How do you solve #1/v+(3v+12)/(v^2-5v)=(7v-56)/(v^2-5v)# and check for extraneous solutions?

1 Answer
Sep 15, 2016

Answer:

#v=21#

Explanation:

#1/v+(3v+12)/(v^2-5v)=(7v-56)/(v^2-5v)#

#1/v+(3v+12)/(v^2-5v)-(7v-56)/(v^2-5v)=0#

The common denominator is #v^2-5v=v(v-5)#

#(v-5+3v+12-(7v-56))/(v^2-5v)=0#

#(v-5+3v+12-7v+56)/(v^2-5v)=0#

#(v+3v-7v-5+12+56)/(v^2-5v)=0#

#(-3v+63)/(v^2-5v)=0#

#-3v+63=0#
#-3v=-63#
#v=(-63)/(-3)#
#v=21#