algebra
posted 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

algebra 
Henry
Eq1: L*W = 425Ft^2.
Eq2: 2L + 2W = 84Ft.
L + W = 42.
L e 42W,
In Eq1, substitute 42W for L:
(42W)W = 425,
42W  W^2  425 = 0,
W^2 42W + 425 = o,
(W25)(W17) = 0,
W25 = 0,
W = 25.
W17 = 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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