2x/x^2-1 - 1/x+1
i got x+1 as my final answer
How have you got $\displaystyle x + 1 $? I don't understand what you're trying to achieve.
I'll presume the denominator on the left side is $\displaystyle x^2 - 1 $
$\displaystyle \frac {2x}{(x^2 - 1)} - \frac{1}{(x + 1)} $
$\displaystyle \frac{(2x)(x + 1) - (1)(x^2 - 1)}{(x^2 - 1)(x + 1)} $
$\displaystyle \frac{(2x^2 + 2x) - (x^2 - 1)}{(x^2 - 1)(x + 1)} $
$\displaystyle \frac{x^2 + 2x + 1}{x^3 + x^2 - x -1}$
Wouldn't it be better to use the least common denominator?
$\displaystyle \frac{2x}{x^2- 1}- \frac{1}{x+1}$
Since $\displaystyle x^2- 1= (x+ 1)(x- 1)$, that is the least common denominator and we can just multiply numerator and denominator of the second fraction by x- 1:
$\displaystyle \frac{2x}{x^2- 1}- \frac{x- 1}{(x+1)(x-1)}= \frac{2x}{x^2- 1}- \frac{x- 1}{x^2- 1}= \frac{x+ 1}{x^2- 1}= \frac{1}{x- 1}$
AStartledDeer, your $\displaystyle \frac{x^2+ 2x+ 1}{x^3+ x^2- x- 1}= \frac{(x+ 1)^2}{(x+1)(x^2- 1)}= \frac{(x+1)^2}{(x+1)^2(x- 1)}= \frac{1}{x- 1}$
That is still not "x+1".
"that is the least common denominator and we can just multiply numerator and denominator of the second fraction by x- 1:"
why did you only multiple the second fraction? can you even do that?
what i did was mutiple both fractions by (x+1)(x-1) THUS making 2x - x-1= x-1
i'm aware i originally said my answer was x+1. this has been revised to x-1
The most efficient method is to find the LOWEST common denominator. To do this you need to factorise each denominator and see which factors are missing from each.
In this case we have $\displaystyle \displaystyle \begin{align*}x^2 - 1 = (x - 1)(x + 1) \end{align*}$ as the first denominator, and $\displaystyle \displaystyle \begin{align*} x + 1 \end{align*}$ as the second denominator. Can you see that you already have a factor of $\displaystyle \displaystyle \begin{align*} x + 1 \end{align*}$ in your first denominator? So all you need to get the lowest common denominator is that extra factor of $\displaystyle \displaystyle \begin{align*} x - 1 \end{align*}$.
It is basically the same thing you learned in arithmetic: in order to add or subtract fractions, you need to get the same denominator.
If, for example, you want to add $\displaystyle \frac{1}{3}$ and $\displaystyle \frac{1}{6}$ you could get a common denominator by multiplying, both numerator and denominator of the first fraction by 6, and of the second fraction by 3: $\displaystyle \frac{1}{3}+\frac{1}{6}= \frac{6}{18}+ \frac{3}{18}= \frac{9}{18}$ which reduces to $\displaystyle \frac{1}{2}$.
But since 6= 2(3), we can get the "least common denominator" of 6 by multiplying numerator and denominator of the first fraction by 2: $\displaystyle \frac{1}{3}+ \frac{1}{6}= \frac{2}{6}+ \frac{1}{6}= \frac{3}{6}= \frac{1}{2}$.