(1/3)-(2/3y)=(7/y)

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Solution for (1/3)-(2/3y)=(7/y) equation: 


D( y )

y = 0

y = 0

y = 0

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

1/3-((2/3)*y) = 7/y // - 7/y

1/3-((2/3)*y)-(7/y) = 0

(-2/3)*y-7*y^-1+1/3 = 0

1/3*y^0-2/3*y^1-7*y^-1 = 0

(1/3*y^1-2/3*y^2-7*y^0)/(y^1) = 0 // * y^2

y^1*(1/3*y^1-2/3*y^2-7*y^0) = 0

y^1

(-2/3)*y^2+(1/3)*y-7 = 0

(-2/3)*y^2+(1/3)*y-7 = 0

DELTA = (1/3)^2-(-7*4*(-2/3))

DELTA = -167/9

DELTA < 0

y in { } 

y belongs to the empty set

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