How do you solve #abs(6 - 2c) - 2 > c + 10#?

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Answer 1 · 2015-07-09T20:56:11

#c<-2#
#c>18#

#abs(6-2c)-2>c+10#

Subtract #2# from both sides of the inequality.

#abs(6-2c)>c+12#

Separate the inequality into 2 other inequalities.

#6-2c>c+12# and #-(6-2c)>c+12#

First Inequality: #6-2c>c+12#

Subtract #6# from both sides.

#-2c>c+12-6# =

#-2c>c+6#

Subtract #c# from both sides.

#-2c-c>6#

#-3c>6#

Divide both sides by #-3#. This will reverse the inequality.

#c<6/(-3)#

#c<-2#

Second Inequality: #-(6-2c)>12#

#-6+2c>c+12#

Add #6# to both sides.

#2c>c+12+6# =

#2c>c+18#

Subtract #c# from both sides.

#2c-c>18#

#c>18#

Answer 2 · 2015-07-09T20:57:41

#c \in ]-\infty, -2[ \cup [18, +\infty]#

#|6−2c|−2>c+10#

#|6−2c|>c+12#

#|a| > b \Rightarrow a > b or a < -b#

#6−2c>c+12 or 6−2c < -c-12#

#-6>3c or 18 < c#

#c < -2 or c > 18#