Asked by Lucia
                the area of a rectangle is 425 square feet. if the perimeter is 84 feet, find the length and the width of the rectangle
            
            
        Answers
                    Answered by
            Henry
            
    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.
    
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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