How do you solve #(x+2)/(x+3)-(x^2)/(x^2-9)=1-((x-1)/(3-x))#?

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Answer 1 · 2017-11-22T04:49:57

#x=(-3-sqrt57)/4# or #(-3+sqrt57)/4#

#(x+2)/(x+3)-(x^2)/(x^2-9)=1-((x-1)/(3-x))#

or #(x+2)/(x+3)-(x^2)/((x+3)(x-3))=1+(x-1)/(x-3)#

Observe that we cannot have #x=+-3#. Now multiplying each term by #(x+3)(x-3)#, we get

#(x+2)(x-3)-x^2=(x-3)(x+3)+(x-1)(x+3)#

or #x^2-x-6-x^2=x^2-9+x^2+2x-3#

or #-x-6=2x^2+2x-12#

or #2x^2+3x-6=0#

or #x=(-3+-sqrt(3^2-4*2*(-6)))/(2*2)#

= #(-3+-sqrt(9+48))/4#

i.e. #x=(-3-sqrt57)/4# or #(-3+sqrt57)/4#