Q:

The perimeter of a rectangle is 86 cm. twice the width exceeds the length by 2cm. write a system and solve to find the dimensions of the rectangle.

Accepted Solution

A:
There are two equation of the system
(1) The perimeter
     2(w + l) = p
     2(w + l) = 86
(2) The length and the width
      2w = l + 2

From the second equation
2w = l + 2
l + 2 = 2w
l = 2w - 2

Subtitute l with 2w - 2 in the first equation to find the width
2(w + l) = 86
2(w + 2w - 2) = 86
2(3w - 2) = 86
6w - 4 = 86
6w = 90
w = 15

Now, find the length by subtituting w with 15
l = 2w - 2
l = 2(15) - 2
l = 30 - 2
l = 28

The dimension of the rectangle
length = 28 cm
width = 15 cm