|5x-2|>= 13

Lösung

∣ 5 x - 2 ∣ \geq 13

x \leq - \frac{11}{5} or x \geq 3

Intervall-Notation

(- \infty, - \frac{11}{5}] \cup [3, \infty)

Dezimale

x \leq - 2.2 or x \geq 3

Schritte zur Lösung

∣ 5 x - 2 ∣ \geq 13

Wende Absoluteregel an: Wenn

∣ u ∣ \geq a, a > 0

dann

u \leq - a

u \geq a

5 x - 2 \leq - 13 or 5 x - 2 \geq 13

5 x - 2 \leq - 13 or 5 x - 2 \geq 13

x \leq - \frac{11}{5} or x \geq 3

(7 x - 5)^{2} = 25

2 x^{2} - 10 x + 12 = 0

6 h + 16 = 46

4 z^{2} + z + 2 = 0

s im pl i f y \frac{5}{9}