Here's the question you clicked on:
puddlebunny
I want to check myself on this one: a/2 + 1/4 divided by a/3 - 1/2 you need same denominator so you multiply a/2 by 2 which gives you 2a/4 + 1/4 = 3a/4. Next part a/3 - 1/2 so you multiply a/3 by 2 to get 2a/6 then 1/2 you multiply by 3 to get 3/6 so 2a/6 - 3/6 = 2a-3/6. Now 3a/4 divided by 2a-3/6 then you use the reciprocal and multiply 3a/4 * 6/3 to get a/-2 (after you reduce and cancel out like terms)?
your 1st step has an error:\[\frac{a}{2}+\frac{1}{4}=\frac{2a}{4}+\frac{1}{4}\ne\frac{3a}{4}\]
do you understand why?
Not sure .. now I'm confused LOL
look at the numerators of that fraction: 2a + 1 = ?
it would just be 2a + 1 / 4
because 2a and 1 are separate
So then it would be 2a + 1/4 divided by 2a - 3 / 6
your 2nd step was correct:\[\frac{a}{3}-\frac{1}{2}=\frac{2a-3}{6}\]
your 3rd step has an error
then 2a + 1 / 4 * 6 / 2a - 3
what you want is:\[\frac{2a+1}{4}\div\frac{2a-3}{6}=\frac{2a+1}{4}\times\frac{6}{2a-3}\]yes - you have it right now :)
yea I saw my error from the first part and realized I made the same mistake and then corrected it
ok - so you know what to do now and understand the process?
Now I am unsure of the rest ...
:) np - just let me know what part is troubling you...
you cannot do anything with the 2a + 1 or the 2a - 3 because the 2a's are attached to something, so the only thing you can mess with are the 4 and 6, correct?
(2a + 1)*3 / 2*(2a - 3) = 6a + 3 / 4a - 6 ?
that is correct. although I would leave it in the form:\[\frac{3(2a+1)}{2(2a-3)}\]
haha yea i just noticed it was already reduced and doing what I did, UN-reduced it LOL
gr8! the good thing is that you recognised your mistakes - which is vital if you want to improve! :D
I second guess myself all the time and change things when they don't need changed or over-think it!
he he - don't worry too much about it - the main thing here is that you have a good "hunger" for knowledge which means you will go far my friend! :D
thanks for your help! (I haven't "math-ed" in about 10 years gosh, that probably gives away my age) I'm a college student again and I'm taking Technical Algebra ... I actually like math now! =)
I left college many many moons ago - oh no! - now I've given away my age as well :) Like you, I still have a thirst for knowledge. :D
You're as "young" as you feel! take care and have a blessed day =) =)
thx - you too my friend! :)