2/x <= 1/(x+6)

Lösung

\frac{2}{x} \leq \frac{1}{x + 6}

x \leq - 12 or - 6 < x < 0

Intervall-Notation

(- \infty, - 12] \cup (- 6, 0)

Schritte zur Lösung

\frac{2}{x} \leq \frac{1}{x + 6}

Rewrite in standard form

\frac{x + 12}{x ( x + 6 )} \leq 0

Identifiziere die Intervalle

x \leq - 12 or - 6 < x < 0

∣ m + 5 ∣ + 9 \leq 30

- 5 x + 6 > - 4

13 = \frac{3}{4} (5 x + 8)

- y^{2} + y + 6 = 0

∣ 3 x - 7 ∣ = 4 x