Question 648064
Average speed on trip home is
( distance in miles ) / ( time in hrs )
Let {{{ t }}} = the time in hours going home
{{{ t + 10/60 }}} = time to go to work
{{{ 56 / ( t + 1/6 ) }}} = speed going to work
{{{ 56 / t }}} = speed going home
-------------------
given:
{{{ 56 / t = 56 / ( t + 1/6 ) + 8 }}}
Multiply both sides by {{{ t*( t + 1/6 ) }}}
{{{ 56*( t + 1/6 ) = 56t + 8t*( t + 1/6 ) }}}
{{{ 56t + 28/3 = 56t + 8t^2 + 4t/3 }}}
Subtract {{{ 56t }}} from both sides
and multiply both sides by {{{ 3 }}}
{{{ 28 = 24t^2+ 4t }}}
{{{ 6t^2 + t - 7 = 0 }}}
Use quadratic formula
{{{ t = (-b +- sqrt( b^2-4*a*c )) / (2*a) }}}
{{{ a = 6 }}}
{{{ b = 1 }}}
{{{ c = -7 }}}
{{{ t = (-1 +- sqrt( 1^2-4*6*(-7) )) / (2*6) }}}
{{{ t = (-1 +- sqrt( 1 + 168 )) / 12 }}}
{{{ t = (-1 +- sqrt( 169 )) / 12 }}}
{{{ t = ( -1 + 13 ) / 12 }}}
{{{ t = 1 }}}
{{{ 56 / t }}} = average speed going home
{{{ 56/t = 56 }}} 
The average speed going home is 56 mi/hr
check:
{{{ 56 / ( t + 1/6 ) }}} = speed going to work
{{{ 56 / (7/6) = 48 }}} mi/hr
rate x time = {{{ 48*( t + 1/6 ) }}}
{{{ 48*(7/6) = 56 }}} mi
OK