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Popular >

|q+4|<= 8

Solution

∣ q + 4 ∣ \leq 8

Solution

- 12 \leq q \leq 4

+1

Interval Notation

[- 12, 4]

Solution steps

∣ q + 4 ∣ \leq 8

Apply absolute rule: If

∣ u ∣ \leq a, a > 0

then

- a \leq u \leq a

- 8 \leq q + 4 \leq 8

q + 4 \geq - 8 and q + 4 \leq 8

q \geq - 12 and q \leq 4

Merge Overlapping Intervals

- 12 \leq q \leq 4

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\frac{x ^{- 2}}{x ^{2}}

2 x (x + 1) = 1

3 n^{2} - 8 n - 3 = 0

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9 m^{3} + 6 m

Frequently Asked Questions (FAQ)

What is |q+4|<= 8 ?
The solution to |q+4|<= 8 is -12<= q<= 4