How do you solve #abs(( 2 - 5x) / 4 ) >= 2/3#?

1 Answer
Alan P.
May 26, 2015

One way to approach this kind of problem is to split it into two cases:
one when the argument of the absolute value is #<0#
and the other when the argument of the absolute value is #>=0#

Given #abs((2-5x)/4) >= 2/3#

Case 1: #(2-5x)/4 < 0#
Note that this implies #x>2/5#

#abs((2-5x)/4) >= 2/3#
becomes equivalent to
#(5x-2)/4 >= 2/3#

#5x-2 >= 8/3#

#5x >= 14/3#

#x>= 14/15#
Note that if #x>=14/15# the restriction #x>2/5# becomes irrelevant.

Case 2: #(2-5x)/4 >=0#
Note this implies #x<=2/5#

#abs((2-5x)/4) >= 2/3#
becomes equivalent to
#(2-5x)/4>= 2/3#

#2-5x >= 8/3#

#2>= 8/3+5x#

#-2/3>=5x#

#-10/3 >= x#
Again, note that #x<=-10/3# implies #x<=2/5# so the restriction that #x<=2/5# is irrelevant.

Combining Case 1 and Case 2:
#abs((2-5x)/4) >= 2/3#
if
#x>=14/15# or #x<=-10/3#

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