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18jonea

  • one year ago

find the domain and simplify

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  1. 18jonea
    • one year ago
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    \[\frac{ x+1 }{ 3x-3 } - \frac{ 4 }{ x^2-3x+2 }\]

  2. Michele_Laino
    • one year ago
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    we have to factorize both denominators

  3. 18jonea
    • one year ago
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    i know and i did the first but dont know how to do the decond first one is 3(x-1)

  4. Michele_Laino
    • one year ago
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    for the second one, you have to search for two numbers such that their sum is 3, and their product is 2

  5. 18jonea
    • one year ago
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    1 and 2

  6. Michele_Laino
    • one year ago
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    correct!

  7. Michele_Laino
    • one year ago
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    so we can write this: \[\Large {x^2} - 3x + 2 = \left( {x - 1} \right)\left( {x - 2} \right)\]

  8. 18jonea
    • one year ago
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    now we have to find the domain

  9. Michele_Laino
    • one year ago
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    therefore: \[\Large \begin{gathered} \frac{{x + 1}}{{3x - 3}} - \frac{4}{{{x^2} - 3x + 2}} = \hfill \\ \hfill \\ = \frac{{x + 1}}{{3\left( {x - 1} \right)}} - \frac{4}{{\left( {x - 1} \right)\left( {x - 2} \right)}} \hfill \\ \end{gathered} \]

  10. Michele_Laino
    • one year ago
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    yes! we have to find the domain

  11. Michele_Laino
    • one year ago
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    in order to do that, we note that we can not divide by zero

  12. 18jonea
    • one year ago
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    Domain is all real numbers except when x = 1 or 2

  13. 18jonea
    • one year ago
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    is that right?

  14. Michele_Laino
    • one year ago
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    so we have to request that: \[\Large \begin{gathered} x - 1 \ne 0 \hfill \\ \left( {x - 1} \right)\left( {x - 2} \right) \ne 0 \hfill \\ \end{gathered} \] yes! correct!

  15. 18jonea
    • one year ago
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    ok now we have to find the common denominator

  16. Michele_Laino
    • one year ago
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    right! we have to find the common denominator between the two denominators: \[\Large \begin{gathered} 3\left( {x - 1} \right) \hfill \\ \left( {x - 1} \right)\left( {x - 2} \right) \hfill \\ \end{gathered} \]

  17. 18jonea
    • one year ago
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    3?

  18. Michele_Laino
    • one year ago
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    no, we have to take all uncommon and common factors with the highest exponent

  19. 18jonea
    • one year ago
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    ok so how do we do that

  20. Michele_Laino
    • one year ago
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    it is simple, x-1 is a common factor, whereas 3 and x-2 are uncommon factor, so the common denominator is: 3*(x-1)*(x-2)

  21. Michele_Laino
    • one year ago
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    and hence, we can write this: \[\Large \begin{gathered} \frac{{x + 1}}{{3x - 3}} - \frac{4}{{{x^2} - 3x + 2}} = \hfill \\ \hfill \\ = \frac{{x + 1}}{{3\left( {x - 1} \right)}} - \frac{4}{{\left( {x - 1} \right)\left( {x - 2} \right)}} = \hfill \\ \hfill \\ = \frac{{...}}{{3\left( {x - 1} \right)\left( {x - 2} \right)}} \hfill \\ \end{gathered} \]

  22. 18jonea
    • one year ago
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    so what is the common denominator

  23. Michele_Laino
    • one year ago
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    it is: \[\Large {3\left( {x - 1} \right)\left( {x - 2} \right)}\]

  24. Michele_Laino
    • one year ago
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    now we have to divide the common denominator by the first numerator, namely what is: \[\Large \frac{{3\left( {x - 1} \right)\left( {x - 2} \right)}}{{3\left( {x - 1} \right)}} = ...?\]

  25. Michele_Laino
    • one year ago
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    you can cancel out common factors

