I was hoping someone could give a little more insight, or perhaps enlighten me to a better way of approaching solving these seemingly simple Algebra 2 inequalities. I did some google searching but I was not able to find the answers I seek. The problem came up when a friend of mine had an equation that looked like this: |x-3| > -2 + |2x + 8| Question 1: What is the generally accepted best way to solve this inequality. My method of solving involved setting up 4 equalities: 1) x-3 = -2 + 2x + 8 2) -(x-3) = -2 + 2x + 8 3) x-3 = -2 - (2x + 8) 4) -(x-3) = -2 - (2x + 8) These yielded the following values for x: 1) x= -9 2) x = -1 3) x = -7/3 4) x = -13 With each of these x values I then went back and plugged into the original inequality. My goal was to check for equality. Only 2) and 4) resulted in equality so I threw the other answers out. Once I had these two solutions I tested the points 0, -5, and -20 in the original inequality. And since -5 was the only value that yielded a true statement, I declared the solution: (technically I had checked -7/3 and -9, checking -5 was unnecessary I realize now) -13 < x < -1 Question 2: Can someone give me more information on why there are extraneous solutions? As I understand things, when you have: |x| = A You need to solve x = A and x = -A When you have: |x| = |y| You need to solve x = y, x = -y, -x = y and -x = -y, but because x = -y, -x = y are the same equation and x = y, -x = -y are the same equation you only need to solve two equations. When you have: |x| = A + |y| You need to solve x = A + y, x = A - y, -x = A + y, and -x = A - y, since these are all different, you need to solve all four. Essentially the prior case (|x| = |y|) is a special case of this, just two of the equations you need to solve happen to be the same. Question 3: Is there a more rigorous way of handling these equations? Should I be using |x| = Sqrt(x^2) here? Thank you in advance.