Jamierox4ev3r
  • Jamierox4ev3r
@nnesha
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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Nnesha
  • Nnesha
\(\Large\frac{x^{2}}{x^{2}-4}\) - \(\Large\frac{x+1}{x+2}\) here is ur question
Jamierox4ev3r
  • Jamierox4ev3r
Oml thank you xD
Nnesha
  • Nnesha
:P

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Nnesha
  • Nnesha
alright 1st) try to factor out the equation 2nd) find common denominator
Jamierox4ev3r
  • Jamierox4ev3r
alright. So if I recall correctly, first we factor the bottom right? So it would look like this: \[\frac{ x ^{2} }{ (x-2)(x+2) }-\frac{ x+1 }{ x+2 }\]
Nnesha
  • Nnesha
correct.
Jamierox4ev3r
  • Jamierox4ev3r
so to make a common denominator, would you just multiply by x-2? :ooo so it would look like this: \(\huge\frac{x^{2}}{(x-2)(x+2)} - \huge\frac{(x+1)(x-2)}{(x+2)(x-2)}\)
Jamierox4ev3r
  • Jamierox4ev3r
if this is correct, I think I'm finally starting to get it O_O
Nnesha
  • Nnesha
yes right common denominator: (x-2)(x+2)
Jamierox4ev3r
  • Jamierox4ev3r
so that is a thing. oh my giblets. Ahh i forget. Now what do I do from here?
Nnesha
  • Nnesha
now multiply numerator of 1st fraction by the denominator of 2nd fraction multiply numerator of 2nd fraction by the denominator of 1st fraction
Jamierox4ev3r
  • Jamierox4ev3r
oh, so would you just cross multiply?
Nnesha
  • Nnesha
yes but don't forget the sign
Jamierox4ev3r
  • Jamierox4ev3r
ah yes, the subtraction sign. thanks bbg
Nnesha
  • Nnesha
\(\color{blue}{\text{Originally Posted by}}\) @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
Jamierox4ev3r
  • Jamierox4ev3r
oh wait. wasn't that a part of making it have a common denominator? XDD so NOW what do I do?
Nnesha
  • Nnesha
\(\huge\frac{x^{2}}{(x-2)(x+2)} - \huge\frac{(x+1)(x-2)}{(x+2)(x-2)}\) 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
Jamierox4ev3r
  • Jamierox4ev3r
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 \(-x^{2}+x+2\)
Nnesha
  • Nnesha
looks good!
Jamierox4ev3r
  • Jamierox4ev3r
So then you would have this: \(\huge\frac{x^{2}-x^{2}+x+2}{(x+2)(x-2)}\)
Nnesha
  • Nnesha
yeah:P
Jamierox4ev3r
  • Jamierox4ev3r
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: \(\huge\frac{1}{x-2}\)
Jamierox4ev3r
  • Jamierox4ev3r
OH MY GOD I GET IT
Nnesha
  • Nnesha
yayaY!
Jamierox4ev3r
  • Jamierox4ev3r
you are my goddess ily
Jamierox4ev3r
  • Jamierox4ev3r
seriously, thank you soo much. You deserve all 4 of those medals XD
Nnesha
  • Nnesha
thanks <3
Nnesha
  • Nnesha
just pray for me hahah medals aren't important :=)
Nnesha
  • Nnesha
you're too smart!
Jamierox4ev3r
  • Jamierox4ev3r
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.
Nnesha
  • Nnesha
thanks my friend!!
Nnesha
  • Nnesha
and yes YOU!
Jamierox4ev3r
  • Jamierox4ev3r
:')
Nnesha
  • Nnesha
o^_^o

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