How do you solve #6/(x^2-8x) = 1/x + 3/(x^2-8x)# and find any extraneous solutions?

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Answer 1 · 2017-07-19T10:41:36

Solution: #x=11#

# 6/(x^2-8x) = 1/x + 3/(x^2-8x) or #

# 6/(x^2-8x) - 3/(x^2-8x) = 1/x or#

# (6-3)/(x^2-8x) = 1/x or 3/(x^2-8x) = 1/x# or

#x^2-8x=3x or x^2-11x =0# or

#x(x-11)=0 #. Eeither #x=0 or x-11= 0 :. x=11#

#1/x or 3/(x^2-8x) # is undefined for #x=0 :. x != 0#

Solution: #x=11# [Ans]