Solve the following :
(a) `1 le |x-2| le 3 " (b) "0 lt |x-3|le 5`
(c ) `|x-2|+|2x-3|=|x-1| " (d) " |(x-3)/(x+1)| le1`

Answer

(a) `[-1,1]cup [3,5] " (b) " [-2,8]-{3} " (c ) " [3//2,2] " (d) "[1,oo)`

Explanation

(a) ` 1 le |x-2| le 3`
`implies-3 le x-2 le -1 " or " 1 le x-2 le 3`
`implies -1 le x le 1 " or "3 le x le 5`
(b) ` 0 lt |x-3| le 5`
`implies -5 le x-3 le 5, x-3 ne 0`
`implies -2 le x le 8, x ne 3`
(c ) `|x-2| +|2x-3|=|x-1|`
`implies |x-2|+|2x-3|=|(2x-3)-(x-2)|`
`implies (x-2) (2x-3) le 0`
`implies 3//2 le x le 2`
(d) `|(x-3)/(x+1)| le 1`
`implies -1 le (x-3)/(x+1) le 1`
`implies 0 le (x-3)/(x+1)+1 " and " (x-3)/(x+1)-1 le 0`
`implies0 le (2x-2)/(x+1) " and " (-4)/(x+1) le 0`
`implies x le -1 " or " x ge 1 " and " x gt -1`
`implies x ge 1`