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anonymous

  • one year ago

What is the solution to the rational equation x over x squared minus 9 plus 1 over x plus 3 equals 1 over 4x minus 12 ?

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  1. Nnesha
    • one year ago
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    \[\huge\rm \frac{ x }{ x^2-9}+\frac{ 1 }{ x+3}=\frac{ 1 }{ 4x-12 }\] like this ?

  2. anonymous
    • one year ago
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    yes

  3. Nnesha
    • one year ago
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    alright first solve left side apply difference of square method to factor x^2-9 \[\rm a^2-b^2=(a+b)(a-b)\]

  4. Nnesha
    • one year ago
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    take square of both terms write ur answer as (sqrt of 1st term + sqrt of 2nd term) (sqrt of 1st term - sqrt of 2nd term) here is an example \[\rm x^2-16=(x+4)(x-4)\]

  5. anonymous
    • one year ago
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    x = -9 over 7 x = 15 x = 15 over 7 x = 9

  6. anonymous
    • one year ago
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    these are the answers i have

  7. Nnesha
    • one year ago
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    ok. so what are the factors of x^2-9 ?

  8. anonymous
    • one year ago
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    (x-9)(x+9)

  9. Nnesha
    • one year ago
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    hmm not what is the square root of 9 ?

  10. anonymous
    • one year ago
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    3

  11. Nnesha
    • one year ago
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    |dw:1440351548850:dw|

  12. Nnesha
    • one year ago
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    yes right so it should be (x-3)(x+3)

  13. Nnesha
    • one year ago
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    \[\huge\rm \frac{ x }{ (x-3)(x+3)}+\frac{ 1 }{ x+3}=\frac{ 1 }{ 4x-12 }\] now find common denominator what is the common denominator at left side ?

  14. anonymous
    • one year ago
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    so 4(3)-12?

  15. Nnesha
    • one year ago
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    no.. common denominator\[\huge\rm \frac{ x }{ \color{Red}{(x-3)(x+3)}}+\frac{ 1 }{\color{Red}{ x+3}}=\frac{ 1 }{ 4x-12 }\] just look at red part what is common ?

  16. anonymous
    • one year ago
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    3

  17. Nnesha
    • one year ago
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    its not just 3 (x-3) is one term so x+3 is common

  18. Nnesha
    • one year ago
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    .. \[\huge\rm \frac{ x }{ \color{Red}{(x-3)(x+3)}}+\frac{ 1 }{\color{Red}{ x+3}}=\frac{ 1 }{ 4x-12 }\] \[\huge\rm \frac{ x +1(x-3)}{ (x-3)(x+3) }\] when you find common denominator you should multiply `numerator` of `1st` fraction by the `denominator` of `2nd` fraction and multiply numerator of `2nd` fraction by the denomiantor of 1st fraction

  19. anonymous
    • one year ago
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    \[-\frac{ 9 }{ 7 }\]

  20. Nnesha
    • one year ago
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    combine like terms .. \[\huge\rm \frac{ x +1x-3}{ (x-3)(x+3) }\] x+1x-3= ??

  21. anonymous
    • one year ago
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    ? is this right

  22. Nnesha
    • one year ago
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    what ??

  23. Nnesha
    • one year ago
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    no that's not right

  24. Nnesha
    • one year ago
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    what are `like terms` in this equation x+x-3 ?

  25. anonymous
    • one year ago
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    x

  26. Nnesha
    • one year ago
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    yes right so combine them x+x = ?

  27. anonymous
    • one year ago
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    x^2?

  28. Nnesha
    • one year ago
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    no when you multiply same bases then you should add their exponents no just add the coefficient 1x+1x=(1+1)x = ?

  29. anonymous
    • one year ago
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    im confused

  30. Nnesha
    • one year ago
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    okay so here we need to `combine` x+x which is same as 1x+1x one is invisible we here is an example 3y+5y = (3+5)y = 8y just add the coefficient \[\rm x \times x=x^2\] when you multiply them `then` you should add their exponents

  31. Nnesha
    • one year ago
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    |dw:1440353412609:dw| leading coefficient a number front of the variable

  32. Nnesha
    • one year ago
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    1x+1x = ?

  33. anonymous
    • one year ago
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    1x

  34. Nnesha
    • one year ago
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    add their coefficient \[1x+1x=(1+1)x= ??x\]

  35. anonymous
    • one year ago
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    2x

  36. Nnesha
    • one year ago
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    yes right .. \[\huge\rm \frac{ x }{ \color{Red}{(x-3)(x+3)}}+\frac{ 1 }{\color{Red}{ x+3}}=\frac{ 1 }{ 4x-12 }\] \[\huge\rm \frac{ 2x-3}{ (x-3)(x+3) }=\frac{1}{4x-12}\] now foil this (x-3)(x+3) are you familiar with the foil method ?

  37. anonymous
    • one year ago
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    x = 9 is this right?

  38. Nnesha
    • one year ago
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    no... show ur work so i'll look ovr to find mistakes :=)

  39. anonymous
    • one year ago
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    idk i dont understand

  40. Nnesha
    • one year ago
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    so you don't know how to foil two parentheses ?

  41. anonymous
    • one year ago
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    ido x^2 +3x-3x-9?

  42. Nnesha
    • one year ago
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    yes right which is same as x^2-9 \[\frac{ 2x-3 }{ x^2-9 }=\frac{ 1 }{ 4x+12 }\]now cross multiply

  43. Nnesha
    • one year ago
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    |dw:1440355272983:dw|

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