How do you solve #(x-5)/(x-8)+(x-6)/(x-7)=(x+97)/(x^2-15x+56)# and check for extraneous solutions?

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Answer 1 · 2017-09-06T09:16:27

#x=-1/2# or #x=14#

In #(x-5)/(x-8)+(x-6)/(x-7)=(x+97)/(x^2-15x+56)#, observe that as #x^2-15x+56=(x-8)(x-7)#, we cannot have #x=8# or #x=7#.

#(x-5)/(x-8)+(x-6)/(x-7)=(x+97)/(x^2-15x+56)#

#hArr((x-5)(x-7)+(x-6)(x-8))/(x^2-15x+56)=(x+97)/(x^2-15x+56)#

or #(x-5)(x-7)+(x-6)(x-8)=x+97#

or #x^2-12x+35+x^2-14x+48=x+97#

or #2x^2-26x+83-x-97=0#

or #2x^2-27x-14=0#

or #2x^2-28x+x-14=0#

or #2x(x-14)+1(x-14)=0#

i.e. #(2x+1)(x-14)=0#

and #x=-1/2# or #x=14#

and there is no extraneous solution as none of these is #7# or #8#