How do you solve $abs(2+3x)=abs(4-2x)$ ?

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Answer 1 · 2017-10-19T14:43:19

See below.

The solutions for $abs(2+3x)=abs(4-2x)$ or equivalently

$sqrt((2+3x)^2)=sqrt((4-2x)^2)$ are included into the solutions for

$(2+3x)^2=(4-2x)^2$ or

$(2+3x)^2-(4-2x)^2=0$ or

$(2+3x+4-2x)(2+3x-4+2x)=0$

or

$(x+6)(5x-2)=0$

or $x = {-6,2/5}$ and both solutions are feasible.

Answer 2 · 2017-10-19T14:45:14

$x_1=-6$ and $x_2=2/5$

$abs(2+3x)=abs(4-2x)$

$(2+3x)^2=(4-2x)^2$

$9x^2+12x+4=4x^2-16x+16$

$5x^2+28x-12=0$

$5x^2+30x-2x-12=0$

$5x*(x+6)-2*(x+6)=0$

$(5x-2)*(x+6)=0$

Hence $x_1=-6$ and $x_2=2/5$

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