Two airplanes left the same airport and arrived at the same destination at the same time. the first airplane left at 8: 00 a. m. and traveled at an average rate of 496 miles per hour. the second airplane left at 8: 30 a. m. and traveled at an average rate of 558 miles per hour. let x represent the number of hours that the first plane traveled. how many hours did it take the first plane to travel to the destination? enter an equation that can be used to solve this problem in the first box. solve for x and enter the number of hours in the second box.

In this question, you are given the plane 1 speed(496miles/h), plane1 depart time(8:00), plane 2 speed(558miles/h) and plane2 depart time(8:30 or 8.5 hour). Both of the planes left from the same airport so their traveling distance should be same. They also arrive in same time. Then the equation would be:

duration = arriving time - depart time
duration1= x= arriving time - 8h
duration2= y= arriving time -8.5h= (arriving time - 8h) -0.5h= x-0.5h 

speed1 * duration1 = speed2 * duration2
496 * x = 558* (x-0.5)
496x = 558x - 279
279= 558x- 496x
279= 62x
x= 4.5hour 

Ithink it is b and d