[ (x + 3)/4] = (x - 2)/3
(x - 2)/3 = k -> x-2 = 3k -> x = 3k + 2
k <= (x+3)/4 < k + 1
k <= (3k +5)/4 < k + 1
4k <= 3k + 5 -> k <= 5 -> , k € (- infinit , 5]
(3k + 5)/4 < (k + 1 ) -> 3k + 5 < 4k + 4 , -k < -1 , k> 1
k € ( 1 , + infinit)
k€ { 2 , 3 ,4 , 5}
k = 2 , x = 8
k = 3 , x = 11
k = 4 , x = 14
k = 5 , x = 17