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

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


D( y )

y = 0

y = 0

y = 0

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

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

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

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

-2/3*y^1-1*y^-1-1*y^0 = 0

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

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

y^1

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

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

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

DELTA = -5/3

DELTA < 0

y in { } 

y belongs to the empty set

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