SOLUTION: Smith bicycled 45 miles going east from Durango, and Jones bicycled 70 miles. Jones averaged 5 miles per hour more than Smit, and his trip took one-half hour longer than Smith's. H

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: Smith bicycled 45 miles going east from Durango, and Jones bicycled 70 miles. Jones averaged 5 miles per hour more than Smit, and his trip took one-half hour longer than Smith's. H      Log On


   

Question 121027: Smith bicycled 45 miles going east from Durango, and Jones bicycled 70 miles. Jones averaged 5 miles per hour more than Smit, and his trip took one-half hour longer than Smith's. How fast was each one treveling?
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
Smith bicycled 45 miles going east from Durango, and Jones bicycled 70 miles. Jones averaged 5 miles per hour more than Smith, and his trip took one-half hour longer than Smith's. How fast was each one traveling?
:
Let s = Smith's speed
then
(s+5) = Jone's speed
:
Write a time equation: Time = Distance/speed
:
Smiths time = Jones time - a half hour
45%2Fs = 70%2F%28%28s%2B5%29%29 - 1%2F2
:
Multiply equation by 2s(s+5)
2s(s+5)*45%2Fs = 2s(s+5)*70%2F%28%28s%2B5%29%29 - 2s(s+5)*1%2F2
:
Cancel out the denominators and you have:
:
2(s+5)*45 = 2s(70) - s(s+5)
:
90s + 450 = 140s - s^2 - 5s
:
+s^2 + 90s - 140s + 5s + 450 = 0
:
s^2 - 45s + 450 = 0
:
This factors to
(s-15)(s-30) = 0
:
It's interesting that we have two valid solutions
s = 15 mph or s = 30 mph is Smith speed
:
Both values will satisfy the problem, although in real life it is not likely that Jone's can maintain 35 mph for 70 mile. 20 mph is more likely for him.