1/3 wins and 2/3 losses
given 2 and 5 = 2/6 = 1/3
1, 3, 4, and 6 = 4/6 = 2/3
Both the x and y have to be greater than zero so you need to add two inequalities, x>0, y>0
No, you should have reduced each base to...
3^(3*2x) = 3^(2*(x-3))
and then you can cancel each base
3*2x = 2*(x-3)
6x = 2x-6
4x= -6
x = -3/2
