Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
1  
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 at 22:59
    
I almost always forget that. Thank you for reminding! –  user160137 Jul 15 at 11:41

3 Answers 3

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.

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

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

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

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.

share|improve this answer
    
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 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 at 14:30

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

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

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.