Question 121027
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?
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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/s}}} = {{{70/((s+5))}}} - {{{1/2}}}
:
Multiply equation by 2s(s+5)
2s(s+5)*{{{45/s}}} = 2s(s+5)*{{{70/((s+5))}}} - 2s(s+5)*{{{1/2}}}
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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
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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 
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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.