How do you solve #-3abs(6-x)<-15#?

1 Answer
Aug 14, 2015

Answer:

#x in (-oo, 1) uu (11, + oo)#

Explanation:

First, isolate the modulus on one side of the inequality. Notice that you can do that by dividing both sides of the inequality by #-3# - do not forget that the sign of the inequality changes when you miltiply or divide by a negative number.

#(color(red)(cancel(color(black)(-3))) * |6-x|)/color(red)(cancel(color(black)(-3))) color(blue)(>) (-15)/(-3)#

#|6 - x| > 5#

Next, use the fact that the expression inside the modulus can be positive or negative to find the solution intervals that would satisfy these possibilities.

  • #6-x>0 implies |6-x| = 6-x#

When the expression inside the modulus is positive, you get

#6 -x > 5#

#-x > -1 implies x < 1#

  • #6-x < 0 implies |6-x| = -(6 - x)#

When the expression inside the modulus is negative, you have

#-(6 - ) > 5#

#-6 + x > 5#

#x > 11#

So, the absolute value inequality will be true for any value of #x# that is smaller than #1# or for any value of #x# that is bigger than #11#.

This means that the solution set for this inequality will be #x in (-oo, 1) uu (11, + oo)#.