How do you solve #abs(x-2)<4#?
1 Answer
Aug 28, 2015
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
The solution set will thus be