# SOLUTION: The length of a rectangle is 2in. less than 7 times its width. the area of the rectangle is 57 sq. in. What are the dimentions of the rectangle?

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 Click here to see ALL problems on Miscellaneous Word Problems Question 385309: The length of a rectangle is 2in. less than 7 times its width. the area of the rectangle is 57 sq. in. What are the dimentions of the rectangle?Found 2 solutions by texttutoring, mananth:Answer by texttutoring(324)   (Show Source): You can put this solution on YOUR website!Let w = width Let L = length Length is 2 less than 7 times the width: L = 7w - 2 Formula for area of a rectangle is A = w*L, and we know that A = 57 Substitute the formula for L into the equation, and solve for w: A= w*L 57 = w(7w-2) 57 = 7w^2 - 2w 0 = 7w^2 - 2w -57 Use the quadratic equation (or WolframAlpha.com) to solve. You should find that w=3 This means that L is: L=7w-2 L=7(3)-2 L=21-2 L=19 So the dimensions are w=3, L=19. Answer by mananth(12270)   (Show Source): You can put this solution on YOUR website!width = x length = 7x-2 .. L*W= Area ... x(7x-2)=57 7x^2-2x=57 7x^2-2x-57=0 7x^2-21x+19x-57=0 7x(x-3)+19(x-3)=0 (x-3)(7x+19)=0 x= 3 which is positive Width = 3 in length = 7x-2 = 21-2=19 in. ... m.ananth@hotmail.ca