9x<=-27\lor 4x>= 36

Lösung

9 x \leq - 27 or 4 x \geq 36

x \leq - 3 or x \geq 9

Intervall-Notation

(- \infty, - 3] \cup [9, \infty)

Schritte zur Lösung

9 x \leq - 27 or 4 x \geq 36

9 x \leq - 27 : x \leq - 3

4 x \geq 36 : x \geq 9

Kombiniere die Bereiche

x \leq - 3 or x \geq 9

(x - 4) (x + 5) = 10

x^{2} - 3 > 6

\frac{1}{5} w - 12 = 20

∣ 2 x ∣ \geq 7

16 t^{2} + 7 t - 78 = 0