The length of a rectangle is 4 yd longer than its width. if the perimeter of the rectangle is 32 yd, find its area.

Hello: 
let : x the length  and : y  the width
2(x+y) the perimeter and ; xy  the area.
x = y +4...(1)
2(x+y) = 32 ...(2)
subsct x by (1) in (2) :
2(y+4 +y) = 32
2y +8 +2y = 32
4y =24
y = 24/4 = 6 yd
x= 6+4=10 yd 
the area is : 60 yd²