Discussion

Explanation:

Case 1: x, y > 0
⇒ 3x + 3y = 7 and 2x + 3y = 1
Solving these two equations we get, 
x = 6 and y = -11/3
This is rejected as y should be positive.

Case 2: x > 0, y < 0
⇒ 3x - y = 7 and 2x + 3y = 1
Solving these two equations we get, 
x = 2 and y = -1
This is accepted.
∴ x + 2y = 2 + 2 × -1 = 0

Case 3: x < 0, y > 0
⇒ 3x + 3y = 7 and 3y = 1
Solving these two equations we get, 
x = 2 and y = 1/3
This is rejected as x should be negative.

Case 4: x, y < 0
⇒ 3x - 3y = 7 and 3y = 1
Solving these two equations we get, 
x = 8/3 and y = 1/3
This is rejected as and y should be negative.

Hence, option (c).

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