SOLUTION: Smith bicycled 45 miles going east from Durango, and Jones bicycled 70 miles. Jones averaged 5mph more than Smith, and his trip took one-half hour longer that Smith's. how fast was

Algebra ->  Algebra  -> Customizable Word Problem Solvers  -> Travel -> SOLUTION: Smith bicycled 45 miles going east from Durango, and Jones bicycled 70 miles. Jones averaged 5mph more than Smith, and his trip took one-half hour longer that Smith's. how fast was      Log On

Ad: Over 600 Algebra Word Problems at edhelper.com
Ad: Algebra Solved!™: algebra software solves algebra homework problems with step-by-step help!
Ad: Algebrator™ solves your algebra problems and provides step-by-step explanations!

   


Question 191433: Smith bicycled 45 miles going east from Durango, and Jones bicycled 70 miles. Jones averaged 5mph more than Smith, and his trip took one-half hour longer that Smith's. how fast was each one traveling?
Answer by stanbon(48551) 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 5mph more than Smith, and his trip took one-half hour longer that Smith's. how fast was each one traveling?
-----------------------------------
Smith DATA:
distance = 45 mi ; time = x hrs ; rate = d/t = 45/x mph
------------------------------------------------------------
Jones DATA:
distance = 70 mi ; time (x+0.5) hrs ; rate = d/t = 70/(x+(1/2)) = mph
-------------------------------------------------------------
Equation:
Jones' rate - Smith's rate = 5
(45/x) - (70/((2x+1)/2) = 5
(9/x) - 28/(2x+1) = 1
9(2x+1) -28x = x(2x+1)
2x^2 + 30x = 18x+9
2x^2 + 12x - 9 = 0
x = [-12 +- sqrt(144 -4*2*-9)]/4
x = [-12 + 14.7]
x = 2.7 hrs
-----------------
Smith rate = 45/x = 45/2.7 = 16 2/3 mph
Jones rate = 16 2/3 + 5 = 21 2/3 mph
-----------------
Cheers,
Stan H.