How do you solve #| 2x + 1 | <| 3x - 2 | #?

1 Answer
Mar 13, 2016

Answer:

#x in ]-oo, 1/5 [ uu ] 3, +oo[#

Explanation:

First find the zeros of each expression in the modules:

#2x+1=0 =>x-=-1/2#

#3x-2=0 =>x-=2/3#

Now we must divide the inequation in three parts:

1) When x>=2/3, both expression are larger than 0, therefore:

#|2x+1|<|3x-2|#

is equivalent to:

#2x+1<3x-2 and x>=2/3 #

#-x<-3 and x>=2/3 #

#x>3 and x>=2/3 #

#x>3#


2)if #-1/2<= x <2/3#:

3x-2 will be <=0, therefore #|3x-2|=-3x+2#:

#2x+1<-3x+2 and -1/2<= x <=2/3#:

#5x<1 and -1/2<= x <=2/3#:

#x<1/5and -1/2<= x <=2/3#:

# -1/2<= x <1/5#


3)if #x<=-1/2#

both expressions are negative so we must get the inverse for both ones:

#-2x-1<-3x+2 and x<=-1/2 #

#x<3 and x<-1/2 #

#x<=-1/2 #

Now the solution is the union of the three solutions:

#x in ]-oo, 1/2]uu[1/2, 1/5 [ uu ] 3, +oo[#

#x in ]-oo, 1/5 [ uu ] 3, +oo[#