Question 29463
SEE BELOW MY COMMENTS......
I am not sure how to set this up. An express and a local train leave Seattle at 3pm and head for Arlington 50 miles away. The express travels twice as fast as the local and arrives 1 hour ahead of the local. Find the speed of the train.
I have tried this several ways.... trying to use elimination, or substitution. I just can't get it. Help!
50 = x + y......FIRST ASSUME LOCAL TRIN SPEED =X.MPH..HENCE EXPRESS SPEED IS =2X..MPH

DISTANCE TO BE COVERED =50 MILES....
TIME TAKEN BY LOCAL =DISTANCE/SPEED = 50/X................................I
TIME TAKEN BY EXPRESS = 50/2X =25/X....................................II
DIFFERENCE = (50/X) - (25/X) = (50-25)/X = 25/X....BUT THIS IS 1 HOUR..SO
25/X=1...OR ......X=25 MPH
SO SPEED OF LOCAL = 25 MPH
SPEED OF EXPRESS = 2*25 = 50 MPH...  
50 = 2y + x Is this even close? Thanks for your help!