Question 1036324
Let {{{ t }}} = time in hrs for 1st part of trip
{{{ 12 - t }}} = his time in hrs for 2nd part of trip
Let {{{ s }}} = his speed in mi/hr for 1st part of trip
{{{ s + 9 }}} = his speed in mi/hr for 2nd part of trip
-------------------------------------------
Equation for 1st part:
(1) {{{ 273 = s*t }}}
Equation for 2nd part:
(2) {{{ 240 = ( s + 9 )*( 12 - t ) }}}
----------------------------
(1) {{{ t = 273/s }}}
(2) {{{ 240 = 12s + 108 - s*t - 9t }}}
(2) {{{ 240 = 12s + 108 - t*( s + 9 ) }}}
(2) {{{ 240 = 12s + 108 - (273/s)*( s + 9 ) }}}
(2) {{{ 240 = 12s + 108 - 273 - 2457/s }}}
(2) {{{ 240 = 12s - 165 - 2457/s }}}
(2) {{{ 2457/s  + 405 = 12s }}}
(2) {{{ 2457 + 405s = 12s^2 }}}
(2) {{{ 12s^2 - 405s - 2457 = 0 }}}
(2) {{{ 4s^2 - 135s - 819 = 0 }}}
----------------------------
Use quadratic formula
{{{ s = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
{{{ a = 4 }}}
{{{ b = -135 }}}
{{{ c = -819 }}}
{{{ s = ( -(-135) +- sqrt( (-135)^2-4*4*(-819) ))/(2*4) }}}
{{{ s = ( 135 +- sqrt( 18225 + 13104 )) / 8 }}}
{{{ s = ( 135 +- sqrt( 31329 )) / 8 }}}
{{{ s = ( 135 + 177 ) / 8 }}}
{{{ s = 312/8 }}}
{{{ s = 39 }}}
and
{{{ s + 9 = 48 }}}
If I had used the negative square root of {{{ 31329 }}}, then
time would be negative, which is not allowed
------------------------
The speed for 1st part is 39 mi/hr
The speed for the 2nd part is 48 mi/hr
---------------------------------
check:
(1) {{{ 273 = s*t }}}
(1) {{{ 273 = 39*t }}}
(1) {{{ t = 7 }}} hrs
and
(2) {{{ 240 = ( s + 9 )*( 12 - t ) }}}
(2) {{{ 240 = ( 39 + 9 )*( 12 - t ) }}}
(2) {{{ 240 = 48*( 12 - t ) }}}
(2) {{{ 240 = 576 - 48t }}} 
(2) {{{ 48t = 336 }}}
(2) {{{ t = 7 }}} hrs
OK