clock related challenge

$\def\deg{{^\circ}}$ The key thing with these clock puzzles is that the hands are not independent; when the minute hand goes forward by some angle $x$, the hour hand also goes forward by the amount $x\over 12$. The hour hand turns $360\deg$ in 12 hours, so $30\deg$ in one hour, and that is $\frac12\deg$ per minute. The minute hand turns $6\deg$ in one minute.

If the time is h:mm, where $0\le mm \lt 60$ and $0\le h \lt 12$, then the minute hand is at position $6\deg\cdot mm$ degrees and the hour hand is at position $h\cdot30\deg+mm\cdot{1\over 2}\deg$ degrees—the $h\cdot30\deg$ term is how far it has turned to get the the beginning of hour $h$, and then the $mm\cdot{1\over 2}\deg$ is how many degrees it has turned in the $mm$ minutes since the beginning of the hour.

Now we need two times, 4:mm and 7:nn, where the hand positions are reversed. The first has the minute hand at $6\deg mm$ and the hour hand at $120\deg + mm\cdot{1\over 2}\deg$. The second has the minute hand at $6\deg nn$ and the hour hand at $210\deg + nn\cdot{1\over 2}\deg$. So we need:

$$ \begin{eqnarray} 6\deg mm & = & 210\deg + nn\cdot{1\over 2}\deg \\ 6\deg nn & = & 120\deg + mm\cdot{1\over 2}\deg \end{eqnarray} $$

Dividing through by $6\deg$ gives:

$$ \begin{eqnarray} mm & = & 35 + {nn\over 12} \\ nn & = & 20 + {mm\over 12} \end{eqnarray} $$

By substitution we get $mm = 36+\frac{132}{143}$ minutes, and $nn = 23+\frac{11}{143}$ minutes. The two times are therefore approximately 4:36:55.359 and 7:23:04.615.