|-2x+17|>= 5

Lösung

∣ - 2 x + 17 ∣ \geq 5

x \leq 6 or x \geq 11

Intervall-Notation

(- \infty, 6] \cup [11, \infty)

Schritte zur Lösung

∣ - 2 x + 17 ∣ \geq 5

Wende Absoluteregel an: Wenn

∣ u ∣ \geq a, a > 0

dann

u \leq - a

u \geq a

- 2 x + 17 \leq - 5 or - 2 x + 17 \geq 5

- 2 x + 17 \leq - 5 or - 2 x + 17 \geq 5

x \leq 6 or x \geq 11

\frac{x - 7.8}{0.36} = 2.33

- x = 13

4 y^{2} + 16 y + 7 = 0

2 x^{2} - 8 x - 42 = 0

6 + 2 x^{2} - 3 x = 8 x^{2}