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Solve |1+ (3)/(x)| gt 2...

Solve `|1+ (3)/(x)| gt 2`

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