Let $r \in \mathbb{R}$ such that $-82 < r < 82$.
Let:
$p_1 = r/44$
$p_2 = r/42.8$
$m_1 = 21p_1$
$m_2 = 21p_2$
Given $ p_1 \pmod 1 $, $ m_1 \pmod 1$, $ m_2 \pmod 1 $ solve for $r$. As pointed out below, $ x \pmod 1 $ in this context is the same as $ x - \lfloor x \rfloor$. For clarity, I'm looking for a solution in the form below:
$r = f(p_1 \pmod 1,m_1 \pmod 1,m_2 \pmod 1) = $
I've written a MATLAB script which shows that there is one solution for $r$ when $r$ is bounded by $-82 < r < 82$, but I can't determine $r$ given only these results.
Let $\{x\}=x-\lfloor x\rfloor$. Also, let $a=\{p_1\}, b=\{m_1\},c=\{m_2\}$.
Then, we want to solve the following system for $r$ when we are given $a,b,c$ :
$$\begin{cases}a=\{\frac{r}{44}\}\\\\ b=\{\frac{21r}{44}\} \\\\c=\{\frac{105r}{214}\}\end{cases}$$
Now, since $-82\lt r\lt 82$, we have $\lfloor\frac{r}{44}\rfloor=-2,-1,0,1$.
So, we have $$\frac{r}{44}=a-2,a-1,a,a+1,$$ i.e. $$r=44(a-2),44(a-1),44a,44(a+1)$$
Therefore, the answer is $$r=\begin{cases}44(a-2)&\text{if $\quad b=\{21(a-2)\}\quad $ and $\quad c=\{\frac{2310(a-2)}{107}\}$} \\\\44(a-1)&\text{if $\quad b=\{21(a-1)\}\quad $ and $\quad c=\{\frac{2310(a-1)}{107}\}$} \\\\44a&\text{if $\quad b=\{21a\}\quad $ and $\quad c=\{\frac{2310a}{107}\}$} \\\\44(a+1)&\text{if $\quad b=\{21(a+1)\}\quad $ and $\quad c=\{\frac{2310(a+1)}{107}\}$}\end{cases}$$
where $a=\{p_1\},b=\{p_2\},c=\{m_2\}$ with $\{x\}=x-\lfloor x\rfloor$.