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Math Help - Area Problem

  1. #1
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    Area Problem

    Hello,
    The Area of a rectangle is 360 square cm, if 12 is added to the length of the figure, and 5 is substracted from the width, the resulting Area is the same. What are the dimensions of the first rectangle?

    Tried to solve it using a system of 2 equations but I'm confused about the "square cm".

    Thanks in advance.
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  2. #2
    MHF Contributor alexmahone's Avatar
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    Quote Originally Posted by norifurippu View Post
    Hello,
    The Area of a rectangle is 360 square cm, if 12 is added to the length of the figure, and 5 is substracted from the width, the resulting Area is the same. What are the dimensions of the first rectangle?

    Tried to solve it using a system of 2 equations but I'm confused about the "square cm".

    Thanks in advance.
    Let the length of the first rectangle be x.
    Let the width of the first rectangle be y.

    xy=360

    (x+12)(y-5)=360

    xy-5x+12y-60=360

    -5x+12y-60=0

    5x-12y+60=0

    5x-12*\frac{360}{x}+60=0

    x-12*\frac{72}{x}+12=0

    x^2+12x-864=0

    (x+36)(x-24)=0

    x=-36 or x=24

    Rejecting the negative root,

    x=24 cm

    y=\frac{360}{24}=15 cm

    The dimensions of the first rectangle are 24 cm and 15 cm.
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  3. #3
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    Thanks a lot!
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