Question 786871
keep it separated mentally into:
(1) Equation for passenger train
(2) Equation for freight train
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Let {{{ t }}} = the time in hours for both trains
Let {{{ s }}} = the speed of the freight train
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Equation for passenger train:
(1)  {{{ 340 = ( s + 10 )*t }}}
Equation for freight train:
(2) {{{ 290 = s*t }}}
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(1) {{{ 340 = s*t + 10t }}}
Substitute (2) into (1)
(1) {{{ 340 = 290 + 10t }}}
(1) {{{ 10t = 50 }}}
(1) {{{ t = 5 }}} hrs
and, since
(2) {{{ 290 = s*t }}}
(2) {{{ 290 = s*5 }}}
(2) {{{ s = 290/5 }}}
(2) {{{ s = 58 }}}
{{{ s + 10 = 68 }}}
The speed of the freight train is 58 mi/hr
The speed of the passenger train is 68 mi/hr
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check:
(1)  {{{ 340 = ( 58 + 10 )*5 }}}
(1) {{{ 340 = 68*5 }}}
(1) {{{ 340 = 340 }}}
and
(2) {{{ 290 = 58*5 }}}
(2) {{{ 290 = 290 }}}
OK