What is 1/(3*(x-1))< 1/4*(x+2)< 1/(6*(x+4)) ?

The solution to 1/(3*(x-1))< 1/4*(x+2)< 1/(6*(x+4)) is (-3-sqrt(129))/6 <x<(sqrt(15)-9)/3

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1/(3(x-1))< 1/4(x+2)< 1/(6*(x+4))

\frac{1}{3 \cdot ( x - 1 )} < \frac{1}{4} \cdot (x + 2) < \frac{1}{6 \cdot ( x + 4 )}

\frac{- 3 - 129}{6} < x < \frac{15 - 9}{3}

Interval Notation

(\frac{- 3 - 129}{6}, \frac{15 - 9}{3})

Decimal Notation

- 2.39296 \dots < x < - 1.70900 \dots

Solution steps

\frac{1}{3 ( x - 1 )} < \frac{1}{4} (x + 2) < \frac{1}{6 ( x + 4 )}

a < u < b

then

a < u and u < b

\frac{1}{3 ( x - 1 )} < \frac{1}{4} (x + 2) and \frac{1}{4} (x + 2) < \frac{1}{6 ( x + 4 )}

\frac{1}{3 ( x - 1 )} < \frac{1}{4} (x + 2) : \frac{- 3 - 129}{6} < x < 1 or x > \frac{129 - 3}{6}

\frac{1}{4} (x + 2) < \frac{1}{6 ( x + 4 )} : x < - \frac{9 + 15}{3} or - 4 < x < \frac{15 - 9}{3}

Combine the intervals

(\frac{- 3 - 129}{6} < x < 1 or x > \frac{129 - 3}{6}) and (x < - \frac{9 + 15}{3} or - 4 < x < \frac{15 - 9}{3})

Merge Overlapping Intervals

\frac{- 3 - 129}{6} < x < \frac{15 - 9}{3}

2 \leq - 3 x - 13 < 8

9 < 4 x + 1 \leq 33

- 6 < x + 3 < 2

2 < ∣ x - 1 ∣ < 5

1 < x + 2 < - 4

What is 1/(3*(x-1))< 1/4*(x+2)< 1/(6*(x+4)) ?
The solution to 1/(3*(x-1))< 1/4*(x+2)< 1/(6*(x+4)) is (-3-sqrt(129))/6 <x<(sqrt(15)-9)/3