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

Lösung

\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}

Intervall-Notation

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

Dezimale

- 2.39296 \dots < x < - 1.70900 \dots

Schritte zur Lösung

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

Wenn

a < u < b

dann

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}

Kombiniere die Bereiche

(\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})

Füge die sich überschneidenden Intervalle zusammen

\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