SOLUTION: the length of a rectangle is 7 feet less than twice its width and its perimeter is 160 feet
what is the area in square footage
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-> SOLUTION: the length of a rectangle is 7 feet less than twice its width and its perimeter is 160 feet
what is the area in square footage
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You can put this solution on YOUR website! Let L = the length of the long sides
Let W = the length of the short sides
Let A = the area of the rectangle
Let P = the perimeter of the rectangle
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A = L * W
P = 2L + 2W
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Given:
L = 2W - 7
P = 160
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Substitution allows us to consider our system of 2 equations with two unknowns into 1 equation with one unknown. We already have L given in terms of W, so substitute that into the equation for P:
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P = 160
P = 2L + 2W
160 = 2(2W - 7) + 2W
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Combine like terms and simplify:
160 = 4W - 14 + 2W
160 + 14 = 4W + 2W
174 = 6W
174/6 = W
W = 29
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go back to the equation for L:
L = 2W - 7
L = 2(29) - 7
L = 51
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the dimensions of the rectangle are:
L = 51 ft
W = 29 ft
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The problem asks for the area of the rectangle:
A = 51 ft * 29 ft
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A = 1,479 ft^2
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Check your work:
P = 2L + 2W
P = 160
P = 2(51) + 2(29)
P = 102 + 58
P = 160 checks
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Cheers,
Lee
