Question -
Answer -
|4 тАУ x | + 1 < 3
Let us subtract 1 from both the sides, we get
|4 тАУ x| + 1 тАУ 1 < 3 тАУ 1
|4 тАУ x| < 2
Let тАШrтАЩ be a positive real number and тАШaтАЩ be a fixed real number. Then,
|a тАУ x| < r тЯ║ a тАУ r < x < a + r
Here, a=4 and r=2
4 тАУ 2 < x < 4 + 2
2 < x < 6
тИ┤ x тИИ (2, 6)