|3x+4|>= 5

Lösung

∣ 3 x + 4 ∣ \geq 5

x \leq - 3 or x \geq \frac{1}{3}

Intervall-Notation

(- \infty, - 3] \cup [\frac{1}{3}, \infty)

Dezimale

x \leq - 3 or x \geq 0.33333 \dots

Schritte zur Lösung

∣ 3 x + 4 ∣ \geq 5

Wende Absoluteregel an: Wenn

∣ u ∣ \geq a, a > 0

dann

u \leq - a

u \geq a

3 x + 4 \leq - 5 or 3 x + 4 \geq 5

3 x + 4 \leq - 5 or 3 x + 4 \geq 5

x \leq - 3 or x \geq \frac{1}{3}

15 x^{2} - 5 x + 6 = 9 x^{2}

- 2 \leq x \leq 5

x^{2} - x - 26 = 0

9 x^{2} + 36 x + 37 = 0

4 - y \geq 9