│x-a│； =2m- 1

get

:m = & lt2

And because a = {-3 ≤ x ≤ 4}, b = {x | 2m- 1 ≤ x ≤ m+ 1}, and b is included in a.

That is to say, a contains B.

Therefore, the upper and lower bounds of a must be less than and greater than the upper and lower bounds of b.

So: m+ 1 =

So: m =

So:-1 = < m = < 2