|3x-3|>9

Solution

∣ 3 x - 3 ∣ > 9

x < - 2 or x > 4

Interval Notation

(- \infty, - 2) \cup (4, \infty)

Solution steps

∣ 3 x - 3 ∣ > 9

Apply absolute rule: If

∣ u ∣ > a, a > 0

then

u < - a

u > a

3 x - 3 < - 9 or 3 x - 3 > 9

3 x - 3 < - 9 or 3 x - 3 > 9

x < - 2 or x > 4

\frac{4 + 2 i}{4 - 2 i}

5 (6) + 3 = 5 x + 3

\frac{5}{3} + \frac{8}{18} - \frac{18}{7}

x^{2} - x = 42