│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