Question 567162
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:


*[tex \LARGE |x-2| = 5-3 = 2]


Now we can take positive and negative solutions:


*[tex \LARGE x-2 = 2] or *[tex \LARGE x-2 = -2], 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


*[tex \LARGE |x-2| > 2], this can either be *[tex \LARGE x-2 > 2] or *[tex \LARGE x-2 < 2] (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.