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anonymous

  • one year ago

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  1. anonymous
    • one year ago
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    \[\frac{ x+a }{ ax }=\frac{ b }{ x }\]

  2. anonymous
    • one year ago
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    @Napolions

  3. anonymous
    • one year ago
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    a=2 and b=3

  4. anonymous
    • one year ago
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    what up

  5. anonymous
    • one year ago
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    can i get a picture

  6. anonymous
    • one year ago
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    ther isnt a picture its just the equation.......

  7. anonymous
    • one year ago
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    alright so which part u need elp with 1 or 2

  8. anonymous
    • one year ago
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    help

  9. anonymous
    • one year ago
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    both...

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  10. anonymous
    • one year ago
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    \[\frac{ x+2 }{ 2x }=\frac{ 3 }{ x }\]

  11. anonymous
    • one year ago
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    alright so do you want to find the variable x

  12. anonymous
    • one year ago
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    is it 4?

  13. anonymous
    • one year ago
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    yes nice job

  14. anonymous
    • one year ago
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    it was 3/4

  15. anonymous
    • one year ago
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    wat was 3/4?

  16. anonymous
    • one year ago
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    the answer it was 3 over 4

  17. anonymous
    • one year ago
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    how?....... i got 4.....

  18. anonymous
    • one year ago
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    2+x =3 and 2x =4

  19. anonymous
    • one year ago
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    call my name when you need help

  20. anonymous
    • one year ago
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    wait how did u get that???

  21. anonymous
    • one year ago
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    @Napolions

  22. anonymous
    • one year ago
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    what up

  23. anonymous
    • one year ago
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    adding 1 up top and multiplying 2 at the bottem

  24. anonymous
    • one year ago
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    im confused......... so first u multiply both sides by 2x.......then it gets canceled...... leaving \[x+2=\frac{ 3 }{ x }2x\]

  25. anonymous
    • one year ago
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    which is \[x+2=6\]

  26. anonymous
    • one year ago
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    now your confusing me

  27. anonymous
    • one year ago
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    @Tazmaniadevil

  28. anonymous
    • one year ago
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    and then u subtract 2 on both sides and get 4 \[x=4\]

  29. anonymous
    • one year ago
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    yes

  30. anonymous
    • one year ago
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    so thats the first part right?

  31. anonymous
    • one year ago
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    yes

  32. anonymous
    • one year ago
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    ok wat about the second part???

  33. anonymous
    • one year ago
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    An extraneous solution is a root of a transformed equation that is not a root of the original equation because it was excluded from the domain of the original equation.

  34. anonymous
    • one year ago
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    that is how you identify it

  35. anonymous
    • one year ago
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    @plzzhelpme

  36. anonymous
    • one year ago
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    i didnt get that @Napolions

  37. anonymous
    • one year ago
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    this website should help u Extraneous Solutions - Hotmath hotmath.com/hotmath_help/topics/extraneous-solutions.htm

  38. anonymous
    • one year ago
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    i mean this one Extraneous Solutions - Hotmath hotmath.com/hotmath_help/topics/extraneous-solutions.htm

  39. anonymous
    • one year ago
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    ya that page isnt opening...... @Napolions

  40. anonymous
    • one year ago
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    go to google and type in extraneous solutions and it is under the one that say hot math

  41. anonymous
    • one year ago
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    wait i think i did it wrong........ \[\frac{ x+2 }{ 2x } =\frac{ 3 }{ x }\] \[ 3(2x)= x(x+2)\] \[6x=2x+2x\] \[6x=4x\] @Napolions

  42. anonymous
    • one year ago
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    cross multiplication??

  43. anonymous
    • one year ago
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    think so

  44. anonymous
    • one year ago
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    im totally confused rn.........

  45. anonymous
    • one year ago
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    @Napolions

  46. anonymous
    • one year ago
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    what up

  47. anonymous
    • one year ago
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    wat do i doo???>.........

  48. anonymous
    • one year ago
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    x_x

  49. anonymous
    • one year ago
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    how does this sound??? x(x+1) = 2x x^2 + x = 2x x^2 -x = 0 x(x-1)=0 x=0 or x=1 the x=0 is extraneous, because we are not allowed to divide by 0 (which is what happens in the original equation if x is 0)

  50. anonymous
    • one year ago
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    i made a mistake in my previous problem....i did x*x is 2x butits x^2

  51. anonymous
    • one year ago
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    ok so make the x 1

  52. anonymous
    • one year ago
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    it was x+22x=3x 3(2x)=x(x+2) 6x=x^2+2x \[x^2+2x-\left(6x\right)=6x-\left(6x\right)\]

  53. anonymous
    • one year ago
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    \[x^2-4x=0\]

  54. anonymous
    • one year ago
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    are u checking to see if you are right do you need my help

  55. anonymous
    • one year ago
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    \[x=\frac{-\left(-4\right)+\sqrt{\left(-4\right)^2-4\cdot \:0\cdot \:1}}{2\cdot \:1}=4\]and \[x=\frac{-\left(-4\right)-\sqrt{\left(-4\right)^2-4\cdot \:0\cdot \:1}}{2\cdot \:1}=0\]

  56. anonymous
    • one year ago
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    can u check it once??

  57. anonymous
    • one year ago
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    ok

  58. anonymous
    • one year ago
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    \[x=4,\:x=0\] right...? and x=0 is the extraneous

  59. anonymous
    • one year ago
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    correct

  60. anonymous
    • one year ago
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    oh ok thxxxx.......

  61. anonymous
    • one year ago
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    Medal plzzz

  62. anonymous
    • one year ago
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    u know i did most of the question ^.^

  63. anonymous
    • one year ago
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    lol but watev thx..........

  64. anonymous
    • one year ago
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    yeah but i helped though but yeah you did most

  65. anonymous
    • one year ago
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    lol

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