rational expression algebra 2?

Question

1/y-1 + 2/y+1 = 0


the 1 and 2 are over the equation like a fraction

Answer 1

Should be written as:-
1 / (y - 1) + 2 / (y + 1) = 0
(y + 1) + 2 (y - 1) = 0
y + 1 + 2y - 2 = 0
3y = 1
y = 1 / 3

Answer 2

(1/[y - 1]) + (2/[y + 1]) = 0
1(y + 1) + 2(y - 1) = 0
y + 1 + 2y - 2 = 0
3y - 1 = 0
3y = 1
y = 1/3

Answer: y = 1/3

Proof:
(1/[1/3 - 1]) + (2/[1/3 + 1]) = 0
(1/[- 2/3]) + (2/[1 1/3]) = 0
(1 * - 3/2) + (2 *3/4) = 0
- 3/2 + 6//4 = 0
- 3/2 + 3/2 = 0

Answer 3

(y+1)+2(y-1)=0 if y≠1, -1
3y=1 --> y=1/3

Saludos.