Let the speed of the slower bus be \( x \) mph. Therefore, the speed of the faster bus is \( x + 10 \) mph.
Both buses travel for 12 minutes, which is \( \frac{12}{60} = \frac{1}{5} \) hours.
In that time, the slower bus covers a distance of: \[ \text{Distance of slower bus} = x \times \frac{1}{5} = \frac{x}{5} \text{ miles} \]
The faster bus covers a distance of: \[ \text{Distance of faster bus} = (x + 10) \times \frac{1}{5} = \frac{x + 10}{5} \text{ miles} \]
After 12 minutes, the total distance between the two buses is the sum of the distances they have traveled: \[ \frac{x}{5} + \frac{x + 10}{5} = \frac{x + x + 10}{5} = \frac{2x + 10}{5} \text{ miles} \]
According to the problem, this distance equals 12 miles: \[ \frac{2x + 10}{5} = 12 \]
To eliminate the fraction, multiply both sides by 5: \[ 2x + 10 = 60 \]
Next, subtract 10 from both sides: \[ 2x = 50 \]
Then, divide both sides by 2 to find \( x \): \[ x = 25 \]
Now, determine the speeds of both buses:
- The speed of the slower bus is \( x = 25 \) mph.
- The speed of the faster bus is \( x + 10 = 25 + 10 = 35 \) mph.
Thus, the speeds of the buses are:
- Slower bus: 25 mph
- Faster bus: 35 mph