Create an equation that results in one extraneous solution

⎮x⎮= 8 - 3x


x = 8 - 3x . . . . . . . . (1)
-x = 8 - 3x . . . . . . . .(2)
From (1), 4x = 8
x = 2
From (2), 2x = 8
x = 4

Putting x = 2 in the original equation:
⎮2⎮= 8-3(2)
2 = 8 - 6
2 = 2

Putting x = 4 in the original equation:
⎮4⎮= 8-3(4)
4 = 8 - 12
4 = -4

Therefore, the valid solution is x = 2, and the extraneous solution is x = 4