Question

1/y +2 +1/y + 2-2y=8/y3-4y

Answer 1

1/y +2 +1/y + 2-2y=8/y3-4y

1/y+2+1/y+2+(-2)y = - (2*(y2-2y-1))/2

(8/y3)-4y=(4*(y4-2))/y3

-(2*(y2-2y-1))/2 = (4*(y4-2))/y3

-2y+(2/y)+4 = (8/y3)-4y

2y+(2/y)-(8/y3)+4=0

(2(y4+2y3+y2-4))/y=0 

1/y3=0

y4+2y3+y2-4=0

y2+y+2=0

d=b2-4ac=1^2-4*(1*2)=-7

d<0

y=-2,   y=1 

Answer 2

To solve the given equation:

1/y + 2 + 1/(y + 2 - 2y) = 8/(y^3 - 4y)

First, simplify the equation by combining like terms and simplifying fractions:

1/y + 2 + 1/(2 - y) = 8/(y^3 - 4y)

Now, find a common denominator for the fractions:

[(2 - y) + 2y + y(y + 2 - 2y)] / [y(2 - y)(y + 2 - 2y)] = 8/(y^3 - 4y)

Simplify further:

[(2 - y) + 2y + y(y + 2 - 2y)] / [y(2 - y)(y + 2 - 2y)] = 8/(y^3 - 4y)

[(2 - y) + 2y + y^2 + 2y - 2y^2] / [y(2 - y)(y + 2 - 2y)] = 8/(y^3 - 4y)

Combine like terms:

(2 - y + 2y + y^2 + 2y - 2y^2) / [y(2 - y)(y + 2 - 2y)] = 8/(y^3 - 4y)

(-y^2 + 5y + 2) / [y(2 - y)(y + 2 - 2y)] = 8/(y^3 - 4y)

Multiply both sides of the equation by the common denominator to eliminate the denominators:

(-y^2 + 5y + 2)(y^3 - 4y) = 8y(2 - y)(y + 2 - 2y)

Expand and simplify:

-y^5 + 4y^4 + 5y^3 - 20y^2 + 2y^3 - 8y^2 + 10y - 40y + 16 = 16y(2 - y)(y + 2 - 2y)

Combine like terms:

-y^5 + 4y^4 + 7y^3 - 28y^2 - 30y + 16 = 16y(2 - y)(y + 2 - 2y)

Rearrange the equation to set it equal to zero:

-y^5 + 4y^4 + 7y^3 - 28y^2 - 30y + 16 - 16y(2 - y)(y + 2 - 2y) = 0

At this point, it becomes a fifth-degree polynomial equation. Finding exact solutions for general fifth-degree polynomials can be challenging and typically requires numerical methods or approximation techniques.