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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

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

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    $\begingroup$ multiply both sides by $n + 4$. The first term on the RHS become $y$. You want to confirm that the second term becomes 2 $\endgroup$
    – Bernard W
    Commented Nov 11, 2017 at 3:25
  • $\begingroup$ $\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} $. $\endgroup$
    – fleablood
    Commented Nov 11, 2017 at 3:26

2 Answers 2

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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.

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$$\frac{y+2}{n+4} =\frac{y}{n+4}+\frac{2}{n+4}$$

$$ =\frac{y}{n+4}.\frac{n}{n}+\frac{2+\frac{n}{2}-\frac{n}{2}}{n+4} $$

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

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