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Jamierox4ev3r

  • one year ago

@nnesha

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  1. Nnesha
    • one year ago
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    \(\Large\frac{x^{2}}{x^{2}-4}\) - \(\Large\frac{x+1}{x+2}\) here is ur question

  2. Jamierox4ev3r
    • one year ago
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    Oml thank you xD

  3. Nnesha
    • one year ago
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    :P

  4. Nnesha
    • one year ago
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    alright 1st) try to factor out the equation 2nd) find common denominator

  5. Jamierox4ev3r
    • one year ago
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    alright. So if I recall correctly, first we factor the bottom right? So it would look like this: \[\frac{ x ^{2} }{ (x-2)(x+2) }-\frac{ x+1 }{ x+2 }\]

  6. Nnesha
    • one year ago
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    correct.

  7. Jamierox4ev3r
    • one year ago
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    so to make a common denominator, would you just multiply by x-2? :ooo so it would look like this: \(\huge\frac{x^{2}}{(x-2)(x+2)} - \huge\frac{(x+1)(x-2)}{(x+2)(x-2)}\)

  8. Jamierox4ev3r
    • one year ago
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    if this is correct, I think I'm finally starting to get it O_O

  9. Nnesha
    • one year ago
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    yes right common denominator: (x-2)(x+2)

  10. Jamierox4ev3r
    • one year ago
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    so that is a thing. oh my giblets. Ahh i forget. Now what do I do from here?

  11. Nnesha
    • one year ago
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    now multiply numerator of 1st fraction by the denominator of 2nd fraction multiply numerator of 2nd fraction by the denominator of 1st fraction

  12. Jamierox4ev3r
    • one year ago
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    oh, so would you just cross multiply?

  13. Nnesha
    • one year ago
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    yes but don't forget the sign

  14. Jamierox4ev3r
    • one year ago
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    ah yes, the subtraction sign. thanks bbg

  15. Nnesha
    • one year ago
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    \(\color{blue}{\text{Originally Posted by}}\) @Jamierox4ev3r so to make a common denominator, would you just multiply by x-2? :ooo so it would look like this: \(\huge\frac{x^{2}}{(x-2)(x+2)} - \huge\frac{(x+1)(x-2)}{(x+2)(x-2)}\) \(\color{blue}{\text{End of Quote}}\) ahh i see you already did that part

  16. Jamierox4ev3r
    • one year ago
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    oh wait. wasn't that a part of making it have a common denominator? XDD so NOW what do I do?

  17. Nnesha
    • one year ago
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    \(\huge\frac{x^{2}}{(x-2)(x+2)} - \huge\frac{(x+1)(x-2)}{(x+2)(x-2)}\) now you\[\huge\rm \frac{ x^2 -\color{ReD}{(x-2)(x+1)} }{( x-2)(x+2)}\] can write the numerator under same denominator foil (x-2)(x+1) and then distribute it by - sign

  18. Jamierox4ev3r
    • one year ago
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    ohhhh right, since the denominators are now common. makes sense :o so first foil, then distribute? If you foil (x-2)(x+1), you get \(x^{2}-x-2\) and then if you distribute the negative sign, you get \(-x^{2}+x+2\)

  19. Nnesha
    • one year ago
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    looks good!

  20. Jamierox4ev3r
    • one year ago
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    So then you would have this: \(\huge\frac{x^{2}-x^{2}+x+2}{(x+2)(x-2)}\)

  21. Nnesha
    • one year ago
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    yeah:P

  22. Jamierox4ev3r
    • one year ago
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    alright. I know that x^2 - x^2 = 0 , and x+2 over x+2 equals one. so your final answer, I believe, would be this: \(\huge\frac{1}{x-2}\)

  23. Jamierox4ev3r
    • one year ago
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    OH MY GOD I GET IT

  24. Nnesha
    • one year ago
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    yayaY!

  25. Jamierox4ev3r
    • one year ago
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    you are my goddess ily

  26. Jamierox4ev3r
    • one year ago
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    seriously, thank you soo much. You deserve all 4 of those medals XD

  27. Nnesha
    • one year ago
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    thanks <3

  28. Nnesha
    • one year ago
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    just pray for me hahah medals aren't important :=)

  29. Nnesha
    • one year ago
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    you're too smart!

  30. Jamierox4ev3r
    • one year ago
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    me? d'aww thank you so much. I do tend to have a strong understanding in math, but some of these topics are things that I've forgotten since I've done them so long ago. Kudos to you for being able to explain things so clearly.

  31. Nnesha
    • one year ago
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    thanks my friend!!

  32. Nnesha
    • one year ago
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    and yes YOU!

  33. Jamierox4ev3r
    • one year ago
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    :')

  34. Nnesha
    • one year ago
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    o^_^o

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