anonymous
  • anonymous
Add and simplify... x/(x-1) - 2x/(x^2-1)
Mathematics
schrodinger
  • schrodinger
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SnuggieLad
  • SnuggieLad
Where are you stuck?
SnuggieLad
  • SnuggieLad
\[\frac{ x }{ (x-1) }-\frac{2x}{x^2-1}\]
anonymous
  • anonymous
I'm pretty sure there is some factoring int there, but I'm not sure.

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anonymous
  • anonymous
You will want to factor the denominator of that second term. Then you will want to get the LCM (Least Common Multiple) for both terms.
anonymous
  • anonymous
Right! So what do I multiply to get 1 and adds up to be 1?
anonymous
  • anonymous
Hint: the denominator of the second term is the difference of 2 perfect squares.
anonymous
  • anonymous
Once you factor that denominator of that second term, you will see one of the factors as the denominator in the first term. Take the OTHER factor of the denominator of the second term and make a factor of "1" out of it (by putting the factor in the top and bottom) and multiply the first term by that. Then you will have a common denominator.
anonymous
  • anonymous
So to start, can you factor x^2 - 1 ?
anonymous
  • anonymous
Another hint:\[a ^{2} - b ^{2} = (a + b)(a - b)\]
anonymous
  • anonymous
Think of 1 as 1^2
anonymous
  • anonymous
If you substitute a=x and b=1 into the equation I wrote out, you should be able to factor:\[x ^{2} - 1\]
anonymous
  • anonymous
Okay, so my final answer would be x^2-x/x^2-1?
anonymous
  • anonymous
No, but that's a good intermediate answer. You need to factor out "x-1" from the top and bottom.
anonymous
  • anonymous
uh.. x/x-1?._.
anonymous
  • anonymous
Or would it be +1?
anonymous
  • anonymous
\[\frac{ x }{ x - 1 } - \frac{ 2x }{ x ^{2} - 1 } = \frac{ x }{ x - 1 } - \frac{ 2x }{ (x - 1)(x + 1) } \] \[\frac{ x(x + 1) }{ (x - 1)(x + 1) } - \frac{ 2x }{ (x - 1)(x + 1) } = \frac{ x ^{2} + x - 2x }{ (x - 1)(x + 1) }\] \[\frac{ x ^{2} - x }{ (x - 1)(x + 1) } = \frac{ x\cancel{(x - 1)} }{ \cancel{(x - 1)}{(x + 1)} }\]
anonymous
  • anonymous
Ohhhh! Okay thanks!
anonymous
  • anonymous
uw!
anonymous
  • anonymous
Would you mind helping me with some other problems? My teacher isn't available.
anonymous
  • anonymous
ok, see you in another question!
anonymous
  • anonymous
Okay!

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