Question 516171: The length of a rectangle is 5 more than it's width. the perimeter is 9 times more then the width. what are the dimensions of the rectangle
Found 3 solutions by algebrahouse.com, gc2000, Maths68:
Answer by algebrahouse.com(1659)   (Show Source):
You can put this solution on YOUR website! x = width
x + 5 = length {length is 5 more than width}
9x = perimeter {perimeter is nine times width}
Perimeter of a rectangle is 2(width) + 2(length}
2x + 2(x + 5) = 9x {perimeter is 2(width) + 2(length)}
2x + 2x + 10 = 9x {used distributive property}
4x + 10 = 9x {combined like terms}
10 = 5x {subtracted 4x from both sides}
x = 2 {divided both sides by 5}
x + 5 = 7 {substituted 2, in for x, into x + 5}
width = 2
length = 7
For more help from me, visit: www.algebrahouse.com
Answer by gc2000(22)  (Show Source):
You can put this solution on YOUR website! The length of a rectangle is 5 more than it's width. the perimeter is 9 times more then the width. what are the dimensions of the rectangle?
The key is often to find 2 equations to solve, and to isolate variables
let x=length, w=width, p = perimeter
p = 2x + 2w
x = w + 5
substitute for x
p = 2(w + 5) + 2w
p = 2w + 10 + 2w
p = 4w + 10 [this is the first equation, simplified]
p = 9w [this is the second equation, which was a given]
substitute 9w for p in first equation:
9w = 4w + 10
subtract 4w from both sides
5w = 10
divide both sides by 5
w = 2
therefore since x = w + 5,
x = 2 + 5
x = 7
The dimensions are length = 7, width = 2
plug dimensions back in to check:
length is width plus 5 (yes)
perimeter is 9w (7 + 7 + 2 + 2 = 18 = 9(2), yes)
Answer by Maths68(1474)  (Show Source):
You can put this solution on YOUR website! Length = l=w+5
Width = w=?
Perimeter = p = 9w
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Perimeter of a Rectangle = length + length + width + width
Perimeter of a Rectangle =2 length +2 width
P = 2l + 2w
9w=2(w+5)+ 2w
9w=2w+10+2w
9w=4w+10
9w-4w=10
5w=10
5w/5=10/5
w=2
============================
Dimensions of the Rectangle
Length = l=w+5 = 2+5 =7
Width = w=2
Perimeter = p = 9w = 9*2=18
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