SOLUTION: a car A left city X traveling at an average velocity of 60 miles per hour. Two hours later a car B left city X traveling on the same road at an average velocity of 80 miles per ho

Just for fun let's do it in our head the easy way, and then we'll do it by
algebra.

In 2 hours Car A, doing 60 mph is 120 miles ahead of Car B when Car B
starts. Car B travels 80 mph which is 20 mph faster, so that's how fast
he's approaching Car A at 80-60 or 20 mph.  So to catch up the 120 miles 
head start at 20 mph approach rate will take 6 hours since 120/20 = 6, 
and they will have traveled 80 mph*6 hours or 480 miles.

Your teacher doesn't want you to do it that way, but by algebra. So
here's how to do it by algebra, even though the "head" way is easier for
this problem.
 
Now by algebra:

Make this chart:

        distance    rate    time
Car A    
Car B    

Let x be the time it takes B to catch up to A
Then x+2 will be the time A has traveled since he 
travels for 2 hours longer.  So fill those in and
also the rates at which they travel:
    

        distance    rate    time
Car A                60      x+2 
Car B                80       x

Now use distance = rate × time to fill in the distances:

        distance    rate    time
Car A    60(x+2)     60      x+2 
Car B      80x       80       x

The two distances are equal, so make the equation:

        60(x+2) = 80x

Solve that and get x = 6 hours

Then use either value for the distance and substitute 6 for x

60(x+2) = 60(6+2) = 60(8) = 480 miles

Or you could substitute in the other value for the distance:

80x = 80(6) = 480 miles.

Edwin