SOLUTION: The length of a rectangle is 1 in. more than twice its width. If the perimeter of the rectangle is 26 in., find the width of the rectangle. I have four choices (4 in, 5 in, 6 in,

Algebra ->  Algebra  -> Customizable Word Problem Solvers  -> Finance -> SOLUTION: The length of a rectangle is 1 in. more than twice its width. If the perimeter of the rectangle is 26 in., find the width of the rectangle. I have four choices (4 in, 5 in, 6 in,      Log On

Ad: Over 600 Algebra Word Problems at edhelper.com
Ad: Algebrator™ solves your algebra problems and provides step-by-step explanations!
Ad: Algebra Solved!™: algebra software solves algebra homework problems with step-by-step help!

   

Question 127441: The length of a rectangle is 1 in. more than twice its width. If the perimeter of the rectangle is 26 in., find the width of the rectangle. I have four choices (4 in, 5 in, 6 in, or 3 in.). I came up with 6 inches but this doesn't feel right. Is this correct? Thanks so much.
Answer by Earlsdon(6103) About Me  (Show Source):
You can put this solution on YOUR website!
Let the width be W and the length be L.
The perimeter of a rectangle is given by:
P+=+2L%2B2W and this, you are told, is 26 inches.
You also know that the length L is 1 more than twice the width, W, so you can write:
L+=+2W%2B1 Now substitute this into the formula for the perimeter to get:
26+=+2%282W%2B1%29%2B2W Simplify and solve for W.
26+=+4W%2B2%2B2W
26+=+6W%2B2 Subtract 2 from both sides.
24+=+6W Divide both sides by 6.
4+=+W
The width is 4 inches.
Check:
P+=+2L%2B2W
P+=+2%282W%2B1%29%2B2W Substitute W = 4.
P+=+2%282%284%29%2B1%29%2B2%284%29
P+=+2%289%29%2B8
P+=+18+%2B+8
P+=+26 It checks!