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Adi3

  • one year ago

Will medal please help find the inverse of g(x) = (1)/x-1 + 3

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  1. Adi3
    • one year ago
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    \[\left(\begin{matrix}\\ \frac{ 1}{ x-1 }\end{matrix}\right) + 3\]

  2. Adi3
    • one year ago
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    \[\frac{ 1 }{ x-1} + 3\]

  3. Adi3
    • one year ago
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    if this more clear

  4. Nnesha
    • one year ago
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    g(x) is same as y so you can replace g(x) with y \[\huge\rm \color{ReD}{y}=\frac{ 1 }{ \color{blue}{x}-1 } +3\] to find inverse `switch x and y` \[\huge\rm \color{blue}{x}=\frac{ 1 }{ \color{red}{y}-1 } +3\] now solve for y

  5. Adi3
    • one year ago
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    ok, x - 3 = \[\frac{ 1 }{ y-1 }\]

  6. Adi3
    • one year ago
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    is that right

  7. Adi3
    • one year ago
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    @IrishBoy123

  8. Adi3
    • one year ago
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    @HELPMEPLZ!!

  9. Adi3
    • one year ago
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    still on inverse, this is too much for me

  10. Nnesha
    • one year ago
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    sorry i was afk yes that's right \[\huge\rm x-3=\frac{ 1 }{ y-1}\] now how would you cancel out y-1 from right side ?

  11. Adi3
    • one year ago
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    divde it?

  12. imqwerty
    • one year ago
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    write y in terms of x and then jst switch y and x :) function which u get is inverse

  13. Nnesha
    • one year ago
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    its already divided by 1 (1/y-1 )to cancel it out you should do opposite of division

  14. Adi3
    • one year ago
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    multiply

  15. Nnesha
    • one year ago
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    yes right multiply both sides by y-1

  16. Adi3
    • one year ago
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    so 1x-3

  17. Nnesha
    • one year ago
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    multiply `both sides` by y-1 \[\huge\rm (y-1) (x-3)=\frac{ 1 }{\cancel{y-1} } \times \cancel{(y-1)}\] we can cancel out y-1 \[\large\rm (y-1)(x-3)=1\] now again to solve for y move `x-3` to the right side

  18. Adi3
    • one year ago
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    ohh, then how to get y by itself?

  19. Nnesha
    • one year ago
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    how would you cancel x-3 from left side ?

  20. Adi3
    • one year ago
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    divide

  21. Nnesha
    • one year ago
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    right so divide both sides by x-3 let me know what you get

  22. Nnesha
    • one year ago
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    it's multiplication( y-1 ) times (x-3) so yes you should divide

  23. Adi3
    • one year ago
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    y-1 = 1/x-3

  24. Nnesha
    • one year ago
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    right whats nexxt ?

  25. Adi3
    • one year ago
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    y = 1/x-3 + 1

  26. Nnesha
    • one year ago
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    right now we should get the common denominator |dw:1443366122753:dw|

  27. Adi3
    • one year ago
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    OK, Can i ask a last question

  28. Adi3
    • one year ago
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    pls

  29. imqwerty
    • one year ago
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    yea y nt :)

  30. Nnesha
    • one year ago
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    yes ?

  31. Adi3
    • one year ago
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    ok, thank you all. find the inverse of \[-\frac{ 2 }{ 5}x - 2\]

  32. Adi3
    • one year ago
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    g(x) =

  33. Nnesha
    • one year ago
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    same steps try it first

  34. Adi3
    • one year ago
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    x+2 = -2/5y

  35. imqwerty
    • one year ago
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    yea nxt

  36. Adi3
    • one year ago
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    thts wht idk

  37. imqwerty
    • one year ago
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    think a way to isolate y

  38. Nnesha
    • one year ago
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    now \[x+2 =\frac{ -2 }{ 5 }y\] what would be the nexxt step ?

  39. Adi3
    • one year ago
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    divide?

  40. imqwerty
    • one year ago
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    divide by what..

  41. Nnesha
    • one year ago
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    right divide what ?

  42. Adi3
    • one year ago
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    idk

  43. Adi3
    • one year ago
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    -2/5

  44. Nnesha
    • one year ago
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    well first you need to get rid of the fraction

  45. imqwerty
    • one year ago
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    if u have 2x and u want just x then what number will u divide it by?

  46. Nnesha
    • one year ago
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    \[\huge\rm x+2=\frac{ -2 }{ 5 }y\] to get rid of the fraction at right side what would u do ? remember it's -2 divided by 5

  47. Nnesha
    • one year ago
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    nd ofc y

  48. Adi3
    • one year ago
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    2/5 = 2.5

  49. Adi3
    • one year ago
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    -2.5

  50. Adi3
    • one year ago
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    @imqwerty -2.5

  51. Nnesha
    • one year ago
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    well i'm pretty sure they aren't looking for decimal answer

  52. Adi3
    • one year ago
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    yup

  53. Nnesha
    • one year ago
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    \[\huge\rm x+2=\frac{ -2 }{ 5 }y\] to get rid of the fraction at right side what would u do ? remember it's -2 divided by 5

  54. Nnesha
    • one year ago
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    how would you cancel out 5 from right side ?

  55. Adi3
    • one year ago
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    multiply

  56. imqwerty
    • one year ago
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    no its not -2.5

  57. Nnesha
    • one year ago
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    multiply by what ?

  58. Adi3
    • one year ago
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    ok

  59. Adi3
    • one year ago
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    they do not want a answer in deci

  60. Adi3
    • one year ago
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    5 @Nnesha

  61. imqwerty
    • one year ago
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    see if u got something like this - \[x \times y\]and u wanna get rid of y then what u do is that just multiplay it by 1/y then u get-\[x \times y \times \frac{ 1 }{ y }\] y and y get cut and u get ur x

  62. Adi3
    • one year ago
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    the answer is \[5 - \frac{ 5 }{ 2 }x\]

  63. Nnesha
    • one year ago
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    yes right multiply both sides by 5 \[\huge\rm 5(x+2)=\frac{ -2 }{ 5 }y \times 5\] always first try to get rid of de fraction

  64. Adi3
    • one year ago
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    ok

  65. Nnesha
    • one year ago
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    can you explain plz how did you get -5 /2 and just 5 ?

  66. Adi3
    • one year ago
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    that is the answer, thats part i dont get it

  67. Nnesha
    • one year ago
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    yes right multiply both sides by 5 \[\huge\rm 5(x+2)=\frac{ -2 }{ 5 }y \times 5\] distribute parentheses by 5 5(x+2) = ???

  68. Adi3
    • one year ago
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    5x + 10

  69. Adi3
    • one year ago
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    it is like 11:00 P.M here so i am gonna sleep, bye, cant understand anything right now. bye

  70. Nnesha
    • one year ago
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    right \[\huge\rm 5x+10 =-2y\]last step

  71. Nnesha
    • one year ago
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    there is only one step ........ -2 ` times ` y so to cancel it out divide by sides by -2

  72. Adi3
    • one year ago
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    (5x + 10)/2

  73. Nnesha
    • one year ago
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    -2 not just 2 and (5x+10)/-2 can be written as 5x/-2 + 10/-2 \[\huge\rm \frac{ 5x+10 }{ -2}=\frac{ 5x }{ -2 }+\frac{ 10 }{ -2}\]now jsut simplify

  74. Adi3
    • one year ago
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    ok, thank you

  75. Nnesha
    • one year ago
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    np

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