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How do you solve #abs(x-2)<4#?

Answer 1

#x in (-2, 6)#

You know that the absolute value of a real number is always positive, regardless of the sign of said number.

This means that you have two possibilities to take into account, more specifically if the expression inside the modulus is positive or if it is negative.

The inequality will take the form

#x-2 < 4#

#x < 6#

This time, the inequality will be

#-(x-2) <4#

#-x + 2 <4#

#-x < 2 implies x > -2#

So, you've determined that any value of #x# that is bigger than #(-2)# or smaller than #6# will satisfy this inequality.

The solution set will thus be #x in (-2, 6)#. For any value of #x in (-oo, -2] uu [6, + 00)# the inequality will no longer be true.

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Answer 2

Kennedy Adams

To solve ( |x-2| < 4 ):

Set up two inequalities: ( x - 2 < 4 ) and ( -(x - 2) < 4 ).
Solve each inequality separately.
Combine the solutions.

The solution is ( -2 < x < 6 ).

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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-2)&lt;4#?

How do you solve #abs(x-2)<4#?