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Question 798717: I've tried this so many times and am stuck. The problem must be solved using 2 variables and by finding 2 equations (then using addition or substitution method to solve)
A. Jason and Steve are 45 miles apart on the canal riding toward each other. Steve is twice as fast as Jason. They meet up after 2 hours. How fast is Steve?
B. after they meet up, Jason heads home. 15 minutes later, Steve leaves to catch up. How long was Jason biking when Steve caught up?
I drew a table and figured one equation was x+y=45 but I cannot find the other equation. I also tried using D=RT. Please help!!!
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! A. Jason and Steve are 45 miles apart on the canal riding toward each other. Steve is twice as fast as Jason. They meet up after 2 hours. How fast is Steve?
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Their speeds are added when going in opposite directions.
45 mi/2 hr = 22.5 mi/hr
J = Jason's speed
S = Steve's speed
J + S = 22.5
S = 2J
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J + 2J = 22.5
J = 7.5 mi/hr
S = 15 mi/hr
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B. after they meet up, Jason heads home. 15 minutes later, Steve leaves to catch up. How long was Jason biking when Steve caught up?
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Assuming they go at the same speeds:
In 15 minutes Jason goes 7.5*0.25 miles = 15/8 miles.
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Steve "gains on" him at 7.5 mi/hr (15 - 7.5)
d = r*t
t = d/r = (15/8)/(15/2) = 1/4 hour to catch up.
Jason was biking 15 + 15 = 30 minutes
I drew a table and figured one equation was x+y=45 but I cannot find the other equation. I also tried using D=RT.
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