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

1 Answer
Aug 28, 2015

Answer:

#x in (-2, 6)#

Explanation:

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.

  • #x-2>=0 implies |x-2| = x-2#

The inequality will take the form

#x-2 < 4#

#x < 6#

  • #x-2 <0 implies |x-2| = -(x-2)#

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.