1. ## Simplifying Rational Expressions

I'm doing my victory lap (Grade 13), and I need a credit in Advanced Functions.

I'm already working my butt off to keep up with the review, but I feel that I need a solid foundation of these concepts before I can really excel in this course. So, If anybody could give me a hand...

I looked at (x^2 - 1) as a difference of squares, and then multiplied each side by the other's denominator to achieve a common denominator.

Then I canceled, and ended up with this:

Collected like terms...

Then, I thought - I'm way off track here. The answer is apparently

I'm hoping someone can do the answer, and explain to me where I went wrong. I've tried it several times, but I'm just too rusty, I think. Thanks a million to whoever's up to the task!

- Cameron

2. [quote=Cam5;183971]I'm doing my victory lap (Grade 13), and I need a credit in Advanced Functions.

I'm already working my butt off to keep up with the review, but I feel that I need a solid foundation of these concepts before I can really excel in this course. So, If anybody could give me a hand...

your expression before cancelling; $\displaystyle =\frac{4(x+1)(x-1) + 5(x-1)}{(x-1)(x-1)(x+1)}$

take common (x-1) from numerator, $\displaystyle =\frac{(x-1)[4(x+1) + 5]}{(x-1)(x-1)(x+1)}$

You should cancel one (x-1) from both Numerator and denominator. After cancellation, you should have $\displaystyle =\frac{4(x+1) + 5}{(x-1)(x+1)}$ $\displaystyle =\frac{4x+9}{(x-1)(x+1)}$

$\displaystyle =\frac{4}{x-1}+\frac{5}{x^2-1}$

$\displaystyle =\frac{4}{x-1}+\frac{5}{(x-1)(x+1)}$

$\displaystyle =\frac{4(x+1) + 5}{(x-1)(x+1)}$

$\displaystyle =\frac{4x+4 + 5}{(x-1)(x+1)}$

$\displaystyle =\frac{4x+9}{(x-1)(x+1)}$

3. Thanks a lot Shyam! That clears it up for me.

I'm still rusty on the factoring, as well, so that little gem didn't present itself to me right away, even when you worked it out for me.