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Solve the following inequality: |x + 1| <_3

<_ = greater than or equal to

User Vsxen
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1 Answer

3 votes

Answer:

x <= -4 or x >= 2

Explanation:

so, it actually says

|x+1| >= 3

so, then, this is valid for all x >= 2 (then x+1 is 3 or higher), and for all x <= -4 (then x+1 is -3 or lower, and |x+1| is still 3 or higher)

User Anthonyhumphreys
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