Yes, that's a good one. I've been long thinking of doing one about the absolute value questions but have forgot about it.
Absolute values
The way to solve this kind of questions is to break the equation (inequality) into two parts, one is when the value is non negative, the other is when the value is negative.
For example:
|x-4|<9
You break it into two parts:
If x-4>=0, then x-4<9, solve for both you get x>=4, x<13. So your solution is 4<=x<13.
If x-4<0, then -(x-4)<9, ie x-4>-9. Solve for both you get x<4, x>-5. So your solution for this part is -5<x<4.
Combine the two solutions, you get -5<x<13 as your final solution.
Another example:
|x+4|>4
If x+4>=0, then x+4>4. Solve for both you get x>=-4, x>0. So your solution is x>0.
If x+4<0, then -(x+4)>4, ie. x+4<-4. Solve for both you get x<-4, x<-8. So your solution is x<-8.
You final solution is x>0 or x<-8.
The same strategy can apply to square questions.
For example: (x+4)^2>4
You could solve it this way:
x^2+8x+12>0
(x+2)(x+6)>0
x>-2 or x<-6
Or you can solve it this way:
If x+4>=0 then x+4>2. Solve for them you get x>-2.
If x+4<0 then x+4<-2. Solve for them you get x<-6.
|y|>|y+1|
if y>=0, y+1>=0, y>y+1, no solution.
if y<0, y+1<0, -y>-(y+1), solution is y<-1
if y>=0, y+1<0, y>-(y+1), no solution.
if y<0, y+1>=0, -y>y+1, solution is -1<=y<-1/2
So your final solution is y<-1/2
You could also solve this question by going the square route.
y^2>(y+1)^2
y^2>y^2+2y+1
2y+1<0
y<-1/2