Two buses leave a station at the same time, traveling in opposite directions. the rate of the faster bus is 15 mph faster than that of the slower bus. at the end of 3 hours, they are 345 miles apart. find the rate of each bus.

A set of equation can be set up:
a=b+15
3(a+b)=345
where a is the faster and b is the slower bus.
substitute b+15 into the second equation, so 6b+45=345, and 6b=300, therefore b=50.
we can then figure out that a=65

The faster bus can be expressed with
x+15=y
where x is the rate of the slower bus and y is the rate of the faster bus.
The buses travel for 3 hours, 345 miles apart. 
So it becomes
3y + 3x = 345
Substitute in the value for y with x+15 and you get
3(x+15) + 3x = 345
And afterwards, you just solve for x (WORK)
3x + 45 + 3x = 345 (Distribution)
6x + 45 = 345 (Combine like terms)
6x = 300 (Subtraction)
x = 50 (Division)
The rate of the slower bus is 50 mph, and plug it in for the equation for the faster bus.
x + 15 = y
(50) + 15 = y
y = 65
The rate of the faster bus if 65 mph.

Ican’t sorry i would if i could