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4/(x-1)le3le6/(x+1)(xgt0)...

`4/(x-1)le3le6/(x+1)(xgt0)`

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Consider first two inequallities,
`4/(x+1)le3`
`rArr 4le3(x+1)`
`rArr 4le 3x+3`
`rArr 4-3le3X` [subtracting 3 on both sides]
`rArr 1 le 3x`
`:. Xge1/3` …(i)
and consider last two inequalities,
`3le6/(x+1)`
`rArr 3(x+1)le6`
`rArr 3x+3le 6`
`rArr 3xle6-3` [subtracting 3 to both sides]
`rArr 3xle3` [dividing by 3]
`:. xle1` ....(ii)
From Eqs. (i) and (ii),
`x in[1/3,1]`
`1/3lexle1`
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