SOLUTION: The length of a rectangular piece of property is one meter more than twice the width. If the perimeter of the property is 302 meters, find the length and width

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Question 1134468: The length of a rectangular piece of property is one meter more than twice the width. If the perimeter of the property is 302 meters, find the length and width
Found 2 solutions by rothauserc, Theo:
Answer by rothauserc(4718) (Show Source):
You can put this solution on YOUR website!
let l be the length of the rectangle and w be the width of the rectangle
:
l = 2w +1
:
twice the length +twice the width = perimeter of the rectangle
:
2(2w+1) +2w = 302
:
4w +2 +2w = 302
:
6w = 300
:
w = 50
:
l = 2(50) +1 = 101
:
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length of rectangle is 101 meters
:
width of rectangle is 50 meters
:
check the answer
:
2(101) +2(50) = 302
:
202 +100 = 302
:
302 = 302
:
answer checks
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Answer by Theo(13342) (Show Source):
You can put this solution on YOUR website!
perimeter is equal to 2 times length + 2 times width.

let L = length
let W = width
let P = perimeter

P = 2L + 2W = 2 * (L + W)

you are given that the length is one meter more than twice the width.

that makes L = 2W + 1.

in the formula P = 2 * (L + W), replace L with 2W + 1 to get:

P = 2 * (2W + 1 + W)

simplify to get P = 2 * (3W + 1)

simplify further to get P = 6W + 2

you are given that the perimeter is equal to 302 meters.

that makes P = 302 which makes P = 6W + 2 which becomes 302 = 6W + 2

subtract 2 from both sides of that equation to get 300 = 6W

solve for W to get W = 300 / 6 = 50

since L = 2W + 1, then L = 101.

you have L = 101 and W = 50

P = 2 * (L + W) becomes P = 2 * (151) which becomes P = 302.

solution looks good.

solution is the length is equal to 101 and the width is equal to 50.