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Simplify: $$\frac{\frac{1}{x-1} + \frac{1}{x^2-1}}{x-\frac 2{x + 1}}$$

This is what I did.

Step 1: I expanded $x^2-1$ into: $(x-1)(x+1)$. And got: $\frac{x+1}{(x-1)(x+1)} + \frac{1}{(x-1)(x+1)}$
Step 2: I calculated it into: $\frac{x+2}{(x-1)(x+1)}$
Step 3: I multiplied $x-\frac{2}{x+1}$ by $(x-1)$ as following and I think this part might be wrong:

  • $x(x-1) = x^2-x$. Times $x+1$ cause that's the denominator =
  • $x^3+x^2-x^2-x = x^3-x$.
  • After this I added the $+ 2$
  • $\frac{x^3-x+2}{(x-1)(x+1)}$

Step 4: I canceled out the denominator $(x-1)(x+1)$ on both sides.
Step 5: And I'm left with: $\frac{x+2}{x^3-x+2}$
Step 6: Removed $(x+2)$ from both sides I got my UN-correct answer: $\frac{1}{x^3}$

Please help me. What am I doing wrong?

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Don't forget that for this problem, you have as an assumption $x \ne 1$, $x \ne -1$, and $x - \frac{2}{x+1} \ne 0 \Rightarrow x \ne -2$. So whatever your simplified form is, you need to add $x \not \in \{1, -1, -2\}$, even if your simplified form is defined for those values. – DanielV Jul 14 '14 at 22:59
I almost always forget that. Thank you for reminding! – user160137 Jul 15 '14 at 11:41
up vote 1 down vote accepted

First recall that \[ \frac{a}{b} \pm \frac{c}{d} =\frac{ad \pm cb}{bd} \] And \[ \frac{\frac{a}{b}}{\frac{c}{d}} =\frac{ad}{bc} \]

Now just simplify, no fancy fractions needed: \[ \frac{\frac{1}{x-1}+\frac{1}{x^2-1}}{x-\frac{2}{x+1}} = \frac{\frac{(x^2-1)+(x-1)}{(x-1)(x^2-1)}}{\frac{x(x+1)-2}{x+1}} = \frac{\frac{(x-1)(x+1)+(x-1)}{(x-1)^2(x+1)}}{\frac{(x+2)(x-1)}{x+1}} = \frac{\frac{(x-1)(x+2)}{(x-1)^2(x+1)}}{\frac{(x+2)(x-1)}{x+1}} = \frac{(x-1)(x+2)(x+1)}{(x-1)^3(x+1)(x+2)} = \frac{1}{(x-1)^2} \] I attempted to be as clear as possible. If you'd like me to elaborate further, just let me know.

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I now understand how I can get (x+2)(x-1). Thanks for the clear explanation! – user160137 Jul 15 '14 at 11:35
Your welcome. Glad I could help. – k170 Jul 15 '14 at 15:47

You need to multiply the +2 by (x-1) before you add it to $x^3-x$ because


Also, in step 6, you can't remove (x+2) from top and bottom. Firstly, $x^3-x+2=x^3-(x-2)$, so there is no $x+2$ in the denominator. Secondly, you need (x+2) to be a factor of the whole denominator, not just part.

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That makes sense, since I only multiplied x and the 2 only has (x+1). I'm gonna try to solve it with that. Thanks. – user160137 Jul 14 '14 at 14:22
But If I multiply it with 2 I got 2x-2. If I add in front the x³-x-(2x-2). I get: x³-3x+2. That can probably be factored into x+2. Let me check. – user160137 Jul 14 '14 at 14:30

It simplifies things a lot if you just multiply the numerator and denominator by $(x+1)(x-1)$

$$\frac{\frac{1}{x-1} + \frac{1}{x^2-1}}{x-\frac 2{x + 1}}\cdot\frac{\frac{(x+1)(x-1)}{1}}{\frac{(x+1)(x-1)}{1}} = \frac{(x+1)+1}{x(x+1)(x-1)-2(x-1)}=\frac{x+2}{(x-1)(x(x+1)-2)}$$ $$=\frac{x+2}{(x-1)(x^2+x-2)}=\frac{x+2}{(x-1)(x+2)(x-1)}=\frac{1}{(x-1)^2}$$

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