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Find the equation of a line parallel to 2y = 3(x − 2) and passing through the point (2, 3).

Mathematics

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3 parts
20

C. y=32x\text{C. } y = \frac{3}{2}x

21

B. {0,−1}\text{B. } \{0, -1\}

22

5y2+24y+165y^2 + 24y + 16

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Part 19

The diagram for this question is not visible in the provided image. Therefore, it is not possible to find the value of xx.


Part 20

C. y=32x\text{C. } y = \frac{3}{2}x

  • Given: A line parallel to 2y=3(x−2)2y = 3(x - 2) and passing through the point (2,3)(2, 3).
  • To find: The equation of this line.
  1. Find the slope of the given line.

    First, rewrite the given equation in slope-intercept form (y=mx+cy = mx + c).

    2y=3(x−2)2y = 3(x - 2) 2y=3x−62y = 3x - 6

    Divide by 2:

    y=32x−3y = \frac{3}{2}x - 3

    The slope of this line is m1=32m_1 = \frac{3}{2}.

  2. Determine the slope of the new line.

    Since the new line is parallel to the given line, their slopes are equal.
    So, the slope of the new line is m2=32m_2 = \frac{3}{2}.

  3. Use the point-slope form to find the equation of the new line.

    The new line passes through (x1,y1)=(2,3)(x_1, y_1) = (2, 3) and has a slope m=32m = \frac{3}{2}.
    The point-slope form is y−y1=m(x−x1)y - y_1 = m(x - x_1).

    y−3=32(x−2)y - 3 = \frac{3}{2}(x - 2)
  4. Convert the equation to slope-intercept form.

    y−3=32x−32×2y - 3 = \frac{3}{2}x - \frac{3}{2} \times 2 y−3=32x−3y - 3 = \frac{3}{2}x - 3

    Add 3 to both sides:

    y=32x−3+3y = \frac{3}{2}x - 3 + 3 y=32xy = \frac{3}{2}x
  5. Compare with the given options.

    The calculated equation matches option C.

    The final answer is C. y=32x\boxed{\text{C. } y = \frac{3}{2}x}.


Part 21

B. {0,−1}\text{B. } \{0, -1\}

  • Given: The expression 5x+36x(x+1)\frac{5x+3}{6x(x+1)}.
  • To find: The values of xx for which the expression is undefined.
  1. Identify the condition for an expression to be undefined.

    A rational expression is undefined when its denominator is equal to zero.

  2. Set the denominator to zero and solve for xx.

    The denominator is 6x(x+1)6x(x+1).

    6x(x+1)=06x(x+1) = 0

    For this product to be zero, at least one of the factors must be zero.

    Case 1: Set the first factor to zero.

    6x=06x = 0 x=06x = \frac{0}{6} x=0x = 0

    Case 2: Set the second factor to zero.

    x+1=0x+1 = 0 x=−1x = -1
  3. List the values of xx.

    The expression is undefined when x=0x = 0 or x=−1x = -1.
    The set of values is {0,−1}\{0, -1\}.

  4. Compare with the given options.

    The calculated set matches option B.

    The final answer is B. {0,−1}\boxed{\text{B. } \{0, -1\}}.


Part 22

5y2+24y+165y^2 + 24y + 16

  • Given:
    • A man is five times as old as his son.
    • In four years time, the product of their ages would be 340.
    • The son's current age is yy.
  • To find: Express the product of their ages in terms of yy.
  1. Define current ages in terms of yy.

    Son's current age = yy
    Man's current age = 5y5y

  2. Define ages in four years time.

    Son's age in four years = y+4y + 4
    Man's age in four years = 5y+45y + 4

  3. Express the product of their ages in four years in terms of yy.

    Product of their ages in four years = (y+4)(5y+4)(y + 4)(5y + 4)

  4. Expand the expression.

    (y+4)(5y+4)=y(5y)+y(4)+4(5y)+4(4)(y + 4)(5y + 4) = y(5y) + y(4) + 4(5y) + 4(4) =5y2+4y+20y+16= 5y^2 + 4y + 20y + 16 =5y2+24y+16= 5y^2 + 24y + 16

    The product of their ages in terms of yy is 5y2+24y+165y^2 + 24y + 16. This expression is equal to 340 according to the problem statement.

    The final answer is 5y2+24y+16\boxed{5y^2 + 24y + 16}.

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