What is |x^2+5x|<6,|x+1|<= 1 ?

The solution to |x^2+5x|<6,|x+1|<= 1 is -6<x<-3\lor-2<x<1

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|x^2+5x|<6,|x+1|<= 1

x^{2} + 5 x < 6, ∣ x + 1 ∣ \leq 1

- 6 < x < - 3 or - 2 < x < 1

Interval Notation

(- 6, - 3) \cup (- 2, 1)

Solution steps

x^{2} + 5 x < 6

Apply absolute rule: If

∣ u ∣ < a, a > 0

then

- a < u < a

- 6 < x^{2} + 5 x < 6

x^{2} + 5 x > - 6 and x^{2} + 5 x < 6

(x < - 3 or x > - 2) and - 6 < x < 1

Merge Overlapping Intervals

- 6 < x < - 3 or - 2 < x < 1

∣ 2 x - 1 ∣ > - 3

∣ 3 x - 4 ∣ \leq - 6

∣ 3 x - 2 ∣ > 1

x^{2} - x - 6 < x + 2

x^{2} + x < 2

What is |x^2+5x|<6,|x+1|<= 1 ?
The solution to |x^2+5x|<6,|x+1|<= 1 is -6<x<-3\lor-2<x<1