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March 30, 2015

Posted by **-Untamed-** on Friday, December 23, 2011 at 5:52pm.

4a^2-1/4a^2-16*2-a/2a-1

This is what I did to try and solve it out:

(2a+1)(2a-1)/(2a+4)(2a-4)*2-a/(2a-1)

I cancelled out the (2a-1)'s and don't get how to solve.

I'm supposed to get the answer :

-2a-1/4(a+2)

But I don't get how to get that^.

- Calculus -
**Lindsay**, Friday, December 23, 2011 at 6:05pman (2a-4)=2(a-2) or -2(2-a) which will cancel out your 2-a,

then do the same for the 2a+4, and this should give you your answer

- Calculus -
**-Untamed-**, Friday, December 23, 2011 at 6:09pmThanks so much =)

- Calculus -
**Reiny**, Friday, December 23, 2011 at 6:59pmI read that as

(4a^2-1)/(4a^2-16) * (2-a)/(2a-1) , my brackets are necessary

= (2a+1)(2a-1)/( 4(a+2)(a-2) * (2-a)/(2a-1)

the (2-a)/(a-2) are opposite, they will give you the -1

so..

=**-(2a+1)/(4(a+2))**, again you have to use brackets to write it on here.

- Calculus -
**-Untamed-**, Saturday, December 24, 2011 at 5:17pmThanks Reiny =) But I to get everything clear, the 2-a and the a-2 cancel right? After I cancel those I am left with (2a+1) as the numerator:S Sorry I am just confused. Also did you factor out the 2a+4 and 2a-4 to get 4(a+2)(a-2)? Could you please explain step by step?

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