What is the solution set for #abs(4x – 3) – 1 > 12#?

1 Answer
Aug 14, 2015

Answer:

#x in (-oo, -5/2) uu (4, +oo)#

Explanation:

Start by isolating the modulus on one side of the inequality. This can be done by adding #1# to both sides

#|4x-3| - color(red)(cancel(color(black)(1))) + color(red)(cancel(color(black)(1))) > 12 + 1#

#|4x-3| > 13#

Since you're dealing with absolute value, you need to take into account the fact that the expression inside the modulus can be negative or positive

  • #4x-3>0 implies |4x-3| = 4x-3#

This means that your inequality can be written as

#4x - 3 > 13#

#4x > 16 implies x > 4#

  • #4x-3<0 implies |4x-3| = -(4x-3)#

This time, you have

#-(4x-3) > 13#

#-4x + 3 > 13#

#-4x > 10 implies x < -5/2#

What this tells you is that for any value of #x# that is bigger than #4# or smaller than #-5/2#, this inequality will be true. This means that the solution set will will #x in (-oo, -5/2) uu (4, +oo)#.