#f(x) = abs(x+2)+abs(2x-4)-abs(x-3)=0#
Making #abs(x+2) = y = abs(x-3)-abs(2x-4)=0#
we have #x = pm y-2#.
I) Choosing #x = y-2#
#y = abs(y-5)-abs(2y-8)#
#y-5 = abs(y-5)-abs(2y-8)-5#
#1 = pm1-(abs(2y-8)+5)/(y-5)# for #y ne 5#
Here we have two possibilities:
I-1)
#0 = abs(2y-8)+5# having two outcomes
I-1-a)
#2y-8=-5->y=3/2#
I-1-b)
#2y-8=5->y=13/2#
I-2)
#2=-( abs(2y-8)+5)/(y-5)#
#2(y-5)=-abs(2y-8)-5#
#2y-10=-abs(2y-8)-5#
#2y-8=-abs(2y-8)-3#
#1=pm 1-3/(2y-8)#
with the only posibility
#2 =- 3/(2y-8)->y = 13/4#
II) Choosing #x = -(y+2)#
#-y+abs(2y+8)-abs(y+5)=0#
#-y+2abs(y+4)-abs(y+5)=0#
#-y-4+2abs(y+4)-abs(y+5)+4=0#
#-(y+4)+2abs(y+4)-abs(y+5)+4=0#
#1 pm2 +(abs[y+5]-4) /(y+4) = 0# with two outcomes
II-1)
#3 +(abs[y+5]-4) /(y+4) = 0#
#3(y+4)+abs(y+5)-4=0#
#3(y+5)+abs(y+5)-4-3=0#
#3pm1 -7/(y+5)=0# with two outcomes
II-1-a)
#4(y+5)=7->y=-13/4#
II-1-b)
#2(y+5)=7->y=-3/2#
II-2)
#-1 +(abs[y+5]-4) /(y+4) = 0#
#-(y+4)+abs(y+5)-4=0#
#-(y+5)+abs(y+5)-4+1=0#
#1pm1+3/(y+5)=0#
with the only outcome
#-2(y+5)+3=0->y=-7/2#
Collecting the results obtained for #x = y-2#
#y = {3/2,13/2,13/4}->x={-1/2,9/2,5/4}#
and for #x = -(y+2)#
#y = {-13/4,-3/2,-7/2}->x={5/4,-1/2,3/2}#
Substituting in #f(x)# none of them is solution.
