How do you solve #x/(x-3)=6/(x-3)#?

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Answer 1 · 2016-08-29T08:44:16

#x=+3, or +6#

#x/(x-3)=6/(x-3)#

i.e. #6(x-3)=x(x-3)#

#6x-18=x^2-3x#

#x^2-9x+18=0#

This is a quadratic in #x#, for which we might use the quadratic formula. Alternatively we could factor it to give:

#(x-3)(x-6)=0#

Thus #x# has roots at #+3# or #+6#