Question 1000285
Both tourists meet at point C, located somewhere between A and B.


The first tourist cycles for {{{t}}} hours at an average speed of {{{v[1]}}} km/hour, going from A to C.
So the distance from A to C is
{{{t*v[1]}}} .
After the two tourists meet at C, the second tourist cycles from C to A,
at an average speed of {{{v[2]}}} km/hour, for {{{9}}} hours.
So the distance from C to A is
{{{9*v[2]}}} .
Since the the distance from A to C is the same as the distance from C to A,
{{{t*v[1]=9*v[2]}}} (equation 1).


The second tourist cycled from B to C in {{{t+6}}} hours, at an average speed of {{{v[2]}}} km/hour.
So the distance from B to C is
{{{(t+6)*v[2]}}} .
After the two tourists meet at C, the second tourist cycles from C to B,
at an average speed of {{{v[1]}}} km/hour, for {{{8}}} hours.
So the distance from C to A is
{{{8*v[1]}}} .
Since the the distance from B to C is the same as the distance from C to B,
{{{8*v[1]=(t+6)*v[2]}}} (equation 2).


Dividing  equation 2 by equation 1, we get
{{{8*v[1]/(t*v[1])=(t+6)*v[2]/(9*v[2])}}}--->{{{8/t=(t+6)/9}}}--->{{{t(t+6)=8*9}}}--->{{{t(t+6)=72}}}--->{{{t^2+6t=72}}}--->{{{t^2+6t+9=72+9}}}--->{{{(t+3)^2=81}}}
From there, we conclude that
{{{t+3=9}}}--->{{{t=6}}} , because {{{t+3=-9}}}--->{{{t=-12}}} does not make sense.
Then {{{t+6=6+6=12}}} .


Substituting {{{t=6}}} into equation 1, we get
{{{6*v[1]=9*v[2]}}}--->{{{2v[1]=3v[2]}}} .


Before they met at C,
the second tourist had cycled {{{(t+6)=12v[2]}}} ,while
the first tourist had cycled {{{t*v[1]=6*v[1]}}} .
Since the first tourist had traveled 12 km less than the second tourist,
{{{6*v[1]=12v[2]-12}}}--->{{{v[1]=2v[2]-2}}}--->{{{2v[1]=4v[2]-4}}} .
{{{system(2v[1]=4v[2]-4,2v[1]=3v[2])}}}--->{{{4v[2]-4=3v[2]}}}--->{{{4v[2]-3v[2]-4=0}}}--->{{{v[2]-4=0}}}--->{{{v[2]=4}}} .
Then {{{system(v[1]=2v[2]-2,v[2]=4)}}}--->{{{v[1]=2*4-2}}}--->{{{v[1]=8-2}}}--->{{{highlight(v[1]=6)}}}