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?

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(16946) (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
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