How do you solve #abs(3 + 2x)>abs(4 - x)#?

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Apr 24, 2018

Answer:

The solution is # x in (-oo, -7) uu(1/3, +oo)#

Explanation:

This is an inequality with absolute values

#|3+2x| >|4-x|#

#|3+2x| -|4-x| >0#

Let #f(x)=|3+2x| -|4-x| #

#{(3+2x>=0),(4-x>=0):}#, #<=>#, #{(x>=-3/2),(x<=4):}#

The sign chart is as follows :

#color(white)(aaaa)##x##color(white)(aaaa)##-oo##color(white)(aaaaaaaaa)##-3/2##color(white)(aaaaaaaaaaa)##4##color(white)(aaaaaa)##+oo#

#color(white)(aaaa)##3+2x##color(white)(aaaaa)##-##color(white)(aaaaaaa)##0##color(white)(aaaa)##+##color(white)(aaaaaaaaa)##+#

#color(white)(aaaa)##4-x##color(white)(aaaaaa)##+##color(white)(aaaaaaa)####color(white)(aaaaa)##+##color(white)(aaaaa)##0##color(white)(aaa)##+#

#color(white)(aaaa)##|3+2x|##color(white)(aaaaa)##-3-2x##color(white)(aa)##0##color(white)(aaa)##3+2x##color(white)(aaaaaa)##3+2x#

#color(white)(aaaa)##|4-x|##color(white)(aaaaaaaa)##4-x##color(white)(aaa)####color(white)(aaaa)##4-x##color(white)(aaa)##0##color(white)(aa)##-4+x#

In the interval #(-oo, -3/2)#,

#f(x)=-3-2x-4+x=-7-x#

#f(x)>0#, #=>#, #-7-x>0#, #=>#, #x<-7#

In the interval #[-3/2, 4]#,

#f(x)=3+2x-4+x=-1+3x#

#f(x)>0#, #=>#, #-1+3x>0#, #x>1/3#

In the interval #( 4, +oo)#,

#f(x)=3+2x+4-x=7+x#

#f(x)>0#, #=>#, #7+x>0#, #=>#, #x<-7#

This result is not valid since #x !in #(4, +oo)##

The solution is

# x in (-oo, -7) uu(1/3, +oo)#

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