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

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Answer 1 · 2018-05-20T00:23:42

#y=2+-sqrt14#

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

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

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

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

#y=(-b+-sqrt(b^2-4ac))/(2a)#

#y=(4+-sqrt((-4)^2-4(1)(-10)))/(2times1)#

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

#y=(4+-sqrt56)/2#

#y=(4+-2sqrt14)/2#

#y=2+-sqrt14#