The Question: A train traveled 960 miles from Town A to Town B. If the train had traveled from Town A to Town B at an average speed of 12 miles per hour less, it would have taken 4 more hours. What was the train's average speed during that journey? The explanations in some of those online courses are not so good. Here's the way I would solve it: Make this chart: DISTANCE RATE TIME ACTUAL TRIP HYPOTHETICAL TRIP Let the speed on the actual trip be x. Fill that in for the RATE of the ACTUAL TRIP. DISTANCE RATE TIME ACTUAL TRIP x HYPOTHETICAL TRIP >>...If the train had traveled from Town A to Town B at an average speed of 12 miles per hour less...<< So we subtract 12 from x, getting x-12 and fill that in for the RATE for the HYPOTHETICAL TRIP. DISTANCE RATE TIME ACTUAL TRIP x HYPOTHETICAL TRIP x-12 We are told that the DISTANCE for both the ACTUAL and the HYPOTHETICAL trip is 960 miles, so we fill that in: DISTANCE RATE TIME ACTUAL TRIP 960 x HYPOTHETICAL TRIP 960 x-12 Now we fill in the TIMEs using TIME = DISTANCE/RATE DISTANCE RATE TIME ACTUAL TRIP 960 x 960/x HYPOTHETICAL TRIP 960 x-12 960/(x-12) Now we look for what we haven't used: >>...it [THE HYPOTHETICAL TRIP] would have taken 4 more hours...<< So HYPOTHETICAL TRIP TIME = ACTUAL TRIP TIME + 4 HOURS or 960/(x-12) = 960/x + 4 Divide thru by 4 240/(x-12) = 240/x + 1 Multiply through by LCD = x(x-12) 240x = 240(x-12) + x(x-12) 240x = 240x - 2880 + x˛ - 12x 0 = x˛ - 12x - 2880 Solve this by the quadratic formula x˛ - 12x - 2880 = 0 The quadratic formula is: ______ -b ą Öb˛-4ac x = 2a where a = 1; b = -12; c = -2880 __________________ -(-12) ą Ö(-12)˛-4(1)(-2880) x = 2(1) _________ 12 ą Ö144+11520 x = 2 _____ 12 ą Ö11664 x = 2 12 ą 108 x = 2 Using the + 12 + 108 x = 2 120 x = = 60 mph 2 Using the - 12 - 108 x = 2 -96 x = = -48 mph 2 so we discard that and the only answer is 60 mph. Edwin