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

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Answer 1 · 2016-09-15T16:49:21

#v=21#

#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#