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Math Help - YR10 Simaltaneous Equations

  1. #1
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    YR10 Simaltaneous Equations

    Hello,

    I cant figure out how to how to write an equation for this question.


    -A rectangular house has a perimeter of 40 meters and the length is 4 meters more than the width. What are the dimensions of the house?


    please help, im really confused about this and i have a test on it on Tuesday.

    TIA.

    cheers
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  2. #2
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    Let l = the length, w = the width.
    If the length is 4 more than the width, then l = w+4 | w we do not know.
    The parameter is given to be 40 so, you have 2 lengths and 2 widths.
    w + (w+4) + w + (w+4) = 40
    4w+8 = 40
    4w = 32
    w = 8

    So the width (w) is 8, since the length (l) is 4 more the width (w) it must be 12. 8 + 12 + 8 12 = 40
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  3. #3
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    i was looking for something like this

    Perimeter = 40

    x (length), y (width)
    therefore eqn(1) = 2x +2y = 40

    Length is 4 more than width
    therefore eqn(2) = x = y + 4

    so your 2 eqns are:
    2x +2y = 40
    x = y + 4

    put them together and you should get y=8 and x=12
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  4. #4
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    Quote Originally Posted by ktmboy View Post
    i was looking for something like this

    Perimeter = 40

    x (length), y (width)
    therefore eqn(1) = 2x +2y = 40

    Length is 4 more than width
    therefore eqn(2) = x = y + 4

    so your 2 eqns are:
    2x +2y = 40
    x = y + 4

    put them together and you should get y=8 and x=12
    Why bother asking a question if in the space of 10 minutes you tell someone who has taken the time to show you what to do that you prefer doing it a different way.
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