Question 148759
Smith bicycled 45 miles going east from Durango
Smith DATA:
distance = 45 miles ; rate = x mph : time = d/r = 45/x hrs.
 and Jones bicycled 70 miles.
Jones DATA:
distance = 70 miles ; rate = (x+5) mph ; time = d/r = 70/(x+5) hrs. 
Jones averaged 5 miles per hour more than Smith and his trip took ½ hour longer than Smith’s. 
How fast was each one traveling? 
EQUATION:
Jones time - Smith time = 1/2
45/x - 70/(x+5) = 1/2
90(x+5) - 140x = x^2+5x
-50x + 450 = x^2 + 5x
x^2 + 60x - 450 = 0
x = [-60 +-sqrt(60^2-4*-450)]/2
x = [-60 +- sqrt(5400)]/2
Positive Solution
x = (-60+ 73.48)/2
x = 13.48/2 = 6.74 mph (Smith's rate)
x+5 = 11.74 mph (Jones's rate)
==========================
Cheers,
Stan H.
------------------------