SOLUTION: The length of a rectangle is five more than twice the width. a) If the area is 18 square feet, find the dimensions of the rectangle. b) If the perimeter is 25 meters, find the dime

Algebra ->  Customizable Word Problem Solvers  -> Misc -> SOLUTION: The length of a rectangle is five more than twice the width. a) If the area is 18 square feet, find the dimensions of the rectangle. b) If the perimeter is 25 meters, find the dime      Log On

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Question 1061101: The length of a rectangle is five more than twice the width. a) If the area is 18 square feet, find the dimensions of the rectangle. b) If the perimeter is 25 meters, find the dimensions of the rectangle.
Answer by Cromlix(4381) About Me  (Show Source):
You can put this solution on YOUR website!
Hi there,
Make width = x
Length = 5 + 2x
Area = Length * Width (* = times)
18 = (5 + 2x)x
18 = 5x + 2x^2
Rearrange
2x^2 + 5x - 18 = 0
Factorize
(2x + 9)(x - 2) = 0
2x + 9 = 0
2x = -9
x = -9/2 (Disregard as -ve)
x - 2 = 0
x = 2
Width = 2 ft
Length = 9 ft.
b)Using same dimensions
Make width = x
Length = 5 + 2x
Perimeter = 2 x Length + 2 x Width
25 = 2(5 + 2x) + 2(x)
25 = 10 + 4x + 2x
Rearrange
6x = 25 - 10
6x = 15
x = 15/6
x = 2.5
Width = 2.5 m
Length = 10 m
Hope this helps :-)