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anonymous

  • one year ago

What is the simplified form of the quantity of x plus 7, all over the quantity of x plus 3 + the quantity of x minus 4, all over 3?

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  1. anonymous
    • one year ago
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    |dw:1436202319724:dw|

  2. anonymous
    • one year ago
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    @ganeshie8 @geerky42 @Michele_Laino

  3. Michele_Laino
    • one year ago
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    hint: the least common multiple between x+3 and 3 is: \[3\left( {x + 3} \right)\]

  4. Michele_Laino
    • one year ago
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    am I right?

  5. anonymous
    • one year ago
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    would it be x2++3x-28/3(x+3)

  6. anonymous
    • one year ago
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    Yes you are:)

  7. Michele_Laino
    • one year ago
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    so we can write this: \[\Large \begin{gathered} \frac{{x + 7}}{{x + 3}} + \frac{{x - 4}}{3} = \hfill \\ \hfill \\ = \frac{{3\left( {x + 7} \right) + \left( {x + 3} \right)\left( {x - 4} \right)}}{{3\left( {x + 3} \right)}} = ... \hfill \\ \end{gathered} \] please continue

  8. anonymous
    • one year ago
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    (x+7)(x+3)(x-4)/3(X+3)

  9. Michele_Laino
    • one year ago
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    not exactly, since you have to compute the multiplications first

  10. anonymous
    • one year ago
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    hOW do I do that?

  11. Michele_Laino
    • one year ago
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    hint:

  12. Michele_Laino
    • one year ago
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    \[3\left( {x + 7} \right) = 3x + 3 \times 7 = 3x + 21\]

  13. anonymous
    • one year ago
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    3(x+7)

  14. anonymous
    • one year ago
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    I'm sorry :(

  15. Michele_Laino
    • one year ago
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    \[\Large \left( {x + 3} \right)\left( {x - 4} \right) = x \cdot x - 4x + 3x - 4 \cdot 3 = ...\] here, I have applied the foil procedure

  16. Michele_Laino
    • one year ago
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    please simplify, what do you get?

  17. anonymous
    • one year ago
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    x^2-1x-12

  18. Michele_Laino
    • one year ago
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    ok!

  19. Michele_Laino
    • one year ago
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    so the numerator is: \[\Large 3x + 21 + {x^2} - x - 12 = ...\]

  20. anonymous
    • one year ago
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    2x+x^2+9

  21. Michele_Laino
    • one year ago
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    perfect! so your expression becomes: \[\Large \frac{{x + 7}}{{x + 3}} + \frac{{x - 4}}{3} = \frac{{{x^2} + 2x + 9}}{{3\left( {x + 3} \right)}}\]

  22. anonymous
    • one year ago
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    ohhh! Thank-you! I need one more please?

  23. Michele_Laino
    • one year ago
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    ok!

  24. anonymous
    • one year ago
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    What is the simplified form of the quantity of x plus 9, all over the quantity of 2x plus 3 + the quantity of x plus 4, all over the quantity of x plus 2?

  25. Michele_Laino
    • one year ago
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    please, make a drawing of your expression

  26. anonymous
    • one year ago
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    ok hold on

  27. Michele_Laino
    • one year ago
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    here the least common multiple is: \[\left( {2x + 3} \right)\left( {x + 2} \right)\]

  28. Michele_Laino
    • one year ago
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    is it right?

  29. anonymous
    • one year ago
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    Yes

  30. anonymous
    • one year ago
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    Now what?

  31. Michele_Laino
    • one year ago
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    ok! So we can write: \[\large \frac{{x + 9}}{{2x + 3}} + \frac{{x + 4}}{{x + 2}} = \frac{{\left( {x + 9} \right)\left( {x + 2} \right) + \left( {x + 4} \right)\left( {2x + 3} \right)}}{{\left( {2x + 3} \right)\left( {x + 2} \right)}}\]

  32. Michele_Laino
    • one year ago
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    now you have to apply the foil method in order to evaluate this: \[\Large \left( {x + 9} \right)\left( {x + 2} \right) = ...?\]

  33. Michele_Laino
    • one year ago
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    please continue

  34. anonymous
    • one year ago
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    x2+11x+18

  35. Michele_Laino
    • one year ago
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    ok! Now do the same with this expression: \[\Large {\left( {x + 4} \right)\left( {2x + 3} \right)}\]

  36. anonymous
    • one year ago
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    2x2+11x+12

  37. Michele_Laino
    • one year ago
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    ok! so the numerator is: \[\Large {x^2} + 11x + 18 + 2{x^2} + 11x + 12 = ...\]

  38. anonymous
    • one year ago
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    2x3+22x+30

  39. Michele_Laino
    • one year ago
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    are you sure?

  40. anonymous
    • one year ago
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    yes. wait no

  41. anonymous
    • one year ago
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    2x4+22x+30

  42. Michele_Laino
    • one year ago
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    I got: 3 x^2+...

  43. Michele_Laino
    • one year ago
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    \[\Large {3{x^2} + 22x + 30}\]

  44. Michele_Laino
    • one year ago
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    is it right?

  45. anonymous
    • one year ago
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    yes bcuz 1x2+2x2=3x2. I forgot about the imaginary 1

  46. Michele_Laino
    • one year ago
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    ok! So your original expression, becomes: \[\Large \frac{{x + 9}}{{2x + 3}} + \frac{{x + 4}}{{x + 2}} = \frac{{3{x^2} + 22x + 30}}{{\left( {2x + 3} \right)\left( {x + 2} \right)}}\]

  47. anonymous
    • one year ago
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    That's not an answer choice, so I guess we will have to foil the bottom also

  48. Michele_Laino
    • one year ago
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    ok! then what is: \[\Large \left( {2x + 3} \right)\left( {x + 2} \right) = ...?\]

  49. anonymous
    • one year ago
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    2x2+7x+6

  50. Michele_Laino
    • one year ago
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    so we get: \[\Large \frac{{x + 9}}{{2x + 3}} + \frac{{x + 4}}{{x + 2}} = \frac{{3{x^2} + 22x + 30}}{{2{x^2} + 7x + 6}}\]

  51. Michele_Laino
    • one year ago
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    yes! that's right!

  52. anonymous
    • one year ago
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    Yay! Ok so I want you to keep helping me, and give u medals, so i AM going to close the question and ask another ok?

  53. Michele_Laino
    • one year ago
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    I'm sorry, I have to go to dinner :(

  54. anonymous
    • one year ago
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    :( Ok well thanks anyways!

  55. Michele_Laino
    • one year ago
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    thanks! :)

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