Question 155059
Note: The information " leave Gray’s Lake at 3 P.M" is extra information that is not helpful. So we can ignore the "3 pm"


Let x = speed of local train 


Let's set up the equation for the local train:


{{{d=rt}}} Start with the distance-rate-time equation


{{{50=(x)t}}} Plug in {{{d=50}}} (the distance from the lake to Chicago) and {{{r=x}}}


{{{50/x=t}}} Divide both sides by "x" to solve for "t"


So the time "t" can be represented by {{{t=50/x}}}


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Now let's set up the equation for the express train:


{{{d=rt}}} Go back to the distance-rate-time equation


Since the "express travels twice as fast as the local", this means that the speed of the express train is {{{2x}}} mph. Also, because the express train "arrives 1 hour ahead of" the local train, this means that the time of the express train is {{{t-1}}} hours


{{{50=2x(t-1)}}} Plug in {{{d=50}}}, {{{r=2x}}} and replace "t" with {{{t-1}}}



{{{50=2x(50/x-1)}}} Plug in {{{t=50/x}}}



{{{50=2x(50/x)-2x(1)}}} Distribute



{{{50=100-2x}}} Multiply



{{{-50=-2x}}} Subtract 100 from both sides



{{{25=x}}} Divide both sides by -2 to isolate "x"



So the answer is {{{x=25}}}. 



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Answer:

So the speed of the local train is 25 mph and the speed of the express train is 50 mph (since it is twice the speed of the local train)