New class of pre-calc, and I've forgotten some stuff from Algebra II. We were given a diagnostic test to see how we've changed from last term to this term.

I've been stuck on this problem for a bit, I'll try to format it the best I can.

Solve: [(r + 3) / (r^2 - 1)] + [(r - 3) / (r^2 - r)] = [2r / (r^2 + r)]

I used all the common denominators; r, r+1, and r-1.

This gives me:

[r(r+3)(r+1)(r-1) / (r+1)(r-1)] + [r(r-3)(r+1)(r-1) / r(r-1)] = [2r(r)(r+1)(r-1) / r(r+1)]

Canceling out all the fractions, this leaves me with:

[r(r+3)] + [(r-3)(r+1)] = [2r(r-1)] =

r^2 + 3r + r^2 + r - 3r - 3 = 2r^2 - 2r

Combine/remove;

2r^2 + r - 3 = 2r^2 - 2r

Move to one side: 3r - 3 = 0

3r = 3
r = 1

However, in the first fraction, if r = 1, (1^2 - 1) = 0, and that would leave it indivisible.

Any help would be appreciated, thanks.

Without doing it myself, and reading what you've done

looks like the answer is no soultion.

You've lost me a bit here:

I used all the common denominators; r, r+1, and r-1.


Your demoniators are r^2-1, r^2-r, and r^2 + r, are they not?

Yes, but I factored them to their lowest components.

Please do explain.

If you're talking of factoring out when there are 2x r's present, your r^2-1 can't be factored that way. it would be r(r-1/r)

r^2-r can become r(r-1), but r^2-1 cannot.

? Not sure what you're saying, I factored r^2 - r and r^2 - 1 differently.

r^2 - r = r (r - 1)

r^2 - 1 = (r+1) (r-1)

I see what you did now, I thought you were doing something else, nevermind, I'll take a look at it when I get a chance tommorow.

Nvm the answer is no solution, I thought I did something wrong since I didn't do any algebra in a while and the first solution didn't work.