# 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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