1/x+1+4/3=7/x+x+4/3x+3

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Solution for 1/x+1+4/3=7/x+x+4/3x+3 equation: 


D( x )

x = 0

x = 0

x = 0

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

1/x+4/3+1 = x+(4/3)*x+7/x+3 // - x+(4/3)*x+7/x+3

1/x-x-((4/3)*x)-(7/x)+4/3-3+1 = 0

(-4/3)*x-x+1/x-7*x^-1+4/3-3+1 = 0

-7/3*x^1-6*x^-1-2/3*x^0 = 0

(-7/3*x^2-2/3*x^1-6*x^0)/(x^1) = 0 // * x^2

x^1*(-7/3*x^2-2/3*x^1-6*x^0) = 0

x^1

(-7/3)*x^2+(-2/3)*x-6 = 0

(-7/3)*x^2+(-2/3)*x-6 = 0

DELTA = (-2/3)^2-(-6*4*(-7/3))

DELTA = -500/9

DELTA < 0

x in { } 

x belongs to the empty set

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