You can
put this solution on YOUR website!I don't know, your solution is a bit flawed. You can't really say that |x-2| + 3 is always equal to x+1, since there are negative solutions involved too.
You should move the 3 to the RHS first:
Now we can take positive and negative solutions:

or

, this yields x=4 and x=0. Here, you would change the sign of the 3, but I wouldn't do that; you're more likely to make a mistake this way.
For the inequality, we have

, this can either be

or

(since we can have a negative solution -x-2 > 2, multiplying by -1 reverses the direction of the inequality). The solutions are x > 4 and x < 0.