A car travels 300 mi averaging a certain speed. If the car had gone 10 mph​ faster, the trip would have taken 1 hr less. Find the​ car's average speed.

Answer 1

300 = rt

300 / r = t

300 = (r + 10)(t - 1)

300 / (r + 10) = t - 1

300 / (r + 10) = (300 / r) - 1

300 / (r + 10) = (300 - r) / r

300r = (r + 10)(300 - r)

300r = 300r + 3000 - r² -10r

0 = -r² - 10r + 3000

0 = r² + 10r - 3000

0 = (r + 60)(x - 50)

50 mph

Check:

300 / 50 = 6

300 . 60 = 5

Answer 2

Distance (X)= rate(r)*time(t)

r=300/t

300 = (300/t+10)(t-1) Distribute and simplify

t^2 -t-30=0

(t-6)(t+5)=0

t=6

300/6 = 50.