How to solve $\lvert{x}\rvert - \lvert{2+x}\rvert = x$? How do I solve the following equation?
$$\lvert{x}\rvert- \lvert{2+x}\rvert= x$$
I was thinking about dividing it into 4 cases: plus plus, plus minus, minus plus and minus minus. What is the best way to solve this?
 A: $$|x|-|x+2|=\left\{\begin{matrix}
 -x+(x+2)& \;\;, x\leq -2  \\\\ 
 -x-(x+2)&\;\;, -2<x<0  \\\\ 
x-(x+2) &\;\; ,x\geq 0  & 
\end{matrix}\right.$$
$\bullet\; $ If $x\leq -2\;,$ Then equation convert into $-x+x+2=x\Rightarrow x=2\;\;(\bf{False})$
$\bullet\; $ If $-2<x< 0\;,$ Then equation convert into $\displaystyle -x-x-2=x\Rightarrow x=-\frac{2}{3}\;\;(\bf{True})$
$\bullet\; $ If $x\geq 0\;,$ Then equation convert into $x-x-2=x\Rightarrow x=-2\;\;(\bf{False})$
So our solutions are $$\displaystyle x= -\frac{2}{3}$$
A: Another way is to use $|x| = \sqrt{x^2}$.  If $|x| - |2+x| = x$ then
$$\begin{align*}
\sqrt{x^2} - \sqrt{(x+2)^2} &= x\\
x^2 + (x+2)^2 -2\sqrt{x^2}\sqrt{(x+2)^2} &= x^2\\
(x+2)^2 &= 2\sqrt{x^2} \sqrt{(x+2)^2}\\
(x+2)^4 &= 4x^2(x+2)^2
\end{align*}$$
So either $x= -2$, which doesn't satisfy the original equation, or else
$$(x+2)^2 = 4x^2$$
which is a quadratic equation in $x$. This has the roots $x=2$ and $x=-2/3$. Now you just have to check whether either of these satisfies the original equation, and only $x=-2/3$ does, so this is the only solution.
A: HINT: Consider the 3 cases: 


*

*$x<-2$

*$-2\le x<0$

*$x\ge 0$

A: $$\lvert{x}\rvert + 2 \ge\lvert{2+x}\rvert$$
$$\therefore x = \lvert{x}\rvert- \lvert{2+x}\rvert \ge -2$$
$$\therefore x+2 \ge 0$$
Thus,
$$\lvert{x}\rvert- \lvert{2+x}\rvert= x$$ becomes$$\lvert{x}\rvert = 2x + 2$$
Now, just consider two cases. $x \lt 0$ and $x \ge 0$.
A: If $x<-2$ then $x + 2 < 0$. So $\lvert x \rvert - \lvert 2 + x \rvert = - x + (2+x)$. So you really just want to solve 
$$
-x + (2+x) = x.
$$
Here the solution is $x = 2$. This contradicts $x < -2$. So there are not solutions for $x<-2$.
Now if $-2\leq x < 0$, then $\lvert x \rvert = -x$ and $\lvert 2+x\rvert = 2 + x$.
If $ x \leq 0$, then $\lvert x\rvert = x$ and $\lvert 2 +x\rvert = 2+x $.

All of this comes down to the fact that 
$$
\lvert \text{something}\rvert = \begin{cases}\text{something} & \text{if } \text{something} \geq 0 \\ 
-\text{something} & \text{if } \text{something} < 0
\end{cases}
$$
