Question 361015
The key is that if they will always meet at
the same time on an observers watch.
since they leave at the same time, each 
friend's equation of motion has the same {{{t}}}, elapsed time.
given:
friend 1:
{{{d = 21t}}}
Friend 2:
{{{60 - d = 15t}}}
By substitution:
{{{60 - 21t = 15t}}}
{{{36t = 60}}}
{{{3t = 5}}}
{{{t = 5/3}}} hr
It takes 1 hr and 40 min for them to meet
friend 1:
{{{d = 21*(5/3)}}}
{{{d = 35}}} mi
friend 2:
{{{60 - d = 15*(5/3)}}}
{{{60 - d = 25}}} mi
friend 1 goes 35 mi and friend 2 goes 25 mi