How do you solve # abs(x+5)>12#?

1 Answer
Aug 31, 2015

Answer:

#x in (-oo, -17) uu (7, + oo)#

Explanation:

You're dealing with the absolute value of an expression, which means that you need totake into account the fact that the absolute value of a real number returns a positive value regardless of the sign of said number.

#color(blue)(|x| = {(x", "x >=0),(-x", "x<0) :}, " "(AA)x in RR)#

This means that you have to look at two possible scenarios, one in which the expression inside the modulus is positive and one in which it's negative.

  • #x+5 >0 implies |x+5| = x + 5#

The inequality takes the form

#x + 5 >12 implies x > 7#

  • #x+5<0 implies |x+5| = -(x+5)#

This time, you get

#-(x+5) > 12#

#-x - 5 > 12#

#-x > 17 implies x < -17#

The inequality will be true for any value of #x# that is greater than #7# or smaller than #(-17)#. The solution set will thus be #x in (-oo, -17) uu (7, + oo)#.