Question 273582
Each cyclist must have their own {{{d=r*t}}} equation
1st cyclist:
{{{d[1] = r[1]*t[1]}}}
2nd cyclist:
{{{d[2] = r[2]*t[2]}}}
given:
The 1st cyclist has a head start which is
{{{6}}} mi/hr x {{{3}}} hrs = {{{18}}} mi
{{{r[1] = 6}}} mi/hr
{{{r[2] = 10}}} mi/hr
-----------------------
Assume I have a stopwatch and I start it when
the 2nd cyclist leaves. When I stop the watch
{{{t[1] = t[2]}}} so I'll call them both {{{t}}}
The 2nd cyclist has to make up the head start, so
{{{d[2] = d[1] + 18}}} mi
------------------------
{{{d[1] = r[1]*t}}}
(1) {{{d[1] = 6t}}}
and
{{{d[2] = r[2]*t}}}
(2) {{{d[1] + 18 = 10t}}}
I'll substitute (1) into (2)
{{{6t + 18 = 10t}}}
{{{4t = 18}}}
{{{t = 4.5}}} hrs
This is the elapsed time after the 1st cyclist got 
a {{{3}}} hr head start, so
the 2nd cyclist catches the 1st cyclist {{{4.5 + 3 = 7.5}}} hrs
after the 1st cyclist left
check answer:
{{{d[2] = 10t}}}
{{{d[2] = 10*4.5}}}
{{{d[2] = 45}}} mi
and
{{{d[1] = 6t }}}
and, since 
{{{d[2] = d[1] + 18}}}
{{{d[1] = d[2] - 18}}}
then,
{{{d[2] - 18 = 6*4.5}}}
{{{d[2] - 18 = 27}}}
{{{d[2] = 45}}} mi
OK