Simple algebra derivation I am reading a paper and came across, what the author claims, is simple algebra. I made a few attempts, but have struggled. The equivalence claim is
$$\frac{y+2}{n+4} = \left(\frac{n}{n+4}\right)\frac{y}{n} + \left(1-\frac{n}{n+4}\right)\frac{1}{2}$$
 A: What exactly is your problem with that expression? You can't see why the two statements on the left and right sides of the equals sign are equivalent?
\begin{align}\require{cancel}
\left(\frac{n}{n+4}\right)\frac{y}{n} + \left(1-\frac{n}{n+4}\right)\frac{1}{2}
&=\frac{\cancel{n}y}{\cancel{n}(n+4)} + \left(\frac{n+4}{n+4}-\frac{n}{n+4}\right)\frac{1}{2}\\
&=\frac{y}{n+4} + \left(\frac{n+4 -n}{n+4}\right)\frac{1}{2}\\
&=\frac{y}{n+4} + \left(\frac{\cancel{n}+4 \cancel{-n}}{n+4}\right)\frac{1}{2}\\
&=\frac{y}{n+4} + \frac{4 }{n+4}\cdot\frac{1}{2}\\
&=\frac{y}{n+4} + \frac{2\cdot 2 }{2(n+4)}\\
&=\frac{y}{n+4} + \frac{\cancel{2}\cdot 2 }{\cancel{2}(n+4)}\\
&=\frac{y}{n+4} + \frac{2}{n+4}\\
&=\frac{y+2}{n+4}
\end{align}
If you want to get from $\frac{y+2}{n+4}$ to $\left(\frac{n}{n+4}\right)\frac{y}{n} + \left(1-\frac{n}{n+4}\right)\frac{1}{2}$, just trace the steps backwards.
A: $$\frac{y+2}{n+4} =\frac{y}{n+4}+\frac{2}{n+4}$$
$$ =\frac{y}{n+4}.\frac{n}{n}+\frac{2+\frac{n}{2}-\frac{n}{2}}{n+4} $$
$$=\left(\frac{n}{n+4}\right)\frac{y}{n} +\left(\frac{2+\frac{n}{2}-\frac{n}{2}}{n+4}\right) .\frac{2}{2}$$
$$=\left(\frac{n}{n+4}\right)\frac{y}{n} +\left(\frac{n+4-n}{n+4}\right).\frac{1}{2}$$
$$=\left(\frac{n}{n+4}\right)\frac{y}{n} + \left(1-\frac{n}{n+4}\right)\frac{1}{2}$$
