|4x+10|>2

Solution

∣ 4 x + 10 ∣ > 2

x < - 3 or x > - 2

Interval Notation

(- \infty, - 3) \cup (- 2, \infty)

Solution steps

∣ 4 x + 10 ∣ > 2

Apply absolute rule: If

∣ u ∣ > a, a > 0

then

u < - a

u > a

4 x + 10 < - 2 or 4 x + 10 > 2

4 x + 10 < - 2 or 4 x + 10 > 2

x < - 3 or x > - 2

2 \cdot \frac{2}{2}

\frac{3}{5} + \frac{8}{3} \div (5 - 10 (4 + 7 \div (\frac{1}{2} - \frac{3}{5})))

2.44 (\frac{0.7}{1})^{\frac{1}{3}} (6 \cdot 1 0^{4})

\frac{x ^{7}}{x ^{9}}

- 7 < 3 - 2 x \leq - 2