How do you solve the absolute value #1/2 abs(3c + 5)= 6c +4#?

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Answer 1 · 2015-05-05T00:10:11

#1/2 abs(3c+5) = 6c+4#

implies
#abs(3c+5) = 12c+8#

Case 1: (3c+5<0)
(#rarr c<-5/3 #)
#abs(3c+5) = -3c-5#

#-3c-5=12c+8#

#-15c = 13#

#c=-13/15 " an extraneous solution since "-13/15> -5/3#

Case 2: (3c+5>=0)
(#rarr c>=5/3#)
#abs(3c+5) = 3c+5#

#3c+5 = 12c+8#
#-9c = 3#
#c=- 1/3 #