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]}

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