Need help finding solution to Complex Fraction?

1
2 - -------------
2y +1
___________________________________

1 1
--------------------------- - --------- 1 2y+1
------------- - -----------------
2y^-5y-3 y - 3 (y-3)(2y+1) (y-3)(2y+1)


Denominator= 2y
---------------
(y-3)(2y+1)
Nominator =

2(2y+1)-1 4y+1
------------- = ---------- so skip one step and take the reciprocal,
2y+1 2y+1

4y +1 (y-3)(2y+1) 4y^-11y-3
-------- x -------------- = ------------- I get lost around here?
2y+1 2y 4y^+2y

SCSmith said:

1
2 - -------------
2y +1

Click to expand...

Repost in readable fashion.

As example, if above is 2 minus 1 divided by sum of 2y + 1, type like this:

2 - 1 / (2y + 1)

/ = division; use * for multiplication.

\(\displaystyle \L \frac{2 - \frac{1}{2y+1}}{\frac{1}{2y^2-5y-3} - \frac{1}{y-3}}=\)

\(\displaystyle \L \frac{\frac{2(2y+1)(y-3) - (y-3)}{(2y+1)(y-3)}}{\frac{1 - (2y+1)}{(2y+1)(y-3)}}=\)

\(\displaystyle \L \frac{(y-3)[2(2y+1) - 1]}{-2y} =\)

\(\displaystyle \L \frac{(y-3)(4y+1)}{-2y} =\)

\(\displaystyle \L \frac{(3-y)(4y+1)}{2y}\)

Here's a suggestion, Smitty.
2y^2 - 5y - 3 factors to: (2y +1)(y - 3)
Let a = 2y +1 and b = y - 3
Then your expression becomes less "wieldy":

(2 - 1/a) / [1/(ab) - 1/b]

You can now simplify that, and substitute back.