Just plug it in!
$${\mathtt{6\,872}} = {\mathtt{14\,592}}{\mathtt{\,\times\,}}\left({\mathtt{1}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{r}}}{{\mathtt{1}}}}\right){\mathtt{\,\times\,}}{\mathtt{11}} \Rightarrow {\mathtt{r}} = {\mathtt{\,-\,}}{\frac{{\mathtt{19\,205}}}{{\mathtt{20\,064}}}} \Rightarrow {\mathtt{r}} = -{\mathtt{0.957\: \!187\: \!001\: \!594\: \!896\: \!3}}$$
I'll try to show some work, but gimme a sec.
As far as I know, we need to isolate the 'r'.
First combine like terms.
$${\mathtt{6\,872}} = {\mathtt{160\,512}}{\mathtt{\,\times\,}}\left({\mathtt{1}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{r}}}{{\mathtt{1}}}}\right)$$
Divide 6872 by 160512
$${\frac{{\mathtt{6\,872}}}{{\mathtt{160\,512}}}} = {\mathtt{0.042\: \!812\: \!998\: \!405\: \!103\: \!7}}$$
$${\mathtt{0.042\: \!812\: \!998\: \!405\: \!103\: \!7}} = {\mathtt{1}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{r}}}{{\mathtt{1}}}}$$
The 'r/1' can be just 'r'. All that's needed is to subtract 1 from 0.0428129984051037
$${\mathtt{0.042\: \!812\: \!998\: \!405\: \!103\: \!7}}{\mathtt{\,-\,}}{\mathtt{1}} = -{\mathtt{0.957\: \!187\: \!001\: \!594\: \!896\: \!3}}$$
$$-{\mathtt{0.957\: \!187\: \!001\: \!594\: \!896\: \!3}} = {\mathtt{r}}$$
And thus, the work =)