How do you solve #8+6/(x-5)=(x-13)/(x-5)# and check for extraneous solutions?

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Answer 1 · 2016-09-08T14:43:01

#x = 3#

Check for restrictions on #x# first. The denominator may not be 0.
#x-5 = 0 rarr x = 5" " rarr# So #x!= 5#

The denominators of the fractions are the same, so we can put them on the same side and add them together.

#8 = (x-13)/(x-5) -6/(x-5)#

#8 = (x-13-6)/(x-5)#

#8 = (x-19)/(x-5) " "larr# now cross-multiply

#8x-40 = x-19#

#8x-x = 40-19#

#7x = 21#

#x = 3#