How do you solve #abs((4x-2)/5)=abs((6x+3)/2)#?

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Answer 1 · 2017-01-22T06:01:03

#-11/38 and -19/22#. The values are zeros ( y = 0 ), in the Socratic graph.

It is #|2x-1|=3.75|2x+1|#. Breaking up,

#2x-1=3.75(2x+1)#, when #x > 1/2#,

giving extraneous #x = -19/22#,

#-(2x-1)=3.75(2x+1)#, when #-1/2< x < 1/2#,

giving #x = -11/38 in (-1/2, 1/2)# and

#-(2x-1)=-3.75(2x+1)#, when #x < -1/2#,

giving #x = -19/22 < -1/2#.

graph{|x-1|-2.5|2x+1| [-1, 0, -3.075, 2.075]}