Two cars leave at the same time from towns that are 460 miles apart and travel toward each other. They meet after 4 hours. Find the rate of each car if one car travels 13 miles per hour faster than the other.

x + x + 13 =460/4

2x + 13 = 115

2x + 13 -13 = 115. -13

2x = 102

2x/2 = 102/2

x = 51

x + 13
= 64

define your variables:

let the rate of the the slower car be x mph
then the rate of the faster car is x+13 mph

They both drive for 4 hours

distance covered by slower car = 4x
distance covered by faster car = 4(x+13)

4x + 4(x+13) = 460
8x 408
x = 51

slower car went 51 mph
faster car went 64 mph

To solve this problem, we can apply the concept of relative speed. Let's denote the rate of the slower car as 'r' miles per hour. Since the other car travels 13 miles per hour faster, its rate will be 'r + 13' miles per hour.

Now, the time it takes for both cars to meet is given as 4 hours. During this time, the combined distance covered by both cars will be equal to the total distance between the towns, which is 460 miles.

Using the formula distance = rate × time, we can set up the following equation:

4r + 4(r + 13) = 460

Let's solve for 'r' by simplifying and solving the equation:

4r + 4r + 52 = 460
8r = 408
r = 408/8
r = 51

So, the rate of the slower car is 51 miles per hour.

To find the rate of the faster car, we can substitute the value of 'r' back into the equation:

Rate of faster car = r + 13 = 51 + 13 = 64 miles per hour.

Therefore, the rate of the slower car is 51 miles per hour, while the rate of the faster car is 64 miles per hour.