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