SOLUTION: If you have a problem like |x-2| + 3 = 5 you would solve it with a positive and negative equation changing the signs of both the 5 and the 3. If you had an inequality |x-2| + 3 > 5

Algebra ->  Absolute-value -> SOLUTION: If you have a problem like |x-2| + 3 = 5 you would solve it with a positive and negative equation changing the signs of both the 5 and the 3. If you had an inequality |x-2| + 3 > 5      Log On


   

Question 567162: If you have a problem like |x-2| + 3 = 5 you would solve it with a positive and negative equation changing the signs of both the 5 and the 3. If you had an inequality |x-2| + 3 > 5 your positive and negative equations would not include changing the sign of the 3. Why is that?
|x-2| + 3 = 5
x + 1 = 5
x=4
|x-2| - 3 = -5
x - 5 = -5
x=0
|x-2| + 3 > 5
x + 1 > 5
x>4
|x-2| + 3 < -5
x + 1 < -5
x<-6
Answer by richard1234(7193) About Me  (Show Source):
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.