Question 1031941
Try analyzing and solving the described problem without checking first what David did.


<pre>
             speed           time        distance

WALK          3              {{{t+7/60}}}       d

FASTER        4              {{{t-5/60}}}       d
</pre>

Continue when that data makes sense to you.


{{{system(3(t+7/60)=d,4(t-1/12)=d)}}}


{{{3t+7*3/60=4t-4/12}}}
{{{3t+7/20=4t-1/3}}}
{{{7/20=t-1/3}}}
{{{t=7/20+1/3}}}
{{{t=21/60+20/60}}}
{{{highlight_green(t=41/60)}}} meaning 41 minutes, the time from beginning the walk until train will leave.


{{{d=4(41/60-5/60)}}}, taking the second equation of the system.
{{{d=4(36/60)}}}
{{{d=36/15}}}
{{{d=18/5}}}
{{{highlight(d=2&3/5)}}}---------MILES


You can clearly see that at least one of David's mistakes was to forget the accounting properly for time units.  Seven minutes is  {{{7/60}}} hour, and 5 minutes is  {{{5/60=1/12}}} hour.