Given that,
- $3x + 2|y| + y = 7 \quad \longrightarrow (1) $
- $x + |x| + 3y = 1 \quad \longrightarrow (2) $
We know that, $|x| = \left\{\begin{matrix} x\;; &x \geq 0 \\ -x \;;& x<0 \end{matrix}\right.$
$\textbf{Case 1:} \; x \geq 0\;;\; y\geq 0$
$\begin{array}{} 3x+3y=7 \\ 2x + 3y = 1\\ \; – \qquad – \quad\; – \\\hline \boxed{x=6}, \; \boxed{ {\color{Red}{y= \frac{-11}{3}\; (\text{rejected})}}} \end{array}$
$\textbf{Case 2:} \quad x \geq 0 \; ;\; y< 0$
$\begin{array}{} (3x – y = 7) \times 3 \\ 2x + 3y =1 \\\hline 9x - 3y = 21 \\ 2x + 3y = 1 \\\hline 11x = 22\; (\text{Adding the equations)} \end{array}$
$\boxed{x=2}, \quad \boxed{y= -1}$
$\therefore$ The value of $x + 2y = 2-2 =0.$
$\textbf{Case 3:} \quad x<0 \; ; y\geq 0 $
Now, $3y=1 \Rightarrow \boxed{y = \frac{1}{3}}$
And, $3x + 3y =7$
$\Rightarrow 3x+1=7$
$\Rightarrow \boxed{{\color{Red} {x=2\;\text{(rejected)}}}}$
$\textbf{Case 4:} \quad x<0 \; ; y< 0 $
Now, $3y=1 \Rightarrow \boxed{{\color{Red} {y = \frac{1}{3}\;\text{(rejected)}}}}$
And, $3x - y =7$
$\Rightarrow 3x-\frac{1}{3}=7$
$\Rightarrow 3x=7+\frac{1}{3}$
$\Rightarrow 3x=\frac{22}{3} \Rightarrow \boxed{{\color{Red} {x=\frac{22}{9}\;\text{(rejected)}}}}$
Correct Answer $:\text{D}$