Really easy algebra problem

Question

Can somebody show me how to go about solving the following (really easy) equation for $\alpha$ please? $$\displaystyle 0 = \sum_{n=1}^{N}\left( \frac{y_n}{\alpha} + \frac{1-y_n}{1-\alpha}\right)$$

Answer 1

Multiplying through by $\alpha(1-\alpha)$ you get $$\begin{align*} 0 &= \sum_n\left(\frac{y_n}{\alpha} + \frac{1-y_n}{1-\alpha}\right)\\ 0 &= \sum_{n}\left( (1-\alpha)y_n + \alpha(1-y_n)\right)\\ 0 &= \sum_n \left( y_n + (1-2y_n)\alpha\right)\\ 0&= \sum_n y_n + \sum_n \alpha(1-2y_n)\\ \sum_n \alpha(2y_n -1) &= \sum_n y_n\\ \alpha\sum_n(2y_n-1) &= \sum_n y_n\\ \alpha &= \frac{\displaystyle \sum_n y_n}{\displaystyle\sum_n (2y_n-1)}. \end{align*} $$

Answer 2

$$\displaystyle 0 = \sum_n\left( \frac{y_n}{\alpha} + \frac{1-y_n}{1-\alpha}\right)$$ $$0 = \sum_n\left( \frac{y_n}{\alpha} \right) + \sum_n\left(\frac{1-y_n}{1-\alpha}\right)$$ $$0 = \frac{1}{\alpha} \sum_n y_n + \frac{1}{1-\alpha} \sum_n {(1-y_n)}$$

Then you can solve it.