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#