|((3x+1))/((x-3))|>6

Lösung

\frac{( 3 x + 1 )}{( x - 3 )} > 6

\frac{17}{9} < x < 3 or 3 < x < \frac{19}{3}

Intervall-Notation

(\frac{17}{9}, 3) \cup (3, \frac{19}{3})

Dezimale

1.88888 \dots < x < 3 or 3 < x < 6.33333 \dots

Schritte zur Lösung

\frac{3 x + 1}{x - 3} > 6

Wende Absoluteregel an: Wenn

∣ u ∣ > a, a > 0

dann

u < - a

u > a

\frac{3 x + 1}{x - 3} < - 6 or \frac{3 x + 1}{x - 3} > 6

\frac{3 x + 1}{x - 3} < - 6 or \frac{3 x + 1}{x - 3} > 6

\frac{17}{9} < x < 3 or 3 < x < \frac{19}{3}

∣ z ∣ \leq 7

∣ 2 x + 5 ∣ > 0.5

- 4 x + 14 + 10 ∣ x - 2 ∣ < 12

∣ x + 2 ∣ \leq x + 3

∣ x ∣ > 2 x + 7