f(x)=|x+2|+|2x−4|−|x−3|=0
Making |x+2|=y=|x−3|−|2x−4|=0
we have x=±y−2.
I) Choosing x=y−2
y=|y−5|−|2y−8|
y−5=|y−5|−|2y−8|−5
1=±1−|2y−8|+5y−5 for y≠5
Here we have two possibilities:
I-1)
0=|2y−8|+5 having two outcomes
I-1-a)
2y−8=−5→y=32
I-1-b)
2y−8=5→y=132
I-2)
2=−|2y−8|+5y−5
2(y−5)=−|2y−8|−5
2y−10=−|2y−8|−5
2y−8=−|2y−8|−3
1=±1−32y−8
with the only posibility
2=−32y−8→y=134
II) Choosing x=−(y+2)
−y+|2y+8|−|y+5|=0
−y+2|y+4|−|y+5|=0
−y−4+2|y+4|−|y+5|+4=0
−(y+4)+2|y+4|−|y+5|+4=0
1±2+|y+5|−4y+4=0 with two outcomes
II-1)
3+|y+5|−4y+4=0
3(y+4)+|y+5|−4=0
3(y+5)+|y+5|−4−3=0
3±1−7y+5=0 with two outcomes
II-1-a)
4(y+5)=7→y=−134
II-1-b)
2(y+5)=7→y=−32
II-2)
−1+|y+5|−4y+4=0
−(y+4)+|y+5|−4=0
−(y+5)+|y+5|−4+1=0
1±1+3y+5=0
with the only outcome
−2(y+5)+3=0→y=−72
Collecting the results obtained for x=y−2
y={32,132,134}→x={−12,92,54}
and for x=−(y+2)
y={−134,−32,−72}→x={54,−12,32}
Substituting in f(x) none of them is solution.
