How do you solve #(y+4)/(y-2)+6/(y-2)=1/(y+3)#?

Question

1636 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-12T12:04:27

Solution: #y=-4 ,y= -8#

#(y+4)/(y-2)+6/(y-2)=1/(y+3)#. Multiplying by #(y-2)(y+3)#

on both sides we get, #(y+4)(y+3)+6(y+3)=y-2# or

#y^2+7y+12+6y+18-y+2=0# or

#y^2+12y+32=0# or

#y^2+8y+4y+32=0# or

#y(y+8)+4(y+8)=0# or

#(y+8)(y+4)=0:.# either #y+8=0:. y=-8#

or #y+4=0:. y=-4#

Solution: #y=-4 ,y= -8# [Ans]