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Cubi-Cal

  • one year ago

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

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  1. SnuggieLad
    • one year ago
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    Where are you stuck?

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

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

  4. tcarroll010
    • one year ago
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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.

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

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

  7. tcarroll010
    • one year ago
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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.

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

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

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

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

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

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

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

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

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

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

  18. tcarroll010
    • one year ago
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    uw!

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

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

  21. Cubi-Cal
    • one year ago
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    Okay!

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