okay the problem is 2/y^2-16 = 10 y^2+4y can anyone help me?

Answer 1

Wasn't it 2/y^2-16 = 10/y^2+4y?

If so:

2/y^2 - 16 = 10/y^2 + 4y
10 (y^2 - 16) = 2 (y^2 + 4y)
5 (y^2 - 16) = y^2 + 4y
5y^2 - 80 = y^2 + 4y
5y^2 - y^2 - 4y = 80
4y^2 - 4y = 80
4 (y^2 - y) = 80
y^2 - y = 80/4
y^2 - y = 20
y^2 = 20 + y

so now, we have this:

2/y^2 - 16 = 10/y^2 + 4y (where we can replace y^2 with 20 + y)
10 (y^2 - 16) = 2 (y^2 + 4y)
10 (20 + y - 16) = 2 (20 + y + 4y)
5 (20 + y - 16) = 20 + y + 4y
100 + 5y - 80 = 20 + 5y
20 + 5y = 20 + 5y

Answer 2

Multiply both sides by y^2:
2 - 16y^2 = 10y^4 + 4y^3, then by collecting terms
10y^4 + 4y^3 + 16y^2 - 2 = 0, a polynomial