Question

The perimeter of a rectangular field is 360yd. The length is 20yd longer than the width. Find the dimensions

Answer 1

Write an equation system based on the problem
# The perimeter is 360 yg
2l + 2w = 360
# The length is 20 yd longer than the width
l = 20 + w

Solve the equation system, substitute 20+w as l to the first equation to find the value of w
2l + 2w = 360
2(20+w) + 2w = 360
40 + 2w + 2w = 360
4w + 40 = 360
4w = 360 - 40
4w = 320
  w = 320/4
  w = 80
The width is 80 yd

Find the length by substituting the value of w
l = 20 + w
l = 20 + 80
l = 100
The length is 100 yd

The dimension:
l = 100 yd
w = 80 yd

Answer 2

If P=360, then 360yd / 4 sides = 90yds each side. Now that it is a rectangle, the Length should be more than the width, so if the length is 20yds more than the width, do 90 - 20 two times,  Width= [ (90yds - 20yds)*2 ] 

To sum things up, Length = 110; Width = 70. | 110*2 = 220, 70 * 2 = 140, 220+140 = 360.