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Cubi-Cal

Add and simplify... x/(x-1) - 2x/(x^2-1)

  • 9 months ago
  • 9 months ago

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  1. SnuggieLad
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    Where are you stuck?

    • 9 months ago
  2. SnuggieLad
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    \[\frac{ x }{ (x-1) }-\frac{2x}{x^2-1}\]

    • 9 months ago
  3. Cubi-Cal
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    I'm pretty sure there is some factoring int there, but I'm not sure.

    • 9 months ago
  4. tcarroll010
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    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.

    • 9 months ago
  5. Cubi-Cal
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    Right! So what do I multiply to get 1 and adds up to be 1?

    • 9 months ago
  6. tcarroll010
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    Hint: the denominator of the second term is the difference of 2 perfect squares.

    • 9 months ago
  7. tcarroll010
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    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.

    • 9 months ago
  8. tcarroll010
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    So to start, can you factor x^2 - 1 ?

    • 9 months ago
  9. tcarroll010
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    Another hint:\[a ^{2} - b ^{2} = (a + b)(a - b)\]

    • 9 months ago
  10. tcarroll010
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    Think of 1 as 1^2

    • 9 months ago
  11. tcarroll010
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    If you substitute a=x and b=1 into the equation I wrote out, you should be able to factor:\[x ^{2} - 1\]

    • 9 months ago
  12. Cubi-Cal
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    Okay, so my final answer would be x^2-x/x^2-1?

    • 9 months ago
  13. tcarroll010
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    No, but that's a good intermediate answer. You need to factor out "x-1" from the top and bottom.

    • 9 months ago
  14. Cubi-Cal
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    uh.. x/x-1?._.

    • 9 months ago
  15. Cubi-Cal
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    Or would it be +1?

    • 9 months ago
  16. tcarroll010
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    \[\frac{ x }{ x - 1 } - \frac{ 2x }{ x ^{2} - 1 } = \frac{ x }{ x - 1 } - \frac{ 2x }{ (x - 1)(x + 1) } \] \[\frac{ x(x + 1) }{ (x - 1)(x + 1) } - \frac{ 2x }{ (x - 1)(x + 1) } = \frac{ x ^{2} + x - 2x }{ (x - 1)(x + 1) }\] \[\frac{ x ^{2} - x }{ (x - 1)(x + 1) } = \frac{ x\cancel{(x - 1)} }{ \cancel{(x - 1)}{(x + 1)} }\]

    • 9 months ago
  17. Cubi-Cal
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    Ohhhh! Okay thanks!

    • 9 months ago
  18. tcarroll010
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    uw!

    • 9 months ago
  19. Cubi-Cal
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    Would you mind helping me with some other problems? My teacher isn't available.

    • 9 months ago
  20. tcarroll010
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    ok, see you in another question!

    • 9 months ago
  21. Cubi-Cal
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    Okay!

    • 9 months ago
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