`|(x+3)/(x-1)|gt1`

This will give two initial solutions,

`(x+3)/(x-1) gt 1` or `(x+3)/(x-1) lt-1`

1) `(x+3)/(x-1) gt 1`

`(x+3)/(x-1)-1 gt 0`

`(x+3-(x-1))/(x-1) gt0 `

This gives,

`4/(x-1)gt0`

`1/(x-1) gt0`

Multiplying both denominator and numerator by (x-1),

`(x-1)/(x-1)^2 gt0`

This would make the denominator alway positive, and the requirement for this inequality would be,

`x-1gt0`

`x gt1`

2)`(x+3)/(x-1) lt-1`

`(x+3)/(x-1) +1 lt 0`

This gives,

`(x+3+x-1)/(x-1)lt0`

`(2x+2)/(x-1) lt 0`

`(x+1)/(x-1) lt 0`

Multiplying both denominator and numerator by (x-1),

`((x+1)(x-1))/(x-1)^2 lt0`

This gives,

(x+1)(x-1) < 0

The solution for this is -1<x<1

By combining above two solutions we can get the final solution.

Therefore the required solution for `|(x+3)/(x-1)|gt1` is `xgt -1` .

**The answer is x > -1**