Question 1017189
Let {{{ s }}} = the speed in mi/hr of the 1st train
{{{ s + 56 }}} = the speed in mi/hr of the 2nd train
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What is the head start of the 1st train in miles?
{{{ d[1] = s*4 }}} ( 3PM to 7 PM is 4 hrs )
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Let {{{ d }}} = distance in miles the 2nd train 
travels until it catches up with 1st train
{{{ t }}} = time in hrs from 7 PM to 10 PM, so
{{{ t = 3 }}} hrs
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Equation for 1st train:
(1) {{{ d - 4s = s*3 }}}
Equation for 2nd train:
(2) {{{ d = ( s + 56 )*3 }}}
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(1) {{{ d = 3s + 4s }}}
(1) {{{ d = 7s }}}
and
(2) {{{ d = 3s + 168 }}}
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{{{ 7s = 3s + 168 }}}
{{{ 4s = 168 }}}
{{{ s = 42 }}}
and
{{{ s + 56 = 98 }}}
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The 1st train's speed is 42 mi/hr
The 2nd train's speed is 98 mi/hr
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check:
(1) {{{ d - 4s = s*3 }}}
(1) {{{ d - 4*42 = 42*3 }}}
(1) {{{ d - 168 = 126 }}}
(1) {{{ d = 294 }}} mi
and
(2) {{{ d = ( s + 56 )*3 }}}
(2) {{{ d = ( 42 + 56 )*3 }}}
(2) {{{ d = 294 }}} mi
OK