Absolute value in Linear programming problem I have this problem, where I have to reformulate it as a linear programming problem.  

Minimize   $ 2x_1 +3 \vert {x_2-10}\vert $

subject to $\vert {x_1+2}\vert + \vert {x_2}\vert ≤5  $

The subject is new to me. Can anyone help?
Alternatively, the constraint $\vert {x_1+2}\vert + \vert {x_2}\vert ≤5$ corresponds to:
$$\pm(x_1+2)\pm x_2 \le 5 \iff 
\begin{cases}\ \ \ \ \ x_1+2+x_2\le 5\\
-(x_1+2)+x_2\le 5\\
\ \ \ \ \ x_1+2-x_2\le 5\\
-(x_1+2)-x_2\le 5\\
\end{cases}$$
The feasible region is:

$\hspace{3cm}$

Since $x_2&lt;10$, then the original problem can be reformulated as:
$$\text{Minimize} \ \ 2x_1+3(10-x_2) \ \ \text{subject to:}\\
 \begin{cases}x_1+x_2\le 3\\
x_1-x_2\ge -7\\
x_1-x_2\le 3\\
x_1+x_2\ge -7\\
\end{cases}$$
The answer is: $f(-2,5)=11$. To be verified: Original and Reformulated.
Minimize $2x_1 +3 \vert {x_2-10}\vert$ subject to $\vert {x_1+2}\vert + \vert {x_2}\vert ≤5$.


It can be converted to $6$ linear programming problems:
$$1) \ \begin{cases}2x_1-3(x_2-10) \to \text{min}\\ -(x_1+2) -(x_2)≤5\\
\color{red}{x_1\le -2}\\
x_2\le0\end{cases} \ \ \ \ 
4) \ \begin{cases}2x_1-3(x_2-10) \to \text{min}\\ (x_1+2) -(x_2)≤5\\
\color{blue}{x_1\ge -2}\\
x_2\le0\end{cases} \\ 
2) \ \begin{cases}2x_1-3(x_2-10) \to \text{min}\\ -(x_1+2) +(x_2)≤5\\
\color{red}{x_1\le -2} \\
0\le x_2\le10 \end{cases} \ \ \ \ 
5) \ \begin{cases}2x_1-3(x_2-10) \to \text{min}\\ (x_1+2) +(x_2)≤5\\
\color{blue}{x_1\ge -2}\\
0\le x_2\le10\end{cases}\\\\
3) \ \begin{cases}2x_1+3(x_2-10) \to \text{min}\\ -(x_1+2) +(x_2)≤5\\
\color{red}{x_1\le -2}\\
10\le x_2 \end{cases}\ \ \ \ 
6) \ \begin{cases}2x_1+3(x_2-10) \to \text{min}\\ (x_1+2) +(x_2)≤5\\
\color{blue}{x_1\ge -2}\\
10\le x_2 \end{cases}\\$$