Equation with absolute value and parameter How to solve this equation over real numbers with parameter $p \in \Bbb R$?
$$(1 - p)(\left\lvert x + 2 \right\rvert + \left\lvert x \right\rvert) = 4 - 3p$$ 
I know how to solve absolute value equations without parameter.
 A: We have
$$|x|+|x+2|=\frac{4-3p}{1-p}\tag1$$
Case 1 : For $x\gt 0$, 
$$(1)\iff x+(x+2)=\frac{4-3p}{1-p}\iff x=\frac{2-p}{2(1-p)}$$
Here, $\frac{2-p}{2(1-p)}\gt 0\iff p\lt 1\quad\text{or}\quad p\gt 2$. 
Case 2 : For $-2\lt x\le 0$,
$$(1)\iff -x+(x+2)=\frac{4-3p}{1-p}\iff p=2$$
Case 3 : For $x\le -2$,
$$(1)\iff -x-(x+2)=\frac{4-3p}{1-p}\iff x=\frac{6-5p}{2(p-1)}$$
Here, $\frac{6-5p}{2(p-1)}\le -2\iff p\lt 1\quad\text{or}\quad p\ge 2$. 
So, the answer is the followings : 


*

*For $p\lt 1\ \text{or}\ p\gt 2$, $x=\frac{2-p}{2(1-p)},\frac{6-5p}{2(p-1)}$.

*For $p=2$, $-2\le x\le 0$.

*For $1\le p\lt 2$, there is no such $x$.
A: HINT....use the definition of $|x|$ in the form 
$$|x|= \begin{cases} x &x\geq 0 \\-x & x<0 \end{cases}$$, So that you can split the equation into three separate cases. So, for example, when $x\geq 0$, the equation becomes $$(1-p)(2x+2)=4-3p$$
A: I am sensing a homework question here :P
Anyways.
Just remember,
|x| = x , when x>=0 & 
|x| = -x , when x<0
So , just change the equation accordingly to these cases and solve it.
For x<0 , above equation becomes
(1-p)(-x-2-x) = (4 - 3p) hence, it becomes
(1-p)(-2x - 2) = (4 - 3p)
Now , for the case x>=0 , I am leaving it to you as homework for trying it out yourself. I hope this helps.
