Question 5092
I think your mistake is not putting parentheses around the time (t-1).


As you said, 
let s = speed of the local
2s = speed of the express


t= time of the local
t-1 = time of the express


Distances are equal at 50 miles, so (rate times time) = (rate times time)

s*(t) = 2s *(t-1)


Since s the speed of the train does not equal zero, you can divide both sides by s:
{{{ (s*t)/s = (2s * (t-1))/s }}}

t= 2(t-1)
t=2t -2
-t = -2
t=2 hours for the local
t-1 = 1 hour for the express


After you do that, then remember that D=RT, so {{{R= D/T}}}

Rate for the local = {{{50/2}}}=25 mph
Rate for the express = {{{50/1}}}= 50 mph.


R^2 at SCC