|2x-2|>6

Solution

∣ 2 x - 2 ∣ > 6

x < - 2 or x > 4

Interval Notation

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

Solution steps

∣ 2 x - 2 ∣ > 6

Apply absolute rule: If

∣ u ∣ > a, a > 0

then

u < - a

u > a

2 x - 2 < - 6 or 2 x - 2 > 6

2 x - 2 < - 6 or 2 x - 2 > 6

x < - 2 or x > 4

(3 - 5)^{2}

6 x - 1 = 4 x + 6

\frac{- 5 ( 2 x - 1 )}{3} + 6 (3 - 2 x) - \frac{2 x}{5} = \frac{- 3 ( x + 1 )}{4}

\frac{3}{∣ x + 3 ∣} \geq ∣ x + 2 ∣

\frac{5 3}{4 2}