College Algebra word problem help.

Question

two cars started from the same place at the same time and traveled in opposite directions. at the end of 6 hours, they were 618 miles apart. if the average rate of the slow car exceeded 1/2 of the average rate of the fast car by 16 mph, find the rate of each car, step by step solution

Answer 1

To solve this you are going to need to set up a system of equations and solve. In my system, I will use the variables F and S to denote Fast and Slow cars, respectively. The first equation will tell us about how far apart they are. This is keeping in mind that rate x time = distance:

 

1) 6F + 6S = 618                  6 hours of rate F and 6 hours of rate S added to a total of 618 miles.

 

The second equation relates the two rates to one another:

 

2) S = (1/2)F + 16                The slower car traveled at 16 mph faster than half of the rate of the fast car.

 

Now we plug (2) into (1):

 

6F + 6(0.5F + 16) = 618

6F + 3F + 96 = 618

9F = 522

F = 58 mph

 

Plug into (2) to get S:

 

S = 0.5(58) + 16

S = 29 + 16

S = 45 mph

 

Finally check with equation (1):

 

6(58) + 6(45) = 618

348 + 270 = 618

618 = 618                  YAY!

 

So the fast car travels at 58 mph and the slow car travels at 45 mph. I hope you understand how this was set up and the steps to solving.