How do you solve #(y+2)/y = 1/(y-5)#?

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Answer 1 · 2015-05-27T01:09:31

Provide #y!=0# and #y!=5# (either of which would result in an attempt to divide by zero)

#(y+2)/y = 1/(y-5)#

can be re-written as
#(y+2)(y-5) = y#

#y^2 -3y -10 = y#

#y^2-4y -10 = 0#

using the quadratic formula for roots
#y = (-b+-sqrt(b^2-4ac)/(2a)

#= (4+-sqrt(16+40))/2#

#= 2+- sqrt(14)#