Algebra II question; will award best answer

Can someone please show me the steps to do these 2 problems? I just need to see where I'm making my mistakes.

3/a + 1/2 = 11/5a

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

1) Multiply both sides by a.

a(3/a + 1/2) = a(11/5a)
a * (3/a) + a * (1/2) = a * (11/5a)
3 + a/2 = 11/5

Multiply both sides by 2.

2(3 + a/2) = 2(11/5)
6 + a = 22/5

Multiply both sides by 5.

5(6 + a) = 5(22/5)
30 + 5a = 22
5a = -8
a = -8/5

2) Mulitply both sides by y+2.

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

Multiply both sides by y.

y(4) = y(y + 2 - 8/y)
4y = y^2 + 2y - 8
y^2 - 2y - 8 = 0
(y - 4)(y + 2) = 0
y - 4 = 0 or y + 2 = 0
y = 4 or y = -2

y = -2 forces us to divide by zero so we get rid of this.

y = 4

3/a + 1/2 =

lcd = 2*a

(6 + a)/(2*a)

4/y + 2 =

lcd = y

(4 + 2*y)/y = 2*(2 + y)/y