SOLUTION: a rectangle has a perimeter of 15 inches. The length is 6 more than twice the width. Find the length and width of the rectangle

Question 496384: a rectangle has a perimeter of 15 inches. The length is 6 more than twice the width. Find the length and width of the rectangle

Found 2 solutions by rabrox, cleomenius:
Answer by rabrox(2) (Show Source): You can put this solution on YOUR website!
Okay, so first we have to set up the variables. Since the length is expressed in terms of the width in the problem, we'll keep it like that in solving the problem. So:
let w= the width
let l= 2w+6
Next, we solve for the perimeter of the rectangle as usual: twice the width + twice the length = the perimeter- and let's plug in our length and width.
2w + 2*(2w+6) =15
After you distribute, it should come out to this: 2w+4w+12= 15
Then you combine like terms (add the w's, and subtract 12 from each side of the equation): 6w= 3. Then divide each side by 6, so w= 1/2.
To get the length, plug it in: 2*(1/2)+6= 7
Width= 1/2 inches
length= 7 inches
Answer by cleomenius(959) (Show Source): You can put this solution on YOUR website!
W = x
L = 2x + 6
2x + 4x + 12 = 15
6x = 3
x = 1/2 The width.
Length = 7.
2w = 1 inches.
2L = 14 inches.
This does check.
Cleomenius

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