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Math Help - word problem, using factoring equation with two solving situations.

  1. #1
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    word problem, using factoring equation with two solving situations.

    The perimeter of a rectangle is 32 inches, and the area is 60 sq. inch. Find the length and width of the rectangle.
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  2. #2
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    Let a,b be the sides of the rectangle.

    We know that 2(a+b)=32 and a\cdot b=60

    Now, solve.
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  3. #3
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    Quote Originally Posted by mirdita59 View Post
    The perimeter of a rectangle is 32 inches, and the area is 60 sq. inch. Find the length and width of the rectangle.
    Let L = length of rectangle, in inches
    And, w = width, in inches

    Perimeter, P = 2L +2w
    32 = 2L +2w
    16 = L +w -------------(1)

    Area, A = w*L
    60 = wL ----------------(2)

    There are your two simultaneous equations.

    One way to continue is by substitution,
    there are many ways to do this substitution itself,
    one way is we use (1) to get the value of L in terms of w,
    L = 16-w --------------(1a)

    Substitute that into (2),
    60 = w(16-w)
    60 = 16w -w^2
    w^2 -16w +60 = 0
    Factor that,
    (w-6)(w-10) = 0
    w = 6 or 10

    When w = 6 inches,
    L = 16-w --------------(1a)
    L = 16-6
    L = 10 inches

    When w = 10 inches,
    L = 16-w --------------(1a)
    L = 16-10
    L = 6 inches
    What? L is shorter than w?
    Ignore this.

    Therefore, length = 10 inches and width = 6 inches. ----------answer.
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