1/x+1 + 5x/(x^2-1) = 4/x-1

What on earth is the lcd! I forgot how to do this easy crap! btw thats to the second power minus 1.

The last time I solved a problem like this, Ronald Reagan was president.

Wrong luke

Could you properly parenthesis that up.. I'm too afraid to assume anything, heh.

Yah, I didn't actually read what he needed. I just know Wolfram Alpha is the shit.

1/(x+1) + 5x/(x^2-1) = 4/(x-1)

Start with the left side:

1/(x+1) + 5x/(x^2-1)

Factor the right term's denominator:

5x/(x^2-1) = 5x/[(x+1)(x-1)]

Both terms of the left side share a factor of (x+1) so

Multiply 1/(x+1) by (x-1) and simplify to get:

(6x - 1)/[(x+1)(x-1)] = 4/(x-1)

And I think you can move forward from there.

This is going to be fun.

1/x+1 + 5x/(x^2-1) = 4/x-1

Let's see. You have:

1/(x+1) = (x+1)^-1 (I don't know why I put this here, but it might be handy later)

5x/(x^2-1) //I want to suppose this is 5x/((x?) -1) and not 5x/( x^(2-1) )

So:

5x/(x?-1) = 4/(x-1) - 1/(x+1)

I'll take the right part:

[ 4 (x+1) / (x+1)*(x-1) ] - [ (x-1) / (x+1)*(x-1) ]

[ (4x+4) / ( x? - x + x -1) ] - [ (x-1) / ( x? - x + x -1) ]

[ (4x+4) / (x?-1) ] - [ (x-1) / (x?-1) ]

Join them:

[ (4x+4) - (x-1) ] / (x?-1)

Wait, where was I going?

I forget I barely passed my maths final. I shouldn't be the one doing this.

EDIT: Ninja'd by Nefarious. Better do what he says, he's probably closer to the right procedure for the above reasons.

Thanks, I got it from there nefarious, didn't think to factor the left side. I was looking for an entire gcf for all three variables. WOOT

Now im pretty sure i use (x+1)(X-1) as the gcf, lets see.

Edit: Yea lol x=5/2 thanks for the help guys.