# Homework Help: Solve For X

1. Dec 8, 2004

### aisha

I tried to do this problem soo many times but cant even get close to the answer, I dont know what to do.

[(x)/(x-2)]+2=[(5x)/(x+2)]+[(3x+1)/(x^(2)-4)]

ok I know the LCD is (x+2)(x-2) and that x^(2)-4 is a perfect square that can be written as (x-2) (x+2)

But I cant get the solution it says x=3, 3/2

2. Dec 8, 2004

### marlon

dear aisha

just make sure that the denominator is equal to x²-4 everywhere...
then you can forget about it (ofcourse x can't be 2 or -2)

like this :

[(x)(x+2)+2(x²-4)=[(5x)(x-2)+[(3x+1)]

Then solve for x...

can you go on from here ???

regards
marlon

3. Dec 8, 2004

### Diane_

If you multiply through by the LCD, you'll get:

(x/(x-2))(x-2)(x+2) + 2(x^2-4) = (5x/(x+2))(x-2)(x+2) + ((3x+1)/(x^2-4))(x+2)(x-2)

Simplifying gives you

x(x+2) + 2(x^2 - 4) = 5x(x-2) + (3x+1)

Multiply it out, collect similar terms, and you'll have a quadratic. That can be factored, or you can use the Quadratic Formula. Don't forget to distribute minus signs.

Does that help?

4. Dec 8, 2004

### aisha

YES THAT HELPS, BUT The quadratic formula I got after multiplied out and collected the like terms was -2x^(2)-x+1 and this factored out and x=-1/2 or x=-1 but the solutions were x=3, 3/2 am I wrong or is the solution wrong?

5. Dec 8, 2004

### Diane_

I think you're off somewhere. The quadratic I ended up with was

2x^2 - 9x + 9 = 0

which has the "proper" solutions. It's probably a simple math error. Go back through it and check it.

Unsolicited advice: I have found that students often make those little errors because they're trying to go too fast. I know that the grunt work of algebra can be snore-inducing, but you need to take it slowly at first. Speed comes with experience, and experience is what you're lacking right now. You'll get there, though. All it takes is time. :) </mommy-mode>

6. Dec 8, 2004

### aisha

Thanks Diane I got it FINALLY