SOLUTION: Mrs. Watt traveled between two cities that were 750 kilometers apart. On the return trip, she increased her travel rate of travel by 20 kilometers per hour and made the trip in 10

Algebra ->  Customizable Word Problem Solvers  -> Travel -> SOLUTION: Mrs. Watt traveled between two cities that were 750 kilometers apart. On the return trip, she increased her travel rate of travel by 20 kilometers per hour and made the trip in 10       Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


    

Question 1199184: Mrs. Watt traveled between two cities that were 750 kilometers apart. On the return trip, she increased her travel rate of travel by 20 kilometers per hour and made the trip in 10 hours less time. Find her rate of travel in each direction.
Answer by ikleyn(52776) About Me  (Show Source):
You can put this solution on YOUR website!
.
Mrs. Watt traveled between two cities that were 750 kilometers apart.
On the return trip, she increased her travel rate of travel by
20 kilometers per hour and made the trip in 10 hours less time.
Find her rate of travel in each direction.
~~~~~~~~~~~~~~~ 

x    = slower speed, in km/h.

x+20 = faster speed.


Time equation

   750%2Fx - 750%2F%28x%2B20%29 = 10  hours.    (1)


Cancel 10 in both sides

   75%2Fx - 75%2F%28x%2B20%29 = 1.


Multiply both sides by x*(x+20)

    75(x+20) - 75x = x*(x+20)

    75x + 1500 - 75x = x^2 + 20x

    x^2 + 20x - 1500 = 0

    (x+50)*(x-30) = 0


Choose the positive root  x = 30 and disregard the negative root x = -50.


CHECK  equation (1) :  750%2F30+-+750%2F50 = 25 - 15 = 10 hours.   ! correct !


ANSWER.  The slower rate is 30 km/h;  the faster rate is 50 km/h.

Solved.