Studying for Math Final need some help.

[Plowed]

Been doing this final study guide which has 120 problems. I am extremely rusty on the beginning chapters so I only got 22 problems done in the past 3 hours.
I have a problem here that I can't figure out how to solve. I can't find anything in past exams/homework or notes that is similar to this.

[Marmaladez]

Okay bear with me since I haven't done math since first year university... and that was like 3 and a half years ago. And I am procrastinating studying for my path final soooo here goes!

8/(y+3) - 2/(y-3) = 12/(y^2 - 9)

Seeing that, I know y^2-9 is the same as (y-3)(y+3). So you would then do this:

8(y-3)/(y+3)(y-3) - 2(y+3)/(y-3)(y+3) = 12/(y-3)(y+3)

Now that all the denominators are the same, you can add/subtract (and bring that 12 over!)!

8(y-3) - 2(y+3) - 12 / (y-3)(y+3) = 0

And expand!

8y - 24 - 2y -6 - 12 / (y-3)(y+3) = 0

Group like terms etc.

8y - 2y - 24 - 6 - 12 / (y-3)(y+3) = 0

6y - 42 / (y-3)(y+3) = 0

Now just looking at the numerator:

6y - 42 = 0

6y = 42

y = 7

And there's your answer! y=7! As for the denominator... (y-3)(y+3), I believe (this is vaguely what I remember), it has to do with the asymptotes. I hope that helps! And good luck

[Plowed]

Oh my god... I know what I was doing wrong. I forgot to make the 6 negative when multiplying with the 2. I kept getting 5 and couldn't figure it out. Thank you. I will post up any other problems I have trouble with in here.

[Marmaladez]

Haha! It's those little things that'll get you I used to make those mistakes all the time, so frustrating! I'll probably be floating around so I'll check back now and then if you still need help

[Fuzzzie]

Eh. We like to nudge people in the right direction for homework assignments but we aren't going to do your work for you.