Aswimming pool is 5m longer than it is wide and is surrounded by a deck 2m wide. the area of the pool and deck together is at least 140 m squared greater than the area of the pool alone. what can you conclude about the dimensions of the pool?

Let x be length of pool and y be width of pool.
we can write system of equations

x - y = 5
(4+x)*(4+y) - x*y = 140

16+4y + 4x +xy -xy = 140
4x + 4y = 124
x + y = 31

now we have system:
x-y = 5
x+y = 31

2x = 36
x = 18
y = 13

This is calculated for minimum difference in surfaces. Because of equation 
x+y = 31 which we got from second condition we can see that for greater difference between surfaces  x+y will be greater which means that dimensions of pool are at least 18 and 13

(1/x + 1)^2 + 6 (1/x + 1) +5