Question 749464
Let {{{ s }}} = the 1st speed in mi/hr
{{{ s + 10 }}} = the increased speed in mi/hr
Let {{{ t }}} = the train's time at {{{ s }}} mi/hr
{{{ t - 2 }}} = the train's time at {{{ s + 10 }}} mi/hr
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given:
(1) {{{ 600 = s*t }}}
(2) {{{ 600 = ( s + 10 )*( t - 2 ) }}}
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(2) {{{ 600 = s*t + 10t - 2s - 20 }}}
(2) {{{ 600 = s*( t - 2 ) + 10t - 20 }}}
(2) {{{ s*( t - 2 ) = 620 - 10t }}}
and,
(1) {{{ 600 = s*t }}}
(1) {{{ s = 600/t }}}
(2) {{{ ( 600/t )*( t - 2 ) = 620 - 10t }}}
(2) {{{ 600 - 1200/t = 620 - 10t }}}
(2) {{{ 10t - 1200/t - 20 = 0 }}}
(2) {{{ 10t^2 - 1200 - 20t = 0 }}}
(2) {{{ t^2 - 2t - 120 = 0 }}}
Use quadratic formula
{{{ t = (-b +- sqrt( b^2 - 4*a*c )) / (2*a) }}} 
{{{ a = 1 }}}
{{{ b = -2 }}}
{{{ c = -120 }}}
{{{ t = (-(-2) +- sqrt( (-2)^2 - 4*1*(-120) )) / (2*1) }}} 
{{{ t = ( 2 +- sqrt( 4 + 480 )) / 2 }}} 
{{{ t = ( 2 +- sqrt( 484 )) / 2 }}} 
{{{ t = ( 2 + 22) / 2 }}} ( can't use negative square root )
{{{ t = 24/2 }}}
{{{ t = 12 }}}
and
(1) {{{ s = 600/t }}}
(1) {{{ s = 600/12 }}}
(1) {{{ s = 50 }}}
The speed of the train is 50 mi/hr
check:
(1) {{{ 600 = s*t }}}
(1) {{{ 600 = 50*12 }}} 
{{{ 600 = 600 }}}
and
(2) {{{ 600 = ( s + 10 )*( t - 2 ) }}}
(2) {{{ 600 = ( 50 + 10 )*( 12 - 2 ) }}}
(2) {{{ 600 = 60*10 }}}
(2) {{{ 600 = 600 }}}
OK