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Cubi-Cal Group Title

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

  • one year ago
  • one year ago

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  1. SnuggieLad Group Title
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    Where are you stuck?

    • one year ago
  2. SnuggieLad Group Title
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    \[\frac{ x }{ (x-1) }-\frac{2x}{x^2-1}\]

    • one year ago
  3. Cubi-Cal Group Title
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    I'm pretty sure there is some factoring int there, but I'm not sure.

    • one year ago
  4. tcarroll010 Group Title
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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.

    • one year ago
  5. Cubi-Cal Group Title
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    Right! So what do I multiply to get 1 and adds up to be 1?

    • one year ago
  6. tcarroll010 Group Title
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    Hint: the denominator of the second term is the difference of 2 perfect squares.

    • one year ago
  7. tcarroll010 Group Title
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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.

    • one year ago
  8. tcarroll010 Group Title
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    So to start, can you factor x^2 - 1 ?

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

    • one year ago
  10. tcarroll010 Group Title
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    Think of 1 as 1^2

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

    • one year ago
  12. Cubi-Cal Group Title
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    Okay, so my final answer would be x^2-x/x^2-1?

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

    • one year ago
  14. Cubi-Cal Group Title
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    uh.. x/x-1?._.

    • one year ago
  15. Cubi-Cal Group Title
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    Or would it be +1?

    • one year ago
  16. tcarroll010 Group Title
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    \[\frac{ x }{ x - 1 } - \frac{ 2x }{ x ^{2} - 1 } = \frac{ x }{ x - 1 } - \frac{ 2x }{ (x - 1)(x + 1) } \] \[\frac{ x(x + 1) }{ (x - 1)(x + 1) } - \frac{ 2x }{ (x - 1)(x + 1) } = \frac{ x ^{2} + x - 2x }{ (x - 1)(x + 1) }\] \[\frac{ x ^{2} - x }{ (x - 1)(x + 1) } = \frac{ x\cancel{(x - 1)} }{ \cancel{(x - 1)}{(x + 1)} }\]

    • one year ago
  17. Cubi-Cal Group Title
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    Ohhhh! Okay thanks!

    • one year ago
  18. tcarroll010 Group Title
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    uw!

    • one year ago
  19. Cubi-Cal Group Title
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    Would you mind helping me with some other problems? My teacher isn't available.

    • one year ago
  20. tcarroll010 Group Title
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    ok, see you in another question!

    • one year ago
  21. Cubi-Cal Group Title
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    Okay!

    • one year ago
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