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If $3x + 2|y| + y = 7$ and $x + |x| + 3y = 1,$ then $x + 2y$ is

1. $\frac{8}{3}$
2. $1$
3. $– \frac{4}{3}$
4. $0$

## 1 Answer

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 $:\text{D}$

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