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Math Help - XY variables vary directly........

  1. #1
    Newbie n_duncan2010's Avatar
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    XY variables vary directly........

    once... again... worksheet. The variables x y vary directly. Use the given values to write an equation that relates x and y


    x=8 , y =24

    x=3 , y=36

    I have no idea... how to do this
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by n_duncan2010 View Post
    once... again... worksheet. The variables x y vary directly. Use the given values to write an equation that relates x and y


    x=8 , y =24

    x=3 , y=36

    I have no idea... how to do this
    if x and y vary directly, it means that one is a constant times the other, that is:

    x = ky

    for the first:
    x = 8, y = 24
    => 8 = 24k
    => k = 1/3
    so x = (1/3)y
    or better yet:
    y = 3x

    for the second:
    x = 3, y = 36
    => 3 = 36k
    => k = 1/12
    so x = (1/12)y
    or y = 12x
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  3. #3
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    I interpret the question to mean derive the equation of the line with those two points:

    y = mx +c

    m is the gradient = (y1-y2)/(x1-x2) = (24-36)/(8-3) = -2.4

    y = -2.4x + c

    Substitute in one of the (x,y) pairs to find c:

    24 = -2.4(8) + c => c = 24 + 19.2 = 43.2

    y = 43.2 - 2.4x
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by jl5000 View Post
    I interpret the question to mean derive the equation of the line with those two points:

    y = mx +c

    m is the gradient = (y1-y2)/(x1-x2) = (24-36)/(8-3) = -2.4

    y = -2.4x + c

    Substitute in one of the (x,y) pairs to find c:

    24 = -2.4(8) + c => c = 24 + 19.2 = 43.2

    y = 43.2 - 2.4x
    well, your way seems to make sense, but i believe it was two different questions for the following reason:

    By definition, we say x and y vary directly if:
    x = ky

    I have never seen a definition of direct variation, or varying directly that accounts for the possibility of a lone constant. that is,

    x = ky + c is not direct variation by definition

    and so i assumed it was two separate problems. plus, n_duncan2010 never said anything about me putting down two answers, so i guess it was really two problems
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  5. #5
    Senior Member ecMathGeek's Avatar
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    I would assume that same as you, Jhevon, but n_duncan2010 didn't give very clear instructions.
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