How do you solve #(3x-2)/(x^2-17x+72)-(x-4)/(x^2-20x+96)=(x-6)/(x^2-21x+108)# and check for extraneous solutions?

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Answer 1 · 2018-01-12T16:03:25

#x_1=-4# and #x_2=15#

#(3x-2)/(x^2-17x+72)-(x-4)/(x^2-20x+96)=(x-6)/(x^2-21x+108)#

#(3x-2)/[(x-8)(x-9)]-(x-4)/[(x-8)(x-12)]=(x-6)/[(x-9)(x-12)]#

#[(3x-2)(x-12)-(x-4)(x-9)]/[(x-8)(x-9)(x-12)]=(x-6)/[(x-9)(x-12)]#

#[(3x^2-38x+24)-(x^2-13x+36)]/[(x-8)(x-9)(x-12)]=(x-6)/[(x-9)(x-12)]#

#2x^2-25x-12=(x-6)*(x*8)#

#2x^2-25x-12=x^2-14x+48#

#x^2-11x-60=0#

#(x+4)*(x-15)=0#

Hence #x_1=-4# and #x_2=15#