i need help with this problem 2/y2-16=10/y2+4y?

Question

the problem is 2 over y2-16 = 10 y2+4y

Answer 1

2/(y^2-16)=10/(y^2+4y)
=>2/(y+4)(y-4)=10/y(y+4)
=>1/(y-4)=5 /y
=>5y-20=y
=>5y-y=20
=>4y=20
=>y=20/4=5

Answer 2

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


there you go mate.

Answer 3

21

Answer 4

If doing a power (I saw you write y2) write it like y^2