SOLUTION: The perimeter of a rectangle is 132 meters. The length is 9 meters more than twice the width. What are the dimensions.

Question 452193: The perimeter of a rectangle is 132 meters. The length is 9 meters more than twice the width. What are the dimensions.
Answer by Leaf W.(135) (Show Source): You can put this solution on YOUR website!
width: x
"The length is 9 meters more than twice the width."
"twice the width": 2x
"9 meters more than twice the width": 2x + 9
Therefore, length: 2x + 9
The perimeter or a rectangle is denoted as 2(length) + 2(width). With the afore-said values of length and width, the perimeter is 2(2x + 9) + 2(x). Since the problem states that the perimeter is 132 meters:
2(2x + 9) + 2x = 132
Solve for x. Distribute the 2 into the 2x and 9:
4x + 18 + 2x = 132
Add like terms (4x and 2x):
6x + 18 = 132
Subtract 18 from both sides:
6x = 114
Divide both sides by 6:
x = 19
So, the width is 19 meters. To find the length, plug this value (19) in for x in the expression for length (2x + 9):
2x + 9
2(19) + 9
38 + 9
47
So, the length is 47 meters.
***Therefore, the length of the rectangle is 47 meters and the width of the rectangle is 19 meters.
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