1=(3/(y+2))+(1/(y-2))

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Solution for 1=(3/(y+2))+(1/(y-2)) equation: 


D( y )

y+2 = 0

y-2 = 0

y+2 = 0

y+2 = 0

y+2 = 0 // - 2

y = -2

y-2 = 0

y-2 = 0

y-2 = 0 // + 2

y = 2

y in (-oo:-2) U (-2:2) U (2:+oo)

1 = 3/(y+2)+1/(y-2) // - 3/(y+2)+1/(y-2)

1-(3/(y+2))-(1/(y-2)) = 0

1-3*(y+2)^-1-(y-2)^-1 = 0

1-3/(y+2)-1/(y-2) = 0

(-3*(y-2))/((y+2)*(y-2))+(-1*(y+2))/((y+2)*(y-2))+(1*(y+2)*(y-2))/((y+2)*(y-2)) = 0

1*(y+2)*(y-2)-3*(y-2)-1*(y+2) = 0

y^2-4*y-4+4 = 0

y^2-4*y = 0

y^2-4*y = 0

y*(y-4) = 0

y-4 = 0 // + 4

y = 4

y*(y-4) = 0

(y*(y-4))/((y+2)*(y-2)) = 0

(y*(y-4))/((y+2)*(y-2)) = 0 // * (y+2)*(y-2)

y*(y-4) = 0

( y-4 )

y-4 = 0 // + 4

y = 4

( y )

y = 0

y in { 4, 0 }

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