A rectanglular window has a width 1' less than its length. The perimeter is 18'. Find the length and width.

What I have so far is......

Let x = length

P=lw

18=2(2x-1) -1

am I on the right track? Thanks in advance!

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- Oct 10th 2008, 11:56 AMtpomaAm I on the right track here??
A rectanglular window has a width 1' less than its length. The perimeter is 18'. Find the length and width.

What I have so far is......

Let x = length

P=lw

18=2(2x-1) -1

am I on the right track? Thanks in advance! - Oct 10th 2008, 12:02 PMChop Suey
NO!

Perimeter is the sum of the side lengths of a polygon, and it is not the area.

A rectangle has 4 sides: 2 sides considered as width and 2 sides considered as lengths.

You can then conclude that the perimeter is $\displaystyle P = 2w + 2l$

The problem also mentions that the width is one inch less than length. From this, you can conclude that:

$\displaystyle w = l - 1$

Given P, solve for l, and then find w using the given relation.