algebra

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the area of a rectangle is 425 square feet. if the perimeter is 84 feet, find the length and the width of the rectangle

  • algebra -

    Eq1: L*W = 425Ft^2.

    Eq2: 2L + 2W = 84Ft.

    L + W = 42.
    L e 42-W,

    In Eq1, substitute 42-W for L:
    (42-W)W = 425,
    42W - W^2 - 425 = 0,
    W^2 -42W + 425 = o,
    (W-25)(W-17) = 0,

    W-25 = 0,
    W = 25.

    W-17 = 0,
    W = 17.

    In Eq1 substitute 17 for W:
    L*17 = 425,
    L = 25 Ft., W = 17.

    When L = 17, W = 25.

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