How do you solve # 3/(x-3) + 4/x +1/3 = x/(x-3)# and find any extraneous solutions?

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Answer 1 · 2018-06-01T10:26:26

#x=6#

#3/(x-3)+4/x+1/3=x/(x-3)#

#4/x+1/3=(x-3)/(x-3)#

I assumed #x!=3#, because #x=3# is extraneous solution. Hence,

#4/x+1/3=1#

#4/x=2/3#

#x=3*4/2=6#