Explain how the following expression was derived?

Question

Can someone explain how the author gets to the expression after the words "This leads to:"

Answer 1

If $$\frac \alpha {1-\alpha} = \frac p q,\tag 1 $$ then $$\alpha = \dfrac p {p+q}.$$

To see this, first multiply both sides of $(1)$ by $1-\alpha$: $$ \alpha = (1-\alpha)\frac p q. $$ Expand: $$ \alpha = \frac p q - \alpha \frac p q. $$ Move all $\alpha$s to one side: $$ \alpha + \alpha \frac p q = \frac p q. $$ Factor $$ \alpha \left( 1 + \frac p q \right) = \frac p q. $$ So $$ \alpha = \frac{p/q}{1+(p/q)}. $$ Then multiply the numerator and denominator by $q$: $$ \alpha = \frac p {q+p}. $$

Answer 2

Using algebraic rearrangement

$$\begin{align} \frac{\alpha(x)}{1-\alpha(x)} & =\frac{\alpha\; \mathsf E(g(\mu)\mid x)}{\beta\; \mathsf E(h(\mu)\mid x)} \\[1ex] \alpha(x)\;\beta\; \mathsf E(h(\mu)\mid x) & =\alpha\; \mathsf E(g(\mu)\mid x)\;\big(1-\alpha(x)\big) \\[1ex] \alpha (x) \left(\alpha\; \mathsf E(g(\mu)\mid x)+\beta\; \mathsf E(h(\mu)\mid x)\right) & = \alpha\; \mathsf E(g(\mu)\mid x) \\[1ex] \alpha(x) &=\frac{\alpha\; \mathsf E(g(\mu)\mid x)}{\alpha\; \mathsf E(g(\mu)\mid x)+\beta\; \mathsf E(h(\mu)\mid x)} \end{align}$$

Then as $\alpha = \tfrac 2 3, \beta=\tfrac 1 3$ (the mixture ratio):

$$\begin{align} \alpha(x) &=\frac{2\; \mathsf E(g(\mu)\mid x)}{2\; \mathsf E(g(\mu)\mid x)+ \mathsf E(h(\mu)\mid x)} \\[1ex] & =\frac{2\int f(\mu\mid x)\,g(x)\operatorname d \mu}{2\int f(\mu\mid x)\,g(x)\operatorname d \mu+\int f(\mu\mid x)\,h(x)\operatorname d \mu} \end{align}$$