(1/6x)+(1/7x)=(1/42)-(1/x)

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Solution for (1/6x)+(1/7x)=(1/42)-(1/x) equation: 


D( x )

x = 0

x = 0

x = 0

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

(1/6)*x+(1/7)*x = 1/42-(1/x) // - 1/42-(1/x)

(1/6)*x+(1/7)*x+1/x-(1/42) = 0

(1/6)*x+(1/7)*x+1/x-1/42 = 0

13/42*x^1+1*x^-1-1/42*x^0 = 0

(13/42*x^2-1/42*x^1+1*x^0)/(x^1) = 0 // * x^2

x^1*(13/42*x^2-1/42*x^1+1*x^0) = 0

x^1

(13/42)*x^2+(-1/42)*x+1 = 0

(13/42)*x^2+(-1/42)*x+1 = 0

DELTA = (-1/42)^2-(1*4*(13/42))

DELTA = -2183/1764

DELTA < 0

x in { } 

x belongs to the empty set

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