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

$$\frac{y+2}{n+4} = \left(\frac{n}{n+4}\right)\frac{y}{n} + \left(1-\frac{n}{n+4}\right)\frac{1}{2}$$

• multiply both sides by $n + 4$. The first term on the RHS become $y$. You want to confirm that the second term becomes 2 Commented Nov 11, 2017 at 3:25
• $\frac n {n+4}\frac yn +(1-\frac n {n+4})\frac 12=\frac y {n+4} +(\frac {n+4}{n+4}-\frac n {n+4})\frac 12=\frac y {n+4} +(\frac 4 {n+4})\frac 12=\frac y {n+4}+\frac 2 {n+4}=\frac {y+2}{n+4}$. Commented Nov 11, 2017 at 3:26

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.
$$\frac{y+2}{n+4} =\frac{y}{n+4}+\frac{2}{n+4}$$
$$=\frac{y}{n+4}.\frac{n}{n}+\frac{2+\frac{n}{2}-\frac{n}{2}}{n+4}$$
$$=\left(\frac{n}{n+4}\right)\frac{y}{n} +\left(\frac{2+\frac{n}{2}-\frac{n}{2}}{n+4}\right) .\frac{2}{2}$$ $$=\left(\frac{n}{n+4}\right)\frac{y}{n} +\left(\frac{n+4-n}{n+4}\right).\frac{1}{2}$$ $$=\left(\frac{n}{n+4}\right)\frac{y}{n} + \left(1-\frac{n}{n+4}\right)\frac{1}{2}$$