Solve `(2)/(|x-3|)gt5,x inR.`

Explanation

Clearly, `x-3ne0 "and therefore",xne3`.
We have, `(2)/(|x-3|)gt5" "...(i)`
Since `|x-3|` is positive, we may multiply both sides of (i) by `|x-3|`. This gives
`2gt5|x-3|`
`iff (2)/(5)gt|x-3|`
`iff|x-3|gt(2)/(5)`
`iff (-2)/(5)ltx-3lt(2)/(5)" " [therefore|x|lta iff-altxlta]`
`iff (-2)/(5)ltx-3and x-3lt(2)/(5)`
`iff (-2)/(5)+3ltx and xlt(2)/(5)+3`
`iff (13)/(5)ltx and xlt(17)/(5)`.
`iff (13)/(5)ltxlt(17)/(5)`.
Also, as shown above, `x ne 3`.
`therefore` solution set `={x inR :(13)/(5)ltxlt(17)/(5)}-{3}`
`=(2.6,3.4)-{3}=(2.6,3)uu(3,3.4)`.