Question 1019044
From 1 PM to 5 PM is 4 hrs
Let {{{ s }}} = 1st train's speed in mi/hr
{{{ s + 36 }}} = 2nd train's speed in mi/hr
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Find the 1st train's head start in miles
{{{ d[1] = s*4 }}}
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From 5 PM to 10 PM is 5 hrs
The start time for these equations is 5 PM
Let {{{ d }}} = distance in miles the 2nd train travels
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Equation for 1st train:
(1) {{{ d - 4s = s*5 }}}
Equation for 2nd train:
(2) {{{ d = ( s + 36 )*5 }}}
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Substitute (2) into (1)
(1) {{{( s + 36 )*5 - 4s = s*5 }}}
(1) {{{ 5s + 180 - 4s = 5s }}}
(1) {{{ 4s = 180 }}}
(1) {{{ s = 45 }}}
and
{{{ s + 36 = 81 }}}
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The 1st train's speed is 45 mi/hr
The 2nd train's speed is 81 mi/hr
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check:
(1) {{{ d - 4s = s*5 }}}
(1) {{{ d - 4s = s*5 }}}
(1) {{{ d = 9s }}}
(1) {{{ d = 9*45 }}}
(1) {{{ d = 405 }}} mi
and
(2) {{{ d = ( s + 36 )*5 }}}
(2) {{{ d = ( 45 + 36 )*5 }}}
(2) {{{ d = 81*5 }}}
(2) {{{ d = 405 }}} mi
OK