How do you solve #3 < absx – 5#?

1 Answer
Aug 15, 2015

Answer:

#x in (-oo, -8) uu (8, +oo)#

Explanation:

Start by isolating the modulus on one side of the inequality. You can do this by adding #5# to both sides

#3 + 5 < |x| - color(red)(cancel(color(black)(5))) + color(red)(cancel(color(black)(5)))#

#8 < |x|#

This is of course equivalent to

#|x| > 8#

Now, you need to take into account the fact that #x# can be both positive or negative, which means that you get

  • #x>0 implies |x| = x#

For positive values of #x#, the inequality will be

#x > 8#

  • #x<0 implies |x| = -x#

For negative values of #x#, the inequality will be

#-x > 8#

Multiply both sides by #-1# to get #x# on the left side of the inequality - do not forget that the sign of the inequality changes when you multiply or divide by a negative number

#-1 * (-x) color(red)(<) 8 * (-1)#

#x < -8#

This means that your origininal inequality will be tru for any value of #x# that is smaller than #-8# or bigger than #8#. In other words, you need #x# to belong to two distinct intervals, #x<-8# and #x>8#, which can be written as #x in (-oo, -8) uu (8, +oo)#