We have to go through slowly and case by case in order to get every solution, then check for extraneous ones. First, we take off the outermost set of absolute value bars, leaving the equations $$ \left\textbackslash{\begin{array}{c} x+\textbar x+2\textbar{} x\textbar\textbar=1 \textbackslash\textbackslash{} x+\textbar x+2\textbar{} x\textbar\textbar=-1 \end{array}\right. $$ Now, moving the $x$ over and taking off the outermost absolute value bars on both equations leaves us with the equations $$ \left\textbackslash{\begin{array}{c} x+2\textbar x\textbar=1-x \textbackslash\textbackslash{} x+2\textbar x\textbar=-1+x \textbackslash\textbackslash{} x+2\textbar x\textbar=-1-x \textbackslash\textbackslash{} x+2\textbar x\textbar=1+x \end{array}\right. $$ Moving the $x$ over in every equation gives $$ \left\textbackslash{\begin{array}{c} 2\textbar x\textbar=1-2 x \textbackslash\textbackslash{} 2\textbar x\textbar=-1 \textbackslash\textbackslash{} 2\textbar x\textbar=-1-2 x \textbackslash\textbackslash{} 2\textbar x\textbar=1 \end{array}\right. $$ Obviously, $2\textbar x\textbar=-1$ cannot happen, so we can throw out that equation. Now, removing the absolute value bars on the three remaining equations gives us $$ \left\textbackslash{\begin{array}{c} 2 x=1-2 x \textbackslash\textbackslash{} 2 x=-1+2 x \textbackslash\textbackslash{} 2 x=-1-2 x \textbackslash\textbackslash{} 2 x=1+2 x \textbackslash\textbackslash{} 2 x=1 \textbackslash\textbackslash{} 2 x=-1 \end{array}\right. $$ Obviously, $2 x=-1+2 x$ and $2 x=1+2 x$ are not real equations, so we can throw those out leaving us with four final equations. Solving each one, we get that $x=\frac{1}{2},-\frac{1}{2}, \frac{1}{4}$, and $-\frac{1}{4}$. Plugging in all those values to the original equation tells us that only $\frac{1}{4}$ is a real solution while the others are all extraneous, so there is only one real solution. 
How many real solutions does $\textbar x+\textbar{} x+2\textbar x\textbar\textbar\textbar=1$ have? 

(A) $$1$$ 
(B) $$2$$ 
(C) $$4$$ 
(D) $$8$$ 