Let's check the options
A.
2x - 1 = 2
2x= 3
x= 3/2=1.5
option A has atleast one solution
B
3y+ 1 = 3y
option B has no solution
C.
5p - (3 + p) = 6p + 1
5p - 3 - p = 6p + 1
4p - 6p = 1 + 3
-2p = 4
p =-2
option C has atleast one solution
D.
4/5 m = 1- 1/5 m
4/5 m + 1/5m = 1
1m = 1
m = 1
Option D has atleast one solution
E.
10 + 0.5w = 1/2w - 10
0.5 w - 1/2 w = -10 - 10
option E has no solution
F.
4a + 3(a-2) = 8a - (6+a)
4a +3a - 6 = 8a -6 - a
7a -6 = 7a - 6
option F has many solution. Hence it also has atleast one solution
Therefore;
option A, C, D and F has atleast one solution