1/y-12=1/3y

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Solution for 1/y-12=1/3y equation: 


D( y )

y = 0

y = 0

y = 0

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

1/y-12 = (1/3)*y // - (1/3)*y

1/y-((1/3)*y)-12 = 0

(-1/3)*y+1/y-12 = 0

1*y^-1-1/3*y^1-12*y^0 = 0

(1*y^0-1/3*y^2-12*y^1)/(y^1) = 0 // * y^2

y^1*(1*y^0-1/3*y^2-12*y^1) = 0

y^1

(-1/3)*y^2-12*y+1 = 0

(-1/3)*y^2-12*y+1 = 0

DELTA = (-12)^2-(1*4*(-1/3))

DELTA = 436/3

DELTA > 0

y = ((436/3)^(1/2)+12)/(2*(-1/3)) or y = (12-(436/3)^(1/2))/(2*(-1/3))

y = -3/2*((436/3)^(1/2)+12) or y = -3/2*(12-(436/3)^(1/2))

y in { -3/2*((436/3)^(1/2)+12), -3/2*(12-(436/3)^(1/2))} 

y in { -3/2*((436/3)^(1/2)+12), -3/2*(12-(436/3)^(1/2)) }

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