Question 192618
Let {{{s}}}= her average speed on the trip to Richmond
Let {{{t[1]}}} = time to get to Richmond
Let {{{t[2]}}} = time for return trip
given:
{{{d = 600}}} mi is the one-way distance
{{{t[1] + t[2] = 22}}} hr
-------------------
{{{d = s*t[1]}}}
{{{600 = s*t[1]}}}
and
{{{d = (s - 10)*(22 - t[1])}}}
{{{600 = (s - 10)*(22 - t[1])}}}
There are 2 equations with 2 unknowns, so it's solvable
{{{600 = 22s - 220 - s*t[1] + 10t[1]}}}
and, since {{{600 = s*t[1]}}}, 
{{{600 = 22s - 220 - 600 + 10t[1]}}}
{{{1420 = 22s + 10*(600/s)}}}
{{{1420s = 22s^2 + 6000}}}
{{{22s^2 - 1420s + 6000 = 0}}}
{{{11s^2 - 710s + 3000 = 0}}}
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} 
{{{a = 11}}}
{{{b = -710}}}
{{{c = 3000}}}
{{{x = (-(-710) +- sqrt( (-710)^2-4*11*3000 ))/(2*11) }}}
{{{x = ( 710 +- sqrt( 504100-132000 ))/ 22 }}}
{{{x = ( 710 +- sqrt( 372100 ))/ 22 }}}
{{{x = ( 710 +- 610 ))/ 22 }}} 
{{{s = 1320/22}}}
{{{s = 60}}} mi/hr
{{{s - 10 = 50}}} mi/hr
the trip to Richmond was made at 60 mi/hr
The trip back at 50 mi/hr
check answer 
{{{600 = (s - 10)*(22 - t[1])}}}
{{{600 = 50*(22 - t[1])}}}
{{{600 = 1100 - 50t[1]}}}
{{{50t[1] = 1100 - 600}}}
{{{50t[1] = 500}}}
{{{t[1] = 10}}} hr
{{{t[2] = 22 - 10}}}
{{{t[2] = 12}}} hr
{{{600 = s*t[1]}}}
{{{600 = 60*10}}}
{{{600 = 600}}}
OK