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)
( $rarr c>=5/3$ )
$abs(3c+5) = 3c+5$

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

How do you solve the absolute value 1/2 abs(3c + 5)= 6c +41/2 abs(3c + 5)= 6c +4?