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OpenStudy (anonymous):
Do you mean 2/(a^2+4a+4) + 9/(a^2-4)
OpenStudy (anonymous):
yes
OpenStudy (anonymous):
The two denominators can be factored. Give it a try.
OpenStudy (anonymous):
2/(a+4) (a+1) + 9/(a+4)(a-4)
OpenStudy (anonymous):
Close. Did you actually multiply them out to see if you get back the original expression?
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OpenStudy (anonymous):
u r doin silly mistake
OpenStudy (anonymous):
i figured out what i did wrong but i am having trouble finding the lcd
OpenStudy (anonymous):
u can... . try...
OpenStudy (anonymous):
ur tryin has much more importance
OpenStudy (anonymous):
i think its (a+2)(a+2) (a+4)(a-4)
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OpenStudy (anonymous):
no....
OpenStudy (anonymous):
second part is wrong
OpenStudy (anonymous):
can u just tell me how to do it then
OpenStudy (anonymous):
First correctly factor the denominators in
2/(a^2+4a+4) + 9/(a^2-4)
OpenStudy (anonymous):
Ooops, missed an earlier post of yours.
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OpenStudy (anonymous):
(a+4)(a-4) is not correct.
OpenStudy (anonymous):
ya.. just examine that part
OpenStudy (anonymous):
i no i fixed it
OpenStudy (anonymous):
(a+2) (a+2)
OpenStudy (anonymous):
u can write (a^2-4)
as (a^2-2^2)
=(a-2)(a+2)
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OpenStudy (anonymous):
got it
?????
OpenStudy (anonymous):
OK
OpenStudy (anonymous):
is this right so far
2a-4/(a+2) (a+2) (a-2) + 9a+18/(a+2 (a+2) (a-2)
OpenStudy (anonymous):
rite
OpenStudy (anonymous):
2/(a-2)(a-2) + 9/(a+2)(a-2)
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OpenStudy (anonymous):
a^2+4a+4 =(a+2)(a+2)
OpenStudy (anonymous):
its +4 my friend
OpenStudy (anonymous):
correction : 2/(a+2)(a_2) + 9/(a+2)(a-2)
OpenStudy (anonymous):
Dumb computer : 2/(a+2)(a+2) + 9/(a+2)(a-2)
OpenStudy (anonymous):
now its rite
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OpenStudy (anonymous):
Yes, finally. lol
OpenStudy (anonymous):
final answer
7a+14/(a+2)a+2)(a-2)
OpenStudy (anonymous):
2a+9a=11a
OpenStudy (anonymous):
i got anther problem
3/a^2+12a+36 + 5/a^2-36
OpenStudy (anonymous):
same type as before
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OpenStudy (anonymous):
No, final answer is not correct.
OpenStudy (anonymous):
3/(a+6)(a+6) + 5 /(a+6)(a-6)
OpenStudy (anonymous):
Working with this latest problem, what is the LCD?
OpenStudy (anonymous):
with the problem im doing now
OpenStudy (anonymous):
Yes.
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OpenStudy (anonymous):
i have (a+6)(a+6)(a-6)
OpenStudy (anonymous):
Yes. So now change the two fractions so they both have this same denominator.
OpenStudy (anonymous):
so i have 3a-18/(a+6)(a+6)(a-6) + 5a+30/(a+6)(a+6)(a-6)
OpenStudy (anonymous):
That looks correct. Now combine the numerators.
OpenStudy (anonymous):
8a+12
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OpenStudy (anonymous):
Yes. Of course, that goes over the denominator. 8a +12 does not factor in any way that can cancel with something in the denominator, so that should be the final answer.