Question 230819
Let {{{s}}} = the speed of the local in mi/hr
Then {{{2s}}} = speed of the express  
Let {{{t}}} = time in hrs it takes for local to arrive in Chicago
Then {{{t - 1}}} time in hrs for express to arrive in Chicago
They both travel the same distance, {{{d = 50}}} mi
Now I can write an equation for each train
For local:
(1) {{{50 = st}}}
For express:
(2) {{{50 = 2s*(t - 1)}}}
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From (2)
{{{50 = 2st - 2s}}}
Substitute from (1)
{{{50 = 2*50 - 2s}}}
{{{2s = 100 - 50}}}
{{{2s = 50}}}
{{{s = 25}}}
and
{{{2s = 50}}}
The speed of the local is 25 mi/hr and 
the speed of the express is 50 mi/hr
check answer:
(1) {{{50 = st}}}
{{{50 = 25t}}}
{{{t = 2}}}
(2) {{{50 = 2s*(t - 1)}}}
{{{50 = 2*25*(2-1)}}}
{{{50 = 50*1}}}
{{{50 = 50}}}
OK