Пусть х - количество марок в начале
х - 6 - 0.25(х - 6) - 2 = (2/3)х
х - 8 - 0.25х - 1.5 = (2/3)х
0.75х - 9.5 = (2/3)х
(3/4)х - (2/3)х = 9.5
(1/12)х = 9.5
х = 114
Пусть х - количество марок в начале
х - 6 - 0.25(х - 6) - 2 = (2/3)х
х - 8 - 0.25х - 1.5 = (2/3)х
0.75х - 9.5 = (2/3)х
(3/4)х - (2/3)х = 9.5
(1/12)х = 9.5
х = 114