SOLUTION: Two cars leave town A at the same time and travel 480 miles (each) to town B. The speed of the slower car is 20 mph less than the speed of the faster one. If the slower car arrives

Question 934100: Two cars leave town A at the same time and travel 480 miles (each) to town B. The speed of the slower car is 20 mph less than the speed of the faster one. If the slower car arrives at Town B two hours after the faster car, the what is the speed ( in mph) of the faster car?
Answer by mananth(16946) (Show Source): You can put this solution on YOUR website!
Car I speed x
Car II speed x + -20

Distance = same 480 miles 1
Car II time - Car I time = 2 hours
t=d/t

480 / x - 480 / ( x + -20 ) = 2.00
LCD= x ( x + -20 )
multiply by LCD
480 ( x + -20 ) - 480 x = 2 x ( x + -20 )
480 x + -9600 - 480 x = 2 X^2 + -40 x
-9600 = 2 X^2 + -40 x
2 X^2 -40 x - 9600 = 0
/2
x^2-20x-4800=0
x^2-20x+100= 4800+100
(x-10)^2= 4900
x-10 = 70 ( taking positive)
x= 80
Car I ------80 mph
Car II-------60 mph

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