algebra
posted by Alex .
The length of a rectangular field is 18 m longer than the width. The field is enclosed with fencing and divided into two parts with a fence parallel to the shorter sides. If 216 m of fencing are required, what are the dimensions of the outside rectangle?

From your description, I am not sure that the fencing is around the whole perimeter of the field. If it is:
L = W + 18
2L + 2W = 216
2(W + 18) + 2W = 216
Solve for W, then put value in the first equation to solve for L. Check by inserting both values into the second equation. 
A rectangular field is 3 yards longer than twice its width. If the perimeter is 156 yards, how wide is the field?