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?

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)