Question 562875
Let {{{ t }}} = time for the trip in hours
at speed {{{ s }}} in mi/hr
given:
(1) {{{ 1500 =s*t }}}
and
(2) {{{ 1500 = ( s + 10 )*( t - 5 ) }}}
(2) {{{ 1500 = s*t + 10t - 5s - 50 }}}
Substitute (1) into (2)
(2) {{{ 1500 = 1500 + 10t - 5s - 50 }}}
(2) {{{ 10t - 5s - 50 = 0 }}}
and
(1) {{{ t = 1500/s }}}
(2) {{{ 10*(1500/s) - 5s - 50 = 0 }}}
(2) {{{ 15000 - 5s^2 - 50s = 0 }}}
(2) {{{ 5s^2 + 50s - 15000 = 0 }}}
(2) {{{ s^2 + 10s - 3000 = 0 }}}
Use quadratic equation
{{{ s = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} 
{{{ a = 1 }}}
{{{ b = 10 }}}
{{{ c = -3000  }}}
{{{ s = (-10 +- sqrt( 10^2-4*1*(-3000) )) / (2*1) }}} 
{{{ s = (-10 +- sqrt( 100 + 12000 )) / 2 }}} 
{{{ s = (-10 +- sqrt( 12100 )) / 2 }}} 
{{{ s = (-10 + 110) / 2 }}} 
{{{ s = 100/2 }}}
{{{ s = 50 }}} ( I can ignore the (-) root )
The speed is 50 mi/hr
check answer:
(1) {{{ 1500 =s*t }}}
(1) {{{ 1500 = 50t }}}
(1) {{{ t = 30 }}}
and
(2) {{{ 1500 = ( 50 + 10 )*( 30 - 5 ) }}}
(2) {{{ 1500 = 60*25 }}}
(2) {{{ 1500 = 1500 }}}
OK