Steve traveled 600 miles at a certain speed. Had he gone 20mph faster, the trip would have taken 1 hour less. Find the speed of his vehicle. Thank you in advance for the help! I'll do it a different way from Stanbon: Let r be the actual rate and t be the actual time. Then the rate for the hypothetical trip would be r+20 and the time for the hypothetical trip would be t-1. So put the information into a DRT chart: DISTANCE RATE TIME Actual trip | 600 | r | t Hypothetical trip| 600 | r+20 | t-1 Now use DISTANCE = (RATE)(TIME) 600 = rt 600 = (r+20)(t-1) FOIL out the right side of the second equation: 600 = rt - r + 20t - 20 Multiply the 1st equation by -1 and add it to the second equation 600 = rt - r + 20t - 20 -600 = -rt --------------------------- 0 = -r + 20t - 20 r = 20t - 20 Substitute in the first equation 600 = rt 600 = (20t - 20)t 600 = 20tē - 20t 0 = 20tē - 20t - 600 Divide every term through by 20 0 = tē - t - 30 0 = (t - 6)(t + 5) t - 6 = 0 t + 5 = 0 t = 6 hrs t = -5 hrs We discard the negative answer. To find r, substitute in 600 = rt 600 = r(6) 600 = 6r 100 = r So he traveled at 100 mi/hr. His vehicle must have been a small plane. Edwin