(3/y^2-9)+(4/y+3)=1

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Solution for (3/y^2-9)+(4/y+3)=1 equation: 


D( y )

y = 0

y^2 = 0

y = 0

y = 0

y^2 = 0

y^2 = 0

1*y^2 = 0 // : 1

y^2 = 0

y = 0

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

4/y+3/(y^2)-9+3 = 1 // - 1

4/y+3/(y^2)-9-1+3 = 0

4*y^-1+3*y^-2-7 = 0

t_1 = y^-1

3*t_1^2+4*t_1^1-7 = 0

3*t_1^2+4*t_1-7 = 0

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

DELTA = 100

DELTA > 0

t_1 = (100^(1/2)-4)/(2*3) or t_1 = (-100^(1/2)-4)/(2*3)

t_1 = 1 or t_1 = -7/3

t_1 = -7/3

y^-1+7/3 = 0

1*y^-1 = -7/3 // : 1

y^-1 = -7/3

-1 < 0

1/(y^1) = -7/3 // * y^1

1 = -7/3*y^1 // : -7/3

-3/7 = y^1

y = -3/7

t_1 = 1

y^-1-1 = 0

1*y^-1 = 1 // : 1

y^-1 = 1

-1 < 0

1/(y^1) = 1 // * y^1

1 = 1*y^1 // : 1

1 = y^1

y = 1

y in { -3/7, 1 }

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