How do you solve #-2abs(x-3)+6 < -4#?

1 Answer
May 16, 2015

There are two possibilities: #x-3# (without the absolutes) is positive or negative (the 'switch point is #x=3#)

If it is positive, we can leave out the absolute brackets, is it is negative, we 'turn around' to #-(x-3)=3-x#

(1) #x-3>=0->x>=3#
#-2(x-3)+6<-4->-2x+6+6<-4->#
#-2x<-16->x>8#

(2)#x-3<0->x<3#
#-2(3-x)+6<-4-> -6+2x+6<-4->#
#2x<-4->x<-2#

Conclusion: #x<-2orx>8#