How do you graph and solve #2x-1-abs[9-3x]<abs[3x]#?

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Answer 1 · 2017-03-21T10:57:39

See below.

#2x-1-3abs(x-3) < 3absx# now supposing #x ne 0#

#2x/absx -1/absx-3abs(x-3)/absx < 3# but

#x/abs x = pm 1# so

#pm2-3<1/absx+3abs(x-3)/absx# or

#abs x max(-5,-2) < 1+abs(x-3)# or

#-2absx < 1 +abs(x-3)# or

#2abs x + abs(x-3) + 1 > 0#

So it is true for all #x#