SOLUTION: Two cars start together and travel in the same direction, one going twice as fast as the other. At the end of 3 hours, they are 96 miles apart. How fast is each car traveling?

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Question 30904: Two cars start together and travel in the same direction, one going twice as fast as the other. At the end of 3 hours, they are 96 miles apart. How fast is each car traveling?
Found 2 solutions by aussie7us, mbarugel:
Answer by aussie7us(3) About Me  (Show Source):
You can put this solution on YOUR website!
This can be solved algebrically. You have two equations.
1- 2X=Y where x is the speed of the first car, and y is the speed of the second (faster) car.
2- 3x%2B3y=96 96 is their distance away from each other, and the 3's stand for the hours. car X has traveled for three hours, and so has car Y.
Next work out the equation. You can use substitution.
+3x%2B3%282Y%29=96 distribute
then
3x%2B6x=96 add like terms
then
9x=96 divide both sides by 9 9x%2F9=96%2F9
x=10%2B2%2F3 so car x (slow car) is traveling at 10 and 2/3 miles an hour.
Now, plug the X value into equation 1 2X=Y
The fast car is travelling at 21 and 1/3 miles an hour.
Y=10%2B2%2F3
You can also check your answer. Plug in both numbers into equation 2. You get
3%2810%2B2%2F3%29%2A+%2821%2B1%2F3%29=96%29 if this equation is true, then your answers are correct.
(it is true)
HOPE THIS HELPS!

Answer by mbarugel(146) About Me  (Show Source):
You can put this solution on YOUR website!
Hello!
The previous solution you got for this question ("The fast car is travelling at 21 and 1/3 miles an hour", etc) is actually incorrect.
Let's call X to the the speed of the faster car, and Y to the speed of the other one, both measured in miles per hour. After 3 hours, the fast car has traveled a distance of 3X miles, and the slow car has traveled a distance of 3Y miles.
Now, since they are 96 miles apart, the equation that describes this is:
3X+-+3Y+=+96+
The other equation ("one going twice as fast as the other") is:
X+=+2Y
[recall that X is the fast car]
So we have the system:
system%283X+-+3Y+=+96+%2CX+=+2Y%29
Substituting the 2nd equation into the 1st one, we get:
3%282Y%29+-+3Y+=+96
6Y+-+3Y+=+96
3Y+=+96
Y+=+32
So the slow car is traveling at 32 mph, and the fast car is traveling at 64 mph.

I hope this helps!
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