**Re: chart problem help**
santasad13 said:

at noon a plane left austin for los angeles, 2100 km away, flying at 500 km per hour.one hour later a jet left los angeles for austin at 700 km per hour. at what time did they pass each other?

Let t = time for the plane which left from Austin

The rate for that plane is 500 km/hr. So, in t hours, the plane travels 500t km (because distance= rate * time)

The second plane left LA one hour later. So, the time traveled by the second plane is t - 1 hours.

The rate for the second plane is 700 km/hr. That plane traveled a distance of 700(t - 1) km.

Now, one plane left LA and the other plane left Austin. When they PASS each other, the distances traveled by the two planes must add up to the total distance between the two cities, or 2100 km. So,

Distance for first plane + distance for second plane = distance between LA and Ausin

500t + 700(t - 1) = 2100

Ok...solve that for t. And that will be the time traveled by the first plane; when the first plane has traveled t hours, the two planes will be passing each other.