|u+7|<= 3

Solution

∣ u + 7 ∣ \leq 3

- 10 \leq u \leq - 4

Interval Notation

[- 10, - 4]

Solution steps

∣ u + 7 ∣ \leq 3

Apply absolute rule: If

∣ u ∣ \leq a, a > 0

then

- a \leq u \leq a

- 3 \leq u + 7 \leq 3

u + 7 \geq - 3 and u + 7 \leq 3

u \geq - 10 and u \leq - 4

Merge Overlapping Intervals

- 10 \leq u \leq - 4

\frac{x}{5} = - 2

- 3 \cdot - 48

(5^{- 8}) (5^{- 10})

(u^{\frac{1}{3}})^{\frac{4}{5}}

x - 3 > 5 or x + 3 < - 2