Question 203472
 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 = D's speed
then
(s+5) = J's speed
:
Write a time equation: Time = {{{dist/speed}}}
:
Smith's time = Jones' time - .5 hrs
{{{45/s}}} = {{{70/((s+5))}}} - .5
Multiply equation by s(s+5)
s(s+5)*{{{45/s}}} = s(s+5)*{{{70/((s+5))}}} - .5s(s+5)
Cancel out the denominators and you have:
45(s+5) = 70s  - .5s^2 - 2.5s
:
45s + 225 = 70s - .5s^2 - 2.5s
arrange as a quadratic equation on the left
.5s^2 + 2.5s + 45s - 70s + 225 = 0
:
.5s^2 - 22.5s + 225 = 0
Multiply equation by 2, get rid of the decimals
s^2 - 45s + 450 = 0
Factors to:
(s - 15)(s - 30) = 0
Two good solutions
s = 15 mph is Smith's speed, then 20 mph is Jones' speed
and
s = 30 mph is Smith's speed, then 35 mph is Jones' speed
:
:
Check solutions, find the time for each trip
Smith: 45/15 = 3 hr
Jones: 70/20 = 3.5 hr
:
You can check the s=30 solution