Posted by -Untamed- on Friday, December 23, 2011 at 5:52pm.
Simplify. State the nonpermissible values.
This is what I did to try and solve it out:
I cancelled out the (2a-1)'s and don't get how to solve.
I'm supposed to get the answer :
But I don't get how to get that^.
Calculus - Lindsay, Friday, December 23, 2011 at 6:05pm
an (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 6:09pm
Thanks so much =)
Calculus - Reiny, Friday, December 23, 2011 at 6:59pm
I 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
= -(2a+1)/(4(a+2)) , again you have to use brackets to write it on here.
Calculus - -Untamed-, Saturday, December 24, 2011 at 5:17pm
Thanks 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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