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#