  26. Michele_Laino
    • one year ago
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    |dw:1440013189699:dw|

  27. 18jonea
    • one year ago
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    why are you cancelling

  28. Michele_Laino
    • one year ago
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    because one factor is at denominator, and the same factor is at denominator

  29. Michele_Laino
    • one year ago
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    please wait a moment

  30. Michele_Laino
    • one year ago
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    the result is x-2, so the next step is: \[\Large \begin{gathered} \frac{{x + 1}}{{3x - 3}} - \frac{4}{{{x^2} - 3x + 2}} = \hfill \\ \hfill \\ = \frac{{x + 1}}{{3\left( {x - 1} \right)}} - \frac{4}{{\left( {x - 1} \right)\left( {x - 2} \right)}} = \hfill \\ \hfill \\ = \frac{{\left( {x + 1} \right)\left( {x - 2} \right) - 4 \cdot ....}}{{3\left( {x - 1} \right)\left( {x - 2} \right)}} \hfill \\ \end{gathered} \]

  31. Michele_Laino
    • one year ago
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    now we have to divide the common denominator by the second denominator, namely: \[\Large \frac{{3\left( {x - 1} \right)\left( {x - 2} \right)}}{{\left( {x - 1} \right)\left( {x - 2} \right)}} = ...\]

  32. 18jonea
    • one year ago
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    i dont think this is right

  33. Michele_Laino
    • one year ago
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    yes! it is right :)

  34. 18jonea
    • one year ago
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    you have to do the common denominator for each part of the fraction

  35. Michele_Laino
    • one year ago
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    the common denominator is only one

  36. 18jonea
    • one year ago
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    so for the first part wouldnt it be \[\frac{ x+1(x-2) }{ 3(x-1)(x-2)}\]

  37. Michele_Laino
    • one year ago
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    that's right!

  38. Michele_Laino
    • one year ago
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    now, we have this: \[\Large \frac{{3\left( {x - 1} \right)\left( {x - 2} \right)}}{{\left( {x - 1} \right)\left( {x - 2} \right)}} = 3\]

  39. Michele_Laino
    • one year ago
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    so we can write this: \[\Large \begin{gathered} \frac{{x + 1}}{{3x - 3}} - \frac{4}{{{x^2} - 3x + 2}} = \hfill \\ \hfill \\ = \frac{{x + 1}}{{3\left( {x - 1} \right)}} - \frac{4}{{\left( {x - 1} \right)\left( {x - 2} \right)}} = \hfill \\ \hfill \\ = \frac{{\left( {x + 1} \right)\left( {x - 2} \right) - 4 \cdot 3}}{{3\left( {x - 1} \right)\left( {x - 2} \right)}} \hfill \\ \end{gathered} \]

  40. Michele_Laino
    • one year ago
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    we have to simplify the numerator: \[\Large \begin{gathered} \frac{{x + 1}}{{3x - 3}} - \frac{4}{{{x^2} - 3x + 2}} = \hfill \\ \hfill \\ = \frac{{x + 1}}{{3\left( {x - 1} \right)}} - \frac{4}{{\left( {x - 1} \right)\left( {x - 2} \right)}} = \hfill \\ \hfill \\ = \frac{{\left( {x + 1} \right)\left( {x - 2} \right) - 4 \cdot 3}}{{3\left( {x - 1} \right)\left( {x - 2} \right)}} = \hfill \\ \hfill \\ = \frac{{{x^2} - x - 2 - 12}}{{3\left( {x - 1} \right)\left( {x - 2} \right)}} = \frac{{{x^2} - x - 14}}{{3\left( {x - 1} \right)\left( {x - 2} \right)}} \hfill \\ \end{gathered} \]

  41. 18jonea
    • one year ago
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    and then that would be the final answer?

  42. Michele_Laino
    • one year ago
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    yes!

  43. 18jonea
    • one year ago
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    Thank you

  44. Michele_Laino
    • one year ago
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    :)

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