Question

@nnesha

Answer 1

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

Answer 2

Oml thank you xD

Answer 3

:P

Answer 4

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

Answer 5

alright. So if I recall correctly, first we factor the bottom right?

So it would look like this:

\[\frac{ x ^{2} }{ (x-2)(x+2) }-\frac{ x+1 }{ x+2 }\]

Answer 6

correct.

Answer 7

so to make a common denominator, would you just multiply by x-2? :ooo

so it would look like this:

\(\huge\frac{x^{2}}{(x-2)(x+2)} - \huge\frac{(x+1)(x-2)}{(x+2)(x-2)}\)

Answer 8

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

Answer 9

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

Answer 10

so that is a thing. oh my giblets.

Ahh i forget. Now what do I do from here?

Answer 11

now
multiply numerator of 1st fraction by the denominator of 2nd fraction
multiply numerator of 2nd fraction by the denominator of 1st fraction

Answer 12

oh, so would you just cross multiply?

Answer 13

yes but don't forget the sign

Answer 14

ah yes, the subtraction sign. thanks bbg

Answer 15

\(\color{blue}{\text{Originally Posted by}}\) @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

Answer 16

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

Answer 17

\(\huge\frac{x^{2}}{(x-2)(x+2)} - \huge\frac{(x+1)(x-2)}{(x+2)(x-2)}\)
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

Answer 18

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 \(-x^{2}+x+2\)

Answer 19

looks good!

Answer 20

So then you would have this:

\(\huge\frac{x^{2}-x^{2}+x+2}{(x+2)(x-2)}\)

Answer 21

yeah:P

Answer 22

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: \(\huge\frac{1}{x-2}\)

Answer 23

OH MY GOD I GET IT

Answer 24

yayaY!

Answer 25

you are my goddess ily

Answer 26

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

Answer 27

thanks <3

Answer 28

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

Answer 29

you're too smart!

Answer 30

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.

Answer 31

thanks my friend!!

Answer 32

and yes YOU!

Answer 33

:')

Answer 34

o^_^o