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| As we have |x - 1| in the equation so we will have two cases | |
| -Case 1: x - 1 ≥ 0, x ≥ 1 => |x - 1| = x - 1 => x - 1 = 6 => x = 6 + 1 = 7 But condition was x ≥ 1 and 7 ≥ 1 => x = 7 is a SOLUTION | -Case 2: x - 1 ≤ 0 => x ≤ 1 => |x - 1| = -(x - 1) = 1 - x => 1 - x = 6 => x = 1 - 6 = -5 But condition was x ≤ 1 and - 5 ≤ 1 => x = -5 is a SOLUTION |
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