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

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


D( y )

y = 0

y = 0

y = 0

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

3*y+1/y-1+3 = y+2/y+1 // - y+2/y+1

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

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

2*y^1-1*y^-1+1*y^0 = 0

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

y^1*(2*y^2+1*y^1-1*y^0) = 0

y^1

2*y^2+y-1 = 0

2*y^2+y-1 = 0

DELTA = 1^2-(-1*2*4)

DELTA = 9

DELTA > 0

y = (9^(1/2)-1)/(2*2) or y = (-9^(1/2)-1)/(2*2)

y = 1/2 or y = -1

y in { -1, 1/2} 

y in { -1, 1/2 }

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