|x+7|<= 2

Solution

∣ x + 7 ∣ \leq 2

- 9 \leq x \leq - 5

Interval Notation

[- 9, - 5]

Solution steps

∣ x + 7 ∣ \leq 2

Apply absolute rule: If

∣ u ∣ \leq a, a > 0

then

- a \leq u \leq a

- 2 \leq x + 7 \leq 2

x + 7 \geq - 2 and x + 7 \leq 2

x \geq - 9 and x \leq - 5

Merge Overlapping Intervals

- 9 \leq x \leq - 5

4 d - 7 = 21

\frac{5 \cdot 1 0 ^{9}}{2 \cdot 1 0 ^{5}}

7 - \frac{x}{3} = 12

- \frac{15}{16 ( 4 ) ^{\frac{7}{2}}}

5 (x - 2)^{2} - 5 (x + 3)^{2} + (2 x - 1) (5 x + 2) - 10 x^{2} = 0