SOLUTION: Greg drove at a constant speed in a rainstorm for 258 miles. He took a​ break, and the rain stopped He then drove 208 miles at a speed that was 9 miles per hour faster than

Algebra ->  Radicals -> SOLUTION: Greg drove at a constant speed in a rainstorm for 258 miles. He took a​ break, and the rain stopped He then drove 208 miles at a speed that was 9 miles per hour faster than      Log On


   

Question 1102120: Greg drove at a constant speed in a rainstorm for 258 miles. He took a​ break, and the rain stopped He then drove 208 miles at a speed that was 9 miles per hour faster than his previous speed. If he drove for 10 ​hours, find the​ car's speed for each part of the trip. THANK YOU!!
Found 3 solutions by htmentor, greenestamps, josmiceli:
Answer by htmentor(1343) About Me  (Show Source):
You can put this solution on YOUR website!
Let s = the speed and t = the time for the first part of the trip
Then s + 9 is the speed for the second part of the trip
The total time is 10 h, so the time for the second leg is 10 - t
For the first leg of the trip s = 258/t -> t = 258/s [1]
For the second leg, s + 9 = 208/(10 - t) [2]
We have two equations and two unknowns
Substitute [1] into [2]:
s + 9 = 208/(10 - 258/s) = 208s/(10s - 258)
(s + 9)(10s - 258) = 208s
10s^2 - 258s + 90s - 2322 - 208s = 0
10s^2 - 376s - 2322 = 0
5s^2 - 188s - 1161 = 0
This gives s = -5.4 and 43. Only the positive solution is valid.
Thus the other speed is 43 + 9 = 52
Ans: 43 mph and 52 mph

Answer by greenestamps(13203) About Me  (Show Source):
You can put this solution on YOUR website!


The fastest way to solve this problem is by trial and error, hoping that the speeds are whole numbers.

258+=+2%2A129+=+2%2A3%2A43
So a good guess for the first part of the trip is 6 hours at 43 mph.

That would leave 4 hours for the rest of the trip, which was 208 miles; 208 miles in 4 hours means 52 mph; and indeed 52mph is 9 mph faster than 43 mph.

Algebraically...

Let x and x+9 be the two speeds. Then 258 miles at speed x plus 208 miles at speed x+9 makes a total of 10 hours:

258%2Fx+%2B+208%2F%28x%2B9%29+=+10
258%28x%2B9%29+%2B+208x+=+10x%28x%2B9%29
258x+%2B+2322+%2B+208x+=+10x%5E2%2B90x
10x%5E2+-+370x+-+2580+=+0
x%5E2+-+37x+-+258+=+0
%28x-43%29%28x%2B6%29+=+0

The lower speed is 43 mph; the higher speed is 52 mph.

Note that in the algebraic solution, we had to factor the quadratic x^2-37x-258; to do that, we had to find two numbers whose product was 258.

But that's exactly what we did in the first place. So the algebraic solution didn't make the work any easier; it only made us do more work (a LOT more!) to get to the answer.

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let +t+ = time in hrs for 258 mi trip
+10+-+t+ = time in hrs for 208 mi trip
Let +s+ = speed in mi/hr for 258 mi trip
+s+%2B+9+ = speed in mi/hr for 208 mi trip
---------------------------------------
(1) +258+=+s%2At+
(2) +208+=+%28+s+%2B+9+%29%2A%28+10+-+t+%29+
----------------------------------
(2) +208+=+10s+%2B+90+-+s%2At+-+9t+
(2) +208+=+10s+%2B+90+-+t%28+s+%2B+9+%29+
and
(1) +t+=+258%2Fs+
Plug this into (2)
(2) +208+=+10s+%2B+90+-+%28+258%2Fs+%29%2A%28+s+%2B+9+%29+
(2) +118+=+10s+-+258+-+2322%2Fs+
(2) +376+=++10s+-+2322%2Fs+
(2) +376s+=+10s%5E2+-+2322+
(2) +10s%5E2+-+376s+-+2322+=+0+
(2) +5s%5E2+-+188s+-+1161+=+0+
Use quadratic formula
+s+=+%28+-b+%2B-sqrt%28+b%5E2+-+4%2Aa%2Ac+%29%29+%2F+%28+2a+%29+
+a+=+5+
+b+=+-188+
+c+=+-1161+
+s+=+%28+188+%2B-sqrt%28+35344+-+4%2A5%2A%28+-1161+%29%29%29+%2F+10+
+s+=+%28+188+%2B-+sqrt%28+3544+%2B+23220+%29%29+%2F+10+
+s+=+%28+188+%2B+sqrt%28+26764+%29+%29%2F10+
+s+=+%28+188+%2B+163.6+%29+%2F+10+
+s+=+351.6%2F10+
+s+=+35.16+
and
+s+%2B+9+=+44.16+
--------------------
He drove 35.16 mi/hr for 1st part and
44.16 mi/hr for 2nd part
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Get another opinion unless you know
what answer is. Check math too.