@nnesha

- Jamierox4ev3r

@nnesha

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- katieb

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- Nnesha

\(\Large\frac{x^{2}}{x^{2}-4}\) - \(\Large\frac{x+1}{x+2}\) here is ur question

- Jamierox4ev3r

Oml thank you xD

- Nnesha

:P

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## More answers

- Nnesha

alright
1st) try to factor out the equation
2nd) find common denominator

- 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

correct.

- 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

if this is correct, I think I'm finally starting to get it
O_O

- Nnesha

yes right common denominator: (x-2)(x+2)

- Jamierox4ev3r

so that is a thing. oh my giblets.
Ahh i forget. Now what do I do from here?

- 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

oh, so would you just cross multiply?

- Nnesha

yes but don't forget the sign

- Jamierox4ev3r

ah yes, the subtraction sign. thanks bbg

- 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

oh wait. wasn't that a part of making it have a common denominator? XDD so NOW what do I do?

- 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

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

looks good!

- Jamierox4ev3r

So then you would have this:
\(\huge\frac{x^{2}-x^{2}+x+2}{(x+2)(x-2)}\)

- Nnesha

yeah:P

- 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

OH MY GOD I GET IT

- Nnesha

yayaY!

- Jamierox4ev3r

you are my goddess ily

- Jamierox4ev3r

seriously, thank you soo much. You deserve all 4 of those medals XD

- Nnesha

thanks <3

- Nnesha

just pray for me hahah
medals aren't important :=)

- Nnesha

you're too smart!

- 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

thanks my friend!!

- Nnesha

and yes YOU!

- Jamierox4ev3r

:')

- Nnesha

o^_^o

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