The length is 4 units less than 3 times the width. The perimeter is 22 units more than twice the width.

L = length

w = width

p = perimeter

perimeter = 2L + 2W

L = 3w-4

p = 2w+22

so

2w+22 = 2 (3w-4) + 2w

2w+22 = 6w-8+2w

2w+22 = 8w-8

2w+30 = 8w

30=6w

w = 30/6 = 5

width = 5 units

Length = 3w-4 = 3*5 = 15-4 = 11

perimeter = 2w+22 = 2*5 = 10+22 = 32

check: p = 2L + 2w = 2 (11) + 2 (5) = 22+10 = 32

Let:l = lengthw = widthp = perimeter

Using the information given we can formulate some equations:l = 3w - 4 [1]p = 2w + 22 [2]Because it is a rectangle:p = 2w + 2l [3]

Using the [2] and [3], we can create another equation by setting them equal to each other, we can do this because they are both equivalent to the perimeter (p); So:2w + 22 = 2w + 2lNow, rearange and solve:2l = 22l = 11We can plug this value of l into [1] and solve for w: (11) = 3w - 43w = 15w = 5We can plug this value of w into [2] to get the perimeter:p = 2 (5) + 22 = 10 + 22 = 32

To summarise:

l = 11

w = 5

p = 32