| 1/2+3x|>= 6

Lösung

\frac{1}{2} + 3 x \geq 6

x \leq - \frac{13}{6} or x \geq \frac{11}{6}

Intervall-Notation

(- \infty, - \frac{13}{6}] \cup [\frac{11}{6}, \infty)

Dezimale

x \leq - 2.16666 \dots or x \geq 1.83333 \dots

Schritte zur Lösung

\frac{1}{2} + 3 x \geq 6

Wende Absoluteregel an: Wenn

∣ u ∣ \geq a, a > 0

dann

u \leq - a

u \geq a

\frac{1}{2} + 3 x \leq - 6 or \frac{1}{2} + 3 x \geq 6

\frac{1}{2} + 3 x \leq - 6 or \frac{1}{2} + 3 x \geq 6

x \leq - \frac{13}{6} or x \geq \frac{11}{6}

(x + 2) \leq 0

x + 8 = 3

2 x + 5 = 8 x - 4

12 r - 18 = 13 r + 18

- 5 x < - 10