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15 (a) Multiply out - OCR - GCSE Maths - Question 15 - 2018 - Paper 2

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15 (a) Multiply out. (3x - 2y)(x + y) Give your answer in its simplest form. (b) 3(2x + d) + c(x + 5) = 10x + 17 Work out the value of c and the value of d. (c... show full transcript

Worked Solution & Example Answer:15 (a) Multiply out - OCR - GCSE Maths - Question 15 - 2018 - Paper 2

Step 1

Multiply out. (3x - 2y)(x + y)

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Answer

To multiply the two binomials, we apply the distributive property:

  1. First, multiply the first term of the first binomial by both terms of the second:

    • 3xx=3x23x \cdot x = 3x^2
    • 3xy=3xy3x \cdot y = 3xy
  2. Next, multiply the second term of the first binomial by both terms of the second:

    • 2yx=2xy-2y \cdot x = -2xy
    • 2yy=2y2-2y \cdot y = -2y^2
  3. Now, combine all these results:

    • 3x2+3xy2xy2y23x^2 + 3xy - 2xy - 2y^2
  4. Finally, simplify by combining like terms:

    • 3x2+(3xy2xy)2y2=3x2+xy2y23x^2 + (3xy - 2xy) - 2y^2 = 3x^2 + xy - 2y^2

Thus, the final answer in simplest form is:

3x2+xy2y23x^2 + xy - 2y^2

Step 2

Work out the value of c and the value of d.

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Answer

To solve the equation:

3(2x+d)+c(x+5)=10x+173(2x + d) + c(x + 5) = 10x + 17

  1. Expand the left-hand side:

    • 6x+3d+cx+5c=10x+176x + 3d + cx + 5c = 10x + 17
  2. Combine like terms:

    • Grouping xx terms: (6+c)x+(3d+5c)=10x+17(6 + c)x + (3d + 5c) = 10x + 17
  3. From the equation, we can deduce the coefficients:

    • For the xx terms: 6+c=106 + c = 10 gives us:
      • c=106=4c = 10 - 6 = 4
    • For the constant terms: 3d+5c=173d + 5c = 17:
      • Substitute c=4c = 4:
      • 3d+20=173d + 20 = 17
      • This simplifies to:
      • 3d=1720=33d = 17 - 20 = -3;
      • Therefore, d=1d = -1

Thus, the values are:

  • c=4c = 4
  • d=1d = -1

Step 3

Solve by factorising. x² - 7x + 10 = 0

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Answer

To solve the quadratic equation by factorising, we need to find two numbers that multiply to 10 and add up to -7.

  1. The two numbers that satisfy these conditions are -5 and -2.

  2. Therefore, we can factorise the quadratic as:

    • (x5)(x2)=0(x - 5)(x - 2) = 0
  3. Setting each factor to zero provides the solutions:

    • x5=0x=5x - 5 = 0 \Rightarrow x = 5
    • x2=0x=2x - 2 = 0 \Rightarrow x = 2

Thus, the solutions to the equation are:

  • x=5x = 5 or x=2x = 2

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