈ 5.7 cm