We have two equations involving absolute values: $3x + 2|y| + y = 7$ ... (1) $x + |x| + 3y = 1$ ... (2) When we see absolute values like $|x|$ and $|y|$, we need to consider different cases based on whether these variables are positive or negative.

The absolute value function behaves differently depending on the sign of the variable: If $x \geq 0$, then $|x| = x$ If $x < 0$, then $|x| = -x$ If $y \geq 0$, then $|y| = y$ If $y < 0$, then $|y| = -y$ Since we have both $|x|$ and $|y|$, we need to check all possible sign combinations.

Case I: $x \geq 0$ and $y \geq 0$ $|x| = x$ and $|y| = y$ Our equations become: $3x + 2y + y = 7$ $3x + 3y = 7$ ... (1') $x + x + 3y = 1$ $2x + 3y = 1$ ... (2') $(3x + 3y) - (2x + 3y) = 7 - 1$ $x = 6$ $2(6) + 3y = 1$ $12 + 3y = 1$ $y = -\frac{11}{3}$ We assumed $y \geq 0$, but we got $y = -\frac{11}{3} < 0$. This case is invalid.

Case II: $x \geq 0$ and $y < 0$ $|x| = x$ and $|y| = -y$ Our equations become: $3x + 2(-y) + y = 7$ $3x - 2y + y = 7$ $3x - y = 7$ ... (1') $x + x + 3y = 1$ $2x + 3y = 1$ ... (2') From equation (1'): $y = 3x - 7$ $2x + 3(3x - 7) = 1$ $2x + 9x - 21 = 1$ $11x = 22$ $x = 2$ $y = 3(2) - 7 = -1$ $x = 2 \geq 0$ and $y = -1 < 0$ This case is valid.

Case III: $x < 0$ and $y \geq 0$ $|x| = -x$ and $|y| = y$ Our equations become: $3x + 2y + y = 7$ $3x + 3y = 7$ ... (1') $x + (-x) + 3y = 1$ $3y = 1$ ... (2') From equation (2'): $y = \frac{1}{3}$ $3x + 3(\frac{1}{3}) = 7$ $3x + 1 = 7$ $x = 2$ We assumed $x < 0$, but we got $x = 2 > 0$. This case is invalid.

Case IV: $x < 0$ and $y < 0$ $|x| = -x$ and $|y| = -y$ Our equations become: $3x + 2(-y) + y = 7$ $3x - y = 7$ ... (1') $x + (-x) + 3y = 1$ $3y = 1$ ... (2') From equation (2'): $y = \frac{1}{3}$ We assumed $y < 0$, but we got $y = \frac{1}{3} > 0$. This case is invalid.

Only Case II gives us a valid solution that satisfies both the equations and the sign assumptions. $x = 2$ and $y = -1$ Therefore: $x + 2y = 2 + 2(-1) = 2 - 2 = 0$

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