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A contractor needs 54 square feet of brick to construct a rectangular walkway. The length of the?

A contractor needs 54 square feet of brick to construct a rectangular walkway. The length of the walkway is 15 feet more than the width. Write an equation that could be used to determine the dimensions of the walkway. Solve this equation to find the length and width, in feet, of the walkway.

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1. L = W + 15 L * W = 54 (W + 15) * W = 54 W^2 + 15W = 54 W^2 + 15W - 54 = 0 W = (-15 +/- sqrt(15^2 - 4 * 1 * -54)) / (2 * 1) W = (-15 +/- sqrt(225 + 216)) / 2 W = (-15 +/- sqrt(441)) / 2 W = (-15 +/- 21) / 2 W = 6/2 or W = -36/2 W = 3 or W = -18 Obviously W cannot equal -18, so W = 3 and L = 18.
2. Width = W L = W + 15 Area, A = 54 sq. ft. LW = A (W + 15)W = 54 W² + 15W = 54 W² + 15W - 54 = 0 (W + 18)(W - 3) = 0 The dimensions are 3 ft. x 18 ft. ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
3. length = width + 15 Area = length * width; area is 54 square feet. Two equations are: L = W + 15 L * W = 54 To solve, we substitute first equation into second equation: (W + 15) * W = 54 w^2 + 15w = 54 w^2 + 15w - 54 = 0 (w+18)(w-3) = 0 w = -18 or 3 The width cannot be negative, so 3 is the width and since length is 15 more than width, length is 18: Length = 18 feet Width = 3 feet