Question 980719
Let {{{ s }}} = the speed of the slower train
{{{ s + 20 }}} = the speed of the faster train
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If the trains started out being 500 mi apart,
and ended up 300 mi apart, then the sum of the
distances they traveled must be 200 mi
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let {{{ d }}} = the distance in miles the 
slower train has traveled until they are 300 mi apart
{{{ 200 - d }}} = the distance in miles the slower train 
has traveled until they are 300 mi apart
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Equation for slower train:
(1) {{{ d = s*2 }}}
Equation for faster train:
(2) {{{ 200 - d = ( s+20 )*2 }}}
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Substitute (1) into (2)
(2) {{{ 200 - 2s = ( s+20 )*2 }}}
(2) {{{ 200 - 2s = 2s + 40 }}}
(2) {{{ 4s = 160 }}}
(2) {{{ s = 40 }}}
and
{{{ s+20 = 60 }}}
The faster train's speed is 60 mi/hr
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check:
(1) {{{ d = s*2 }}}
(1) {{{ d = 40*2 }}}
(1) {{{ d = 80 }}}
and
(2) {{{ 200 - d = ( s+20 )*2 }}}
(2) {{{ 200 - d = ( 40+20 )*2 }}}
(2) {{{ 200 - d = 120 }}}
Note that {{{ 80 + 120 = 200 }}} which is the sum
of the distances they traveled
OK