SOLUTION: Smith travels 45 miles going East from the center of the town and jones travels 70 miles going west from the same point. If Jones averages 5 miles per hour more than Smith and his

Algebra ->  Expressions-with-variables -> SOLUTION: Smith travels 45 miles going East from the center of the town and jones travels 70 miles going west from the same point. If Jones averages 5 miles per hour more than Smith and his       Log On


   



Question 302973: Smith travels 45 miles going East from the center of the town and jones travels 70 miles going west from the same point. If Jones averages 5 miles per hour more than Smith and his trip1/2 hour longer than Smith. How fast was each of them traveling.
Found 2 solutions by mananth, checkley77:
Answer by mananth(16946) About Me  (Show Source):
You can put this solution on YOUR website!
Smith travels 45 miles going East from the center of the town and jones travels 70 miles going west from the same point. If Jones averages 5 miles per hour more than Smith and his trip1/2 hour longer than Smith. How fast was each of them traveling.
let smith's rate be x mph
time taken by smith = 45/x hours
jones's rate = x+5 mph
time taken by jones = 70/ x+5
70 /x+5 - 45/x = 1/2
70x-45x-225=1/2 *x(x+5 )
2(25x-225)=x^2+5x
50x-450=x^2+5x
x^2-50x+5x+450=0
x^2-45x+450=0
x^2-30x-15x+450=0
x(x-30)-15(x-30)=0
(x-30)(x-15)=0
x= 30 OR 15 mph
Ananth









Answer by checkley77(12844) About Me  (Show Source):
You can put this solution on YOUR website!
D=RT
45=RT SMITH'S TRIP
T=45/R
70=(R+5)(T+.5) JONES' TRIP
70=(R+5)(45/R+.5)
70=(R+5)(45+.5R)/R
70=(45R+2.5R+225+.5R^2)/R
70R=.5R^2+47.5R+225
.5R^2+47.5R-70R+225=0
.5R^2-22.5R+225=0
(.5R-15)(X-15)=0
.5R-15=0
.5R=15
R=15/.5
R=30 MPH. FOR SMITH'S SPEED.
45=30*T
T=45/30=1.5 HOURS.
JONES' SPEED=30+5=35 MPH.
70=(35)(1.5+.5)
70=35*2
70=70