SOLUTION: Please help me solve this prroblem: A passenger train travels 392mi in the same time that it takes a freight train to travel 322mi. if the passenger train travels 20mi/h faster th

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Question 133198This question is from textbook prentice hall mathematics algebra 2
: Please help me solve this prroblem:
A passenger train travels 392mi in the same time that it takes a freight train to travel 322mi. if the passenger train travels 20mi/h faster than the freight train, find the speed of each train
This question is from textbook prentice hall mathematics algebra 2

Found 2 solutions by edjones, ptaylor:
Answer by edjones(8007) About Me  (Show Source):
You can put this solution on YOUR website!
ts=d, t=time, s=speed, d=distance
Let x be the speed of the freight train.
t=d/s In this problem the times are equal
So, d/s=d/s
322/x=392/(x+20)
322(x+20)=392x Multiply each side by x(x+20) to eliminate fractions.
322x+6440=392x
322x-322x+6440=392x-322x
6440=70x
70x/70=6440/70
x=92 mph
x+20=112 mph
.
Check:
322/92=3.5 hrs
392/112=3.5 hrs

Answer by ptaylor(2198) About Me  (Show Source):
You can put this solution on YOUR website!
Distance(d) equals Rate(r) times Time(t) or d=rt; r=d/t and t=d/r
Let r=rate of freight train
Then r+20=rate of passenger train
Time it takes freight train=322/r
Time it takes passenger train=392/(r+20)
Now we are told that the above two times are equal, so:
322/r=392/(r+20) multiply each side by r(r+20) (or cross multiply)
322(r+20)=392r get rid of parens
322r+6440=392r subtract 322r from each side
322r-322r+6440=392r-322r collect like terms
6440=70r divide both sides by 70
r=92 mph------------------------------speed of freight train
r+20=92+20=112 mph------------------------speed of passenger train
CK
322/92=392/112
3.5=3.5

Hope this helps---ptaylor