How do you solve #abs(4x-3)<7#?

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Answer 1 · 2016-07-28T17:37:36

#-1 < x < 2.5# - i.e. #x# is between #-1# and #2.5#

As #|4x-3|<7#,

either #4x-3<7# i.e, #4x<7+3# or

#4x=10# i.e. #x<10/4# i.e. #x<2.5#

or #-(4x-3)<7# i.e. #-4x+3<7# or

#3-7<4x# or #4x>-4# i.e. #x>-1#

Hence, we have #x<2.5# and #x>-1#, which can be written better as

#-1 < x < 2.5# - i.e. #x# is between #-1# and #2.5#