Solving Equations And Inequations Based On The Absolute Function Today I came across some equations and inequations based on the absolute function.
These were 


*

*$|x^2+4x+3|+2x+6=0$

*$|x^2+6x+7|=|x^2+4x+4|+|2x+3|$

*$1\le |x-1|\le 3$

*$\frac{2}{|x-4|}\gt 1$

*$||x|-1|\le 1$

*$|x+1|\gt |2x-1|$


I am really new to these kinds of problems.So can someone please show me how to solve at least a few of them so that I can learn the method to deal with sums like these ? Thank You.Even hints will be gladly appreciated :-)!!
 A: If you have only one absolute value in the problem you can get it to the form $|a|>b$ or $|a|<b$, these are easy to solve from the definition of absolute value. For example if $|x-4|<2$ (fourth problem) then $-2<x-4<2$ and $2<x<6$.
However, if there are more than one absolute value, for every absolute value $|a|$ there are two options: $a\geq 0 $ and then $|a|=a$ or $a\leq 0 $ and then $|a|=-a$. So basically you ignore the absolute value but remember there are a couple cases.
For example:
$$|x+1|>|2x-1|$$
If $x\geq\frac{1}{2}$ then both $x+1 , 2x-1$ are positives and we get $x+1>2x-1 \Rightarrow x<2 $. So one solution is $ \frac{1}{2} \leq x < 2 $.
If $-1\leq x \leq \frac{1}{2} $ then $x+1$ is positive and $2x-1$ is negative, so we get $x+1>1-2x \Rightarrow 3x>0 \Rightarrow x>0$ and we obtain another solution: $0<x\leq \frac{1}{2} $
If $x\leq -1$ then both are negative and we need to solve $-x-1>1-2x \Rightarrow x>2$ but this is not possible because $x\leq -1$
In conclusion, $ 0< x \leq \frac{1}{2}\ \operatorname{or} \frac{1}{2}\leq x <2   $ which is equivalent to $0<x<2$.
A: Equations and inequations like this are only a few steps more complex than their "absolute"-less counterparts.
For your first example:
$|x^2+4x+3|+2x+6=0$  
You can regard this equation as two different cases:


*

*$x^2+4x+3+2x+6=0$ and  $x^2+4x+3\ge0$  

*$-(x^2+4x+3)+2x+6=0$ and $x^2+4x+3<0$


You solve each of these separately. When you calculate the interval of x, you see if it fits the solution. If it doesn't, discard it. 
The final set of solutions is made of all valid solutions.
Same principle for every similar problem.
