7/x+4/3=7/5-5/3x

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Solution for 7/x+4/3=7/5-5/3x equation: 


D( x )

x = 0

x = 0

x = 0

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

7/x+4/3 = 7/5-((5/3)*x) // - 7/5-((5/3)*x)

(5/3)*x+7/x+4/3-(7/5) = 0

(5/3)*x+7/x+4/3-7/5 = 0

5/3*x^1+7*x^-1-1/15*x^0 = 0

(5/3*x^2-1/15*x^1+7*x^0)/(x^1) = 0 // * x^2

x^1*(5/3*x^2-1/15*x^1+7*x^0) = 0

x^1

(5/3)*x^2+(-1/15)*x+7 = 0

(5/3)*x^2+(-1/15)*x+7 = 0

DELTA = (-1/15)^2-(4*7*(5/3))

DELTA = -10499/225

DELTA < 0

x in { } 

x belongs to the empty set

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