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waheguru Group Title

How Do I solve this

  • 2 years ago
  • 2 years ago

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  1. waheguru Group Title
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    • 2 years ago
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  2. freckles Group Title
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    What do you mean without canceling out?

    • 2 years ago
  3. waheguru Group Title
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    Can you show how to solve this

    • 2 years ago
  4. freckles Group Title
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    I'm not really sure what that means here? You mean without clearing the denominators?

    • 2 years ago
  5. estudier Group Title
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    get the number on one side of the = and the others on the other side so that you can manipulate them directly.

    • 2 years ago
  6. freckles Group Title
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    I would first clear the denominators by multiplying both sides by (x-7)

    • 2 years ago
  7. waheguru Group Title
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    CAn you show me please I tried many times

    • 2 years ago
  8. freckles Group Title
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    Multiply the 2 terms on the left by (x-7) And Multiply the 1 term on the right by (x-7)

    • 2 years ago
  9. freckles Group Title
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    What do you get after performing this?

    • 2 years ago
  10. estudier Group Title
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    No, no........

    • 2 years ago
  11. freckles Group Title
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    What @estudier ?

    • 2 years ago
  12. waheguru Group Title
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    x^2-7x+6x-42=7x-49 right?

    • 2 years ago
  13. waheguru Group Title
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    \[x^2-7x+6x-42=7x-49 \]

    • 2 years ago
  14. freckles Group Title
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    Where does that x^2 come from?

    • 2 years ago
  15. estudier Group Title
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    |dw:1342565026387:dw|

    • 2 years ago
  16. waheguru Group Title
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    Freckles when I timed x(x-7)

    • 2 years ago
  17. freckles Group Title
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    What is \[\frac{x}{x-7} \cdot (x-7)=?\] \[6 \cdot (x-7)=?\] \[\frac{7}{x-7} \cdot (x-7)=?\]

    • 2 years ago
  18. estudier Group Title
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    This equation is inconsistent.

    • 2 years ago
  19. waheguru Group Title
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    the first one \[x^2-7x\]

    • 2 years ago
  20. freckles Group Title
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    no @waheguru \[\frac{a}{a} , \text{ assuming of course } a \neq 0\]

    • 2 years ago
  21. freckles Group Title
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    =1

    • 2 years ago
  22. estudier Group Title
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    Why do I feel like I am talking to myself here?

    • 2 years ago
  23. freckles Group Title
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    left out the a/a=1 part

    • 2 years ago
  24. waheguru Group Title
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    So the x's cancel out and it becomes 7x?

    • 2 years ago
  25. freckles Group Title
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    \[\frac{x(x-7)}{(x-7)}\]

    • 2 years ago
  26. estudier Group Title
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    There is no need to multiply initially that is just a waste of time.

    • 2 years ago
  27. freckles Group Title
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    What factor does the bottom and top have in common?

    • 2 years ago
  28. waheguru Group Title
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    @estudier I know you can cencel out but i wanna try this way

    • 2 years ago
  29. estudier Group Title
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    The equation is inconsistent!1111

    • 2 years ago
  30. waheguru Group Title
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    x @freckles

    • 2 years ago
  31. freckles Group Title
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    The top and bottom both have the factor (x-7) in common

    • 2 years ago
  32. waheguru Group Title
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    Yes

    • 2 years ago
  33. estudier Group Title
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    I am not cancelling anything.......

    • 2 years ago
  34. estudier Group Title
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    x-7/x-7 = -6 is just nonsense

    • 2 years ago
  35. freckles Group Title
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    \[\frac{x \not{(x-7)}}{ \not{(x-7)}}\] Imagine that canceler thingy is over the whole factor (x-7) Couldn't do it in latex so we have the first one x So Now try the other two

    • 2 years ago
  36. estudier Group Title
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    You two evidently have nothing better to do.....:-)

    • 2 years ago
  37. freckles Group Title
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    @estudier I know what you are saying but I just want to convince him

    • 2 years ago
  38. waheguru Group Title
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    @estudier Then How do you think we solve this

    • 2 years ago
  39. estudier Group Title
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    There is no solution, the equation is not consistent (it is a false equation)

    • 2 years ago
  40. waheguru Group Title
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    no its not, there is an asnwer !!!!!!

    • 2 years ago
  41. estudier Group Title
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    It is like writing 1= 2

    • 2 years ago
  42. freckles Group Title
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    Right but we need to show him that...and convince him... If he doesn't understand your way, then there is no hurt in me showing this other way

    • 2 years ago
  43. freckles Group Title
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    @estudier is right there is no solution

    • 2 years ago
  44. waheguru Group Title
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    guys ok thanks

    • 2 years ago
  45. estudier Group Title
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    He doesn't want to understand it, that is a different thing.

    • 2 years ago
  46. freckles Group Title
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    We will get x=7 but that is a contradiction because x cannot be 7

    • 2 years ago
  47. freckles Group Title
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    Doing it the way I was trying to show @waheguru

    • 2 years ago
  48. freckles Group Title
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    So therefore no solution

    • 2 years ago
  49. estudier Group Title
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    When simplifying add or subtract like things first.....

    • 2 years ago
  50. waheguru Group Title
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    guys look at this http://www.youtube.com/watch?v=q7_xdLZHN6g the same question appears at about 5 miutes

    • 2 years ago
  51. waheguru Group Title
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    this is not impossible noting is

    • 2 years ago
  52. waheguru Group Title
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    @estudier @freckles http://www.youtube.com/watch?v=q7_xdLZHN6g at 5 mins

    • 2 years ago
  53. estudier Group Title
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    The equation is inconsistent (I don't care what is on youtube)

    • 2 years ago
  54. waheguru Group Title
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    Ur right i am so sorry

    • 2 years ago
  55. waheguru Group Title
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    I just realized Brother I appoloze

    • 2 years ago
  56. estudier Group Title
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    Accepted.

    • 2 years ago
  57. waheguru Group Title
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    I was just confued

    • 2 years ago
  58. freckles Group Title
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    @waheguru the way the video is doing it was the exact way I was going to show you

    • 2 years ago
  59. freckles Group Title
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    Did you see that she got x=7 but then since we have x-7 on bottom then x cannot be 7 so there is no solution

    • 2 years ago
  60. waheguru Group Title
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    yea

    • 2 years ago
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