Can someone please verify this proof on the solution of rational equations? I have a rational equation in the following form:
$$R(x) = \sum_{i=1}^n\frac{\alpha_ix + \beta_i}{\gamma_i x + \delta_i} = 0$$, where the coefficients may be zero.
I would like to solve for $x$ or determine if it is not possible.
My approach is as follows: multiply by the product of all denominators:
$$R(x) = \prod_{i=1}^n(\gamma_i x + \delta_i)\sum_{i=1}^n(\alpha_i x + \beta_i) = 0$$
Then, expanding the sum and collecting like terms:
$$R(x) = \prod_{i=1}^n(\gamma_i x + \delta_i)\bigg(x\sum_{i=1}^n\alpha_i + \sum_{i=1}^n\beta_i\bigg) = P(x)\bigg(Ax + B\bigg) = 0$$
Therefore, the solution must be $x = \frac{-B}{A}$ provided that this is not a root of P(x).
Is that correct?  Did I miss anything after multiplying by the polynomial?

Note that 
$$\prod_{i=1}^n(\gamma_i x + \delta_i)\sum_{i=1}^n\frac{\alpha_ix + \beta_i}{\gamma_i x + \delta_i} \color{red}{\neq} \prod_{i=1}^n(\gamma_i x + \delta_i)\sum_{i=1}^n(\alpha_i x + \beta_i).$$
Rather
$$\prod_{i=1}^n(\gamma_i x + \delta_i)\sum_{i=1}^n\frac{\alpha_ix + \beta_i}{\gamma_i x + \delta_i} = \sum_{i=1}^n\left((\alpha_i x + \beta_i)\prod_{j=1\\ j\neq i}^n(\gamma_j x + \delta_j)\right).$$
