|x+8|<= 2

Solution

∣ x + 8 ∣ \leq 2

- 10 \leq x \leq - 6

Solution steps

∣ x + 8 ∣ \leq 2

Apply absolute rule: If

∣ u ∣ \leq a, a > 0

then

- a \leq u \leq a

- 2 \leq x + 8 \leq 2

x + 8 \geq - 2 and x + 8 \leq 2

x \geq - 10 and x \leq - 6

Merge Overlapping Intervals

- 10 \leq x \leq - 6

\frac{5 x - 1}{5} + \frac{1 - 2 x}{3} = 2

9 - 6 x \geq 3 x - 8

(\frac{1}{2} x^{4}) (16 x^{5})

\frac{24}{6}

- 32 + 320 - 38