Linear equation | Finding the unknown values I have a set of equations as follow,

*

*$a_1\cdot x_1 + a_2\cdot x_2 + a_3\cdot x_3 + a_4\cdot x_4 + \dots + a_{11} \cdot x_{12} + a_{12}\cdot x_{12}  = 6000$

*$x_1 + x_2 + x_3 + x_4 + \dots x_{11} + x_{12} = 100 $

*$\{a_1,a_2,a_3, \dots , a_{11}, a_{12}\} \leq 60$
How can I find the values for all $a_i, \text{ and }  x_i$,not all values of $a_i$ can be equal to $60$ also if possible integer, else the general solution.

*

*How the question get affected if I change the constrain on $a_i$ to some $X$, say $0 < a_i \leq X$,

*Also if the constrain on $a_i$ is , say $a_i>0$.


If solvable in Matlab please show the code also .

If not solvable please mention why.

What I tried is
$$
\begin{bmatrix}
a_1&a_2 & a_3 &\dots & a_{11} & a_{12} \\
1 & 1 & 1& \dots & 1 & 1
\end{bmatrix}\begin{bmatrix}
x_1 \\ x_2 \\ x_3  \\ \vdots\\ x_{11} \\ x_{12}
\end{bmatrix} = \begin{bmatrix}
6000 \\ 100
\end{bmatrix}
$$
The rank will be almost $2$ so atmost I can get $10$ free variables. but since the values for $a_i$'s are also unknown I don't know how to proceed.
 A: So we have
$$
\left\{ \matrix{
  0 \le a_{\,k}  \le 60\quad \left| {\,1 \le k \le 12} \right. \hfill \cr 
  0 \le x_{\,k}  \le 100\quad \left| {\,1 \le k \le 12} \right. \hfill \cr 
  x_{\,1}  + x_{\,2}  +  \cdots  + x_{\,12}  = 100 \hfill \cr 
  {\bf a} \cdot {\bf x} = 6000 \hfill \cr}  \right.
$$
If we put $b_k = 60 -a_k$ then
$$
\left\{ \matrix{
  b_{\,k}  = 60 - a_{\,k} \quad \left| {\,1 \le k \le 12} \right. \hfill \cr 
  u_{\,k}  = 1\quad \left| {\,1 \le k \le 12} \right. \hfill \cr 
  0 \le x_{\,k}  \le 100 \hfill \cr 
  0 \le b_{\,k}  \le 60\quad \left| {\,1 \le k \le 12} \right. \hfill \cr 
  {\bf u} \cdot {\bf x} = 100 \hfill \cr 
  6000 = {\bf a} \cdot {\bf x} = \left( {60\,{\bf u} - {\bf b}} \right) \cdot {\bf x} =  \hfill \cr 
   = 6000 - {\bf b} \cdot {\bf x}\quad  \Rightarrow \quad {\bf b} \cdot {\bf x} = 0 \hfill \cr}  \right.
$$
and clearly, with ${\bf b}, \, {\bf x}$ having non-negative components we must have
$$
\left\{ \matrix{
  {\bf b} = {\bf 0}\; \Rightarrow \;{\bf a} = 60\,{\bf u} \hfill \cr 
  0 \le x_{\,k}  \le 100 \hfill \cr 
  {\bf u} \cdot {\bf x} = 100 \hfill \cr}  \right.
$$
That is, $\bf x$ are the vectors lying on the diagonal plane of a 12-D hypercube of side  $[0,100]$.
There are  $101^{12}  \approx  10^{24}$ points inside that cube, and those on the diagonal plane
are exactly the number of weak compositions
of $100$ into $12$ parts , i.e. $\binom {100+12-1}{11} \approx 2.8 \cdot 10^{14}$.
So your first care should be how to manage such a huge amount of data, if you want to list all the solutions.
You can somewhat reduce the size by taking advantage of the permutation invariance.
If $(x_1,x_2, \cdots)$ is a solution, then any permutation of the twelve components will be.
So you can limit to list the solutions in order, e.g.  non-decreasing.
The number in this case will be reduced to that of the partitions of $100$ into up to twelve parts
(because the components may be null), which is $ \approx 1.7 \, \cdot 10^7$.
A: Assume that $\exists i\in \{1, 2, ..., 12\}: a_i < 60$, then
$$6000 = a_1 x_1 + ... + a_{12} x_{12} < 60x_1 + ...+ 60x_{12} = 60(x_1 + ...+x_{12}) = 6000 $$
is contradiction. So $\forall i \in \{1, 2, ..., 12\}: a_i \geq 60$, so $a_1 = a_2 = ... = a_{12} = 60$
