1.

Solve `|(x-3)/(x+1)| le 1`

Answer» `|(x-3)/(x+1)| le 1` or `-1 le (x-3)/(x+1) le 1`
`rArr (x-3)/(x+1)-1 le 0 " and " 0 le (x-3)/(x+1)+1`
`rArr (-4)/(x+1) le 0 " and " 0 le (2x -2)/(x+1)`
`rArr x lt -1 " and " { x lt - 1 or x ge 1}`
`rArr x ge 1`


Discussion

No Comment Found