How do you solve $∣ ∣ ∣ \frac{4 x - 2}{5} ∣ ∣ ∣ = ∣ ∣ ∣ \frac{6 x + 3}{2} ∣ ∣ ∣$ $abs((4x-2)/5)=abs((6x+3)/2)$ ?

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Answer 1 · 2017-01-22T06:01:03

$- \frac{11}{38} and - \frac{19}{22}$ $-11/38 and -19/22$ . The values are zeros ( y = 0 ), in the Socratic graph.

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

$2 x - 1 = 3.75 (2 x + 1)$ $2x-1=3.75(2x+1)$ , when $x > \frac{1}{2}$ $x > 1/2$ ,

giving extraneous $x = - \frac{19}{22}$ $x = -19/22$ ,

$- (2 x - 1) = 3.75 (2 x + 1)$ $-(2x-1)=3.75(2x+1)$ , when $- \frac{1}{2} < x < \frac{1}{2}$ $-1/2< x < 1/2$ ,

giving $x = - \frac{11}{38} \in (- \frac{1}{2}, \frac{1}{2})$ $x = -11/38 in (-1/2, 1/2)$ and

$- (2 x - 1) = - 3.75 (2 x + 1)$ $-(2x-1)=-3.75(2x+1)$ , when $x < - \frac{1}{2}$ $x < -1/2$ ,

giving $x = - \frac{19}{22} < - \frac{1}{2}$ $x = -19/22 < -1/2$ .

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

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