Answer 1

2|x + 3| > 1
Dividing by 2 on both sides, we get
`|"x" + 3| > 1/2` Now, |x| > k implies x < – k or x > k

∴ `("x" + 3) < -1/2 or ("x" + 3) > 1/2` Subtracting 3 from both sides, we get

∴ `"x" < (-1)/2 - 3 or "x" > 1/2 - 3`

∴ `"x" < (-1 - 6)/2 or "x" > (1 - 6)/2`

∴ `"x" < (-7)/2 or "x" > (-5)/2`

∴ x can take all real values less `(-7)/2` or it can take values greater than `(-5)/2`

∴ Solution set is `(−∞,-7/2)∪(-5/2,∞)`