How do you solve #abs(4x-5)>16 #?

1 Answer
Jul 7, 2015

Answer:

#abs(4x-5) > 16# when #x<-11/4# and #x>21/4#

enter image source here

Explanation:

The "absolute value" function is defined as :

When #x in RR,#

#|x|=x,ifx≥0#
#|x|=−x,ifx<0#

And #4x-5>=0# when #x>=5/4#

Then, #abs(4x-5)=4x-5, if x>=5/4#
And, #abs(4x-5)=-(4x-5)=-4x+5, if x<5/4#

To resolve the inequality, we have to separate them in two parts :
when #x>=5/4# and when #x<5/4#
# #
If #x>=5/4# :
#abs(4x-5) > 16 <=> 4x-5 > 16 <=> 4x > 21 <=> color(red)(x>21/4)#
# #
# #
If #x<5/4# :
#abs(4x-5) > 16 <=> -4x+5 > 16 <=> -4x > 11 <=> -x>11/4#

Recall : when we multiply or divide each side of an inequality by a negative number, we have to inverse the sign of the inequality

Ex : If #-a>0# then #acolor(red)<0# (multiplied by #-1#)
# #
# #
Then #abs(4x-5) > 16 <=> color(red)(x<-11/4)#
# #
# #
Therefore, #abs(4x-5) > 16# when #x<-11/4# and #x>21/4#
# #
# #
Other notation :
#abs(4x-5) > 16# if #x in ]-oo;-11/4[uu]21/4;+oo[#