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

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


D( y )

y = 0

y+y^-1+1/3 = 0

y = 0

y = 0

y+y^-1+1/3 = 0

y+y^-1+1/3 = 0

1*y^1+1*y^-1+1/3*y^0 = 0

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

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

y^1

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

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

DELTA = (1/3)^2-(1*1*4)

DELTA = -35/9

DELTA < 0

y in { } 

y in (-oo:0) U (0:+oo)

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

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

y+y^-1+1/3 = 0

y+y^-1+1/3 = 0

1*y^1+1*y^-1+1/3*y^0 = 0

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

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

y^1

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

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

DELTA = (1/3)^2-(1*1*4)

DELTA = -35/9

DELTA < 0

y in { } 

1 = 0

y-3+1 = 0

y-3+1 = 0

y-2 = 0

y-2 = 0

y-2 = 0 // + 2

y = 2

y = 2

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