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

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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]

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