Question

A swimming 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?

Answer 1

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
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16+4y + 4x +xy -xy = 140
4x + 4y = 124
x + y = 31
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now we have system:
x-y = 5
x+y = 31
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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