Two automobiles leave the same city simultaneously and both head towards another city. The speed of one is 10 km/hour greater than the speed of the other, and this is why the first automobile arrives at the destination 1 hour before the other. Find the speed of both automobiles knowing that the distance between the two cities is 560 km.

speed for the slower automobile = 70

speed for the faster automobile = 80

Step-by-step explanation:

Speed of slower automobile = s and time taken by the slower automobile = t

Speed for faster automobile = s + 10 and time taken by the faster automobile = t - 1

D = s*t

For the slower automobile

560 = st

For the faster automobile

560 = (s + 10) (t-1)

560 = st - s + 10t - 10

Remember st = 560

560 = 560 - s + 10 (560/s) - 10

560-560 + 10 = - s + 5600/s

10 = - s + 5600/s

10s = - s² + 5600

s² + 10s - 5600 = 0

s = (-b ± (√b²-4ac)) / 2a

s = (-10±√ (10² - 4 (1) (-5600))) / 2

s = (-10 ±√22500) / 2

s = (-10+150) / 2 or (-10 - 150) / 2

s=70 or - 80

therefore speed for the slower automobile = 70

and speed for the faster automobile = 80