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How do you solve #abs(x+1)<3#?

Answer 1

Kennedy Adams

#-4 < x <2#

As #|x+1|<3#, we have

either #x+1<3# i.e. #x<2#

or #-(x×1)<3# i.e. #0 < x+1+3# or #x > -4#

Hence solution is #-4 < x < 2#

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Answer 2

Jace Anderson

To solve the inequality (|x + 1| < 3), you need to consider two cases:

When (x + 1) is positive or zero: In this case, the absolute value of (x + 1) is equal to (x + 1).
When (x + 1) is negative: In this case, the absolute value of (x + 1) is equal to (-(x + 1)).

For case 1: (x + 1 < 3)

For case 2: (-(x + 1) < 3)

Now, solve each case separately:

Case 1: (x + 1 < 3) (x < 3 - 1) (x < 2)

Case 2: (-(x + 1) < 3) (-x - 1 < 3) (-x < 3 + 1) (-x < 4) (x > -4)

So, combining the solutions from both cases, the solution set for (|x + 1| < 3) is (-4 < x < 2).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

How do you solve #abs(x+1)&lt;3#?

How do you solve #abs(x+1)<3#?