How do you solve #2/(v+6) + 1/(v^2+5v-6) = 6/(v^2+5v-6)# and find any extraneous solutions?

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Answer 1 · 2016-06-06T00:15:44

#v = 7/2#

#v^2+5v-6=(v-1)(v+6)#
then the common denominator is #(v+1)(v-6)# so

#2/(v+6) + 1/(v^2+5v-6) = 6/(v^2+5v-6)#

is equivalent to

#(2(v-1))/((v-1)(v+6)) + 1/((v-1)(v+6)) = 6/((v-1)(v+6))#

or

# 2(v-1)+1-6=0 #

so

#v = 7/2#