2x-10>3x+a, -10-a>x.
If -10-a=5, a=-15, so 2(x-5)>3x-15, and x<5. We need x≤5, however.
If -10-a=6, a=-16 and x<-10-a, so x<6 which includes all values of x≤5, as well as 5<x<6. In fact, a<-15 is the general solution for a such that x≤5. For example, if a=-15.01, -10-a=5.01, 5.01>x, or x<5.01 which includes x≤5.