Question 1152267
Let {{{ s }}} = the speed of the passenger train in mi/hr
{{{ s - 20 }}} = the speed of the freight train
{{{ ( s - 20 )*2 }}} = the distance of the freight train's head start 
Let {{{ d }}} = distance the passenger train travels in 3 hrs
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Equation for the passenger train:
(1) {{{ d = s*3 }}}
Equation for the freight train:
(2) {{{ d - ( s - 20 )*2 = ( s - 20 )*3 }}}
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Plug (1) into (2)
(2) {{{ 3s - ( s - 20 )*2 = ( s - 20 )*3 }}}
(2) {{{ 3s - 2s + 40 = 3s - 60 }}}
(2) {{{ 2s = 100 }}}
(2) {{{ s = 50 }}}
and 
{{{ s - 20 = 30 }}}
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The rate of the passenger train is 50 mi/hr
The rate of the freight train is 30 mi/hr
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check:
(1) {{{ d = s*3 }}}
(1) {{{ d = 50*3 }}}
(1) {{{ d = 150 }}} mi
and
(2) {{{ d - ( s - 20 )*2 = ( s - 20 )*3 }}}
(2) {{{ d - 30*2 = 30*3 }}}
(2) {{{ d = 60 = 90 }}}
(2) {{{ d = 150 }}}
and
The head start for freight train is:
{{{ d[1] = 30*2 }}}
{{{ d[1] = 60 }}}
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The time for the remaining 90 miles is:
{{{ t = 90/30 }}}
{{{ t = 3 }}} hrs
OK 
get a 2nd opinion if you need to