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| As we have |3 - x| in the equation so we will have two cases | |
| -Case 1: 3 - x ≥ 0 => x ≤ 3 => |3 - x| = 3 - x => 3 - x < x + 5 => 2x > -2 => x > -1 But condition was x ≤ 3 and x > 3 is in the same range => -1 < x ≤ 3 is the SOLUTION | -Case 2: 3 - x ≤ 0 => x ≥ 3 => |3 - x| = -(3 - x) = x - 3 => x - 3 < x + 5 => 8 > 0 => Which is TRUE ALWAYS But condition was x ≥ 3 and it is TRUE ALWAYS => x ≥ 3 is the SOLUTION |
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