Question

Yeah I don't understand this one either... Help please?
https://media.glynlyon.com/g_alg02_2013/5/175.gif

Answer 1

lol your math teacher must hate you

Answer 2

I'm such a good student, I never did anything wrong lol.

Answer 3

\[\large \frac{12a^2-3}{2}\times (2a-1)^{-2}\left(\frac{6}{2a+1}\right)^{-1}\]

Answer 4

lets get rid of the negative exponents first and rewrite as
and rewrite as
\[\large \frac{12a^2-3}{2}\times \left(\frac{1}{2a-1}\right)^2\times \left(\frac{2a+1}{6}\right)\]

Answer 5

damn type=o !!\[\large \frac{12a^2-3}{2}\times \left(\frac{1}{2a+1}\right)^2\times \left(\frac{2a+1}{6}\right)\]

Answer 6

now cancel the common factor of \(2a+1\) top and bottom and get \[\large \frac{12a^2-3}{2}\times \left(\frac{1}{2a+1}\right)\times \left(\frac{1}{6}\right)\]

Answer 7

more to go though as \(12a^2-3=3(4a^1-1)=3(2a+1)(2a-1)\)

Answer 8

\[\large \frac{12a^2-3}{2}\times \left(\frac{1}{2a+1}\right)\times \left(\frac{1}{6}\right)\]
\[=\large \frac{3(2a+1)(2a-1)}{2}\times \left(\frac{1}{2a+1}\right)\times \left(\frac{1}{6}\right)\]

Answer 9

some more cancellations gives \[=\large \frac{3(2a-1)}{2}\times \left(\frac{1}{2a+1}\right)\times \left(\frac{1}{6}\right)\]

Answer 10

and finally, since \(\frac{3}{12}=\frac{1}{4}\) you are left only with
\[\large\frac{2a-1}{4(2a+1)}\]

Answer 11

you were probably checking instagram in class, which is why your math teacher gives you questions like this

Answer 12

That is a good possibility. Haha i wrote down all the steps and I think I understand it more now. Makes more sense the way you explained it than the way my teacher explained it. :)

Answer 13

thanks
good luck

Answer 14

I'll pay more attention in class next time!