How to solve a system of nonlinear equation, with y known and x or its coefficients unknown? While solving a complex problem I have ended up with this simplified problem:
There are eight straight lines in the plane. They are notated as follows:
\begin{gather}
\tag{1}
\label{1}
y=k_1 x+b_1\\
y=k_1 x+b_2\\
y=k_1 x+b_3\\
y=k_1 x+b_4
\end{gather}
\begin{gather}
\tag{2}
\label{2}
y=k_2 x+b_5\\
y=k_2 x+b_6\\
y=k_2 x+b_7\\
y=k_2 x+b_8
\end{gather}
The slopes are $k_1$ and $k_2$, the intercepts on the $y$ axis are $b_1,b_2,b_3,\dotsc,b_8$, and $k_i,b_j \in (0 ,1)$, $i=1,2$; $j=1,2,\dotsc,8$ and $b_1+b_3=b_2+b_4$, $b_5+b_7=b_6+b_8$.
Suppose there are four vectors of independent variables $X$, denoted as $X_1,X_2,X_3,X_4$, where $X_p=(x_{p1},x_{p2},\dotsc,x_{ps},\dotsc,x_{pn})$ corresponds to a collection of the abscissas of $n$  points, and $X_1+X_3=X_2+X_4$. Substituting $X_1, X_2, X_3, X_4$ into equations \eqref{1} and \eqref{2} in turn to obtain eight vectors of dependent variables $Y$, denoted as $Y_1, Y_2,\dotsc, Y_8$, and $Y_q =(y_{q1},y_{q2},\dotsc,y_{qs},\dotsc,y_{qn})$ corresponds to a collection of the  ordinates of $n$ points, which are notated as follows:
\begin{gather}
\tag{3}
\label{3}
Y_1=k_1 X_1+b_1 I\\
Y_2=k_1 X_2+b_2 I\\
Y_3=k_1 X_3+b_3 I\\
Y_4=k_1 X_4+b_4 I
\end{gather}
\begin{gather}
\tag{4}
\label{4}
Y_5=k_2 X_1+b_5 I\\
Y_6=k_2 X_2+b_6 I\\
Y_7=k_2 X_3+b_7 I\\
Y_8=k_2 X_4+b_8 I.
\end{gather}
where $I$ is $n$-dimensional all-ones vector.
Question:

*

*How to solve these equation to obtain  the coefficients $k_i$, $b_j$, under the condition of $Y$ known and $X$ unknown? Closed form solution or numerical solution?

*What's the minimum  number of $n$ for which these equations are solvable?


ADDED Additional information
$\forall x_{p,s}\in X_p,x_{p,s} \in (0,1)$
$b_q=c_q d_1$ and $b_{q+4}=c_q d_2$,$c_q,d_i \in (0,1)$, $q \in \{1,2,3,4\}$,$i \in \{1,2\}$ .
 A: First we notice that the relations $b_1+b_3=b_2+b_4$, $b_5+b_7=b_6+b_8$, and $X_1+X_3=X_2+X_4$ imply that $Y_1+Y_3=Y_2+Y_4$ and $Y_5+Y_7=Y_6+Y_8$. If the these relations do not hold for given $Y_q$, there are no solutions. So, we assume that they do hold, in which case the fourth ($Y_4=\dots$) and eights ($Y_8=\dots$) equations can be considered redundant, while $b_4$ and $b_8$ are expressed from the others $b_i$ as $b_4 = b_1+b_3 - b_2$ and $b_8 = b_5 + b_7 - b_8$.
So, we need to the remaining 6 equations with respect to $k_1,k_2,b_1,b_2,b_3,b_5,b_6,b_7$. From these equations we obtain:
$$\begin{cases}
k_2 Y_1 - k_1 Y_5 = (k_2 b_1 - k_1 b_5) I, \\
k_2 Y_2 - k_1 Y_6 = (k_2 b_2 - k_1 b_6) I, \\
k_2 Y_3 - k_1 Y_7 = (k_2 b_3 - k_1 b_7) I.
\end{cases}
$$
Denoting $Y'_q := (y_{q,1}-y_{q,n}, y_{q,2}-y_{q,n}, \dots, y_{q,n-1}-y_{q,n})$, we can rewrite the above system in the following equivalent form:
$$(\star)\qquad
\begin{cases}
k_2 Y'_1 - k_1 Y'_5 = 0, \\
k_2 Y'_2 - k_1 Y'_6 = 0, \\
k_2 Y'_3 - k_1 Y'_7 = 0, \\
k_2 y_{1,n} - k_1 y_{5,n} = k_2 b_1 - k_1 b_5, \\
k_2 y_{2,n} - k_1 y_{6,n} = k_2 b_2 - k_1 b_6, \\
k_2 y_{3,n} - k_1 y_{7,n} = k_2 b_3 - k_1 b_7.
\end{cases}
$$
That is, viewing $Y'_q$ as column-vectors, we conclude that
$$M\cdot \begin{bmatrix} k_2 \\ -k_1 \end{bmatrix} = 0,$$
where
$$
M:=\begin{bmatrix}
Y'_1 & Y'_3 \\
Y'_2 & Y'_6 \\
Y'_3 & Y'_7 
\end{bmatrix}
$$
is $3(n-1) \times 2$ matrix. In other words, $\begin{bmatrix} k_2 \\ -k_1 \end{bmatrix}$ belongs to the kernel of $M$.
When $k_1$ and $k_2$ are determined, the last three equations of $(\star)$ represent a linear system with respect to $b_i$, which we can solve. If we are interested in just any solution, we can simply take $b_i = y_{i,n}$.

ADDED. The OP indicated in the comments that the given $Y_q$ are noisy, and thus we need to solve the equation approximately rather than exactly. In this case, it is worth to keep all 8 equations for better accuracy.
First, we can estimate the value of $\alpha:=\frac{k_2}{k_1}$ as the average of
$$\frac{y_{q,i} - y_{q,j}}{y_{q+4,i} - y_{q+4,j}}$$
over all $q\in\{1,2,3,4\}$ and $i\ne j$ from $\{1,2,\dots,n\}$.
Notice there is no way to estimate $k_1$ and $k_2$ individually from the given data, but we can fix any value for $k_1$, say $k_1=1$, to obtain the estimation $k_2=\alpha k_1$.
Then, we can estimate $b_q$ by just averaging $y_{q,i}$ over all $i\in\{1,2,\dots,n\}$.
