Question 953717
Let {{{ s }}} = Dave's speed in km/hr in train
{{{ s + 9 }}} = Dave's speed in km/hr  in car
Let {{{ t }}} = Dave's time in hrs in train
{{{ 14 - t }}} = Dave's time in hrs in car
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Equation for train:
(1) {{{ 216 = s*t }}}
Equation for car:
(2) {{{ 216 = ( s + 9 )*( 14 - t ) }}}
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(2) {{{ 216 = 14s + 126 - s*t - 9t }}}
(2) {{{ 90 = s*( 14 - t ) - 9t }}}
and
(1) {{{ s = 216/t }}}
Substitute (1) into (2)
(2) {{{ 90 = ( 216/t )*( 14 - t )  - 9t }}}
(2) {{{ 90 = 3024/t - 216 - 9t }}}
(2) {{{ 9t + 306 - 3024/t = 0 }}}
Multiply both sides by {{{ t }}}
(2) {{{ 9t^2 + 306t - 3024 = 0 }}}
(2) {{{ t^2 + 34t - 336 = 0 }}}
Use the quadratic formula
{{{ t = ( -b +- sqrt( b^2 - 4*a*c )) / (2*a) }}}
{{{ a = 1 }}}
{{{ b = 34 }}}
{{{ c = -336 }}}
{{{ t = ( -34 +- sqrt( 34^2 - 4*1*(-336) )) / (2*1) }}}
{{{ t = ( -34 +- sqrt( 1156 + 1344 )) / 2 }}}
{{{ t = ( -34 +- sqrt( 2500 )) / 2 }}}
{{{ t = ( -34 + 50) / 2 }}}
{{{ t = 16/2 }}}
{{{ t = 8 }}}
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(1) {{{ s = 216/t }}}
(1) {{{ s = 216/8 }}}
(1) {{{ s = 27 }}}
and
{{{ s + 9 = 36 }}}
{{{ 27 + 9 = 36 }}}
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The speed of the train is 27 km/hr
The speed of the car is 36 km/hr
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check:
(2) {{{ 216 = ( 27 + 9 )*( 14 - 8 ) }}}
(2) {{{ 216 = 36*6 }}}
(2) {{{ 216 = 216 }}}
OK