Let the nomber of 2 pointers be x and that of 3 pointers be y, then
2x + 3y = 30 . . . (1)
x = y + 5 . . . (2)
Putting (2) into (1), gives 2(y + 5) + 3y = 30
2y + 10 + 3y = 30
5y = 30 - 10 = 20
y = 20/5 = 4
x = 4 + 5 = 9
Therefore, there are nine 2-pointers and five 3-pointers
