System of two (quite similar) equations Given the real, natural and binary successions $\{t_1,...,t_N\} \in \mathbb{R}$, $\{n_1,...,n_N\} \in \mathbb{N}$ and $\{E_1,...,E_N\} \in \{0,1\}$ we would like to find the $(x,y)$ that satisfies the following equations
$$\begin{cases} \displaystyle\sum_{i=1}^N \frac{E_i}{x-n_iy} = \displaystyle\sum_{i=1}^N n_i t_i \\ \displaystyle\sum_{i=1}^N \frac{E_in_i}{x-n_iy} = \displaystyle\sum_{i=1}^N n^2_i t_i \end{cases}$$
Both equations are quite similar, and my intuition says that there is an analytical solution, but I cannot find it. 
Would you be able to find the explicit solution? (or prove that there is not solution?)
 A: Given


*

*$b_1, b_2, \dots, b_n \in \{0,1\}$

*$m_1, m_2, \dots, m_n \in \mathbb N$

*$c_1, c_2 \in \mathbb R$


define
$$z_i := \dfrac{1}{x - m_i y}$$
We have a system of $2$ equations in $\mathrm z \in \mathbb R^n$
$$\begin{bmatrix} b_1 & b_2 & \dots & b_n\\ b_1 m_1 & b_2 m_2 & \dots & b_n m_n\end{bmatrix} \begin{bmatrix} z_1\\ z_2\\ \vdots\\ z_n\end{bmatrix} = \begin{bmatrix} c_1\\ c_2\end{bmatrix}$$
or, in a more succinct form, $\mathrm A \mathrm z = \mathrm c$. If $n > 2$, this is an underdetermined system whose least-norm solution is
$$\hat{\mathrm z} := \mathrm A^T (\mathrm A \mathrm A^T)^{-1} \mathrm c$$
If all the entries of $\hat{\mathrm z}$ are nonzero, then we have an overdetermined system of equations
$$\begin{array}{rl} x - m_1 y &= \hat z_1^{-1}\\ x - m_2 y &= \hat z_2^{-1}\\ &\vdots \\ x - m_n y &= \hat z_n^{-1}\end{array}$$
Lastly, we compute the least-squares solution $(\hat x, \hat y)$. If $\hat z_i = 0$, then the $i$-th equation, whose right-hand side is illegal, is simply discarded.
