1.) To solve ||x-2|=5, we need to consider two cases:i) x-2 = 5ii) x-2 = -5
i) x-2 = 5Adding 2 to both sides:x = 7
ii) x-2 = -5Adding 2 to both sides:x = -3
Therefore, the solutions to ||x-2|=5 are x = 7 and x = -3.
2.) To solve ||2x-12|+4|=8, we again consider two cases:i) 2x-12+4 = 8ii) -(2x-12)+4 = 8
i) 2x-8 = 8Adding 8 to both sides:2x = 16Dividing by 2:x = 8
ii) -(2x-12)+4 = 8-2x + 12 + 4 = 8-2x + 16 = 8Subtracting 16 from both sides:-2x = -8Dividing by -2:x = 4
Therefore, the solutions to ||2x-12|+4|=8 are x = 8 and x = 4.
1.) To solve ||x-2|=5, we need to consider two cases:
i) x-2 = 5
ii) x-2 = -5
i) x-2 = 5
Adding 2 to both sides:
x = 7
ii) x-2 = -5
Adding 2 to both sides:
x = -3
Therefore, the solutions to ||x-2|=5 are x = 7 and x = -3.
2.) To solve ||2x-12|+4|=8, we again consider two cases:
i) 2x-12+4 = 8
ii) -(2x-12)+4 = 8
i) 2x-8 = 8
Adding 8 to both sides:
2x = 16
Dividing by 2:
x = 8
ii) -(2x-12)+4 = 8
-2x + 12 + 4 = 8
-2x + 16 = 8
Subtracting 16 from both sides:
-2x = -8
Dividing by -2:
x = 4
Therefore, the solutions to ||2x-12|+4|=8 are x = 8 and x = 4.