If the point m (1 + A, 2A + 2) is in the third quadrant, the value range of a is obtained It's a process

If the point m (1 + A, 2A + 2) is in the third quadrant, the value range of a is obtained It's a process

If yes, please see solution a, if not, please see solution B
Solution A:
∵ about the x-axis
The ordinate is the opposite number of the ordinate of point M
∵ in the third quadrant
∴1-a＜0 -2a-2＜0
The solution is a ＞ 1, a ＞ - 1
∴a＞1
Solution B:
∵ about the x-axis
The ordinate is the opposite number of the ordinate of point M
∵ in the third quadrant
∴1+a＜0 -2a-2＜0
The solution is a ＜ - 1, a ＞ - 1
The original inequality system has no solution
There is no such point, and the value of a does not exist