How do you solve #(x+3)/(x-3 ) + x/(x-5) = ( x+5)/(x-5)# and find any extraneous solutions?

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Answer 1 · 2018-05-29T18:29:48

#x=0# or #x=7#

It must be

#xne 3# and #xne 5#

Multiplying by

#(x-3)(x-5)#

we get

#(x+3)(x-5)+x(x-3)=(x+5)(x-3)#

Multiplying this out

#x^2+3x-5x-15+x^2-3x=x^2+5x-3x-15#

#2x^2-5x-15=x^2+2x-15#

so
#x^2-7x)=0#

#x(x-7)=0#
so
#x=0#
or
#x=7#