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    714_valladolid Posts: 1, Reputation: 1
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    #1

    Jun 4, 2008, 06:13 PM
    Algebra 1
    The perimeter of the deck on the Morton's house is 150 feet. The length of the deck is 15 feet more than the width. What are the dimensions of the deck?
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    ebaines Posts: 12,131, Reputation: 1307
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    #2

    Jun 5, 2008, 07:33 AM
    Set up two equations in two unknowns. Let W = the width of the deck and L equal the length. From what you are given you know that:

    Perimeter is 150 ft, so 2W + 2L = 150.
    Length is 15 feet more than the width, so L = W+15

    Can you solve from here?
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