SOLUTION: if the dimension of a rectangle are such that the length is 3 inches more than the width. if the length were doubled and the width decrease by 1 inch, the area would be 50in square
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Question 366601: if the dimension of a rectangle are such that the length is 3 inches more than the width. if the length were doubled and the width decrease by 1 inch, the area would be 50in square. what is the length and the width Answer by amoresroy(361) (Show Source):
You can put this solution on YOUR website! Let L = the length of rectangle in inches
W = the width of rectangle in inches
L = W + 3
2L(W-1) = 50
Substitute L=W+3 in 2nd equation to express the 2nd equation in terms of W
2(W+3)(W-1) = 50
(2W+6)(W-1) = 50
2W^2-2W+6W-6-50=0
Combine like terms and divide by 2
W^2+2W-28=0
Solve by quadratic formula
W = -2 +/- [4-4(1)(-28)]/2
W = 4.385
L = 4.385 +3 = 7.385
