You have:
. . . . .
The insides of the absolute values change signs at x = -1/2 and at x = 3, so you have three intervals; namely,
and
Consider each interval separately. If x < -1/2, then 2x + 1 < 0 and 3 - x > 0, so you have:
. . . . . . . . . . . . . . . . . . . .
Since x = -4 is less than -1/2, then this solution is valid:
.
If -1/2 < x < 3, then 2x + 1 > 0 and 3 - x > 0, so:
. . . . . . . . . . . . . . .
Then the inequality works for
.
Now check when x > 3:
. . . . . . . . . .
But x cannot be both greater than three and less than minus four, so there is no solution on this interval.
. . . . .
This can be confirmed by graphing, by the way. For further information, try
here.
. . . . .