let 'a' represent the area, 'l' represent the length and 'w' represent the width

area = length * width

a=lw, so l = a/w or w = a/l

a+200 = (l+8)(w+3)

a+255 = (l+3)(w+8)

substitute 'w' with 'a/l':

a+200 = (l+8)(a/l + 3)

solving for 'a' we have:

a = -(l(3l-176))/8

substituting 'a' with "(l(3l-176))/8" and 'w' with 'a/l':

(l(3l-176))/8 + 255 = (l+3)(((l(3l-176))/8)/l + 8)

which solves to l=24

returning to "a+200 = (l+8)(a/l + 3)", we solve for 'a':

a+200 = (24+8)(a/24 + 3)

which solves to a=312

w = 312/24 = 13

for the original rectangle, the width is 13, the length is 24 and the area is 312

It is assumed that we are looking for the original dimensions of the rectangle, which we will call 'w' for width and 'l' for length. From the given, we can write:

(l+8)(w+3)=lw+200

(if the length of a rectangle is increased by 8 feet and its width by 3 feet its area will be increased by 200 square feet.)

(l+3)(w+8)=lw+255

(if its length is increased by 3 feet and its width by 8 feet its area will be increased by 255 square feet.)

These two equations are written using the general formula for area of a rectangle=lw.

Using Algebrator to solve for 'l' and 'w', we get:

l=24 ft

w=13 ft