Final answer:
By setting up an equation where the sum of the distances covered by both planes equals 11648 kilometers and solving for x, we find out that the slower plane's speed is 799 km/h and the faster plane's speed is 865 km/h.
Step-by-step explanation:
To solve the problem where two airplanes leave an airport at the same time and travel in opposite directions with one plane traveling 66 km/h faster than the other, and they are 11648 kilometers apart after 7 hours, we must first understand that the distances traveled by both planes are cumulative when determining how far apart they are. We let the speed of the slower plane be x km/h and the speed of the faster plane be (x + 66) km/h.
In 7 hours, the slower plane covers 7x kilometers while the faster plane covers 7(x + 66) kilometers. Since they are flying in opposite directions, their combined distance is the sum of both distances, which equals 11648 kilometers. We can write the equation:
7x + 7(x + 66) = 11648
Simplifying the equation:
14x + 462 = 11648
14x = 11648 - 462
14x = 11186
x = 799
Hence, the speed of the slower plane is 799 km/h and the faster plane is 865 km/h (799 + 66).