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Destinyyyy

  • one year ago

Help with a few problems..

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  1. Destinyyyy
    • one year ago
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    Solve by any method. (4x+3)(x+2)=3

  2. dinamix
    • one year ago
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    4x^2+8x+3x+6 = 3 4x^2+11x+3=0 \[\Delta = 121-48 =73\]

  3. Destinyyyy
    • one year ago
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    Um your missing some numbers

  4. Destinyyyy
    • one year ago
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    Final answer---> -11+- Square root 73 all over 8

  5. Destinyyyy
    • one year ago
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    Solve by any method. 1/x + 1/x+2 = 1/5

  6. Destinyyyy
    • one year ago
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    @Nnesha

  7. dinamix
    • one year ago
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    u complete it cuz i find u delta

  8. Destinyyyy
    • one year ago
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    What?

  9. dinamix
    • one year ago
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    \[x(1) = \frac{ -b-\sqrt{\Delta} }{ 2a } , x(2) = \frac{ -b+\sqrt{\Delta} }{ 2a }\]

  10. dinamix
    • one year ago
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    u have find solution

  11. Destinyyyy
    • one year ago
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    Um? Which problem are you talking about?

  12. dinamix
    • one year ago
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    (4x+3)(x+2) = 3 @Destinyyyy

  13. dinamix
    • one year ago
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    this i mean

  14. Destinyyyy
    • one year ago
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    Oh. I completed that question.

  15. Destinyyyy
    • one year ago
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    I now need help on this one- Solve by any method. 1/x + 1/x+2 = 1/5

  16. dinamix
    • one year ago
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    \[\frac{ x+2+x }{ (x+2)(x) }=1/5\]

  17. dinamix
    • one year ago
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    u understand that @Destinyyyy how i do it right ?

  18. Destinyyyy
    • one year ago
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    Thats no right... 1/x + 1/x+2 = 1/5

  19. Destinyyyy
    • one year ago
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    not *

  20. dinamix
    • one year ago
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    hmm |dw:1440723213093:dw|

  21. Destinyyyy
    • one year ago
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    Umm I cant see the end part..

  22. Destinyyyy
    • one year ago
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    Um are you trying to say times everything by x+2 ?? Its hard to read

  23. dinamix
    • one year ago
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    \[(\frac{ 1 }{ x } *\frac{ x+2 }{ x+2 }) +(\frac{ 1 }{ x+2}*\frac{ x }{ x })\]

  24. dinamix
    • one year ago
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    \[\frac{ x+2 }{ x(x+2) }+\frac{ x }{ x(x+2) } = 1/5\]

  25. dinamix
    • one year ago
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    \[\frac{ x+2+x }{ x(x+2) } =1/5\]

  26. Destinyyyy
    • one year ago
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    Sorry but im not understanding.. I thought im suppose to get rid of the fractions?

  27. dinamix
    • one year ago
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    \[\frac{ 2x+2 }{ x^2+2x } =1/5\]

  28. Destinyyyy
    • one year ago
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    I dont get how your going from the original problem to the first image

  29. Destinyyyy
    • one year ago
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    Where are you getting x+2/x+2 from? And where did the parentheses come from?

  30. dinamix
    • one year ago
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    u will get fuction now cuz we dont have way we must do that to solve problem

  31. Destinyyyy
    • one year ago
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    Huh?

  32. Destinyyyy
    • one year ago
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    @jim_thompson5910

  33. dinamix
    • one year ago
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    tell to him if my mrthod is correct or no , cuz i am sure about my answer is true

  34. Destinyyyy
    • one year ago
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    Again huh? Can you please type in proper English so I can understand?

  35. Destinyyyy
    • one year ago
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    @.Gjallarhorn. @IrishBoy123

  36. dinamix
    • one year ago
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    we sould have same denominator of this 1/x+(1/x+2)

  37. dinamix
    • one year ago
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    should*

  38. Destinyyyy
    • one year ago
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    Where are you getting parenthesis from??

