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Popular >

|x+7|>= 7

Solution

∣ x + 7 ∣ \geq 7

Solution

x \leq - 14 or x \geq 0

+1

Interval Notation

(- \infty, - 14] \cup [0, \infty)

Solution steps

∣ x + 7 ∣ \geq 7

Apply absolute rule: If

∣ u ∣ \geq a, a > 0

then

u \leq - a

or

u \geq a

x + 7 \leq - 7 or x + 7 \geq 7

x + 7 \leq - 7 or x + 7 \geq 7

x \leq - 14 or x \geq 0

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(\frac{12 p ^{\frac{3}{2}}}{15 p})^{4}

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3 (4 x + 4) = 2 (5 x + 9) - 12

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9 x^{2} - 6 x y - 8 y^{2}

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\frac{3}{2} + \frac{5}{4} + \frac{9}{8} + \frac{17}{16} + \frac{33}{32} + \frac{65}{64} - 7

2 x = 4 x + 4

Frequently Asked Questions (FAQ)

What is |x+7|>= 7 ?
The solution to |x+7|>= 7 is x<=-14\lor x>= 0