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Cubi-Cal

  • 2 years ago

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

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  1. SnuggieLad
    • 2 years ago
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    Where are you stuck?

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

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

  4. tcarroll010
    • 2 years 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
    • 2 years ago
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    Right! So what do I multiply to get 1 and adds up to be 1?

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

  7. tcarroll010
    • 2 years 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
    • 2 years ago
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    So to start, can you factor x^2 - 1 ?

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

  10. tcarroll010
    • 2 years ago
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    Think of 1 as 1^2

  11. tcarroll010
    • 2 years 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
    • 2 years ago
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    Okay, so my final answer would be x^2-x/x^2-1?

  13. tcarroll010
    • 2 years 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
    • 2 years ago
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    uh.. x/x-1?._.

  15. Cubi-Cal
    • 2 years ago
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    Or would it be +1?

  16. tcarroll010
    • 2 years 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
    • 2 years ago
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    Ohhhh! Okay thanks!

  18. tcarroll010
    • 2 years ago
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    uw!

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

  20. tcarroll010
    • 2 years ago
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    ok, see you in another question!

  21. Cubi-Cal
    • 2 years ago
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    Okay!

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