I saw this question
Let $A ∈ M_{m\times n}(\mathbb{Q})$ and $b ∈ \mathbb{Q}^m$. Suppose that the system of linear equations $Ax = b$ has a solution in $\mathbb{R}^n$. Does it necessarily have a solution in $\mathbb{Q}^n$?
and I thought I'd give an interesting, possibly wrong, approach to solving it. I'm not sure if such things can be done, if not maybe you can help me refine.
I considered the form of the equality as
$$ A^{(1)}x_1+\cdots+A^{(n)}x_n=b, $$
where $A^{(i)}$ is a column vector of $A$. I then noticed that for $x_i\in\mathbb{R\setminus Q}=\mathbb{T}$ then, and this is where I think I'm doing something forbidden, each $x$ has the represenation
$$ x_1 = k_{11}\tau_1+\cdots+k_{1p}\tau_p \\ x_2= k_{21}\tau_1+\cdots+k_{2p}\tau_p \\ \vdots \hspace{4cm} \vdots \\ x_n=k_{n1}\tau_1+\cdots+k_{np}\tau_p,$$
where $\tau_i$ is a distinct irrational number, $k_{ij}\in\mathbb{R}$, and $p$ is the number of such distinct irrational numbers. I wound this out, but there may be a discrepancy with $p$ and $m$. I feel this method can lead me to the answer, but I'm not sure where to go from here.
EDIT$^1$:
I end up getting something like this, I believe, after substitution:
$$ A(k^{(1)}\tau_1+\cdots+k^{(p)}\tau_p)=b$$
Here, $k^{(i)}$ is the vector
$$\begin{pmatrix} k_{1i} \\ \vdots \\ k_{ni} \end{pmatrix}.$$
EDIT$^2$:
I think there is no discrepancy with $p$ and $m$ because $A\in M_{m\times n}(\mathbb{Q})$, $K\in M_{n\times p}(\mathbb{R})$, and $\tau\in M_{p\times 1}(\mathbb{T})$, so
$$ (m\times n)\cdot (n\times p)\cdot (p\times 1) = m\times 1. $$