  39. Destinyyyy
    • one year ago
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    Can you show me step by step?

  40. dinamix
    • one year ago
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    just parenthesis for u to understand .

  41. Destinyyyy
    • one year ago
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    Oh.. Well they seem to be confusing me instead

  42. dinamix
    • one year ago
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    i give u example to understand okey

  43. Destinyyyy
    • one year ago
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    Um okay

  44. dinamix
    • one year ago
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    |dw:1440724788988:dw| how collecting this numbers

  45. dinamix
    • one year ago
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    @Destinyyyy

  46. Destinyyyy
    • one year ago
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    Okay?

  47. dinamix
    • one year ago
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    did u see my question or no

  48. dinamix
    • one year ago
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    |dw:1440725128863:dw|

  49. Destinyyyy
    • one year ago
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    Yes I see it but I dont know what you want me to do or say

  50. jim_thompson5910
    • one year ago
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    The original problem is this, right? \[\Large \frac{1}{x} + \frac{1}{x+2}= \frac{1}{5}\]

  51. Destinyyyy
    • one year ago
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    Yes @jim_thompson5910

  52. jim_thompson5910
    • one year ago
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    since we have an equation, we can multiply every fraction by the LCD 5x(x+2) to clear out the fractions \[\large \frac{1}{x} + \frac{1}{x+2}= \frac{1}{5}\] \[\large {\color{red}{5x(x+2)}}*\frac{1}{x} + {\color{red}{5x(x+2)}}*\frac{1}{x+2}= {\color{red}{5x(x+2)}}*\frac{1}{5}\] \[\large {\color{red}{5\cancel{x}(x+2)}}*\frac{1}{\cancel{x}} + {\color{red}{5x*\cancel{(x+2)}}}*\frac{1}{\cancel{x+2}}= {\color{red}{\cancel{5}*x(x+2)}}*\frac{1}{\cancel{5}}\] \[\large 5(x+2)*1 + 5x*1 = x(x+2)*1\] \[\large 5(x+2) + 5x = x(x+2)\] Hopefully you see how I was able to clear out the fractions?

  53. Destinyyyy
    • one year ago
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    Yes I thought thats what I was suppose to do but wasnt entirely sure. Thank you

  54. jim_thompson5910
    • one year ago
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    are you able to solve `5(x+2)+5x=x(x+2)` from here?

  55. Destinyyyy
    • one year ago
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    Yes!

  56. jim_thompson5910
    • one year ago
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    ok great

  57. dinamix
    • one year ago
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    @jim_thompson5910 lol i think my method is easy look first

  58. Destinyyyy
    • one year ago
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    5x+10+5x= x^2+2x 10x+10=x^2+2x 8x+10=x^2 x^2-8x-10=0

  59. jim_thompson5910
    • one year ago
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    now use the quadratic formula to solve `x^2-8x-10=0`

  60. Destinyyyy
    • one year ago
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    8+- square root 104 over 2

  61. Destinyyyy
    • one year ago
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    8+- 2 square root 26 over 2

  62. jim_thompson5910
    • one year ago
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    There's one more step. You're doing great so far

  63. Destinyyyy
    • one year ago
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    Umm.. 8+2 square root 26/2 --> 10 square root 26 over 2 8-2 square root 26/2 --> 6 square root 26 over 2 That doesnt seem right though

  64. jim_thompson5910
    • one year ago
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    |dw:1440726115004:dw|

  65. jim_thompson5910
    • one year ago
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    divide each part (circled) by 2 |dw:1440726151837:dw|

  66. Destinyyyy
    • one year ago
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    Thats what im missing okay..

  67. Destinyyyy
    • one year ago
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    4+- square root 26

  68. jim_thompson5910
    • one year ago
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    correct

  69. Destinyyyy
    • one year ago
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    Thank you.. I have two more. Do you mind helping? Both are word problems

  70. jim_thompson5910
    • one year ago
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    sure, but post them one at a time (one per post) to avoid clutter or lag

  71. Destinyyyy
    • one year ago
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