Remember, the perimeter is just the sum of all the sides of the rectangle.
So in any rectangle, L + W + L + W = P which simplifies to 2L+ 2W = P.
In this rectangle, we know that P = 22, so 2L + 2W = 22.
The problem also tells us that the length is 5 units larger then the width. This means that L = W + 5, so we can replace L in the perimeter equation with W + 5.
2(W+5) + 2W = 22
2W + 10 + 2W = 22
4W + 10 = 22
4W = 12
W = 3
L = 3 + 5 = 8
The dimensions are 3 by 8.