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

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Answer 1 · 2017-11-09T23:07:57

$x=-22/5$

$abs[(2x-5)/3]=abs[(3x+4)/2]$

After squaring both sides,

$(2x-5)^2/9=(3x+4)^2/4$

$(4x^2-20x+25)/9=(9x^2+24x+16)/4$

$4*(4x^2-20x+25)=9*(9x^2+24x+16)$

$16x^2-80x+100=81x^2+216x+144$

$65x^2+296x+44=0$

$65x^2+276x+20x+44=0$

$13x*(5x+22)+4*(5x+22)=0$

$(13x+4)*(5x+22)=0$

Hence $x_1=-22/5$ and $x_2=-4/13$ . But $x=-4/13$ doesn't provide solution for original problem. Thus solution of it is $x=-22/5$

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