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. 0.3 •4 is the same as 4 groups of 0.32. 7•0.8 is the same as 7 groups of 0.83. it increases by 0.5 every time, so 0.5x4 it increases by 0.9 every time, so 0.9x5. 0.9•12=10.86. 0.47•5=2.35i can finish the next ones in a separate post